Mounika Kandukuri, Author at WorkSheets Buddy

Worksheet on Same Base and Same Parallels | Figures having Same Base and Same Parallels Worksheets

January 27, 2021 by Mounika Kandukuri

Students who are interested in learning the complete concept of geometric shapes having the same base and between the same parallels can refer to this page. Go through our Worksheet on Same Base and Same Parallels and prepare well for the competitive exams also. It includes the shortcuts, tricks, and detailed examples that are useful for the school students and one who is preparing for the jobs.

To get the complete knowledge of the same base and same parallels, learn and practice more example questions. Check the important points that say the relationship between the geometric shapes having the same base in the below sections.

Relationship between Shapes Having Same Base and Same Parallels

The following are relations between the areas of few shapes lying on the same base and between the same parallels.

If two parallelograms are on the same base and between the same parallels, those areas are equal.
When a parallelogram and rectangle lies on the same base and between the same parallels, then the area of a parallelogram is equal to the area of the rectangle.
If a parallelogram and triangle having the same base and are between the same parallels, then the area of a parallelogram is twice the area of the triangle.
If a rectangle and a triangle lie on the same base and between the same parallels, then the area of the rectangle is equal to half of the area of the triangle.
When Triangles on the Same Base and between the Same Parallel, then the area of triangles are equal.

1. In the parallelogram ABCD, E, F are any two points on sides AB and BC respectively. Show that area of ∆ ADF and the area of ∆ CDE are equal.

Solution:

Draw a line from F perpendicular to AD and a line from E perpendicular to CD.

Here, the triangle ADF and parallelogram ABCD are on the same base AD and between the same parallels CD and FG

We already know that Triangle and Parallelogram on the Same Base and between Same Parallels then the area of triangle is equal to the half of the area of the parallelogram.

So, the Area of △ ADF = 1/2 x Area of Parallelogram ABCD —- (i)

Here, CD is the common base for △ CDE, parallelogram ABCD, and the same parallels are BC and EH

In the same way, Area of △ CDE = 1/2 x Area of parallelogram ABCD —- (ii)

By observing the equation (i) and (ii) we can say that

Area of △ CDE = Area of △ ADF.

Hence shown.

2. Calculate the area of quadrilateral ABCD?

Solution:

Hint: We can use Heron’s formula to find the area of triangle ABC and triangle ABD, then add those to get the quadrilateral area.

Semiperimeter of △ ABC s = (a + b + c)/ 2

s = 12 + 25 + 17)/2 = 54/2 = 27 units

Area of △ ABC = √[s (s – a) (s – b) (s – c)]

= √[27 (27 – 12) (27 – 25) (27 – 17)]

= √(27 x 15 x 2 x 10)

= √8100 = 90 square units.

Semiperimeter of △ ABD s = (12 + 20 + 28)/2 = 60/2 = 30 units

Area of △ ABD = √[s (s – a) (s – b) (s – c)]

= √[30 (30 – 12) (30 – 20) (30 – 28)]
= √(30 x 18 x 10 x 2)

= √10800 = 60√3 square units.

Area of quadrilateral ABCD = Area of △ ABD + Area of △ ABC

= 90 + 60√3 = 30(3 + 2√3) square units.

3. Parallelogram PQRS and rectangle PQUT are on the same base PQ and between the same parallels PQ and TR. Also, the area of the parallelogram is 108 cm² and the width of the rectangle is 6 cm. Find the length of the rectangle and its area.

Solution:

Given that,

Area of the parallelogram = 108 cm²

Width of the rectangle = 6 cm

Area of the rectangle = length x width

We know that, if the parallelogram and rectangle lie on the same base and between the same parallels, then the area of the parallelogram is equal to the area of the rectangle.

So, Area of the parallelogram = Area of the rectangle

108 = length x width

108 = length x 6

length = 108/6

= 18 cm

Therefore, the rectangle length is 18 cm, and area is 108 cm².

4. Parallelogram PQRS and PQTU are on the same base PQ and between the same parallels PQ and UR. Area of parallelogram PQRS = 144 cm² and the altitude of the parallelogram PQTU = 16 cm. Find the length of the common side of two parallelograms.

Solution:

Given that,

Area of the parallelogram PQRS = 144 cm²

The altitude of the parallelogram PQTU = 16 cm

We have two parallelograms PQRS and PQTU are on the same base PQ and between the same parallels PQ and UR.

So, the area of the parallelogram PQRS is equal to the area of parallelogram PQTU.

Area of the parallelogram PQRS = Area of the parallelogram PQTU

144 = base x altitude

144 = base x 16

base = 144/16

base = 9

Therefore, the length of the common side of two parallelograms is 9 cm.

5. In the adjoining figure, ∆ PQR is right-angled at Q in which QR = 12 cm and PQ = 4 cm. Find the area of ∆QSR; given that PS ∥ QR.

Solution:

Given that,

QR = 12 cm, PQ = 4 cm

The triangles PQR and QRS lie on the same base QR and between the same parallels PS and QR.

Then the area of △ PQR = Area of △ QRS

1/2 x QR x PQ = 1/2 x QR x height

1/2 x 12 x 4 = 1/2 x 12 x height

24 = 6 x height

height of △ QRS = 24/6 = 4 cm

Then, area of ∆ QSR = 1/2 x 12 x 4

= 12 x 2 = 24 cm².

6. Parallelogram PQRS, ∆ PQS, and rectangle PQTU have the same base PQ. If the area of ∆ PSQ = 48 cm², then find the area of parallelogram PQRS and the area of rectangle PQTU.

Solution:

Given that,

The area of ∆ PSQ = 48 cm²

Parallelogram PQRS and Triangle PQS are the same base PQ, and between the same parallels PQ, RS.

Then the area of parallelogram PQRS = Area of ∆ PSQ

Area of the parallelogram PQRS = 48 cm²

The parallelogram PQRS, rectangle PQTU having the same base and between the same parallels PQ and UT.

Then the area of parallelogram PQRS = Area of rectangle PQTU

Area of rectangle PQTU = 48 cm²

Therefore, the area of parallelogram PQRS is 48 cm² and the area of rectangle PQTU is 48 cm².

