## Worksheet on Word Problems on Linear Equation | Linear Equations Word Problems Worksheet

Find the solution for any typical problem of linear equations by practicing our Worksheet on Word Problems on Linear Equation. You can find out different types of equations from simple to complex ones. We are offering solutions for every equation along with an explanation. Practice using the below Linear Equations Word Problems Worksheet and learn how to find a solution for a given linear equation problem quickly.

## Linear Equations Problems with Solutions

1. Convert the following statements into equations?
(a) 5 added to a number is 9.
(b) 3 subtracted from a number is equal to 12.
(c) 5 times a number decreased by 2 is 4.
(d) 2 times the sum of the numbers K and 7 is 13.

Solution:

(a) The given statement is 5 added to a number is 9.
Let’s consider a number as ‘K’.
Then, adding 5 with K should be 9. That is 5 + K = 9.
By converting the given statement into an equation we will get the equation like 5 + K = 9.
(b) The given statement is 3 subtracted from a number is equal to 12.
Let’s consider a number as ‘K’.
Then, subtracting 3 from K should be 12. That is K – 3 = 12.
By converting the given statement into an equation we will get the equation as K – 3 = 12.
(c) The given statement is5 times a number decreased by 2 is 4.
Let’s consider a number as ‘K’.
Five times a number is equal to 5K.
5Kis decreased by 2 and that is 5K – 2.
Then, by decreasing 2 from 5K should be 4. That is5K – 2 = 4.
By converting the given statement into an equation we will get the equation as 5K – 2 = 4.
(d)The given statement is 2 times the sum of the number K and 7 is 13.
Let the sum of the number K and 7 is K + 7.
Now, 2 times the sum of K and 7 is 2 ( K + 7 ) which is equal to 13.
By converting the given statement into an equation we will get the equation as
2 ( K + 7 ) = 13.

2. A number is 12 more than the other. Find the numbers if their sum is 48?

Solution:

The given statement is a number is 12 more than the other and if their sum is 48.
Let’s consider the first number as ‘K’.
If another number is 12 more than K and that is K + 12.
So, the sum of two numbers is
K + K + 12 = 48.
2K + 12 = 48.
2K = 48 – 12 = 36.
K = 36 ÷ 2 = 18
So first number K is 18 then other number is 12 + K = 12 + 18 = 30.
The sum of the two numbers 18 + 30 = 48.

The two numbers are 18 and 30.

3. Twice the number decreased by 22 is 48. Find the number?

Solution:

The given statement is twice the number decreased by 22 is 48.
Let’s consider the number as ‘K’.
Twice the number is 2K.
2K is decreased by 22, that is 2K – 22.
So 2K – 22 = 48.
2k = 48 + 22 = 70.
K = 70 ÷ 2 = 35.

That is the number is K = 35.

4. Seven times the number is 36 less than 10 times the number. Find the number?

Solution:

The given statement is Seven times the number is 36 less than 10 times the number.
Let’s consider a number as ‘K’.
Seven times the number is 7K.
Ten times the number is 10K.
36 less than ten times a number is 10K – 36.
7K should be equal to 10K – 36.
That is 7K = 10K – 36.
36 = 10K – 7K.
36 = 3K.
K = 36 ÷ 3 = 12.

The number is 12.

5. 4/5 of a number is more than 3/4 of the number by 5. Find the number?

Solution:

The given statement is 4/5 of a number is more than 3/4 of the number by 5.
Let’s consider a number as ‘K’.
4 / 5 of a number is 4K / 5.
3 / 4 of a number is 3K / 4.
4K / 5 is more than 3K / 4 by 5.
That is 3K / 4 = 4K / 5 + 5.
By using LCM on RHS.
3K / 4 = ( 4K + 25 ) / 5.
By using cross multiplication on L.H.S and R.H.S for the above equation.
( 3K ) X 5 = ( 4K + 25 ) X 4
15K = 16K + 100.
K = 100.
The number is 100.

6. The sum of two consecutive even numbers is 38. Find the numbers?

Solution:

The given statement is the sum of two consecutive even numbers is 38.
Let’s consider the consecutive even numbers are K + 2 and K + 4.
Sum of two consecutive even numbers is ( K + 2 ) + ( K + 4 ) = 38.
2K + 6 = 38.
2K = 38 – 6 = 32.
K = 32 ÷ 2 = 16.
K + 2 = 16 + 2 = 18.
K + 4 = 16 + 4 = 20
So the two consecutive even numbers are 18 and 20.

7. The sum of three consecutive odd numbers is 51. Find the numbers?

Solution:

The given statement is the sum of three consecutive odd numbers is 51.
Let’s consider the consecutive odd numbers as K + 1, K + 3 and K + 5.
The Sum of three consecutive odd numbers is
( K + 1 ) + ( K + 3 ) + ( K + 5 ) = 51.
3K + 9 = 51.
3K = 51 – 9 = 42.
K = 42 ÷ 3 = 14.
K + 1 = 15.
K + 3 = 17.
K + 5 = 19.
The consecutive odd numbers are 15, 17, and 19.

8. Rene is 6 years older than her younger sister. After 10 years, the sum of their ages will be 50 years. Find their present ages?

Solution:

The given statement Rene is 6 years older than her younger sister. After 1 0 years, the sum of their ages will be 50 years.
Let’s consider younger sister age as ‘K’.
Rene is 6 years older than her younger sister, that is 6 + K.
After 10 years the sum of Rene and Younger sister age will be equal to 50 can be written as
( K + 10 ) + ( K + 6 + 10 ) = 50.
2K + 26 = 50.
2K = 50 – 26 = 24.
K = 24 ÷ 2 = 12.
Younger sister age is 12.
Rene is 6 years older than her younger sister that is 12 + 6 = 18.

9. The length of a rectangle is 10 m more than its breadth. If the perimeter of a rectangle is 80 m, find the dimensions of the rectangle?

Solution:

The given statement is the length of a rectangle is 10m more than its breadth and the perimeter of a rectangle is 80m.
Let’s consider the breadth of a rectangle as ‘K’.
b = K.
The length of a rectangle is 10m more than the breadth that is K + 10.
l = K + 10.
The perimeter of rectangle is 2 ( l + b ) = 80m.
2 ( K + 10 + K ) = 80m.
2K + 10 = 80 ÷ 2 = 40m.
2K = 40 – 10 = 30m
K = 30 ÷ 2 = 15m.
The breadth of a rectangle K = 15m.
Length of a rectangle K + 10 = 15 + 10 = 25m.

10. A 300 m long wire is used to fence a rectangular plot whose length is twice its width. Find the length and breadth of the plot?

Solution:

The given statement is a 300 m long wire is used to fence a rectangular plot whose length is twice its width.
Let’s consider the width of the rectangle as ‘K’.
b = K.
The length of the rectangle is twice its width that is 2K.
l = 2K.
Perimeter of a rectangle is 2 ( l + b ) = 300.
2 ( K + 2K ) = 300m.
3K = 300 ÷ 2 = 150m.
K = 150 ÷ 3 = 50m.
The width of the rectangle is K = 50m.
The length of rectangle is 2K = 2 X 50 = 100m.

11. The denominator of a fraction is greater than the numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is 3/2, find the fraction?

Solution:

The given statement is the denominator of a fraction is greater than the numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is 3/2.
Let’s consider the numerator as ‘K’.
The denominator of the fraction is greater than the numerator by 8 that is K + 8.
So fraction is K / K + 8.
The numerator is increased by 17 and the denominator is decreased by 1 which is
( K + 17 ) / ( K + 8 – 1 ) = 3 / 2.
By applying cross multiplication for above equation we get
2 ( K + 17 ) = 3 ( K + 7 ).
2K + 34 = 3K + 21.
34 – 21 = 3K – 2K.
K = 13.
The numerator of the fraction is K = 13.
Denominator of fraction is K + 8 = 13 + 8 = 21.
So fraction is 13 / 21.

12. A sum of \$2700 is to be given in the form of 63 prizes. If the prize is either \$100 or \$25, find the number of prizes of each type?

Solution:

The given statement is a sum of \$2700 is to be given in the form of 63 prizes. If the prize is either \$100 or \$25.
Let’s consider ‘K’ as the number of the \$100 prizes and ‘L’ as the number of the \$25 prize.
100K + 25L = \$2700 ——– (1).
Total number of prizes is 63 that is K + L = 63 ——- (2).
Multiply the equation (2) with 100 on both sides. That is
100K + 100L = \$6300 ——– (3).
Subtract the equation (1) and (3).
100K + 25L = \$2700.
100K + 100L = \$6300.
(-)—–(-)——-(-)——————-
– 75L = – \$3600.
L = 3600 ÷ 75 = 48.
Substitute the L value in equation number (2).
That is K + 48 = 63.
K = 63 – 48 = 15.
Finally, the number of prizes are 15 and 48.

13. In a class of 42 students, the number of boys is 2/5 of the girls. Find the number of boys and girls in the class?

Solution:

The given statement isa class of 42 students, the number of boys is 2/5 of the girls.
Let’s consider the number of girls in the class is ‘K’.
So, the number of boys is 2 / 5 of the girls. That is 2K/ 5.
The total number of students in the class is 42. That is
2K / 5 + K = 42.
2K + 5K = 42 X 5 = 210.
7K = 210.
K = 210 ÷ 7 = 30.
The total number of girls in the class is K = 30
The total number of boys in the class is 2K / 5 = 2 X 30 / 5 = 2 X 6 = 12.

14. Among the two supplementary angles, the measure of the larger angle is 36° more than the measure of smaller. Find their measures?

Solution:

The given statement is the two supplementary angles, the measure of the larger angle is 36° more than the measure of smaller.
Let’s consider the small-angle as ‘K’ and the larger angle as ‘L’.
‘K’ and ‘L’ are supplementary angles. That is K + L = 180° ———(1).
The larger angle is 36° more than the smaller angle. That is L = K + 36°.
Substitute the larger angle value in equation (1).
Then, K + K + 36° = 180°.
2K = 180° – 36° = 144°.
K = 144° ÷ 2 = 72°.
L = K + 36° = 72° + 36° = 108°.
The measures of the angles are 72° and 108°.

15. My mother is 12 years more than twice my age. After 8 years, my mother’s age will be 20 years less than three times my age. Find my age and my mother’s age?

Solution:

The given statement is mother’s age is 12 years more than twice my age. After 8 years, my mother’s age will be 20 years less than three times my age.
Let’s consider my age as ‘K’.
My mother’s age is 12 years more than twice my age and that is 2K + 12.
After 8 years.
My age is K + 8.
My mother’s age is 2K + 12 + 8.
My mother’s age will be 20 years less than three times my age.
3(K + 8) – 20 = 2K + 20.
3K + 24 – 20 = 2K + 20.
3K + 4 = 2K + 20.
3K – 2K = 20 – 4 = 16.
My age is K = 16 years.
My mother’s age is 2K + 12 = 2(16) + 12 = 44 years.

16. In an isosceles triangle, the base angles are equal and the vertex angle is 80°. Find the measure of the base angles?

Solution:

The given statement is in an isosceles triangle, the base angles are equal and the vertex angle is 80°.
As per the given information, the vertex angle is 80°.
The base angles are equal.
Triangle is K + K + 80° = 180°.
2K = 180° – 80° = 100°.
K = 100 ÷ 2 = 50°.
So, the measures of the base angles are 50°.

17. Adman’s father is 49 years old. He is 5 years older than four times Adman’s age. What is Adman’s age?

Solution:

The given statement is an Adman’s father is 49 years old. He is 5 years older than four times Adman’s age.
Let’s consider an Adman’s age as ‘K’
Adman’s father is 49 years old.
Adman’s father is 5 years older than four times Adman’s age. That is
4K + 5 = 49.
4K = 49 – 5 = 44.
K = 44 ÷ 4 = 11 years
So, an Adman’s age is 11 years.

18. The cost of a pencil is 25 cents more than the cost of an eraser. If the cost of 8 pencils and 10 erasers is \$12.80, find the cost of each?

Solution:

The given statement is the cost of a pencil is 25 cents more than the cost of an eraser. If the cost of 8 pencils and 10 erasers is \$12.80.
Let’s consider the cost of an eraser as ‘K’.
The cost of a pencil is 25 cents more than the cost of an eraser. That is K + 25.
The cost of 8 pencils and 10 erasers is \$12.80.
1 dollar = 100 cents. So, \$12.80 is 1280 cents.
8(K + 25) + 10K = 1280 cents.
8K + 200 + 10K = 1280 cents.
18K = 1280 – 200 = 1080 cents.
K = 1080 ÷ 18 = 60 cents.
The cost of an eraser is 60 cents.
Cost of pencils is K + 25 = 60 + 25 = 85 cents.

19. Divide 36 into two parts in such a way that 1/5 of one part is equal to 1/7 of the other?

Solution:

The given statement is Divide 36 into two parts in such a way that 1/5 of one part is equal to 1/7 of the other.
By dividing the 36 into two parts
K / 5 = (36 – K) / 7.
By cross multiplying the above terms, we will get
7K = (36 – K)5.
7K = 36 X 5 – 5K.
7K + 5K = 180.
12K = 180.
K = 180 ÷ 12 = 15.
The first part is 15 and the second part is 36 – 15 = 21.

20. The length of the rectangle exceeds its breadth by 3 cm. If the length and breadth are each increased by 2 cm, then the area of the new rectangle will be 70 sq. cm more than that of the given rectangle. Find the length and breadth of the given rectangle?

Solution:

The given statement is the length of the rectangle exceeds its breadth by 3 cm. If the length and breadth are each increased by 2 cm, then the area of the new rectangle will be 70 sq. cm more than that of the given rectangle.
Let’s consider the breadth of the rectangle as ‘K’.
b = K.
Then, the length of the rectangle exceeds its breadth by 3 cm. That is K + 3
l = K + 3.
If the length and breadth are each increased by 2cm is K + 2 and K + 3 + 2.
The area of a rectangle is 70sq. cm
Area of the rectangle is (l X b) = 70 sq. cm.
(K + 2) X (K + 5) = K(K + 3) + 70.
K^2 + 5K + 2K + 10 = K^2 + 3K + 70.
7K + 10 = 3K + 70.
4K = 60.
K = 15.
L = K + 3 = 15 + 3 = 18.
The length of the rectangle is 18 and the breadth of the rectangle is 15.

## Worksheet on Significant Figures | Significant Figures Practice Worksheets

Take the guidance on the concept of Significant Figures by taking the help of Worksheet on Significant Figures. Practice all the problems present in the Significant Figures Worksheets and get a good grip on the topic. Test your subject knowledge by taking the practice tests available on our website. Also, you can easily enhance your problem-solving skills by solving Worksheet on Significant Figures on a daily basis. Assess your preparation standards and concentrate on the areas you are facing difficulty.

## Significant Figures Examples with Answers

Check different problems impose on Significant Figures and get a grip on every concept available in the Significant Figures.

1. Find the Number of a Significant Figure in Each of the Following
(a) 8.4
(b) 173.6 m
(c) 407 g
(d) 4.68 m
(e) 8.1165 kg
(f) 0.054 km
(g) 0.00343 l
(h) 93.040 mg
(i) 30.030300 g
(j) 40.00 km
(k) 3.600 ml
(l) 70.002

Solution:

(a) There are 2 significant figures available in 8.4. They are 8, 4.
(b) There are 4 significant figures available in 173.6 m. They are 1, 7, 3, and 4.
(c) There are 3 significant figures available in 407 g. They are 4, 0, and 7.
(d) There are 3 significant figures available in 4.68 m. They are 4, 6, and 8.
(e) There are 5 significant figures available in 8.1165 kg. They are 8, 1, 1, 6, and 5.
(f) There are 2 significant figures available in 0.054 km. They are 5, and 4.
(g) There are 3 significant figures available in 0.00343 l. They are 3, 4, and 3.
(h) There are 5 significant figures available in 93.040 mg. They are 9, 3, 0, 4, and 0.
(i) There are 8 significant figures available in 30.030300 g. They are 3, 0, 0, 3, 0, 3, 0, and 0.
(j) There are 4 significant figures available in 40.00 km. They are 4, 0, 0, and 0.
(k) There are 4 significant figures available in 3.600 ml. They are 3, 6, 0, and 0.
(l) There are 5 significant figures available in 70.002. They are 7, 0, 0, 0, and 2.

2. Round off each of the following correct up to 3 significant figures
(a) 57.3628 g
(b) 6.31874 kg
(c) 44.422 km
(d) 60.001 cm
(e) 0.0023596 m
(f) 0.0024030 l

Solution:

(a) Given that 57.3628 g. It has 6 significant figures. To round off the given number into 3 significant digits, we need to round it off to 1 place after the decimal.
The digit 6 is greater than 5. So, the digit 3 becomes 4, and the digits 6, 2, and 8 disappear.
Therefore, 57.3628 g = 57.4 g rounded off to 3 significant figures.
(b) Given that 6.31874 kg. It has 6 significant figures. To round off the given number into 3 significant digits, we need to round it off to 2 places after the decimal.
The digit 8 is greater than 5. So, the digit 1 becomes 2, and the digits 8, 7, and 4 disappear.
Therefore, 6.31874 kg = 6.32 kg rounded off to 3 significant figures.
(c) Given that 44.422 km. It has 5 significant figures. To round off the given number into 3 significant digits, we need to round it off to 1 place after the decimal.
The digit 2 is less than 5. So, the digit 4 remains 4, and the digits 2, and 2 disappear.
Therefore, 44.422 km = 44.4 km rounded off to 3 significant figures.
(d) Given that 60.001 cm. It has 5 significant figures. To round off the given number into 3 significant digits, we need to round it off to 1 place after the decimal.
The digit 0 is less than 5. So, the digit 0 remains 0, and the digits 0, and 1 disappear.
Therefore, 60.001 cm = 60.0 cm rounded off to 3 significant figures.
(e) Given that 0.0023596 m. It has 5 significant figures. To round off the given number to 3 significant digits, we need to round it off to 5 places after the decimal.
The digit 9 is greater than 5. So, the digit 5 becomes 6, and the digits 9, and 6 disappear.
Therefore, 0.0023596 m = 0.00236 m rounded off to 3 significant figures.
(f) Given that 0.0024030 l. It has 5 significant figures. To round off the given number to 3 significant digits, we need to round it off to 5 places after the decimal.
The digit 3 is less than 5. So, the digit 0 remains 0, and the digits 3, and 0 disappear.
Therefore, 0.0024030 l = 0.00240 l rounded off to 3 significant figures.

3. Round off
(a) 16.367 g correct to the three significant figures.
(b) 0.00949 mg correct to the two significant figures.
(c) 0.005618 g correct to the one significant figures.
(d) 28.303 mm correct to the four significant figures.
(e) 33.422 km correct to the four significant figures.
(f) 4.0832 kg correct to the two significant figures.
(g) 0.004628 mm correct to the one significant figures.

Solution:

(a) Given that 16.367 g. It has 5 significant figures. To round off the given number into 3 significant digits, we need to round it off to 1 place after the decimal.
The digit 6 is greater than 5. So, digit 3 becomes 4, and the digits 6, and 7 disappear.
Therefore, 16.367 g = 16.4 g rounded off to 3 significant figures.
(b) Given that 0.00949 mg. It has 3 significant figures. To round off the given number into 2 significant digits, we need to round it off to 4 places after the decimal.
The digit 9 is greater than 5. So, the digit 4 becomes 5, and the digit 9 disappear.
Therefore, 0.00949 mg = 0.0095 mg rounded off to 2 significant figures.
(c) Given that 0.005618 g. It has 4 significant figures. To round off the given number into 1 significant digit, we need to round it off to 3 places after the decimal.
The digit 6 is greater than 5. So, the digit 5 becomes 6, and the digits 6, 1, and 8 disappear.
Therefore, 0.005618 g = 0.006 g rounded off to 3 significant figures.
(d) Given that 28.303 mm. It has 5 significant figures. To round off the given number into 4 significant digits, we need to round it off to 2 places after the decimal.
The digit 3 is less than 5. So, the digit 0 remains 0, and the digits 3 disappear.
Therefore, 28.303 mm = 28.30 mm rounded off to 4 significant figures.
(e) Given that 33.422 km. It has 5 significant figures. To round off the given number to 4 significant digits, we need to round it off to 2 places after the decimal.
The digit 2 is less than 5. So, digit 2 remains 2, and the digits 2 disappear.
Therefore, 33.422 km = 33.42 km rounded off to 4 significant figures.
(f) Given that 4.0832 kg. It has 5 significant figures. To round off the given number to 2 significant digits, we need to round it off to 1 place after the decimal.
The digit 8 is greater than 5. So, the digit 0 becomes 1, and the digits 8, 3, and 2 disappear.
Therefore, 4.0832 kg = 4.1 kg rounded off to 2 significant figures.
(g) Given that 0.004628 mm. It has 4 significant figures. To round off the given number to 1 significant digit, we need to round it off to 3 places after the decimal.
The digit 6 is greater than 5. So, the digit 4 becomes 5, and the digits 6, 2, and 8 disappear.
Therefore, 0.004628 mm = 0.005 mm rounded off to 1 significant figure.

4. Round off
(a) \$ 3067.665 to the nearest cents.
(b) 0.00588 m to the nearest cm.
(c) 18.0333 kg to the nearest g.
(d) \$ 49.63 to the nearest dollar.

Solution:

(a) Given that \$ 3067.665. It has 7 significant figures. To round off the \$ 3067.665 to the nearest cents, we need to round it off to 3 places after the decimal.
The digit 5 is equal to 5. So, digit 6 becomes 7, and the digit 5 disappear.
Therefore, \$ 3067.665 = \$ 3067.67
(b) Given that 0.00588 m. It has 3 significant figures. To round off the 0.00588 m to the nearest cm, we need to round it off to 3 places after the decimal.
The digit 5 is equal to 5. So, the digit 5 becomes 6, and the digit 8, and 8 disappear.
Therefore, 0.00588 m = 0.01 m.
(c) Given that 18.0333 kg. It has 6 significant figures. To round off the 18.0333 kg to the nearest g, we need to round it off to 3 places after the decimal.
The digit 3 is less than 5. So, the digit 3 remains 3, and the digits 3 disappear.
Therefore, 18.0333 kg = 18.033 kg.
(d) Given that \$ 49.63. It has 4 significant figures. To round off the \$ 49.63 to the nearest dollar, we need to round it off to the nearest number.
The digit \$ 49.63 is closer to \$ 50.
Therefore, \$ 49.63 = \$ 50.

## Worksheet on Three Dimensional Figures | Three Dimensional Shapes Practice Worksheets

We are here to help all the people who wish to learn the Three Dimensional Figures concepts. We have given detailed explanations and also different problems on 3-Dimensional objects along with explanations. Also, students can easily improve their preparation level by practicing with Worksheet on Three Dimensional Figures. Use the Three Dimensional Figures Worksheets and try to answer the questions on your own. Solving the 3D Figures Problems regularly helps you to improve your accuracy and speed in the final exams.

## Objective Questions on Three Dimensional Figures

1. Mention the number of faces for each of the given figures

(i) Cuboid
(ii) Tetrahedron
(iii) Triangular prism
(iv) Cube
(v) Square pyramid

Solution:

1. (i) 6
(ii) 4
(iii) 5
(iv) 6
(v) 5
Explanation:
(i) A Cuboid is made up of six rectangles, each of the rectangles is called the face.
(ii) A tetrahedron has 4 faces all are triangles.
(iii) A triangular prism consists of 2 triangular faces and 3 rectangular faces. Therefore, total the triangular prism has 5 faces.
(iv) A cube consists of 6 faces.
(v) A square pyramid consists of faces one of which is a square face and the rest four are triangular faces.

2. Write down the number of edges of each of the following figures
(i) Tetrahedron
(ii) Triangular prism
(iii) Cube
(iv) Rectangular pyramid

Solution:

1. (i) 6
(ii) 9
(iii) 12
(iv) 8
Explanation:
(i) A tetrahedron has 6 edges.
(ii) A triangular prism consists of 9 edges. If PQRSTV is a triangular prism, then the 9 edges of the triangular prism are PQ, QR, RP, ST, TV, VS, PS, QT, RV.
(iii) A cube consists of 12 edges. If PQRS is a cube, then the 12 edges of the cube are PQ, QR, RS, SP, EF, FG, GH, HE, PE, SH, QF, RG.
(iv) A rectangular pyramid consists of 8 edges. If the rectangular pyramid OPQRS has 8 edges, then PQ, QR, RS, SP, OP, OQ, OR, and OS are edges.

3. Write down the number of vertices of each of the following figures
(i) Tetrahedron
(ii) Square pyramid
(iii) Cuboid
(iv) Triangular prism

Solution:

1. (i) 4
(ii) 5
(iii) 8
(iv) 6
Explanation:
(i) A tetrahedron has 4 vertices.
(ii) A square pyramid consists of 5 vertices. If OPQRS is a square pyramid having O, P, Q, R, S as its vertices.
(iii) A cuboid has 8 vertices. If PQRS is a cuboid, then the 8 vertices of the cuboid are P, Q, R, S, E, F, G, H.
(iv) A triangular prism consists of 6 vertices. If PQRS is a triangular prism, then the 6 vertices of the triangular prism are P, Q, R, S, T, V.

4. Fill in the blanks
(i) A triangular pyramid is called a …………….. .
(ii) The point at which three faces of a figure meet is known as its ……………..
(iii) A cube has …………….. vertices, …………….. edges and …………….. faces.
(iv) A cuboid is also known as a rectangular …………….

Solution:

1. (i) tetrahedron
(ii) vertex
(iii) 8, 12, 6
(iv) prism

5. Which of the following has a single vertex?
(i) Sphere and Cylinder
(ii) Sphere
(iii) Cylinder
(iv) Cone

Solution:

The answer is Cone. A cone has only a single vertex. The cone decreases smoothly from the circular flat base to the top point.

6. Which of the below statements are not correct?
(i) Each face of a polyhedron is a polygon.
(ii) The faces of a polyhedron are all flat.
(iii) A polyhedron is a three-dimensional figure.
(iv) A cube and a cone are both polyhedrons.

Solution:

(iv) A cube and a cone are both polyhedrons.

7. A ………………… is a space figure with six faces, 8 vertices, and its opposite sides are parallel.
(i) Triangular Prism
(ii) Pyramid
(iii) Rectangular Prism
(iv) Cylinder

Solution:

(iii) Rectangular Prism

A Rectangular Prism is a space figure with six faces, 8 vertices, and its opposite sides are parallel.

8. How many faces does a triangular prism have?
(i) Five
(ii) Four
(iii) Three
(iv) Two

Solution:

(i) Five
A triangular prism has Five faces.

## Worksheet on Different Types of Quadrilaterals | Types of Quadrilaterals Worksheets

Solve all problems available on Worksheet on Different Types of Quadrilaterals to get good marks in the exam. You can learn different methods to solve a single problem on the Quadrilateral Practice Worksheets. Check out Free Quadrilateral Worksheets on our website and get a grip on complete Quadrilateral concepts in minutes. Practice using the Types of Quadrilaterals Worksheets available and be familiar with various questions. For your convenience, we even provided step by step solutions to all the problems making it easy for you to understand the problems.

## Solved Problems on Different Types of Quadrilaterals

1. Construct a parallelogram PQRS in which PQ = 5.2 cm, QR = 4.7 cm and PR = 7.6 cm.

Solution:

Steps of Construction:
Given that a parallelogram PQRS in which PQ = 5.2 cm, QR = 4.7 cm and PR = 7.6 cm.
1. Draw a line segment of length 5.2 cm and mark the ends as P and Q.
2. Take the point P as a center and draw an arc by taking the radius 7.6 cm.
3. Next, take point Q as a center and draw an arc by taking the radius 4.7 cm. Mark the point as R where the two arcs cross each other. Join the points Q and R as well as P and R.
Note: A parallelogram is a simple quadrilateral with two pairs of parallel sides. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure.
4. By taking the point P as a center, draw an arc with a radius of 4.7 cm.
5. By taking the point R as a center, draw an arc with a radius of 5.2 cm.
6. Mark the point as S where the two arcs cross each other. Join the points R and S as well as P and S.

PQRS is a required parallelogram.