7. LM is median of ∆ JKL, JK = 15 cm, and the altitude of the ∆ JKL = 5 cm, find the area of ∆ JLM and area of ∆ LMK.

Solution:

Given that,

JK = 15 cm, altitude of the triangle JKL = 5 cm

Area of ∆ JKL = 1/2 x base x altitude of the triangle JKL

= 1/2 x 15 x 5 = 1/2 x 75

= 37.5 cm²

Triangle JKL and Triangle JLM lie on the same base JL and between the same parallel.

So, area of ∆ JKL = Area of ∆ JLM

Area of ∆ JLM = 37.5 cm²

Triangle JLM, triangle LMK is having the same base LM and between the same parallel

So, area of ∆ LMK = Area of ∆ JLM

Area of LMK = 37.5 cm²

Therefore, the area of ∆ JLM is 37.5 cm², and the area of ∆ LMK is 37.5 cm².

8. ∆ PQR is an isosceles triangle with ST//QR. The medians SR and QT intersect each other at O.

Prove that area of (1) ∆ QTS = ∆ RST

(2) ∆ QOS = ∆ ROT

(3) ∆ PQT = ∆ PRS

Solution:

(1)

Given triangles QTS and RST are on the same base TS and between the same parallels.

The area of △ QTS = 1/2 x TS x OS

Area of △ RST = 1/2 x ST x OT

From the given figure, OT = OS

Area of △ RST = 1/2 x ST x OS

So, Area of △ RST = Area of △ QTS

Hence proved.

(2)

Triangle QOS and triangle ROT.

From the figure, we can say that OT = OS

Area of triangle QOS = 1/2 x OS x altitude of △ QOS

Area of triangle ROT = 1/2 x OT x altitude of △ ROT

= 1/2 x OS x altitude of △ ROT

the altitude of △ ROT = altitude of △ QOS

Therefore, Area of triangle QOS = Area of triangle ROT

Hence proved.

(3)

Area of triangle PQT = Area of triangle QST + Area of triangle PST

Area of triangle QST = 1/2 x ST x OS

Area of triangle PRS = Area of triangle RST + Area of triangle PST

Area of triangle RST = 1/2 x ST x OS

Area of triangle PRS = 1/2 x ST x OS + Area of trinagle PST

Area of triangle PQT = 1/2 x ST x OS + Area of triangle PST

Therefore, Area of triangle PRS = Area of triangle PQT

Hence proved.

9. Parallelogram PQRS, ∆PQS, and rectangle PQTU have the same base PQ. If the area of ∆PSQ = 56 cm², then find the area of parallelogram PQRS and the area of rectangle PQTU.

Solution:

Given that,

The area of the triangle PSQ = 56 cm²

Parallelogram PQRS and Triangle PQS are the same base PQ, and between the same parallels PQ, RS.

Then the area of parallelogram PQRS = Area of ∆ PSQ

Area of the parallelogram PQRS = 56 cm²

The parallelogram PQRS, rectangle PQTU having the same base and between the same parallels PQ and UT.

Then the area of parallelogram PQRS = Area of rectangle PQTU

Area of rectangle PQTU = 56 cm²

Therefore, the area of parallelogram PQRS is 56 cm² and the area of rectangle PQTU is 56 cm².

10. Parallelogram PQRS and rectangle PQUT are on the same base PQ and between the same parallels PQ and TR. Also, the area of the parallelogram is 85 cm² and the width of the rectangle is 5 cm. Find the length of the rectangle and its area.

Solution:

Given that,

Area of the parallelogram = 85 cm²

Width of the rectangle = 5 cm

Area of the rectangle = length x width

We know that, if the parallelogram and rectangle lie on the same base and between the same parallels, then the area of the parallelogram is equal to the area of the rectangle.

So, Area of the parallelogram = Area of the rectangle

85 = length x width

85 = length x 5

length = 85/5

= 17 cm

Therefore, the rectangle length is 17 cm, and the area is 85 cm².

Worksheet on Area of a Polygon | Questions on Polygon Area | Area of a Polygon Worksheet

January 25, 2021 by Mounika Kandukuri

Worksheet on Area of a Polygon is helpful to the students who are willing to solve the questions on area of the pentagon, square, hexagon, octagon, and n-sided polygons. In case the students are preparing for any kind of test, then they can start preparation from this Area of the Polygon Worksheet. Utilize the Polygon Area Worksheets and score better grades in the exam.

In this Worksheet on Polygon Area, all types of questions are covered with hints. Learn How to Find the Area of a Polygon with the Problems available here. So, Practice problems from the Worksheet on Area of a Polygon as many times as possible so that you will understand the concept behind them.

1. Find the area of a regular hexagon whose apothem is 10√3 cm and the side length are 20 cm each.

Solution:

Given that,

Hexagon apothem = 10√3 cm

Side length of regular hexagon = 20 cm

Perimeter of the hexagon = (20 + 20 + 20 + 20 + 20 + 20)

= 120 cm

Area of a regular hexagon = ½ x perimeter x apothem

= ½ x 120 x 10√3

= 60 x 10√3 = 600√3

Therefore, area of a regular hexagon is 600√3 cm².

2. What is the area of a regular octagon with a side length of 12 m.

Solution:

Given that,

Side length of regular octagon = 12 m

Area of regular octagon = 2(1 + √2) a²

= 2(1 + √2) 12²

= 2(1 + √2) 144

= 288 (1 + √2) = 288(2.414)

= 695.293 cm²

Therefore, the area of a regular octagon is 695.293 cm².

3. Find the area of a regular pentagon of side:

(a) 5 cm

(b) 12 cm

Solution:

(a) 5 cm

Side of a regular pentagon a = 5 cm

Area of a regular pentagon = 1/4 ((√(5 (5 + 2 √5) a²))

= 1/4 ((√(5 (5 + 2 √5) 5²))

= 1/4 ((√(5 (5 + 2 √5) 25))

= 1/4 (6.88) (25)

= 172.04/4 = 43.01 cm²

Therefore, the pentagon area is 43.01 cm².