2. Construct a parallelogram PQRS in which PQ = 4.3 cm, PS = 4 cm and QS = 6.8 cm.

Solution:

Steps of Construction:
Given that a parallelogram PQRS in which PQ = 4.3 cm, PS = 4 cm and QS = 6.8 cm.
1. Draw a line segment of length 4.3 cm and mark the ends as P and Q.
2. Take the point P as a center and draw an arc by taking the radius 4 cm.
3. Next, take point Q as a center and draw an arc by taking the radius 6.8 cm. Mark the point as S where the two arcs cross each other. Join the points Q and S as well as P and S.
Note: A parallelogram is a simple quadrilateral with two pairs of parallel sides. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure.
4. By taking the point Q as a center, draw an arc with a radius of 4 cm.
5. By taking the point S as a center, draw an arc with a radius of 4.3 cm.
6. Mark the point as R where the two arcs cross each other. Join the points R and S as well as R and Q.

PQRS is a required parallelogram.

3. Construct a parallelogram ABCD in which BC = 6 cm, AB = 4 cm and ∠ABC = 60°.

Solution:

Steps of Construction:
Given that a parallelogram ABCD in which BC = 6 cm, AB = 4 cm and ∠ABC = 60°.
1. Draw a line segment of length 4 cm and mark the ends as A and B.
2. Take point B as a center and make a point by taking 60º using a protector.
3. Next, take point B as a center and draw an arc by taking the radius 6 cm. Mark the point as C where the point and arc cross each other. Join the points C and B.
Note: A parallelogram is a simple quadrilateral with two pairs of parallel sides. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure.
4. By taking the point C as a center, draw an arc with a radius of 4 cm.
5. By taking point A as a center, draw an arc with a radius of 6 cm.
6. Mark the point as D where the two arcs cross each other. Join the points D and C as well as D and A.

ABCD is a required parallelogram.

4. Construct a parallelogram PQRS in which QR = 5 cm, ∠PQR = 120° and RS = 4.8 cm.

Solution:

Steps of Construction:
Given that a parallelogram PQRS in which QR = 5 cm, ∠PQR = 120° and RS = 4.8 cm.
1. Draw a line segment of length 5 cm and mark the ends as Q and R.
2. Take point R as a center and make a point by taking 120º using a protector.
3. Next, take point R as a center and draw an arc by taking the radius 4.8 cm. Mark the point as S where the point and arc cross each other. Join the points S and R.
Note: A parallelogram is a simple quadrilateral with two pairs of parallel sides. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure.
4. By taking the point S as a center, draw an arc with a radius of 5 cm.
5. By taking point Q as a center, draw an arc with a radius of 4.8 cm.
6. Mark the point as P where the two arcs cross each other. Join the points P and S as well as P and Q.

PQRS is a required parallelogram.

5. Construct a parallelogram PQRS, one of whose sides is 4.4 cm and whose diagonals are 5.6 cm and 7 cm. Measure the other side?

Solution:

Steps of Construction:
Given that a parallelogram PQRS, one of whose sides is 4.4 cm and whose diagonals are 5.6 cm and 7 cm.
1. Draw a line segment of length 4.4 cm and mark the ends as P and Q.
2. Make the diagonals half to get the exact vertices of a parallelogram. Take point P as a center and draw an arc by taking the radius 2.8 cm.
3. Next, take point Q as a center and draw an arc by taking the radius 3.5 cm. Mark the point as O where the point and arc cross each other. Join the points P and O, Q and O.
4. Extend the line PO with a radius of 3.5 cm and mark it as R.
5. Extend the line QO with a radius of 2.8 cm and mark it as S.
6. Join the points P and S as well as Q and R, R and S.

PQRS is a required parallelogram.

6. Construct a parallelogram PQRS in which PQ = 6.5 cm, PR = 3.4 cm, and the altitude PL from P is 2.5 cm. Draw the altitude from R and measure it?

Solution:

Steps of Construction:
Given that a parallelogram PQRS in which PQ = 6.5 cm, PR = 3.4 cm, and the altitude PL from P is 2.5 cm.
1. Draw a line segment of length 6.5 cm and mark the ends as P and Q.
2. Take point P as a center and draw a perpendicular line PX by taking the radius 2.5 cm and mention that point as L. Draw a parallel line to PQ at a point L.
3. Next, take point P as a center and draw an arc by taking the radius of 3.4 cm. Mark the point as R where the point and arc cross each other. Join the points P and R, Q and R.
4. Take point R as a center and draw an arc by taking the radius of 6.5 cm. Mark the point as S where the point and arc cross each other. Join the points R and S, P and S.

PQRS is a required parallelogram.

7. Construct a parallelogram PQRS, in which diagonal PR = 3.8 cm, diagonal QS = 4.6 cm, and the angle between PR and QS is 60°.

Solution:

Steps of Construction:
Given that a parallelogram PQRS, in which diagonal PR = 3.8 cm, diagonal QS = 4.6 cm, and the angle between PR and QS is 60°.
1. Draw a line segment of length 3.8 cm and mark the ends as P and R.
2. Bisect the line PR and point it as O.
3. Next, take point O and make a point by taking the angle 60° by taking the O as a center.
4. Take point O as a center and draw an arc by taking the radius of 2.3 cm on both sides of O and name them as Q and S. Join the points PQ, QR, RS, SP.

PQRS is a required parallelogram.

8. Construct a rectangle PQRS whose adjacent sides are 11 cm and. 8.5 cm.

Solution:

Steps of Construction:
Given that a rectangle PQRS whose adjacent sides are 11 cm and. 8.5 cm.
1. Draw a line segment of length 11 cm and mark the ends as P and Q.
2. Take point P as a center and make a point by taking 90º using a protector and make the point as E.
Rectangle: All angles of a rectangle are 90º.
3. Next, take point P as a center and draw an arc by taking the radius 8.5 cm. Mark the point as S where the point and arc cross each other. Join the points S and P.
4. Take point Q as a center and draw an arc by taking the radius of 8.5 cm.
5. Also, take point S as a center and draw an arc by taking the radius of 11 cm.
Mark the point as R where the two arcs cross each other. Join the points S and R, Q and R.

PQRS is a required rectangle.

9. Construct a square, each of whose sides measures 5.4 cm.

Solution:

Steps of Construction:
Given that a square, each of whose sides measures 5.4 cm.
1. Draw a line segment of length 5.4 cm and mark the ends as P and Q.
2. Take point P as a center and make a point by taking 90º using a protector and make the point as E.
Square: All angles of a square are 90º.
3. Next, take point P as a center and draw an arc by taking the radius 5.4 cm. Mark the point as S where the point and arc cross each other. Join the points S and P.
4. Take point Q as a center and make a point by taking 90º using a protector. Take point Q as a center and draw an arc by taking the radius of 5.4 cm.
5. Also, take point S as a center and draw an arc by taking the radius of 5.4 cm. Mark the point as R where the two arcs cross each other. Join the points S and R, Q and R.

PQRS is a required square.

10. Construct a square, each of whose diagonals measures 5.6 cm.

Solution:

Steps of Construction:
Given that a square, each of whose diagonals measures 5.6 cm.
1. Draw a line segment of length 5.6 cm and mark the ends as P and R.
2. Bisect the line PR and take half of its radius 2.8 cm. Take point P as a center and draw an arc by taking the radius 2.8 cm.
3. Next, take point R as a center and draw an arc by taking the radius 2.8 cm. Mark the point as S where the point and arc cross each other. Join the points S and P, S and R.
4. Similarly, draw two arcs with 2.8 cms and make the point as Q.
5. Join the points P and Q, S and Q.

PQRS is a required square.

11. Construct a rectangle ABCD in which BD = 3.6 cm and diagonal AD = 6 cm. Measure the other side of the rectangle.

Solution:

Steps of Construction:
Given that a rectangle ABCD in which BD = 3.6 cm and diagonal AD = 6 cm.
1. Draw a line segment of length 3.6 cm and mark the ends as B and D.
2. Take point B as a center and make a point by taking 90º using a protector and make the point as E.
Rectangle: All angles of a rectangle are 90º.
3. Next, take point C as a center and draw an arc by taking the radius 6 cm. Mark the point as A where the point and arc cross each other. Join the points A and B, A and D.
4. Similarly, draw two arcs and make the point as C.
5. Join the points A and C, D and C.

ABCD is a required rectangle.

12. Construct a rhombus PQRS when the length measures of the diagonals are 8 cm and 6 cm.

Solution:

Steps of Construction:
Given that a rhombus PQRS when the length measures of the diagonals are 8 cm and 6 cm.
1. Draw a line segment of length 8 cm and mark the ends as P and R.
2. Draw perpendicular bisector XY of PR meeting PR at O.
3. Next, From O cut off OS = 1/2 × 6 cm = 3 cm along OX and OQ = 1/2 × 6 cm =3 cm along OY.
4. Join PQ, QR, RS, and SP.

PQRS is a required rhombus.

13. Construct a rhombus PQRS in which PQ = 4 cm and diagonal PR is 6.5 cm.

Solution:

Steps of Construction:
Given that a rhombus PQRS in which PQ = 4 cm and diagonal PR is 6.5 cm.
1. Draw a line segment of length 4 cm and mark the ends as P and Q.
2. Take the point Q as a center and draw an arc by taking the radius 4 cm.
3. Take the point P as a center and draw an arc by taking the radius 6.5 cm and name it as R.
4. Join PR and QR.
5. Take the point R as a center and draw an arc by taking the radius 4 cm.
6. Take the point P as a center and draw an arc by taking the radius 4 cm. Mark the point as S where the point and arc cross each other. Join the points S and P, S and R.

PQRS is a required rhombus.

14. Draw a rhombus ABCD whose side is 7.2 cm and one angle is 60°.

Solution:

Steps of Construction:
Given that a rhombus ABCD whose side is 7.2 cm and one angle is 60°.
1. Draw a line segment of length 7.2 cm and mark the ends as A and B.
2. Take point A as a center and make a point by taking 60º using a protector.
3. Next, take point A as a center and draw an arc by taking the radius 7.2 cm. Mark the point as D where the point and arc cross each other. Join the points D and A.
4. Take point D as a center and draw an arc by taking the radius 7.2 cm. Take point B as a center and draw an arc by taking the radius of 7.2 cm.
5. Mark the point as C where the two arcs cross each other. Join the points C and B, C and D.

ABCD is a required rhombus.

15. Construct a trapezium PQRS in which PQ = 6 cm, QR = 4 cm, RS = 3.2 cm, ∠Q = 75° and SR ∥ PQ.

Solution:

Steps of Construction:
Given that a trapezium PQRS in which PQ = 6 cm, QR = 4 cm, RS = 3.2 cm, ∠Q = 75° and SR ∥ PQ.
1. Draw a line segment of length 6 cm and mark the ends as P and Q.
2. Take point Q as a center and make a point by taking 75º using a protector.
3. Next, take point Q as a center and draw an arc by taking the radius 4 cm. Mark the point as R where the point and arc cross each other. Join the points Q and R.
4. RS || PQ, so angle Q + angle R = 180º, angle C = 105º as they are interior angles.
5. Take point R as a center and make a point by taking 105º using a protector.
6. Next, take point R as a center and draw an arc by taking the radius 3.2 cm. Mark the point as S where the point and arc cross each other. Join the points R and S, S and P.

PQRS is a required rhombus.

16. Draw a trapezium PQRS in which PQ ∥ SR, PQ = 7 cm, QR = 5 cm, PS = 6.5 cm and ∠Q = 60°.

Solution:

Steps of Construction:
Given that a trapezium PQRS in which PQ ∥ SR, PQ = 7 cm, QR = 5 cm, PS = 6.5 cm and ∠Q = 60°.
1. Draw a line segment of length 7 cm and mark the ends as P and Q.
2. Take point Q as a center and make a point by taking 60º using a protector.
3. Next, take point Q as a center and draw an arc by taking the radius 5 cm. Mark the point as R where the point and arc cross each other. Join the points Q and R.
4. RS || PQ, so angle Q + angle R = 180º, angle C = 120º as they are interior angles.
5. Take point R as a center and make a point by taking 120º using a protector.
6. Next, take point R as a center and draw an arc by taking the radius 6.5 cm. Mark the point as S where the point and arc cross each other. Join the points R and S, S and P.

PQRS is a required rhombus.

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We have provided various problems according to the new updated syllabus in the following sections. Solve all of them and check your answers to clear your doubts. For a better understanding of the concept, we have given the step by step explanation for each question.

Check out the below list to know the depth concepts available in Quadrilateral concepts.

### Solved Problems on Quadrilateral Worksheet

1. Fill in the blanks:
(i) A quadrilateral has …………… diagonals.
(ii) A quadrilateral has …………… angles.
(iii) How many sides present in a quadrilateral?
(iv) A quadrilateral has …………… vertices, no three of which are……………….
(v) A diagonal of a quadrilateral is a line segment that joins two ……………… vertices of the quadrilateral.
(vi) The sum of the angles of a quadrilateral is ……………….

Solution:

(i) Two
(ii) Four
(iii) Four
(iv) four, collinear
(v) Opposite
(vi) 360°

(i) How many pairs of opposite sides are there? Name them.
(ii) How many pairs of adjacent sides are there? Name them.
(iii) Also, find how many pairs of adjacent angles are there? Name them.
(iv) How many diagonals are there? Name them.
(v) How many pairs of opposite angles are there? Name them.

Solution:

(i) two; (PQ, SR), (PS, QR)
(ii) four; (PQ, QR), (QR, RS), (RS, SP), (SP, PQ)
(iii) four; (∠P, ∠Q), (∠Q, ∠R), (∠R, ∠S), (∠S, ∠P)
(iv) two; (PR, QS)
(v) two; (∠P, ∠R), (∠Q, ∠S)

3. Prove that the sum of the angles of a quadrilateral is 360°.

Solution:

• ∠PQR, ∠QRS, ∠RSP, and ∠SPQ are the internal angles.
• PR is a diagonal
• PR divides the quadrilateral into two triangles, ∆PQR and ∆PSR

We know that the sum of internal angles of a quadrilateral is 360°, that is, ∠PQR + ∠QRS + ∠RSP + ∠SPQ = 360°.

let’s prove that the sum of all the four angles of a quadrilateral is 360 degrees.
The sum of angles in a triangle is 180°. Now consider triangle PSR,
∠S + ∠SPR + ∠SRP = 180° (Sum of angles in a triangle)
Now consider triangle PQR,
∠Q + ∠QPR + ∠QRP = 180° (Sum of angles in a triangle)
On adding both the equations obtained above we have,
(∠S + ∠SPR + ∠SRP) + (∠Q + ∠QPR + ∠QRP) = 180° + 180°
∠S + (∠SPR + ∠QPR) + (∠QRP + ∠SRP) + ∠Q = 360°
We see that (∠SPR + ∠QPR) = ∠SPQ and (∠QRP + ∠SRP) = ∠QRS.
Replacing them we have, ∠S + ∠SPQ + ∠QRS + ∠Q = 360°
That is, ∠S + ∠P + ∠R + ∠Q = 360°.