(b) 12 cm

Side of a regular pentagon a = 12 cm

Area of a regular pentagon = 1/4 ((√(5 (5 + 2 √5) a²))

= 1/4 ((√(5 (5 + 2 √5) 12²))

= 1/4 ((√(5 (5 + 2 √5) 144))

= (144/4) (6.88)

= 36 x 6.88 = 247.68 cm²

Therefore, area of a pentagon is 43.01 cm².

4. Find the area of a regular pentagon whose each side is 12 cm long and the radius of the inscribed circle is 6.5 cm.

Solution:

Given that,

The side length of pentagon = 12 cm

The radius of the inscribed circle = 6.5 cm

In ΔOMC, by Pythagoras theorem

OC² = OM² + MC²

6.5² = h² + 6²

42.25 = h² + 36

h² = 42.25 – 36

h² = 6.25

h = √(6.25)

= 2.5

The area of triangle ODC = 1/2 x DC x OM

= 1/2 x 12 x 2.5 = 6 x 2.5 = 15 cm²

The pentagon includes 5 triangles

So the area of the Pentagon is five times the area of the triangle

Pentagon area = 5 x area of one triangle

= 5 x 15 = 75 cm²

Therefore, the area of the pentagon is 75 cm².

5. Find the area of a regular pentagon whose each side is 5 cm long and the radius of the whose circumscribed circle is 3.5 cm.

Solution:

Given that,

Side length of regular pentagon a = 5 cm

The radius of circumscribed circle R = 3.5 cm

Area of the pentagon = n/2 × a × √(R² – a²/4)

= 5/2 x 5 x √(3.5² – 5²/4)

= 5/2 x 5 x √(12.25 – 25/4)

= 5/2 x 5 x √(12.25 – 6.25)

= 5/2 x 5 x √(6)

= 25/2 x 2.44

= 25 x 1.224 = 30.618 cm²

Therefore, the pentagon area is 30.618 cm².

6. What is the area of a parallelogram that has a base of 12 ¾ in and a height of 2 ½ in?

Solution:

Given that,

Base = 12 ¾ inches = 51/4 in

Height = 2 ½ inches = 5/2 in

Area of the parallelogram = base x height

= (51/4) x (5/2)

= 255/8 square inches.

Therefore, the parallelogram area is 255/8 sq inches.

7. The diagonals of a rhombus are 12 m and 26 m. What is the area of the rhombus?

Solution:

Given that,

Diagonals of a rhombus are a = 12 m, b = 26 m

Area of the rhombus = 1/2 (a x b)

= 1/2 (12 x 26) = 6 x 26 = 156 m²

Therefore, the area of rhombus is 156 m².

8. Find the area of polygon ABCDEFG.

Solution:

The polygon can be split into two trapeziums and a triangle.

The area of polygon ABCDEFG is given by the sum of the area of trapezium ABCG and CDFG and the area of triangle DEF.

Height of trapezium ABCG = 3 cm

Height of trapezium CDFG = (6 – 3) = 3 cm

Height of triangle DEF = (8 – 6) = 2 cm

Area of trapezium ABCG = (sum of parallel sides) × height/2

= (4 + 7)×3/2

= 33/2 = 16.5 cm²

Area of trapezium CDFG = (7 + 4) ×3/2

= 33/2 = 16.5 cm²

Area of triangle DEF = (base ×height)/2

= (4 × 2)/2 = 8/2

= 4 cm²

So, area of polygon ABCDEFG = area of ABCG + area of CDFG + area of DEF

= 16.5 + 16.5 + 4 = 37 cm²

9. Find the area of 15 sided polygon having a side length of 2 cm.

Solution:

Given that,

Number of sides of regular polygon = 15

Side length of polygon = 2 cm

Area of regular polygon A = [n * a² * cot(π/n)]/4

A = (15 * 2² * cot(π/15))/4

= (15 x 4 x 4.704)/4

= 282.24/4 = 70.56

Therefore, area of 15 sided polygon having 2 cm length is 70.56 cm².

10. Find the area, the perimeter of the regular heptagon with side length 15 cm and apothem is 25√3 cm.

Solution:

Given that,

The apothem of regular heptagon a = 25√3 cm

Side length = 15 cm

Perimeter of regular heptagon = 15 + 15 + 15 + 15 + 15 + 15 + 15

= 105 cm

The regular heptagon area = ½ x perimeter x apothem

= ½ x 105 x 25√3 = 2625√3/2 = 1312.5√3 cm²

Therefore, the perimeter and area of the heptagon is 105 cm, 1312.5√3 cm².

11. Find the regular hexagon area whose distance from the center of sides to the hexagon center is 32 cm.

Solution:

Given that,

The distance from the center of sides to the hexagon center = 32 cm

Regular hexagon area = (3√3)/2 × distance from the center of sides to the center of the hexagon

= (3√3)/2 x 32 = (3√3) x 16

= 48√3 cm²

Therefore, the hexagon area is 48√3 cm².

Worksheet on Trapezium | Trapezium Problems and Solutions

January 22, 2021 by Mounika Kandukuri

If you need help solving trapezium questions you can take the help of worksheets available here. Have a glance at the Worksheet on Trapezium during your practice and take your preparation to the next level. Identify the knowledge gap and allot time to the areas you are lagging. For complete guidance make use of the Trapezium Worksheets and clear all your doubts related to them.

We have given various Area of Trapezium Practice Questions and Perimeter of Trapezium Questions 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 included the step by step explanation for each question.

1. The length of the parallel sides of a trapezium is in the ratio 10: 13 and the distance between them is 25 cm. If the area of trapezium is 655 cm², find the length of the parallel sides.

Solution:

Given that,

The ratio of parallel sides of the trapezium = 10: 13.

Distance between them = 25 cm

Area of trapezium = 655 cm²

Let the common ratio be x,

Then, the parallel sides are 10x, 13x

Area of trapezium formula is 1/2(a + b)h

Area of trapezium = 655 cm²

1/2(10x + 13x)25 = 655

23x . 25 = 655 . 2

575x = 1310

x = 1310/575

x = 2.278

The parallel sides of the trapezium are 10 x 2.278 = 22.78, 13 x 2.278 = 29.617

Therefore, the length of parallel sides of a trapezium are 22.78 cm, 29.617 cm.

2. Calculate the area of a trapezium whose parallel sides are 15 cm, 22 cm, and the distance between them is 26 cm.

Solution:

Given that,

The parallel sides of the trapezium are a = 15 cm, b = 22 cm

Distance between sides h = 26 cm

Area of trapezium = ½(a + b)h

A = ½(15 + 22) x 26

= 37 x 13 = 481 cm²

Therefore, the area of a trapezium is 481 cm².

3. Find the area of a trapezium whose parallel sides are 32 cm and 27 cm, and the distance between them is 15 cm.

Solution:

Given that,

The parallel sides of the trapezium are a = 32 cm, b = 27 cm

The distance between parallel sides h = 15 cm

Area of trapezium = ½(a + b)h

= ½(32 + 27)15

= ½(59) 15

= 885/2 = 442.5 cm²

Therefore, the area of a trapezium is 481 cm².

4. The area of a trapezium is 459 cm². Its parallel sides are in the ratio 7 : 10 and the perpendicular distance between them is 9 cm. Find the length of each of the parallel sides.

Solution:

Given that,

Area of a trapezium = 459 cm²

The ratio of parallel sides of the trapezium = 7 : 10

The perpendicular distance between parallel sides = 9 cm

Let the common ratio be x,

Then, the parallel sides are 7x, 10x

The area of the trapezium formula is 1/2(a + b)h

Area of trapezium = 459 cm²

1/2(a + b)h = 459

1/2(7x + 10x) 9 = 459

1/2(17x) 9 = 459

153x/2 = 459

153x = 459 x 2

153x = 918

x = 918/153

x = 6

The parallel sides of the trapezium are 7 x 6 = 42 cm, 10 x 6 = 60 cm.

Therefore, the length of each parallel side is 42 cm, 60 cm.

5. The area of a trapezium is 150 cm² and its height is 5 cm. If one of the parallel sides is longer than the other by 6 cm, find the two parallel sides.

Solution:

Given that,

The area of trapezium = 150 cm²

Height h = 5 cm

Let the parallel sides of the trapezium are a, b.

One of the parallel sides is longer than the other by 6 cm.

a = b + 6

The area of the trapezium formula is 1/2(a + b)h

Area of trapezium = 150 cm²

1/2(b + 6 + b)5 = 150

1/2(2b + 6)5 = 150

1/2(2b + 6) = 150/5

1/2(2b + 6) = 30

2b + 6 = 30 x 2

2b + 6 = 60

2b = 60 – 6

2b = 54

b = 54/2

b = 27.5 cm

a = 27.5 + 6 = 33.5 cm

Therefore, two parallel sides are 33.5 cm, 27.5 cm.

6. The area of a trapezium is 180 cm² and its height is 8 cm. If one of the parallel sides is 1.5 times the other side, find the two parallel sides.

Solution:

Given that,

Trapezium area = 180 cm²

Height h = 8 cm

Let a, b are the parallel sides of the trapezium.

One of the parallel sides is 1.5 times the other side

a = 1.5b

Area of trapezium = 1/2(a + b)h

180 = 1/2(1.5b + b)8

180 = 4(2.5b)

10b = 180

b = 180/10

b = 18

a = 1.5 x 18 = 27

Therefore, the length of parallel sides of a trapezium is 18 cm, 27 cm.

7. The area of a trapezium is 560 cm². If the lengths of its parallel sides are 25 cm and 32 cm, find the distance between them.

Solution:

Given that,

The area of trapezium = 560

Length of parallel sides a = 25 cm, b = 32 cm

Area of trapezium = 1/2(a + b)h

560 = 1/2(25 + 32)h

57h/2 = 560

57h = 560 x 2

57h = 1120

h = 1120/57

h = 19.64 cm

Therefore, the trapezium height is 19.64 cm.

8. The area of a trapezium is 1200 cm² and the distance between its parallel sides is 12 cm. If one of the parallel sides is 84 cm, find the other.

Solution:

Given that,

Trapezium area = 1200

The distance between parallel sides h = 12 cm

Parallel side a = 84 cm

Area of trapezium = 1/2(a + b)h

1200 = 1/2(84 + b)12

1200 = (84 + b)6

1200/6 = 84 + b

200 = 84 + b

b = 200 – 84

b = 116

Therefore, other parallel side of a trapezium = 116 cm.

9. The parallel sides of a trapezium are 35 cm and 20 cm. Its nonparallel sides are both equal, each being 15 cm. Find the area of the trapezium.

Solution:

ABCD is a trapezium

AB = 35 cm, CD = AE = 20 cm, BC = 15 cm and AD = 15 cm

Through C, draw CE || AD, meeting AB at E

Draw CF ⊥ AB

Now, EB = (AB – AE) = (AB – DC)

EB = 35 – 20 = 15 cm

CE = AD = 15 cm

Now, in ∆EBC, we have CE = BC = 15 cm

It is an isosceles triangle.

Also, CF ⊥ AB

So, F is the midpoint of EB.

Therefore, EF = ¹/₂ × EB = 15/2 = 7.5 cm

Thus, in right-angled ∆CFE, we have CE = 15 cm, EF = 7.5 cm

By Pythagoras’ theorem, we have

CF = [√CE² – EF²]

= √(15² – 7.5²) = √(225 – 56.25) = √(168.75)

= 12.99

Thus, the distance between the parallel sides is 12.99 cm.

Area of trapezium ABCD = 1/2 × (sum of parallel sides) × (distance between them)

= 1/2 x (35 + 20) x 12.99

= 1/2 x (55 x 12.99)

= 714.45/2 = 357.225

Hence, the area of a trapezium = 357.225 cm².

10. ABCD is a trapezium in which AB ∥ CD, AD ⊥ DC, AB = 16 cm, BC = 12 cm and DC = 22 cm. Find the area of the trapezium.