Or, the sum of angles of a quadrilateral is 360°. This is the angle sum property of quadrilaterals.

4. The three angles of a quadrilateral are 74°, 56°, and 106°. Find the measure fourth angle?

Solution:

Given that the angles of a quadrilateral are 74°, 56°, and 106°.
The sum of the angles of a quadrilateral is 360°.
The quadrilateral consists of four angles.
Let the fourth angle is x.
x + 74° + 56° + 106° = 360°
x = 360° – 74° – 56° – 106°
x = 124°

The fourth angle is 124°.

5. The angles of a quadrilateral are in the ratio 2 : 4 : 6 : 8. Find the measure of each of these angles?

Solution:

Given that the angles of a quadrilateral are in the ratio 2 : 4 : 6 : 8.
Let the common ratio is x.
2x + 4x + 6x + 8x = 360°
20x = 360°
Divide the above equation by 20 on both sides.
20x/20 = 360°/20
x = 18°
2x = 2 . 18° = 36°
4x = 4 . 18° = 72°
6x = 6 . 18° = 108°
8x = 8 . 18° = 144°
The measure of each of these angles is 36°, 72°, 108°, 144°.

6. A quadrilateral has three acute angles, each measuring 70°. Find the measure of the fourth angle?

Solution:

Given that a quadrilateral has three acute angles, each measuring 70°.
Sum of four angles in any Quadrilateral =360°
Three angles each =70°
70° + 70° + 70° + fourth angle =360°
Fourth angle =360° – 210°
= 150°

Therefore, the fourth angle is 150°.

7. Three angles of a quadrilateral are equal and the measure of the fourth angle is 90°. Find the measure of each of the equal angles?

Solution:

Given that three angles of a quadrilateral are equal and the measure of the fourth angle is 90°.
Let each unknown angle is x. The sum of the angles of a quadrilateral are equal to 360°
x + x + x + 90° = 360°
3x + 90° = 360°
3x = 360° – 90°
3x = 270°
Divide the above equation by 3 on both sides.
3x/3 = 270°/3
x = 90°

The three angles are 90°, 90°, and 90°.

8. Two angles of a quadrilateral measure 65° and 55° respectively. The other two angles equal. Find the measure of each of these equal angles?

Solution:

Given that two angles of a quadrilateral measure 65° and 55° respectively. The other two angles equal.
The sum of the angles of a quadrilateral is equal to 360 degrees.
Let the other two angles be x.
So, x + x + 65° + 55° = 360°
2x + 120° = 360°
2x = 360 – 120° = 240°
x = 240°/2
x = 120 degrees

Hence, the Other two angles are 120 degrees each.

9. Two angles of a quadrilateral measure 45° and 75° respectively. The other two angles equal. Find the measure of each of these equal angles?

Solution:

Given that two angles of a quadrilateral measure 45° and 75° respectively. The other two angles equal.
The sum of the angles of a quadrilateral is equal to 360 degrees.
Let the other two angles be x.
So, x + x + 45° + 75° = 360°
2x + 120° = 360°
2x = 360 – 120° = 240°
x = 240°/2
x = 120 degrees

Hence, the Other two angles are 120 degrees each.

Worksheet on Construction on Quadrilateral is the first source for every student who wishes to learn complete concepts of Construction of Quadrilateral. You can see different problems along with answers, and explanations where you can find different processes to solve one problem. Practice all the problems and improve your preparation level easily. Quadrilateral Worksheet is included on our website with practice questions and solved problems.

1. Construct a quadrilateral PQRS in which PQ = 4.4 cm, QR = 6.2 cm, RS = 5.4 cm, SP = 5.2 cm And PR = 8.2 cm.

Solution:

Steps of Construction:
Firstly, draw a rough figure of the quadrilateral with the given dimensions.

1. Draw a line segment of length 4.4 cm and mark the ends as P and Q.
2. Take the point P as a center and draw an arc by taking the radius 8.2 cm.
3. Next, take point Q as a center and draw an arc by taking the radius 6.2 cm. Mark the point as R where the two arcs cross each other. Join the points Q and R as well as P and R.
4. By taking the point P as a center, draw an arc with a radius of 5.2 cm.
5. By taking the point R as a center, draw an arc with a radius of 5.4 cm.
6. Mark the point as S where the two arcs cross each other. Join the points R and S as well as P and S.

The final result is the required quadrilateral.

2. Construct a quadrilateral ABCD in which AB = 5.4 cm, BC = 4.6 cm, CD = 4.3 cm, DA = 3.5 cm and diagonal AC = 5.6 cm

Solution:

Steps of Construction:
Given that a quadrilateral ABCD in which AB = 5.4 cm, BC = 4.6 cm, CD = 4.3 cm, DA = 3.5 cm and diagonal AC = 5.6 cm
1. Draw a line segment of length 5.6 cm and mark the ends as A and C.
2. Take point A as a center and draw an arc by taking the radius 5.4 cm above the diagonal.
3. Next, take point C as a center and draw an arc by taking the radius 4.6 cm above the diagonal. Mark the point as B where the two arcs cross each other. Join the points A and B as well as B and C.
4. By taking point A as a center, draw an arc with a radius of 3.5 cm below the diagonal.
5. By taking the point C as a center, draw an arc with a radius of 4.3 cm below the diagonal.
6. Mark the point as D where the two arcs cross each other. Join the points A and D as well as C and D.

The final result is the required quadrilateral.

3. Construct a quadrilateral PQRS in which PQ = 3.5 cm, QR = 3.8 cm, RS = SP = 4.5 cm and diagonal QR = 5.6 cm.

Solution:

Steps of Construction:
Given that a quadrilateral PQRS in which PQ = 3.5 cm, QR = 3.8 cm, RS = SP = 4.5 cm and diagonal QR = 5.6 cm.
1. Draw a line segment of length 5.6 cm and mark the ends as S and R.
2. Take point S as a center and draw an arc by taking the radius 5.6 cm.
3. Next, take point R as a center and draw an arc by taking the radius 3.8 cm. Mark the point as Q where the two arcs cross each other. Join the points S and Q as well as R and Q.
4. By taking point S as a center, draw an arc with a radius of 4.5 cm.
5. By taking the point Q as a center, draw an arc with a radius of 3.5 cm.
6. Mark the point as P where the two arcs cross each other. Join the points Q and P as well as S and P.

The final result is the required quadrilateral.

4. Construct a quadrilateral PQRS in which PQ = 3.6 cm, QR = 3.3 cm, PS = 2.7 cm, diagonal PR = 4.6 cm and diagonal QS = 4 cm.

Solution:

Steps of Construction:
Given that a quadrilateral PQRS in which PQ = 3.5 cm, QR = 3.8 cm, RS = SP = 4.5 cm and diagonal QR = 5.6 cm.
1. Draw a line segment of length 3.6 cm and mark the ends as P and Q.
2. Take point P as a center and draw an arc by taking the radius 4.6 cm.
3. Next, take point Q as a center and draw an arc by taking the radius 3.3 cm. Mark the point as R where the two arcs cross each other. Join the points P and R as well as R and Q.
4. By taking point P as a center, draw an arc with a radius of 2.7 cm.
5. By taking the point Q as a center, draw an arc with a radius of 4 cm.
6. Mark the point as S where the two arcs cross each other. Join the points Q and S as well as S and P.

The final result is the required quadrilateral.

5. Construct a quadrilateral ABCD in which AC = AD = 6 cm, BC = 7.5 cm , BD = 10 cm and CD = 5 cm. Measure the remaining side.

Solution:

Steps of Construction:
Given that a quadrilateral ABCD in which AC = AD = 6 cm, BC = 7.5 cm , BD = 10 cm and CD = 5 cm. Measure the remaining side.
1. Draw a line segment of length 6 cm and mark the ends as A and D.
2. Take point A as a center and draw an arc by taking the radius 6 cm.
3. Next, take point D as a center and draw an arc by taking the radius 5 cm. Mark the point as C where the two arcs cross each other. Join the points D and C.
4. By taking point D as a center, draw an arc with a radius of 10 cm.
5. By taking the point C as a center, draw an arc with a radius of 7.5 cm.
6. Mark the point as B where the two arcs cross each other. Join the points B and C as well as B and D.

The final result is the required quadrilateral.

6. Construct a quadrilateral PQRS in which PQ = 3.4 cm, RS = 3 cm, SP = 5.7 cm, PR = 8 cm and QS = 4 cm.

Solution:

Steps of Construction:
Given that a quadrilateral PQRS in which PQ = 3.4 cm, RS = 3 cm, SP = 5.7 cm, PR = 8 cm and QS = 4 cm.
1. Draw a line segment of length 3.4 cm and mark the ends as P and Q.
2. Take the point Q as a center and draw an arc by taking the radius 4 cm.
3. Next, take point P as a center and draw an arc by taking the radius 5.7 cm. Mark the point as S where the two arcs cross each other. Join the points Q and S as well as P and S.
4. By taking the point P as a center, draw an arc with a radius of 8 cm.
5. By taking the point S as a center, draw an arc with a radius of 3 cm.
6. Mark the point as R where the two arcs cross each other. Join the points R and S as well as P and R.

The final result is the required quadrilateral.

7. Construct a quadrilateral PQRS in which PQ = QR = 3.5 cm, PS = RS = 5.2 cm and ∠PQR = 120°.

Solution:

Steps of Construction:
Given that PQ = QR = 3.5 cm, PS = RS = 5.2 cm and ∠PQR = 120°.
1. Draw a line segment of length 3.5 cm and mark the ends as P and Q.
2. Take point Q as a center and make a point by taking 120º using a protector.
3. Next, take point Q as a center and draw an arc with a radius of 3.5 cm. Mark the point as R where the two points are meet at a point. Join the points Q and R.
4. By taking the point P as a center, draw an arc with a radius of 5.2 cm.
5. By taking the point R as a center, draw an arc with a radius of 5.2 cm.
6. Mark the point as S where the two arcs cross each other. Join the points P and S, R and S.

The final result is the required quadrilateral.

8. Construct a quadrilateral PQRS in which PQ = 2.9 cm, QR = 3.2 cm, RS = 2.7 cm, SP = 3.4 cm and ∠P = 70°.

Solution:

Steps of Construction:
Given that a quadrilateral PQRS in which PQ = 2.9 cm, QR = 3.2 cm, RS = 2.7 cm, SP = 3.4 cm and ∠P = 70°.
1. Draw a line segment of length 2.9 cm and mark the ends as P and Q.
2. Take point P as a center and make a point by taking 70º using a protector.
3. Next, take point P as a center and draw an arc with a radius of 3.4 cm. Mark the point as S where the two points are meet at a point. Join the points P and S.
4. By taking the point S as a center, draw an arc with a radius of 2.7 cm.
5. By taking the point Q as a center, draw an arc with a radius of 3.2 cm.
6. Mark the point as R where the two arcs cross each other. Join the points Q and S, R and S.

The final result is the required quadrilateral.

9. Construct a quadrilateral PQRS in which PQ = 3.5 cm, QR = 5 cm, RS = 4.6 cm, ∠Q = 125° and ∠R = 60°.

Solution:

Steps of Construction:
Given that a quadrilateral PQRS in which PQ = 3.5 cm, QR = 5 cm, RS = 4.6 cm, ∠Q = 125° and ∠R = 60°.
1. Draw a line segment of length 5 cm and mark the ends as Q and R.
2. Take point Q as a center and make a point by taking 125º using a protector.
3. Next, take point R as a center and make a point by taking 60º using a protector.
4. By taking the point Q as a center, draw an arc with a radius of 3.5 cm.
5. Mark the point as P where the two points are meet at a point. Join the points Q and P.
6. By taking the point R as a center, draw an arc with a radius of 4.6 cm.
7. Mark the point as S where the point and arc cross each other. Join the points P and S.

The final result is the required quadrilateral.

10. Construct a quadrilateral ABCD in which AB = 6 cm, BC = 5.6 cm, CD = 2.7 cm, ∠B = 45° and ∠C = 90°.

Solution:

Steps of Construction:
Given that a quadrilateral ABCD in which AB = 6 cm, BC = 5.6 cm, CD = 2.7 cm, ∠B = 45° and ∠C = 90°.
1. Draw a line segment of length 6 cm and mark the ends as A and B.
2. Take point B as a center and make a point by taking 45º using a protector.
3. By taking the point B as a center, draw an arc with a radius of 5.6 cm.
5. Mark the point as C where the point and arc are meet at a point. Join the points B and C.
6. By taking the point C as a center, draw an arc with a radius of 2.7 cm.
7. Take point C as a center and make a point by taking 90º using a protector.
7. Mark the point as D where the point and arc cross each other. Join the points D and A, D and C.

The final result is the required quadrilateral.

11. Construct a quadrilateral PQRS in which PQ = 5.6 cm, QR = 4 cm, ∠P = 50°, ∠Q = 105° and ∠S = 80°.

Solution:

Steps of Construction:
Given that a quadrilateral PQRS in which PQ = 5.6 cm, QR = 4 cm, ∠P = 50°, ∠Q = 105° and ∠S = 80°.
We know that ∠P + ∠Q + ∠R + ∠S = 360°
So, 50° + 105° + ∠R + 80° = 360°
∠R = 125°
1. Draw a line segment of length 5.6 cm and mark the ends as P and Q.
2. Take point P as a center and make a point by taking 50º using a protector. Also, take point Q as a center and make a point by taking 150º using a protector.
3. By taking the point Q as a center, draw an arc with a radius of 4 cm.
5. Mark the point as R where the point and arc are meet at a point. Join the points Q and R.
6. By taking the point R as a center, make a point by taking 125º using a protector.
7. Mark the point as S where the point and arc cross each other. Join the points S and P, S and R.

The final result is the required quadrilateral.