Solution:

From B draw BP perpendicular DC

Therefore, AB = DP = 16 cm

So, PC = DC – DP

= (22 – 16) cm

= 6 cm

Now, area of trapezium ABCD = Area of rectangle ABPD + Area of △ BPC

△BPC is right-angled at ∠BPC

Therefore, using Pythagoras theorem,

BC² = BP² + PC²

12² = BP² + 6²

144 = BP² + 36

BP² = 144 – 36

BP² = 108

BP = √(108)

= 10.39 cm

Now, area of trapezium ABCD = Area of rectangle ABPD + Area of ∆BPC

= AB × BP + 1/2 × PC × BP

= 16 × 10.39 + 1/2 × 6 × 10.39

= 166.24 + 1/2 x 62.34

= 166.24 + 31.17

= 197.41

Therefore, area of trapezium is 197.41 cm².

11. ABCD is a trapezium in which AB is parallel to DC and angle A is equal to angle B = 40 degrees find angle C and angle D are these angles equal?

Solution:

In trapezium ABCD,

AB is parallel to DC and AD and BC are trans

∠A + ∠D = 180°

40° + ∠D = 180°

∠D = 180° – 40°

= 140°

∠C and ∠D are equal.

Yes, ∠C and ∠D are equal and its value is 140°.

12. The parallel sides of a trapezium are 25 cm and 11 cm, while its nonparallel sides are 15 cm and 13 cm. Find the area of the trapezium.

Solution:

ABCD is a trapezium whose sides are

AB = 25 cm, BC = 13 cm, CD = 11 cm and DA = 15 cm

Now, draw a line DE which is parallel to BC to form a parallelogram

Hence, DE = BC

To find the area of trapezium we will add an area of triangle DAE and the area of a parallelogram

The area of △ DAE is

AE = AB – CD = 25 – 11 = 14

s = (AE + DE + DA)/2 = (14 + 13 + 15)/2

= 42/2 = 21

Area = √[s (s – a) (s – b) (s – c)]

= √[21 (21 – 14) (21 – 13) (21 – 15)]

= √[21 (7) (8) (6)]

= √(7056) = 84 cm²

Now, DF is perpendicular to AE

Area of triangle DAE = 1/2 x base x height

84 = 1/2 x 14 x h

84 = 7h

h = 84/7

h = 12

Now, area of parallelogram = 11 x 12 = 132

Area of trapezium = Area of triangle + Area of the parallelogram

= 132 + 84 = 216

Therefore, the trapezium area is 216 cm².

13. The dimensions of an isosceles trapezium are AB = 6 cm, CD = 4 cm, AD = 14 cm. Find its area.

Solution:

Given that,

AB = a = 6 cm, CD = b = 4 cm, AD = c = 14 cm

Area of an isosceles trapezium formula is 1/2 [√(c² – (a – b)²) (a + b)]

Trapezium area = 1/2 [√(c² – (a – b)²) (a + b)]

= 1/2 [√(14² – (6 – 4)²) (6 + 4)]

= 1/2 [√(196 – (-2)²) (10)]

= 1/2 [√(196 – 4 x 10)]

= 1/2(√196 – 40)

= 1/2 (√156)

= 1/2(12.48)

= 6.24 cm²

Therefore, area of an isosceles trapezium is 6.24 cm².

Worksheet on Area and Perimeter of Triangle | Area and Perimeter of Triangle Worksheet with Answers

January 20, 2021 by Mounika Kandukuri

Students looking for different problems on Perimeter and Area of the Triangle can get them in one place. Use the Worksheet on Area and Perimeter of Triangle and kick start your preparation. Students can assess their strengths and weaknesses by solving all questions from Triangle Area and Perimeter Worksheet. Make use of these worksheets and understand different formulas and ways of solving isosceles triangle, equilateral triangle, scalene, and right-angled triangle areas and perimeter.

The Questions covered in the Area and Perimeter of Triangle Worksheet include various triangles area and perimeter. Get those formulas and step by step process to solve all questions. Here, we are offering a detailed solution for each and every problem for a better understanding of concepts.

1. Find the Area, Perimeter of the following triangles:

(a)

(b)

(c)

Solution:

(a)

Sides of the triangle are a = 12, b = 16, c = 20

s = (a + b + c)/2

s = (12 + 16 + 20)/2 = 48/2

s = 24

Area of the triangle formula is A = √[s (s – a) (s – b) (s – c)]

A = √[24 (24 – 12) (24 – 16) (24 – 20)]

= √[24 (12) (8) (4)]

= √(9216) = 96 sq units.

Perimeter of the traingle = (a + b + c)

= 12 + 16 + 20 = 48 units

∴ The area, perimeter of the given triangle is 96 sq units, 48 units.

(b)

The sides of the given triangle are a = 5, b = 13, c = 12

Semiperimeter of the traingle is s = (a + b + c)/2

s = (5 + 13 + 12)/2 = 30/2 = 15

Area of the triangle formula is A = √[s (s – a) (s – b) (s – c)]

A = √[15 (15 – 5) (15 – 13) (15 – 12)]

= √[15 (10) (2) (3)]

= √[900] = 30

Perimeter of the triangle formula is P = (a + b + c)

P = (5 + 13 + 12) = 30

∴ The perimeter and area of the triangle is 30 units, 30 sq units.

(c)

Sides of the triangle are a = 30 cm, b = 35 cm, c = 55 cm

Perimeter of the triangle = (a + b + c)

= (30 + 35 + 55) = 120 cm

Semiperimeter of the triangle s = (a + b + c)/2

= (30 + 35 + 55) = 120/2 = 60 cm

Area of the triangle formula is A = √[s (s – a) (s – b) (s – c)]

A = √[60 (60 – 30) (60 – 35) (60 – 55)]

= √[60 (30) (25) (5)]

= √[225,000] = 474.34 sq. cm

2. Find the area of the triangle whose dimensions are:

(i) base = 20 cm, height = 5 cm

(ii) base = 6.5 m, height = 8 m

(iii) base = 28 m, height = 7 m

Solution:

(i)

Given that,

base = 20 cm, height = 5 cm

Area of the triangle formula is A = ½ x base x height

= ½ x 20 x 5

Area = 10 x 5 = 50 cm²

(ii)