12. Construct a quadrilateral ABCD in which AB = 5 cm, BC = 6.5 cm, ∠A = ∠C = 100° and ∠D = 75°.

Solution:

Steps of Construction:
Given that a quadrilateral ABCD in which AB = 5 cm, BC = 6.5 cm, ∠A = ∠C = 100° and ∠D = 75°.
We know that ∠A + ∠B + ∠C + ∠D = 360°
So, 100° + ∠B + 100° + 75° = 360°
∠B = 125°
1. Draw a line segment of length 5 cm and mark the ends as A and B.
2. Take point A as a center and make a point by taking 100º using a protector. Also, take point B as a center and make a point by taking 125º using a protector.
3. By taking the point B as a center, draw an arc with a radius of 6.5 cm.
5. Mark the point as C where the point and arc are meet at a point. Join the points B and C.
6. By taking the point C as a center, make a point by taking 100º using a protector.
7. Mark the point as D where the point and arc cross each other. Join the points D and A, D and C.

The final result is the required quadrilateral.

13. Construct a quadrilateral PQRS in which PQ = 4 cm, PR = 5 cm, PS = 5.5 cm and ∠PQR = ∠PRS = 90°.

Solution:

Steps of Construction:
Given that a quadrilateral PQRS in which PQ = 4 cm, PR = 5 cm, PS = 5.5 cm and ∠PQR = ∠PRS = 90°.
1. Draw a line segment of length 4 cm and mark the ends as P and Q.
2. Take point Q as a center and make a point by taking 90º using a protector. Also, take point P as a center and make a point by taking 90º using a protector.
3. By taking the point P as a center, draw an arc with a radius of 5 cm.
5. Mark the point as R where the point and arc are meet at a point. Join the points Q and R.
6. By taking the point P as a center, draw an arc with a radius of 5.5 cm.
7. Mark the point as S where the point and arc cross each other. Join the points S and P, S and R.

The final result is the required quadrilateral.

## Worksheet on Parallelogram | Parallelogram Questions and Answers

Check the problems available on the given Worksheet on Parallelogram. We have included different problems according to the new updated syllabus. Improve your preparation level by solving every problem available here. Access all the Parallelogram problems for free of cost on our website. A clear explanation is provided for every problem here. You can assess your preparation standard using the Questions on Parallelogram available and learn the methods on how to solve related problems.

## Questions on Parallelogram

1. PQRS is a parallelogram in which ∠P = 110°. Find the measure of each of the angles ∠Q, ∠R, and ∠S.

Solution:

In a parallelogram PQRS, if ∠P = 110°, then ∠R = 110°.
The Sum of adjacent angles is supplementary in a parallelogram.
∠P + ∠Q = 180º.
Substitute ∠P value in the above equation.
110º + ∠Q = 180º
∠Q = 180º – 110º
∠Q = 70º
As the Opposite angles are equal in a parallelogram,
∠Q = ∠S = 70º

The final Answer is ∠Q = ∠S = 70º, ∠R = 110°.

2. Two adjacent angles of a parallelogram are equal. What is the measure of each of these angles?

Solution:

We know that Two adjacent angles of a parallelogram are equal.
Let the angle be m.
Adjacent angles of a parallelogram = 180
m + m = 180
2m = 180
m = 180/2
m = 90.

The final answer is every angle is 90º.

3. The ratio of two sides of a parallelogram is 10 : 8. If its perimeter is 108 cm, find the lengths of its sides?

Solution:

Let the lengths of two sides of the parallelogram be 10a cm and 8a cm respectively.
Find the perimeter using given values.
Then, its perimeter = 2(10a + 8a) cm = 2 (18a) cm = 36a cm.
Therefore, 36a = 108 ⇔ a = 108/36 = 3.

Therefore, one side = (10 × 3) cm = 30 cm and other side = (8 × 3) cm = 24 cm.

4. Two adjacent angles of a parallelogram are (6x + 15)° and (2x – 5)°. What is the value of x?

Solution:

The sum of the adjacent angles of parallelogram = 180
6x + 15 + 2x – 5 = 180
8x + 10 = 180
8x = 180 – 10
8x = 170
x = 170/8

The final answer is x = 21.25.

5. The length and breadth of a rectangle are in the ratio 16 : 12. If the diagonal measures 100 cm. Find the perimeter of a rectangle?

Solution:

Let m be the common multiple.
Length = 16m
According to Pythagoras theorem,
(16m)² + (12m)²=(100)²
256m²+ 144m² = 10,000
400m² = 10,000
m² = 10,000/400
m = 25
So, Length = 16m = 400 cm
Breadth = 12m = 300 cm
Perimeter = 2 (l × b)
= 2 (400 + 300)
= 1400 cm

So, perimeter of rectangle is 1400 cm.

6. A rhombus has diagonals of 32 cm and 24 cm. Find the length of each side?

Solution:

Given that One diagonal is 32 and another 24 then half of both is 16 and 12. Diagonal of a rhombus bisect at 90º.
By pythogaurus theorem
h² = (16)² + (12)²
h² = 256 + 144 = 400
h = √400 = 20

Side = 20.

7. The sum of two opposite angles of a parallelogram PQRS is 130º find the measure of each of its angles.

Solution:

Given that the sum of two opposite angles of a parallelogram PQRS is 130º.
Let the angle be a. So, because the sum of two opposite angles is 130º.
a + a = 130
2a = 130
a = 65.
So, Again assume that angle is a
65 + 65 + a + a = 360
2a = 360 – 130
2a = 230
a = 115
So, 1st angle is 65
2nd angle is 65
3rd angle is 115
4th angle is 115.

8. In the adjacent figure, PQRS is a parallelogram and line segments PT and RV bisect the angles P and R respectively. Show that PT ∥ RV.

Solution:

In triangles PST and RQV, we have PS = QR, ∠Q = ∠S, and ∠SPT = ∠QRV.
[Since, ∠P = ∠R = ¹/₂∠P = ¹/₂∠R, ie., ∠SPT = ∠QRV]
Therefore, ∆ PST ≅ ∆RQV. And therefore, RS – ST = PQ – QV.
So, RT = PV.
Therefore, PTQV is a parallelogram.

Hence, PT ∥ RV.

9. The sides of a rectangle are in the ratio 5 : 4 and its perimeter is 90 cm. Find its length and breadth.

Solution:

The sides of a rectangle are in the ratio 5 : 4
Let the ratio be m.
so, Length = 5m
Write the formula of the Perimeter of a rectangle and substitute the values in it.
Perimeter of rectangle = 2 (l + b) = 2(5m + 4m) = 2(9m) = 18m
Now we are given that its perimeter is 90 cm
So, 2 (9m) = 18m
18m = 90
m = 90/18 = 5
So, Length = 5m = 5 . 5 = 25
Breadth = 4m = 4 . 5 = 20

Hence its length and breadth are 25 cm and 20 cm respectively.

10. The perimeter of a parallelogram is 120 cm. If one of the sides is longer than the other by 10 cm, find the length of each of its a side

Solution:

Let the first side is a
then the second side becomes (a + 10)
According to the question 2(a + a + 10) = 120
2(2a + 10) = 120
2a + 10 = 60
2a = 60 – 10
2a = 50
a = 25

First side = a = 25cm
second side = a + 10 = 35cm

11. Name each of the following parallelograms.
(i) Name the parallelogram its diagonals are equal and the adjacent sides are unequal.
(ii) The diagonals are equal and the adjacent sides are equal.
(iii) Name the parallelogram which consists of all the sides that are equal and one angle is 90°.
(iv) All the sides are equal and one angle is 60°.
(v) The diagonals are unequal and the adjacent sides are equal.
(vi) All the angles are equal and the adjacent sides are unequal.

Solution:

(i) rectangle
(ii) square
(iii) square
(iv) rhombus
(v) rhombus
(vi) rectangle

12. State the below statements are true or false.

(i) The diagonals of a parallelogram are equal.
(ii) Every rhombus is a kite.
(iii) Every rectangle is a square.
(iv) The diagonals of a rectangle are perpendicular to each other.
(v) Every square is a rhombus.
(vi) Every square is a parallelogram.
(vii) The diagonals of a rhombus are equal.
(viii) Every rectangle is a parallelogram.
(ix) Every parallelogram is a rectangle.
(x) Check: Every rhombus is a parallelogram.

Solution:

(i) False
(ii) False
(iii) False
(iv) False
(v) True
(vi) True
(vii) False
(viii) True
(ix) False
(x) True

## Worksheet on Factoring Quadratic Trinomials | Factoring Trinomials Worksheet with Answers

The best resource to learn Factorization of Quadratic Trinomials Problems is Worksheet on Factoring Quadratic Trinomials. We have given different problems according to the updated syllabus on Factoring and Solving Quadratic Equations Worksheets. Solve all problems to get a complete grip on the Factorization of Quadratic Trinomials problems. Also, to learn complete factorization problems, check Factorization Worksheets, and improve your preparation level.

(i) a2 + 5a + 6
(ii) a2 + 10a + 24
(iii) a2 + 12a + 27
(iv) a2 + 15a + 56
(v) a2 + 19a + 60
(vi) a2 + 13a + 40
(vii) a2 – 10a + 24
(viii) a2 – 23a + 42
(ix) a2 – 17a + 16
(x) a2 – 21a + 90

Solution:

(i) The Given expression is a2 + 5a + 6.
By comparing the given expression a2 + 5a + 6 with the basic expression x^2 + ax + b.
Here, a = 1, b = 5, and c = 6.
The sum of two numbers is m + n = b = 5 = 3 + 2.
The product of two number is m * n = a * c = 1 * (6) = 6 = 3 * 2
From the above two instructions, we can write the values of two numbers m and n as 3 and 2.
Then, a2 + 5a + 6 = a2 + 3a + 2a + 6.
= a (a+ 3) + 2(a + 3).
Factor out the common terms.
(a + 3) (a + 2)

Then, a2 + 5a + 6  = (a + 3) (a + 2).

(ii) The Given expression is a2 + 10a + 24.
By comparing the given expression a2 + 5a + 6 with the basic expression x^2 + ax + b.
Here, a = 1, b = 10, and c = 24.
The sum of two numbers is m + n = b = 10 = 6 + 4.
The product of two number is m * n = a * c = 1 * (24) = 24 = 6 * 4
From the above two instructions, we can write the values of two numbers m and n as 6 and 4.
Then, a2 + 10a + 24 = a2 + 6a + 4a + 24.
= a (a+ 6) + 4(a + 6).
Factor out the common terms.
(a + 6) (a + 4)

Then, a2 + 10a + 24  = (a + 6) (a + 4).

(iii) The Given expression is a2 + 12a + 27.
By comparing the given expression a2 + 12a + 27 with the basic expression x^2 + ax + b.
Here, a = 1, b = 12, and c = 27.
The sum of two numbers is m + n = b = 12 = 9 + 3.
The product of two number is m * n = a * c = 1 * (27) = 27 = 9 * 3
From the above two instructions, we can write the values of two numbers m and n as 9 and 3.
Then, a2 + 12a + 27 = a2 + 9a + 3a + 27.
= a (a+ 9) + 3(a + 9).
Factor out the common terms.
(a + 9) (a + 3)

Then, a2 + 12a + 27  = (a + 9) (a + 3).

(iv) The Given expression is a2 + 15a + 56.
By comparing the given expression a2 + 15a + 56 with the basic expression x^2 + ax + b.
Here, a = 1, b = 15, and c = 56.
The sum of two numbers is m + n = b = 15 = 8 + 7.
The product of two number is m * n = a * c = 1 * (56) = 56 = 8 * 7
From the above two instructions, we can write the values of two numbers m and n as 8 and 7.
Then, a2 + 15a + 56 = a2 + 8a + 7a + 56.
= a (a+ 8) + 7(a + 8).
Factor out the common terms.
(a + 8) (a + 7)

Then, a2 + 15a + 56  = (a + 8) (a + 7).

(v) The Given expression is a2 + 19a + 60.
By comparing the given expression a2 + 19a + 60 with the basic expression x^2 + ax + b.
Here, a = 1, b = 19, and c = 60.
The sum of two numbers is m + n = b = 19 = 15 + 4.
The product of two number is m * n = a * c = 1 * (60) = 60 = 15 * 4
From the above two instructions, we can write the values of two numbers m and n as 15 and 4.
Then, a2 + 19a + 60 = a2 + 15a + 4a + 60.
= a (a+ 15) + 4(a + 15).
Factor out the common terms.
(a + 15) (a + 4)

Then, a2 + 19a + 60 = (a + 15) (a + 4).

(vi) The Given expression is a2 + 13a + 40.
By comparing the given expression a2 + 13a + 40 with the basic expression x^2 + ax + b.
Here, a = 1, b = 13, and c = 40.
The sum of two numbers is m + n = b = 13 = 8 + 5.
The product of two number is m * n = a * c = 1 * (40) = 40 = 8 * 5
From the above two instructions, we can write the values of two numbers m and n as 8 and 5.
Then, a2 + 13a + 40 = a2 + 8a + 5a + 40.
= a (a + 8) + 5(a + 8).
Factor out the common terms.
(a + 8) (a + 5)

Then, a2 + 13a + 40 = (a + 8) (a + 5).

(vii) The Given expression is a2 – 10a + 24.
By comparing the given expression a2 – 10a + 24 with the basic expression x^2 + ax + b.
Here, a = 1, b = -10, and c = 24.
The sum of two numbers is m + n = b = -10 = -6 – 4.
The product of two number is m * n = a * c = 1 * (24) = 24 = -6 * -4
From the above two instructions, we can write the values of two numbers m and n as -6 and -4.
Then, a2 – 10a + 24 = a2 – 6a – 4a + 24.
= a (a – 6) – 4(a – 6).
Factor out the common terms.
(a – 6) (a – 4)

Then, a2 – 10a + 24 = (a – 6) (a – 4).

(viii) The Given expression is a2 – 23a + 42.
By comparing the given expression a2 – 23a + 42 with the basic expression x^2 + ax + b.
Here, a = 1, b = -23, and c = 42.
The sum of two numbers is m + n = b = -23 = -21 – 2.
The product of two number is m * n = a * c = 1 * (42) = 42 = -21 * -2
From the above two instructions, we can write the values of two numbers m and n as -21 and -2.
Then, a2 – 23a + 42 = a2 – 21a – 2a + 42.
= a (a – 21) – 2(a – 21).
Factor out the common terms.
(a – 21) (a – 2)

Then, a2 – 23a + 42 = (a – 21) (a – 2).