Given that,

base = 6.5 m, height = 8 m

Area of the triangle formula is A = ½ x base x height

A = ½ x 6.5 x 8

Area = 6.5 x 4 = 26 m²

(iii)

Given that,

base = 28 m, height = 7 m

The area of the triangle formula is A = ½ x base x height

= ½ x 28 x 7 = 14 x 7

Area = 98 m²

3. Find the base of a triangle whose

(i) Area = 612 cm², height = 6.2 cm

(ii) Area = 528 cm², height = 10 cm

(iii) Area = 126 m², height = 8 m

Solution:

(i)

Given that,

Area = 612 cm², height = 6.2 cm

Area A = ½ x base x height

612 = ½ x base x 6.2

base = (2 x 612)/6.2 = 1224/6.2

base = 197.41 cm

(ii)

Given that,

Area = 528 cm², height = 10 cm

Area A = ½ x base x height

528 = ½ x base x 10

base = 528/5 = 105.6 cm

(iii)

Given that,

Area = 126 m², height = 8 m

Area A = ½ x base x height

126 = ½ x base x 8

base = 126/4 = 31.5 m

4. Find the height of the triangle, whose

(i) Area = 1500 m², Base = 7.5 m

(ii) Area = 480 cm², Base = 8 cm

(iii) Area = 2580 cm², Base = 12 cm

Solution:

(i) Given that,

Area = 1500 m², Base = 7.5 m

Area A = ½ x base x height

1500 = ½ x 7.5 x height

height = 3000/7.5

height = 400 m

(ii) Given that,

Area = 480 cm², Base = 8 cm

Area A = ½ x base x height

480 = ½ x 8 x height

height = 480/4

height = 120 cm

(iii) Given that,

Area = 2580 cm², Base = 12 cm

Area A = ½ x base x height

2580 = ½ x 12 x height

height = 2580/6

height = 430 cm.

5. The three sides of the triangle are 10 cm, 14 cm, 18 cm. Find its semi perimeter, perimeter, and area.

Solution:

Given that,

Sides of the triangle are a = 10 cm, b = 14 c, c = 18 cm

Perimeter = (a + b + c)

= 10 + 14 + 18 = 42 cm

Semi perimeter s = (a + b + c)/2

= (10 + 14 + 18)/2 = 42/2

= 21 cm

Area of the triangle = √[s (s – a) (s – b) (s – c)]

= √[21 (21 – 10) (21 – 14) (21 – 18]

= √[21 (11) (7) (3)]

= √[4851] = 69.64 cm²

6. Find the area, the perimeter of the triangle, whose sides are 12 m, 15 m, and the semiperimeter is 17.5 m.

Solution:

Given that,

The sides of the triangle are a = 12 m, b = 15 m

Triangle semiperimeter s = 17.5

s = (a + b + c)/2

17.5 = (12 + 15 + c)/2

35 = 27 + c

c = 35 – 27

c = 8 m

Perimeter of the triangle = (a + b + c)

= 12 + 15 + 8 = 35 m

Triangle area = √[s (s – a) (s – b) (s – c)]

= √[17.5 (17.5 – 12) (17.5 – 15) (17.5 – 8)]

= √[17.5 (5.5) (2.5) (9.5)] = √[2,285.9375]

= 47.81 m².

∴ The Triangle perimeter is 35 m, another side is 8 m, and the area is 47.81 m².

7. The three sides of the triangle are in the ratio 5: 2: 6 and its perimeter is 78 units. Find its area?

Solution:

Given that,

The ratio of three sides of the triangle = 5: 2: 6

Triangle perimeter = 78 units

Let us take x as the side of the triangle.

Then, the three sides of a triangle are 5x, 2x, 6x.

Triangle perimeter = 78 units

5x + 2x + 6x = 78

13x = 78

x = 78/13

x = 6 units

So, three sides of the triangle are 30, 12, 36

Semi perimeter s = (30 + 12 + 36)/2 = 78/2 = 39

Triangle area = √[s (s – a) (s – b) (s – c)]

= √[39 (39 – 30) (39 – 12) (39 – 36)]

= √[39 (9) (27) (3)]

= √[28,431] = 168.61 sq units.

8. Find the base of the triangle, whose area is 120 m², altitude is 15 m?

Solution:

Given that,

Altitude = 15 m

Triangle area A = 120 m²

½ x base x height = 120

½ x base x 15 = 120

base = 240/15 = 16 m

9. Find the height of the triangle, whose base is 9 units, and the area is 144 sq units?

Solution:

Given that,

Base = 9 units

Triangle Area = 144 sq units

½ x base x height = 144

½ x 9 x height = 144

height = 288/9 = 32 units.

10. Find the area of an equilateral triangle, the length of whose sides is 11 cm?

Solution:

Given that,

Side of the equilateral triangle a = 11 cm

Area of triangle = (√3/4) a²

Area = (√3/4) 11² = (√3/4) x 11 x 11

= 52.39 cm²

∴ The area of an equilateral triangle is 52.39 cm².

11. The base and height of a triangle are in the ratio 15: 14 and its area is 320 m². Find the height and base of the triangle?

Solution:

Given that,

The ratio of base and height of triangle = 15: 14

Let us take 15x, 14x as the triangle base and height.

Area of triangle = 320

½ x base x height = 320

½ x 15x x 14x = 320

105x² = 320

x² = 320/105

x = √(3.047) = 1.74 m

Height = 14 x 1.74 = 24.44 m

Base = 15 x 1.74 = 26.1 m

12. Find the area of an isosceles right-angled triangle of equal sides 15 cm each?

Solution:

Given that,

Right-angles triangle sides = 15 cm

Triangle area = ½ x side x side

= ½ x 15 x 15 = 225/2 = 112.5 cm

∴ The area of the triangle is 112.5 cm.

13. Find the area of a triangle, whose sides are 5 m, 4 m and angle is 120°.

Solution:

Given that,

The sides of a triangle are a = 5 m, b = 4 m

Angle c = 180°.

Area of the triangle = ½ab sin (A)

= ½ x 5 x 4 x sin(120°)

= 10 x (√3/2)

= 5√3

∴ Area of the triangle = 5√3 sq units.

14. Find the other side and perimeter of the right-angled triangle, whose sides are 8 cm, 15 cm?

Solution:

Given that,

The sides of the triangle are a = 8 cm, b = 15 cm

As per the Pythagorean theorem,

c² = a² + b²

c² = 8² + 15²

c² = 64 + 225

c² = 289

c = √289 = 17 cm

Perimeter = (8 + 15 + 17) = 40

∴ Another side, the perimeter of the right-angled triangle is 17 cm, 40 cm.

15. Find the area of a right-angled triangle whose hypotenuse is 13 cm and one of its sides containing the right angle is 12 cm. Find the length of the other side and perimeter.

Solution:

Given that,

Hypotenuse = 13 cm

Side = 12 cm

Hypotenuse² = Side² + Side²

13² = 12² + Side²

Side² = 169 – 144

Side² = 25

Side = √25 = 5 cm

Semi perimeter s = (13 + 5 + 12)/2 = 30/2 = 15

Perimeter = (13 + 5 + 12) = 30 cm

Triangle Area = √[s (s – a) (s – b) (s – c)]

= √[15 (15 – 5) (15 – 12) (15 – 13)]

= √[15 (10) (3) (2)] = √[900]

= 30 cm²

∴ Area is 30 cm², other side is 5 cm.

16. The area of the triangle is equal to that of the square whose each side measures 30 cm. Find the side of the triangle whose corresponding altitude is 36 cm.

Solution:

Given that,

Area of Triangle = Area of square

Side of square = 30 cm

Altitude of triangle = 36 cm

Area of square = 30 x 30 = 900 cm²

Area of Triangle = Area of square

Area of Triangle = 900

½ x base x height = 900

½ x base x 36 = 900

base = 1800/36 = 50 cm

∴ Side of the triangle is 50 cm.

17. The length of one of the diagonals of a field in the form of a quadrilateral is 52 m. The perpendicular distance of the other two vertices from the diagonal is 15 m and 10 m, find the area of the field.

Solution:

Given that,

Diagonal = 52 m

Altitudes h1 = 15, h2 = 10

Area of quadrilateral = ½ x diagonal (h1 + h2)

= ½ x 52 x (15 + 10)

= 26(25)

= 650 m²

Therefore, the quadrilateral area is 650 m².

18. The area of a right triangle is 156 sq units and one of its legs is 13 units, find the other leg?

Solution:

Given that,

Base = 13 units

Area of triangle = 156 sq units

½ x base x height = 156

½ x 13 x height = 156

height = 312/13 = 24.

19. ∆ ABC is isosceles with AB = AC = 18 cm, BC = 12 cm. The height AD from A to BC is 7 cm. Find the area, perimeter of ∆ABC.

Solution:

Given that,

Sides of the triangle are AB = AC = 18 cm, BC = 12 cm

Height AD = 7 cm

∆ABD + ∆ADC = ∆ABC

Area of ∆ABD = ½bh

= ½ x 18 x 7

= 9 x 7 = 63 cm²

Area of ∆ABC = Area of ∆ABD + ∆ADC

= 2 x 63 = 126 cm².

20. ∆ ABC is an equilateral triangle with sides 28 cm, Find its perimeter and area?

Solution:

Given that,

Side of triangle = 28 cm

Perimeter = (28 + 28 + 28) = 84 cm

Area = (√3/4)a²

= (√3/4) x 28² = (√3/4) x 784

= √3 x 196 = 339.4

Therefore, perimeter, area of the equilateral trinagle is 84 cm, 339.4 cm².

Worksheet on Circumference and Area of Circle | Circle Area, Circumference Worksheet with Answers

January 20, 2021 by Mounika Kandukuri

Worksheet on Circumference and Area of Circle available on this page is a great resource for students. The Area and Circumference of a Circle Worksheet are quite comfortable to use and is available for free. Students can simply solve the various questions available in the Circumference and Area Word Problems Worksheet and learn how to solve problems on the same. To know many other mensuration topic worksheets, stay in touch with our site.

Students will get the questions as per the latest syllabus that helps them to attend the exam with more confidence. So, prepare all the questions in the Area and Circumference of a Circle Worksheet with Solutions and try to attempt questions.

1. Find the area, circumference, and diameter of the circles whose radius is

(a) 10 cm

(b) 6.5 cm

Solution:

(a) 10 cm

Given radius r = 10 cm

The circle formulas are

The diameter of the circle d = 2r

d = 2 x 10

d = 20 cm

Circumference of the circle C = 2πr

C = 2 x (22/7) x 10

C = 440/7

C = 62.85 cm

Area of the circle A = πr²

A = π x (10)² = π x 10 x 10

A = 100 x (22/7) = 2200 / 7

A = 314.285 cm²

∴ Area of circle is 314.285 cm², circumference is 62.85 cm, and diameter is 20 cm.

(b) 6.5 cm

Given radius r = 6.5 cm

The diameter of the circle d = 2r

d = 2 x 6.5

d = 13 cm

Circumference of the circle C = 2πr

C = 2 x (22/7) x 6.5

C = 286/7 = 40.85 cm

Area of the circle A = πr²

A = (22/7) x (6.5)²

= (22/7) x 42.25

= 132.78 cm²

∴ Area of circle is 314.285 cm², circumference is 40.85 cm, and diameter is 13 cm.

Given radius r = 8 cm

The diameter of the circle d = 2r

d = 2 x 8

d = 16 cm

Circumference of the circle C = 2πr

C = 2 x (22/7) x 8

= 352/7 = 50.28 cm

Area of the circle A = πr²

A = (22/7) x (8)²

= (22/7) x 64

= 1408/7

= 201.14 cm²

∴ Area of circle is 201.14 cm², the circumference is 50.28 cm, and diameter is 16 cm.

2. Find the area, circumference of the circle whose diameter is

(a) 12 cm

(b) 18 cm

Solution:

(a) 12 cm

Given diameter d = 12 cm

The Radius of the circle r = d/2

r = 12/2

r = 6 cm

Circumference of the circle C = 2πr

C = 2 x 3.14 x 6

C = 37.68 cm

Area of the circle A = πr²

A = 3.14 x (6)²

A = 3.14 x 36

A = 113.04 cm²

∴ Radius, circumference, and area of the circle is 6 cm, 37.68 cm, 113.04 cm².