(ix) The Given expression is a2 – 17a + 16.
By comparing the given expression a2 – 17a + 16 with the basic expression x^2 + ax + b.
Here, a = 1, b = -17, and c = 16.
The sum of two numbers is m + n = b = -17 = -16 – 1.
The product of two number is m * n = a * c = 1 * (16) = 16 = -16 * -1
From the above two instructions, we can write the values of two numbers m and n as -16 and -1.
Then, a2 – 17a + 16 = a2 – 16a -a + 16.
= a (a – 16) – 1(a – 16).
Factor out the common terms.
(a – 16) (a – 1)

Then, a2 – 17a + 16 = (a – 16) (a – 1).

(x) The Given expression is a2 – 21a + 90.
By comparing the given expression a2 – 21a + 90 with the basic expression x^2 + ax + b.
Here, a = 1, b = -21, and c = 90.
The sum of two numbers is m + n = b = -21 = -15 – 6.
The product of two number is m * n = a * c = 1 * (90) = 90 = -15 * -6
From the above two instructions, we can write the values of two numbers m and n as -15 and -6.
Then, a2 – 21a + 90 = a2 – 15a – 6a + 90.
= a (a – 15) – 6(a – 15).
Factor out the common terms.
(a – 15) (a – 6)

Then, a2 – 21a + 90 = (a – 15) (a – 6).

2. Factorize the expressions completely

(i) a2 – 22a + 117
(ii) a2 – 9a + 20
(iii) a2 + a – 132
(iv) a2 + 5a – 104
(v) b2 + 7b – 144
(vi) c2 + 19c – 150
(vii) b2 + b – 72
(viii) a2 + 6a – 91
(ix) a2 – 4a -77
(x) a2 – 6a – 135

Solution:

(i) The Given expression is a2 – 22a + 117.
By comparing the given expression a2 – 22a + 117 with the basic expression x^2 + ax + b.
Here, a = 1, b = -22, and c = 117.
The sum of two numbers is m + n = b = -22 = -13 – 9.
The product of two number is m * n = a * c = 1 * (117) = 117 = -13 * -9
From the above two instructions, we can write the values of two numbers m and n as -13 and -9.
Then, a2 – 22a + 117 = a2 – 13a – 9a + 117.
= a (a – 13) – 9(a – 13).
Factor out the common terms.
(a – 13) (a – 9)

Then, a2 – 22a + 117 = (a – 13) (a – 9).

(ii) The Given expression is a2 – 9a + 20.
By comparing the given expression a2 – 9a + 20 with the basic expression x^2 + ax + b.
Here, a = 1, b = -9, and c = 20.
The sum of two numbers is m + n = b = -9 = -5 – 4.
The product of two number is m * n = a * c = 1 * (20) = 20 = -5 * -4
From the above two instructions, we can write the values of two numbers m and n as -5 and -4.
Then, a2 – 9a + 20 = a2 – 5a -4a + 20.
= a (a – 5) – 4(a – 5).
Factor out the common terms.
(a – 5) (a – 4)

Then, a2 – 9a + 20 = (a – 5) (a – 4).

(iii) The Given expression is a2 + a – 132.
By comparing the given expression a2 + a – 132 with the basic expression x^2 + ax + b.
Here, a = 1, b = 1, and c = -132.
The sum of two numbers is m + n = b = 1 = 12 – 11.
The product of two number is m * n = a * c = 1 * (-132) = -132 = 12 * -11
From the above two instructions, we can write the values of two numbers m and n as 12 and -11.
Then, a2 + a – 132 = a2 + 12a – 11a – 132.
= a (a + 12) – 11(a + 12).
Factor out the common terms.
(a + 12) (a – 11)

Then, a2 + a – 132 = (a + 12) (a – 11).

(iv) The Given expression is a2 + 5a – 104.
By comparing the given expression a2 + 5a – 104 with the basic expression x^2 + ax + b.
Here, a = 1, b = 5, and c = -104.
The sum of two numbers is m + n = b = 5 = 13 – 8.
The product of two number is m * n = a * c = 1 * (-104) = -104 = 13 * -8
From the above two instructions, we can write the values of two numbers m and n as 13 and -8.
Then, a2 + 5a – 104 = a2 + 13a – 8a – 104.
= a (a + 13) – 8(a + 13).
Factor out the common terms.
(a + 13) (a – 8)

Then, a2 + 5a – 104 = (a + 13) (a – 8).

(v) The Given expression is b2 + 7b – 144.
By comparing the given expression b2 + 7b – 144 with the basic expression x^2 + ax + b.
Here, a = 1, b = 7, and c = -144.
The sum of two numbers is m + n = b = 7 = 16 – 9.
The product of two number is m * n = a * c = 1 * (-144) = -144 = 16 * -9
From the above two instructions, we can write the values of two numbers m and n as 16 and -9.
Then, b2 + 7b – 144 = b2 + 16b – 9b – 144.
= b (b + 16) – 9(b + 16).
Factor out the common terms.
(b + 16) (b – 9)

Then, b2 + 7b – 144 = (b + 16) (b – 9).

(vi) The Given expression is c2 + 19c – 150.
By comparing the given expression c2 + 19c – 150 with the basic expression x^2 + ax + b.
Here, a = 1, b = 19, and c = -150.
The sum of two numbers is m + n = b = 19 = 25 – 6.
The product of two number is m * n = a * c = 1 * (-150) = -150 = 25 * -6
From the above two instructions, we can write the values of two numbers m and n as 25 and -6.
Then, c2 + 19c – 150 = c2 + 25c – 6c – 150.
= c (c + 25) – 6(c + 25).
Factor out the common terms.
(c + 25) (c – 6)

Then, c2 + 19c – 150 = (c + 25) (c – 6).

(vii) The Given expression is b2 + b – 72.
By comparing the given expression b2 + b – 72 with the basic expression x^2 + ax + b.
Here, a = 1, b = 1, and c = -72.
The sum of two numbers is m + n = b = 1 =  9 – 8.
The product of two number is m * n = a * c = 1 * (-72) = -72 = 9 * -8
From the above two instructions, we can write the values of two numbers m and n as 9 and -8.
Then, b2 + b – 72 = b2 + 9b – 8b – 72.
= b (b + 9) – 8(b + 9).
Factor out the common terms.
(b + 9) (b – 8)

Then, b2 + b – 72 = (b + 9) (b – 8).

(viii) The Given expression is a2 + 6a – 91.
By comparing the given expression a2 + 6a – 91 with the basic expression x^2 + ax + b.
Here, a = 1, b = 6, and c = -91.
The sum of two numbers is m + n = b = 6 =  13 – 7.
The product of two number is m * n = a * c = 1 * (-91) = -91 = 13 * -7
From the above two instructions, we can write the values of two numbers m and n as 13 and -7.
Then, a2 + 6a – 91 = a2 + 13a – 7a – 91.
= a (a + 13) – 7(a + 13).
Factor out the common terms.
(a + 13) (a – 7)

Then, a2 + 6a – 91 = (a + 13) (a – 7).

(ix) The Given expression is a2 – 4a -77.
By comparing the given expression a2 – 4a -77 with the basic expression x^2 + ax + b.
Here, a = 1, b = -4, and c = -77.
The sum of two numbers is m + n = b = -4 =  -11 + 7.
The product of two number is m * n = a * c = 1 * (-77) = -77 = -11 * 7
From the above two instructions, we can write the values of two numbers m and n as -11 and 7.
Then, a2 – 4a -77 = a2 – 11a + 7a -77.
= a (a – 11) + 7(a – 11).
Factor out the common terms.
(a – 11) (a + 7)

Then, a2 – 4a -77 = (a – 11) (a + 7).

(x) The Given expression is a2 – 6a – 135.
By comparing the given expression a2 – 6a – 135 with the basic expression x^2 + ax + b.
Here, a = 1, b = -6, and c = -135.
The sum of two numbers is m + n = b = -6 =  -15 + 9.
The product of two number is m * n = a * c = 1 * (-135) = -135 = -15 * 9
From the above two instructions, we can write the values of two numbers m and n as -15 and 9.
Then, a2 – 6a – 135 = a2 – 15a + 9a – 135.
= a (a – 15) + 9(a – 15).
Factor out the common terms.
(a – 15) (a + 9)

Then, a2 – 6a – 135 = (a – 15) (a + 9).

3. Factor by splitting the middle term

(i) a2 – 11a – 42
(ii) a2 – 12a – 45
(iii) a2 – 7a – 30
(iv) a2 – 5a – 24
(v) 3a2 + 10a + 8
(vi) 3a2 + 14a + 8
(vii) 2a2 + a – 45
(viii) 6a2 + 11a – 10
(ix) 3a2 – 10a + 8
(x) 2a2 – 17a – 30

Solution:

(i) The Given expression is a2 – 11a – 42.
By comparing the given expression a2 – 11a – 42 with the basic expression x^2 + ax + b.
Here, a = 1, b = -11, and c = -42.
The sum of two numbers is m + n = b = -11 =  -14 + 3.
The product of two number is m * n = a * c = 1 * (-42) = -42 = -14 * 3
From the above two instructions, we can write the values of two numbers m and n as -14 and 3.
Then, a2 – 11a – 42 = a2 – 14a + 3a- 42.
= a (a – 14) + 3(a – 14).
Factor out the common terms.
(a – 14) (a + 3)

Then, a2 – 11a – 42 = (a – 14) (a + 3).

(ii) The Given expression is a2 – 12a – 45.
By comparing the given expression a2 – 12a – 45 with the basic expression x^2 + ax + b.
Here, a = 1, b = -12, and c = -45.
The sum of two numbers is m + n = b = -12 =  -15 + 3.
The product of two number is m * n = a * c = 1 * (-45) = -45 = -15 * 3
From the above two instructions, we can write the values of two numbers m and n as -15 and 3.
Then, a2 – 12a – 45 = a2 – 15a + 3a- 45.
= a (a – 15) + 3(a – 15).
Factor out the common terms.
(a – 15) (a + 3)

Then, a2 – 12a – 45 = (a – 15) (a + 3).

(iii) The Given expression is a2 – 7a – 30.
By comparing the given expression a2 – 7a – 30 with the basic expression x^2 + ax + b.
Here, a = 1, b = -7, and c = -30.
The sum of two numbers is m + n = b = -7 =  -10 + 3.
The product of two number is m * n = a * c = 1 * (-30) = -30 = -10 * 3
From the above two instructions, we can write the values of two numbers m and n as -10 and 3.
Then, a2 – 7a – 30 = a2 – 10a + 3a – 30.
= a (a – 10) + 3(a – 10).
Factor out the common terms.
(a – 10) (a + 3)

Then, a2 – 7a – 30 = (a – 10) (a + 3).

(iv) The Given expression is a2 – 5a – 24.
By comparing the given expression a2 – 5a – 24 with the basic expression x^2 + ax + b.
Here, a = 1, b = -5, and c = -24.
The sum of two numbers is m + n = b = -5 =  -8 + 3.
The product of two number is m * n = a * c = 1 * (-24) = -24 = -8 * 3
From the above two instructions, we can write the values of two numbers m and n as -8 and 3.
Then, a2 – 5a – 24 = a2 – 8a + 3a – 24.
= a (a – 8) + 3(a – 8).
Factor out the common terms.
(a – 8) (a + 3)

Then, a2 – 5a – 24 = (a – 8) (a + 3).

(v) The Given expression is 3a2 + 10a + 8.
By comparing the given expression 3a2 + 10a + 8 with the basic expression ax^2 + bx + c.
Here, a = 3, b = 10, and c = 8.
The sum of two numbers is m + n = b = 10 =  6 + 4.
The product of two number is m * n = a * c = 3 * (8) = 24 = 6 * 4
From the above two instructions, we can write the values of two numbers m and n as 6 and 4.
Then, 3a2 + 10a + 8 = 3a2 + 6a + 4a + 8.
= 3a (a + 2) + 4(a + 2).
Factor out the common terms.
(a + 2) (3a + 4)

Then, 3a2 + 10a + 8 = (a + 2) (3a + 4).

(vi) The Given expression is 3a2 + 14a + 8.
By comparing the given expression 3a2 + 14a + 8 with the basic expression ax^2 + bx + b.
Here, a = 3, b = 14, and c = 8.
The sum of two numbers is m + n = b = 14 = 12 + 2.
The product of two number is m * n = a * c = 3 * (8) = 24 = 8 * 3
From the above two instructions, we can write the values of two numbers m and n as 8 and 3.
Then, 3a2 + 14a + 8 = 3a2 + 8a + 3a + 8.
= a (3a + 8) + 1 (3a + 8).
Factor out the common terms.
(3a + 8) (a + 1)

Then, 3a2 + 14a + 8 = (3a + 8) (a + 1).

(vii) The Given expression is 2a2 + a – 45.
By comparing the given expression 2a2 + a – 45 with the basic expression ax^2 + bx + b.
Here, a = 2, b = 1, and c = -45.
The sum of two numbers is m + n = b = 1 = 10 – 9.
The product of two number is m * n = a * c = 2 * (-45) = -90 = 10 * -9
From the above two instructions, we can write the values of two numbers m and n as 10 and -9.
Then, 2a2 + a – 45 = 2a2 + 10a – 9a – 45.
= 2a (a + 5) – 9 (a + 5).
Factor out the common terms.
(a + 5) (2a – 9)

Then, 2a2 + a – 45 = (a + 5) (2a – 9).

(viii) The Given expression is 6a2 + 11a – 10.
By comparing the given expression 6a2 + 11a – 10 with the basic expression ax^2 + bx + b.
Here, a = 6, b = 11, and c = -10.
The sum of two numbers is m + n = b = 11 = 15 – 4.
The product of two number is m * n = a * c = 6 * (-10) = -60 = 15 * -4
From the above two instructions, we can write the values of two numbers m and n as 15 and -4.
Then, 6a2 + 11a – 10 = 6a2 + 15a – 4a – 10.
= 3a (2a + 5) – 2 (2a + 5).
Factor out the common terms.
(2a + 5) (3a – 2)

Then, 6a2 + 11a – 10 = (2a + 5) (3a – 2).

(ix) The Given expression is 3a2 – 10a + 8.
By comparing the given expression 3a2 – 10a + 8 with the basic expression ax^2 + bx + b.
Here, a = 3, b = -10, and c = 8.
The sum of two numbers is m + n = b = -10 = -6 – 4.
The product of two number is m * n = a * c = 3 * (8) = 24 = -6 * -4
From the above two instructions, we can write the values of two numbers m and n as -6 and -4.
Then, 3a2 – 10a + 8 = 3a2 – 6a – 4a + 8.
= 3a (a – 2) – 4 (a – 2).
Factor out the common terms.
(a – 2) (3a – 4)

Then, 3a2 – 10a + 8 = (a – 2) (3a – 4).

(x) The Given expression is 2a2 – 17a – 30.
By comparing the given expression 2a2 – 17a – 30 with the basic expression ax^2 + bx + b.
Here, a = 2, b = -17, and c = -30.
The sum of two numbers is m + n = b = -17 = -20 + 3.
The product of two number is m * n = a * c = 2 * (-30) = -60 = -20 * 3
From the above two instructions, we can write the values of two numbers m and n as -20 and 3.
Then, 2a2 – 17a – 30 = 2a2 – 20a + 3a – 30.
= 2a (a – 10) + 3 (a – 10).
Factor out the common terms.
(a – 10) (2a + 3)

Then, 2a2 – 17a – 30 = (a – 10) (2a + 3).

## Worksheet on Factoring Trinomials by Substitution | Solving Trinomials Worksheets

For a clear understanding of the factorization trinomials by substitution, take the help of our worksheets. Follow the questions on Worksheet on Factoring Trinomials by Substitution for reference and grade up your skills level on the concept. Clear all your doubts on the factorization concept by following the below-solved examples of factoring trinomials by substitution.

We know the factorization process for x^2 + ax + b or ax^2 + bx + c. But, for the trinomial expression, we need to substitute the terms in the place of common factors. So that, we will get the expression in the form of x^2 + ax + b or ax^2 + bx + c. Check Factorization Worksheets to understand the complete factorization concept.

## Solved Examples of Factoring Trinomials by Substitution

1. Factor the following trinomials using the substitution method

(i) 4(x – y)^2 – 14(x – y) – 8.
(ii) 3(3a + 2)^2 + 5(3a + 2) – 2.
(iii) 2(x + 2y)^2 + (x + 2y) – 1.
(iv) (x + y)^2 – (x + y) – 6.
(v) (a2 – 3a)^2 – 38(a^2 -3a) – 80.
(vi) 6(a – b)^2 – a + b – 15.
(vii) (x^2 – 3y^2)^2 – 16(x^2 – 3y^2) + 63.
(viii) (x^2 + 2x)^2 – 22(x^2 + 2x) + 72.
(ix) (a^2 – 8a)^2 – 29(a^2 – 8a) + 180.
(x) (p + q)^2 – 8p – 8q + 7.

Solution:

(i) The given expression is 4(x – y)^2 – 14(x – y) – 8.
Replace the (x – y) with p. That is 4p^2 – 14p – 8.
The above expression matches with the basic expression ax^2 + bx + c.
Here, a = 4, b = – 14, c = – 8.
a*c = 4 * ( – 8) = – 32 and b = – 14.
So, – 32 = – 16 * 2 and – 14 = – 16 + 2.
Then, 4p^2 – 16 p + 2p – 8 = 4p(p – 4) + 2(p – 4).
Factor out the common terms from the above expression. That is,
(p – 4) (4p + 2).
Now, replace the term p with the (x – y). That is,
(x – y – 4) (4(x – y) + 2) = (x – y – 4) 2(2x – 2y + 1).

Then, 4(x – y)^2 – 14(x – y) – 8 is equal to 2(x – y – 4) (2x – 2y + 1).

(ii) The given expression is 3(3a + 2)^2 + 5(3a + 2) – 2.
Replace the (3a + 2) with p. That is,
3p^2 + 5p – 2.
The above expression matches with the basic expression ax^2 + bx + c.
Here, a = 3, b = 5, c = – 2.
a*c = 3 * ( – 2) = – 6 and b = 5.
So, – 6 = 6 * (- 1) and 5= 6 – 1.
Then, 3p^2 + 5p – 2 = 3p^2 + 6p – p – 2.
= 3p(p + 2) – (p + 2).
Factor out the common terms from the above expression. That is,
(3p – 1)(p + 2).
Now, replace the term p with the (3a + 2). That is,
(3(3a + 2) – 1) (3a + 2 + 2) = (9a + 6 – 1) (3a + 4)
= (9a + 5) (3a + 4)

Then, 3(3a + 2)^2 + 5(3a + 2) – 2 is equal to (9a + 5) (3a + 4).

(iii) The given expression is 2(x + 2y)^2 + (x + 2y) – 1.
Replace the (x + 2y) with p. That is,
2p^2 + p – 1.
The above expression matches with the basic expression ax^2 + bx + c.
Here, a = 2, b = 1, c = – 1.
a*c = 2 * ( – 1) = – 2 and b = 1.
So, – 2 = 2 * ( – 1) and 1 = 2 – 1.
Then, 2p^2 + p – 1 = 2p^2 + 2p – p – 1.
= 2p(p + 1) – (p + 1).
Factor out the common terms from the above expression. That is,
(2p – 1) (p + 1).
Now, replace the term p with the (x + 2y). That is,
(2p – 1) (p + 1) = (2(x + 2y) – 1) (x + 2y + 1).
= (2x + 4y – 1) (x + 2y + 1).

Then, 2(x + 2y)^2 + (x + 2y) – 1 is equal to (2x + 4y – 1) (x + 2y + 1).

(iv) The given expression is (x + y)^2 – (x + y) – 6.
Replace the (x + y) with p. That is,
p^2 – p – 6.
The above expression matches with the basic expression ax^2 + bx + c.
Here, a = 1, b = – 1, c = – 6.
a*c = 1 * ( – 6) = – 6 and b = – 1.
So, – 6 = – 3 * 2 and – 1 = – 3 + 2.
Then, p^2 – p – 6 = p^2 – 3p + 2p – 6.
= p(p – 3) + 2(p – 3).
Factor out the common terms from the above expression. That is,
(p – 3) (p + 2).
Now, replace the term p with the (x + y). That is,
(p – 3) (p + 2) = (x + y – 3) (x + y + 2).

Then, (x + y)^2 – (x + y) – 6 is equal to (x + y – 3) (x + y + 2).

(v) The given expression is (a^2 – 3a)^2 – 38(a^2 -3a) – 80.
Replace the (a^2 – 3a) with p. That is,
p^2 – 38p – 80.
The above expression matches with the basic expression ax^2 + bx + c.
Here, a = 1, b = – 38, c = – 80.
a*c = 1 * ( – 80) = – 80 and b = – 38.
So, – 80 = – 40 * 2 and – 38 = – 40 + 2.
Then, p^2 – 38p – 80 = p^2 – 40p + 2p – 80.
= p(p – 40) + 2(p – 40).
Factor out the common terms from the above expression. That is,
(p – 40) (p + 2).
Now, replace the term p with the (a^2 – 3a). That is,
(p – 40) (p + 2) = (a^2 – 3a – 40) ( a^2 – 3a + 2).

Then, (a2 – 3a)^2 – 38(a^2 -3a) – 80 is equal to (a^2 – 3a – 40) ( a^2 – 3a + 2).

(vi) The given expression is 6(a – b)^2 – a + b – 15.
We can write it as 6(a – b)^2 – (a – b) – 15.
Replace the (a – b) with p. That is,
6p^2 – p – 15.
The above expression matches with the basic expression ax^2 + bx + c.
Here, a = 6, b = – 1, c = – 15.
a*c = 6 * ( – 15) = – 90 and b = – 1.
So, – 90 = – 10 * 9 and – 1 = – 10 + 9.
Then, 6p^2 – p – 15 = 6p^2 – 10p + 9p – 15.
= 2p(3p – 5) + 3(3p – 5).
Factor out the common terms from the above expression. That is,
(3p – 5) (2p + 3).
Now, replace the term p with the (a – b). That is,
(3p – 5) (2p + 3) = (3(a – b) – 5) (2(a – b) + 3).

Then, 6(a – b)^2 – a + b – 15 is equal to (3(a – b) – 5) (2(a – b) + 3).

(vii) The given expression is (x^2 – 3y^2)^2 – 16(x^2 – 3y^2) + 63.
Replace the (x^2 – 3y^2) with p. That is,
p^2 – 16p + 63.
The above expression matches with the basic expression ax^2 + bx + c.
Here, a = 1, b = – 16, c = 63.
a*c = 1 * 63 = 63 and b = – 16.
So, 63 = – 7 * ( – 9) and – 16 = – 7 – 9.
Then, p^2 – 16p + 63 = p^2 – 9p – 7p + 63.
= p(p – 9) – 7(p – 9).
Factor out the common terms from the above expression. That is,
(p – 9) (p – 7).
Now, replace the term p with the (x^2 – 3y^2). That is,
(p – 9) (p – 7) = (x^2 – 3y^2 – 9) (x^2 – 3y^2 – 7).

Then, (x^2 – 3y^2)^2 – 16(x^2 – 3y^2) + 63 is equal to (x^2 – 3y^2 – 9) (x^2 – 3y^2 – 7).

(viii) The given expression is (x^2 + 2x)^2 – 22(x^2 + 2x) + 72.
Replace the (x^2 + 2x) with p. That is,
p^2 – 22p + 72.
The above expression matches with the basic expression ax^2 + bx + c.
Here, a = 1, b = – 22, c = 72.
a*c = 1 * 72 = 72 and b = – 22.
So, 72 = – 18 * ( – 4) and – 22 = – 18 + ( – 4).
Then, p^2 – 22p + 72 = p^2 – 18p – 4p + 72.
= p(p – 18) – 4(p – 18).
Factor out the common terms from the above expression. That is,
(p – 18) (p – 4).
Now, replace the term p with the (x^2 + 2x). That is,
(p – 18) (p – 4) = (x^2 + 2x – 18) (x^2 + 2x – 4).

Then, (x^2 + 2x)^2 – 22(x^2 + 2x) + 72 is equal to (x^2 + 2x – 18) (x^2 + 2x – 4).

(ix) The given expression is (a^2 – 8a)^2 – 29(a^2 – 8a) + 180.
Replace the (a^2 – 8a) with p. That is,
p^2 – 29p + 180.
The above expression matches with the basic expression ax^2 + bx + c.
Here, a = 1, b = – 29, c = 180.
a*c = 1 * 180 = 180 and b = – 29.
So, 180 = – 20 * (- 9) and – 29 = – 20 – 9.
Then, p^2 – 29p + 180 = p^2 – 9p – 20p + 180.
= p(p – 9) – 20(p – 9).
Factor out the common terms from the above expression. That is,
(p – 9) (p – 20).
Now, replace the term p with the (a^2 – 8a). That is,
(p – 9) (p – 20) = (a^2 – 8a – 9) (a^2 – 8a – 20).

Then, (a^2 – 8a)^2 – 29(a^2 – 8a) + 180 is equal to (a^2 – 8a – 9) (a^2 – 8a – 20).

(x) (p + q)2 – 8p – 8q + 7.
Solution: The given expression is
(p + q)2 – 8p – 8q + 7.
We can write it as (p + q)^2 – 8(p + q) + 7.
Replace the (p + q) with x. That is,
x^2 – 8x + 7.
The above expression matches with the basic expression ax^2 + bx + c.
Here, a = 1, b = – 8, c = 7.
a*c = – 1 * (-7) = 7 and b = – 8.
So, 7 = – 7 * – 1and – 8 = – 7 – 1.
Then, x^2 – 8x + 7 = x^2 – 7x – x + 7.
= x(x – 7) – (x – 7).
Factor out the common terms from the above expression. That is,
(x – 7) (x – 1).
Now, replace the term x with the (p + q). That is,
(x – 7) (x – 1) = (p + q – 7) (p + q – 1).
Then, (p + q)2 – 8p – 8q + 7is equal to (p + q – 7) (p + q – 1).

2. Factor trinomials using substitution

(i) (a – 2b)^2 + 7(a – 2b) – 18
(ii) 3(x – y)^2 – (x – y) – 44
(iii) (5x – 3y)^2 + 8(5x – 3y) + 16
(iv) (a – 4b)^2 – 10(a – 4b) + 25
(v) (3x – 4)^2 – 4(3x – 4) – 12
(vi) (7x – 1)^2 + 12(7x – 1) – 45

Solution:

(i) The given expression is (a – 2b)^2 + 7(a – 2b) – 18.
Replace the (a – 2b) with x. That is,
x^2 + 7x – 18.
The above expression matches with the basic expression ax^2 + bx + c.
Here, a = 1, b = 7, c = – 18.
a*c = 1 * (- 18) = – 18 and b = 7.
So, – 18 = 9 * ( -2) and 7 = 9 + ( – 2).
Then, x^2 + 7x – 18 = x^2 – 2x + 9x – 18.
= x(x – 2) + 9(x – 2).
Factor out the common terms from the above expression. That is,
(x – 2) (x + 9).
Now, replace the term x with the (a – 2b). That is,
(x – 2) (x + 9) = (a – 2b – 2) (a – 2b+ 9).

Then, (a – 2b)^2 + 7(a – 2b) – 18 is equal to (a – 2b – 2) (a – 2b + 9).

(ii) The given expression is 3(x – y)^2 – (x – y) – 44.
Replace the (x – y) with p. That is,
3p^2 – p – 44.
The above expression matches with the basic expression ax^2 + bx + c.
Here, a = 3, b = – 1, c = – 44.
a*c = 3 * (- 44) = – 132 and b = – 1.
So, – 132 = – 12 * 11and – 1 = – 12 + 11.
Then, 3p^2 – p – 44 = 3p^2 – 12p + 11p – 44.
= 3p(p – 4) + 11(p – 4)
Factor out the common terms from the above expression. That is,
(p – 4) (3p + 11).
Now, replace the term p with the (x – y). That is,
(p – 4) (3p + 11) = (x – y – 4) (3(x – y) + 11).

Then, 3(x – y)^2 – (x – y) – 44 is equal to (x – y – 4) (3(x – y) + 11).

(iii) The given expression is (5x – 3y)^2 + 8(5x – 3y) + 16.
Replace the (5x – 3y) with p. That is,
p^2 + 8p + 16.
The above expression matches with the basic expression ax^2 + bx + c.
Here, a = 1, b = 8, c = 16.
a*c = 1 * 16 = 16 and b = 8.
So, 16 = 4 * 4and 8 = 4 + 4.
Then, p^2 + 8p + 16 = p^2 + 4p + 4p + 16.
= p(p + 4) + 4(p + 4).
Factor out the common terms from the above expression. That is,
(p + 4) (p + 4).
Now, replace the term p with the (5x – 3y). That is,
(p +4) (p + 4) = (5x – 3y + 4) (5x – 3y + 4).

Then, (5x – 3y)^2 + 8(5x – 3y) + 16 is equal to (5x – 3y + 4) (5x – 3y + 4).

(iv) The given expression is (a – 4b)^2 – 10(a – 4b) + 25.
Replace the (a – 4b) with p. That is,
p^2 – 10p + 25.
The above expression matches with the basic expression ax^2 + bx + c.
Here, a = 1, b = – 10, c = 25.
a*c = 1 * 25 = 25 and b = – 10.
So, 25 = – 5 * ( – 5) and – 10 = – 5 + ( – 5).
Then, p^2 – 10p + 25 = p^2 – 5p – 5p + 25.
= p(p – 5) – 5(p – 5).
Factor out the common terms from the above expression. That is,
(p – 5) (p – 5).
Now, replace the term p with the (a – 4b). That is,
(p – 5) (p – 5) = (a – 4b – 5) (a – 4b – 5).

Then, (a – 4b)^2 – 10(a – 4b) + 25 is equal to (a – 4b – 5) (a – 4b – 5).

(v) The given expression is (3x – 4)^2 – 4(3x – 4) – 12.
Replace the (3x – 4) with p. That is,
p^2 – 4p – 12.
The above expression matches with the basic expression ax^2 + bx + c.
Here, a = 1, b = – 4, c = – 12.
a*c = 1 * ( – 12) = – 12 and b = – 4.
So, – 12 = – 6 * 2 and – 4 = – 6 + 2.
Then, p^2 – 4p – 12 = p^2 – 6p + 2p – 12.
= p(p – 6) + 2(p – 6).
Factor out the common terms from the above expression. That is,
(p – 6) (p + 2).
Now, replace the term p with the (3x – 4). That is,
(p – 6) (p + 2) = (3x – 4 – 6) (3x – 4 + 2).
= (3x – 10) (3x- 2).

Then, (3x – 4)^2 – 4(3x – 4) – 12 is equal to (3x – 10) (3x- 2).

(vi) The given expression is (p + q)^2 – 8p – 8q + 7.
Replace the (7x – 1) with p. That is,
p^2 + 12p – 45.
The above expression matches with the basic expression ax^2 + bx + c.
Here, a = 1, b = 12, c = – 45.
a*c = 1 * ( – 45) = – 45 and b = 12.
So, – 45 = 15 * ( – 3) and 12 = 15 + ( – 3).
Then, p^2 + 12p – 45 = p^2 + 15p – 3p – 45.
= p(p + 15) – 3(p + 15).
Factor out the common terms from the above expression. That is,
(p + 15) (p – 3).
Now, replace the term p with the (7x – 1). That is,
(p + 15) (p – 3) = (7x – 1 + 15) (7x – 1 – 3).
= (7x + 14) (7x – 4).

Then, (p + q)^2 – 8p – 8q + 7 is equal to (7x + 14) (7x – 4).

## Worksheet on Factorization | Factorisation Worksheet with Answers

Learn the complete concept of factorization by having the best material with you. Follow our Worksheet on Factorization and get a grip on the entire concept. To get the complete knowledge of the factorization process, learn and practice our Factorization Worksheets more and more. We are providing the worksheets with a number of factorization questions and have solutions explained in detail.

1. Factorize each of the following expression

(i) – 5a^2 + 5ab – 52a
(ii) x^2yz + xy^2z + xyz^2
(iii) – 4x^5 – 16x^3y – 20x^2y^2
(iv) x^3yz + 4xy^3 + 14x^3
(v) – a^2 + 3a + a – p – 2
(vi) x^2p + y^2p + z^2p – x^2q – y^2q – z^2q

Solution:

(i) The given expression is – 5a^2 + 5ab – 52a.
Factor out the common term from the above expression.That is,
a(- 5a + 5b – 52).

– 5a^2 + 5ab – 52a is equal to a(- 5a + 5b – 52).

(ii) The given expression is x^2yz + xy^2z + xyz^2
Factor out the common term from the above expression.That is,
xyz(x + y + z).

x^2yz + xy^2z + xyz^2 is equal to xyz(x + y + z).

(iii) The given expression is – 4x^5 – 16x^3y – 20x^2y^2.
Factor out the common term from the above expression. That is,
– 4x^2(x^3 + 4xy + 5y^2).

– 4x^5 – 16x^3y – 20x^2y^2 is equal to 4x^2(x^3 + 4xy + 5y^2).

(iv) The given expression is x^3yz + 4xy^3 + 14x^3
Factor out the common term from the above expression.That is,
x(x^2yz + 4y^2 + 14x^2).

x^3yz + 4xy^3 + 14x^3is equal to x(x^2yz + 4y^2 + 14x^2).

(v) The given expression is – a^2 + 3a + a – p – 2.
Factor out the common term from the above expression.That is,
a(-a + 3 + 1) – (p + 2).
(a – 1) (p – a + 2).

– a^2 + 3a + a – p – 2 is equal to (a – 1) (p – a + 2).

(vi) The given expression is x^2p + y^2p + z^2p – x^2q – y^2q – z^2q.
Factor out the common term from the above expression.That is,
x^2(p – q) + y^2(p – q) + z^2(p – q) = (p – q) (x^2 + y^2 + z^2).

x^2p + y^2p + z^2p – x^2q – y^2q – z^2q is equal to (p – q) (x^2 + y^2 + z^2).

2. Factorize using the formula of the difference of two squares

(i) 25a^2 – 36a^2b^2
(ii) 49x^2 – y^4
(iii) 81x^4 – y^2
(iv) a^3 – 25a.
(v) 32x^2y – 200y^3
(vi) 4x^2 + 12xy + 9y^2 – 9.

Solution:

(i) The given expression is 25a^2 – 36a^2b^2.
We can write it as (5a)^2 – (6ab)^2.
Factor out the common term from the above expression. That is,
a^2[(5)^2 – (6b)^2].
By comparing the above expression, it matches with the basic expression (a)^2 – (b)^2 = (a + b) (a – b).
So, a^2[(5)^2 – (6b)^2] = a^2(5 + 6b) (5 – 6b).

25a^2 – 36a^2b^2 is equal to a^2(5 + 6b) (5 – 6b).

(ii) The given expression is 49x^2 – y^4.
We can write it as (7x)^2 – (y^2)^2.
By comparing the above expression, it matches with the basic expression (a)^2 – (b)^2 = (a + b) (a – b).
So, (7x)^2 – (y^2)^2= (7x + y^2) (7x – y^2).

49x^2 – y^4 is equal to (7x + y^2) (7x – y^2).

(iii) The given expression is 81x^4 – y^2.
We can write it as (9x^2)^2 – (y)^2.
By comparing the above expression, it matches with the basic expression (a)^2 – (b)^2 = (a + b) (a – b).
So, (9x^2)^2 – (y)^2 = (9x^2 + y) (9x^2 – y).

81x^4 – y^2 is equal to (9x^2 + y) (9x^2 – y).

(iv) The given expression is a^3 – 25a
factor out the common terms from the above expression. That is,
a (a^2 – 25).
We can write it as a (a^2 – 5^2).
By comparing the above expression, it matches with the expression a^2 – b^2 = (a + b) (a – b).
So, a (a^2 – 5^2) = a (a + 5) (a – 5).

a^3 – 25a is equal to a(a + 5) (a – 5).

(v) The given expression is 32x^2y – 200y^3.
Factor out the common terms from the above expression. That is,
8y (4x^2 – 25y^2).
We can write it as 8y ((2x)^2 – (5y)^2).
By comparing the above expression, it matches the expression a^2 – b^2 = (a + b) (a – b).
So,8y ((2x)^2 – (5y)^2) = 8y (2x + 5y) (2x – 5y).

32x^2y – 200y^3 is equal to 8y(2x + 5y) (2x – 5y).

(vi) The given expression is 4x^2 + 12xy + 9y^2 – 9.
We can write it as (2x)^2 + 2(2x) (3y) + (3y)^2 – 9.
By comparing the above expression. It matches with the expression a^2 + 2ab + b^2 = (a + b)^2.
Then, (2x)^2 + 2(2x) (3y) + (3y)^2 – 9 = (2x + 3y)^2 – 9.
(2x + 3y)^2 – 9 = (2x + 3y)^2 – (3)^2
By comparing the above expression, it matches the expression a^2 – b^2 = (a + b) (a – b).
So, (2x + 3y)^2 – (3)^2 = (2x + 3y + 3) (2x + 3y – 3).

4x^2 + 12xy + 9y^2 – 9 is equal to (2x + 3y + 3) (2x + 3y – 3).

3. Find the values of

(i) (8)^2 – (6)^2
(ii) (27^2/3)^2 – (8 1/3)^2
(iii) (42)^2 – (14)^2
(iv) (10003)^2 – (9997)^2
(v) (9.2)^2 – (0.8)^2

Solution:

(i) The given expression is (8)^2 – (6)^2.
By comparing the above equation, it matches with the basic expression
(a)^2 – (b)^2 = (a + b) (a – b).
(8)^2 – (6)^2 = (8 + 6) (8 – 6).
= (14) (2) = 28.

(8)^2 – (6)^2 = 28.

(ii) The given expression is (27^2 / 3)^2 – (8 ^1/3)^2.
We can write it as
(3^3 * 2 / 3)^2 – (2^3 * 1 / 3)^2 = (3^2)^2 – (2)^2.
By comparing the above equation, it matches with the basic expression
(a)^2 – (b)^2 = (a + b) (a – b).
(9)^2 – (2)^2 = (9 + 2) (9 – 2) = (11) (7)= 77.

(27^2 / 3)^2 – (8 ^1/3)^2 is equal to 77

(iii) The given expression is (42)^2 – (14)^2.
By comparing the above expression with the basic expressions, it matches with the expression a^2 – b^2 = (a + b) (a – b).
(42)^2 – (14)^2 = (42 + 14) (42 – 14) = (56) (28) = 1568.

(42)^2 – (14)^2 is equal to 1568.

(iv) The given expression is (10003)^2 – (9997)^2.
By comparing the above expression with the basic expressions, it matches with the expression a^2 – b^2 = (a + b) (a – b).
(10003)^2 – (9997)^2 = (10003 + 9997) (10003 – 9997).
= (20,000) (6) = 1,20,000.

(10003)^2 – (9997)^2 is equal to 1,20,000.

(v) The given expression is (9.2)^2 – (0.8)^2.
By comparing the above expression with the basic expressions, it matches the expression a^2 – b^2 = (a + b) (a – b).
(9.2)^2 – (0.8)^2 = (9.2 + 0.8) (9.2 – 0.8) = (10) (8.4) = 84.

(9.2)^2 – (0.8)^2 = 84.

4. Factorization of trinomial

(i) 10x^2y^2 – 11xy + 3
(ii) 12a^2 + 11a – 5
(iii) 15x^2y^2 – 21xy + 6
(iv) a^4 – 10a^2 + 9

Solution:

(i) The given expression is 10x^2y^2 – 11xy + 3.
If xy = a, then 10a^2 – 11a + 3.
By comparing the above expression, it matches with the expression ax^2 + bx + c.
Here, a = 10, b = – 11, c = 3
a * c = 10 * 3 = 30, b = – 11.
30 = – 6 * ( – 5) and – 11 = – 6 + ( – 5).
So, 10a^2 – 11a + 3 = 10a^2 – 5a – 6a + 3.
10a^2 – 5a – 6a + 3 = 5a(2a – 1) – 3(2a – 1).
Factor out the common terms from the above expression. That is,
(2a – 1) (5a – 3).
Here, replace a with the xy, then
(2xy – 1) (5xy – 3).

10x^2y^2 – 11xy + 3 is equal to (2xy – 1) (5xy – 3).

(ii) The given expression is 12a^2 + 11a – 5.
By comparing the above expression, it matches with the expression ax^2 + bx + c.
Here, a = 12, b = 11, c = – 5.
a * c = 12 * (- 5) = – 60, b = 11.
-60 = 15 * ( – 4) and 11 = 15 + ( – 4).
So, 12a^2 + 11a – 5= 12a^2 – 4a + 15a – 5.
= 4a(3a – 1) + 5(3a – 1).
Factor out the common terms from the above expression. That is,
(3a – 1) (4a + 5).

12a^2 + 11a – 5 is equal to (3a – 1) (4a + 5).

(iii) The given expression is 15x^2y^2 – 21xy + 6.
If xy = a, then 15a^2 – 21a + 6.
By comparing the above expression, it matches with the expression ax^2 + bx + c.
Here, a = 15, b = – 21, c = 6.
a * c = 15 * 6 = 90 and b = – 21.
90 = – 15 * ( – 6) and – 21 = – 15 + (- 6).
15a^2 – 21a + 6 = 15a^2 – 15a – 6a + 6.
= 15a(a – 1) – 6(a – 1).
Factor out the common terms from the above expression. That is,
(a – 1) (15a – 6).
Replace a with the xy. That is,
(xy – 1) (15xy – 6).

15x^2y^2 – 21xy + 6 is equal to (xy – 1) (15xy – 6).

(iv) The given expression is a^4 – 10a^2 + 9.
We can write it as (a^2)^2 – 10a^2 + 9.
If a^2 = x, then x^2 – 10x + 9.
By comparing the above expression with the basic expressions, it matches with the ax^2 + bx + c.
Here, a = 1, b = – 10, c = 9.
a * c = 1 * 9 = 9 and b = – 10.
9 = – 9 * ( – 1) and – 10 = – 9 – 1.
x^2 – 9x – x + 9 = x(x – 9) – (x – 9).
Factoring out common terms from the above expression. That is,
(x – 9) (x – 1).
Replace the x with a^2. That is,
(a^2 – 9) (a^2 – 1).

a^4 – 10a^2 + 9 is equal to (a^2 – 9) (a^2 – 1).