(b) 18 cm

Given diameter d = 18 cm

The Radius of the circle r = d/2

r = 18/2

r = 9 cm

Circumference of the circle C = 2πr

C = 2 x 3.14 x 9

C = 56.52 cm

Area of the circle A = πr²

A = 3.14 x 9²

A = 3.14 x 81

A = 254.34 cm²

∴ Radius, circumference, and area of the circle is 9 cm, 56.52 cm, 254.34 cm².

Given diameter d = 15 cm

The Radius of the circle r = d/2

r = 15/2

r = 7.5 cm

Circumference of the circle C = 2πr

C = 2 x 3.14 x 7.5

C = 47.1 cm

Area of the circle A = πr²

A = 3.14 x (7.5)²

= 3.14 x 56.25

= 176.62

∴ The radius, circumference, and area of the circle is 7.5 cm, 47.1 cm, 176.62 cm².

3. Calculate the circumference of the circle whose area is 212 cm².

Solution:

Given that,

Area of the circle A = 212 cm²

πr² = 212

r² = 212/π

r² = (212/22) x 7

r² = 9.636 x 7

r = √(67.45)

r = 8.21 cm

Circumference of the Circle C = 2πr

C = 2 x (22/7) x 8.21

C = 361.24/7

C = 51.6 cm

∴ Circle circumference is 51.6 cm.

4. If the circumference of a circular sheet is 47 cm, find its area.

Solution:

Given that,

Circle circumference C = 47 cm

2πr = 47

r = 47/2π

r = 47/(2 x 3.14)

r = 47/6.28

= 7.48 cm

Area of the circle A = πr²

A = 3.14 x (7.48)²

A = 3.14 x 56 = 175.8 cm²

∴ The circle area is 175.8 cm².

5. A circular swimming pool is 25 feet in diameter. How many feet around is the pool?

Solution:

Given that,

The diameter of the circular swimming pool = 25 feet

Radius = 25/2 = 12.5 feet

Circumference of the swimming pool = 2πr

= 2 x 3.14 x 12.5 = 78.5 feet

∴ 78.5 feet is around the circular swimming pool.

6. The diameter of a bangle is 26 cm. How many times the bangle will revolve in order to travel a distance of 500 cm.

Solution:

Given that,

The diameter of the bangle d = 26 cm

Distance covered in 1 rotation = Circumference of the bangle

So, circumference C = πd

C = 3.14 x 26

C = 81.64 cm

Number of times bangle should revolve = 500/81.64

= 6.12

∴ 6 times the bangle should revolve to cover a distance of 500 cm.

7. From a circular sheet of a radius 7 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet.

Solution:

Given that,

The radius of the outer circle R = 7 cm

Radius of inner circle r = 3 cm

Area of the Outer Circle A = πr²

A = 3.14 x 7²

= 3.14 x 49

= 153.86 cm²

Area of the inner circle a = πr²

a = 3.14 x 3²

= 3.14 x 9

= 28.26 cm²

Area of the remaining sheet = Area of the outer circle – Area of the inner circle

= 153.86 – 28.26 = 125.6 cm².

8. If the circumference of a circle is 86 cm, find its area, diameter, and radius.

Solution:

Given that,

Circumference of the circle C = 86 cm

2πr = 86

r = 86/2π

r = 86/(2 x 3.14) = 86/6.28

r = 13.69 cm

The diameter of the circle d = 2r

d = 13.69 x 2 = 27.38 cm

Area of the circle A = πr²

A = 3.14 x (13.69)² = 3.14 x 13.69 x 13.69

A = 588.48 cm²

∴ Circle radius is 13.69 cm, the area is 588.48 cm², and diameter is 27.38 cm.

9. Find the perimeter of the semicircle whose diameter is 32 m.

Solution:

Given that,

Semicircle diameter = 32 cm

Semicircle radius = diameter/2

= 32/2 = 16 cm

The perimeter of the semicircle = πr

= 3.14 x 16

= 50.24 cm

∴ The perimeter of the semicircle is 50.24 cm.

10. The ratio of the radii of two wheels is 10: 13. Find the ratio of their circumference and areas.

Solution:

Given that,

The ratio of radii of two wheels is 10: 13.

The ratio of wheels circumferences = 2πr : 2πR

= r : R = 10 : 13

The ratio of wheels areas = πr² : πR²

= r² : R²

= 10² : 13²

= 100 : 169

∴ The ratio of wheels circumferences is 10: 13 and areas is 100 : 169.

11. The radius of a cycle wheel is 25 cm. Find the number of turns required to cover a distance of 1585 m.

Solution:

Given that,

The radius of the cycle r = 25 cm

Distance covered in 1 turn = Circumference of the cycle wheel

So, circumference C = 2πr

C = 2 x 3.14 x 25

C = 157 cm

Number of turns cycle wheel should revolve = (1585 x 100)/157

= 1,009.55

∴ 1,009.55 number of turns required to cover a distance of 1585 m.

12. A girl wants to make a square-shaped figure from a circular wire of a radius 49 cm. Determine the sides of a square.

Solution:

Given that,

radius r = 49 cm

Length of the wire = Circumference of the circle = 2πr

= 2 x (22/7) x 49 = 2156/7

= 308 cm

Let the side of the square be ‘s’.

The perimeter of the square = Length of the wire = 4s

s = 308/4

s = 77 cm

Therefore, the sides of the square is 77 cm.

13. To cover a distance of 5 km a wheel rotates 2500 times. Find the radius of the wheel?

Solution:

Given that,

Number of rotations = 2500

The total distance covered = 5 km

Circumference of the wheel = Distance covered in 1 rotation = 2πr

In 2500 rotations, The distance covered = 5 km = 500000 cm

Hence, in 1 rotation, the distance covered = 500000/2500

= 250 cm

But this is equal to the circumference. Hence, 2πr = 250 cm

r = 250/2π

= 250/(2 x 3.14)

= 250/6.28 = 39.8 cm

Therefore, the radius of the wheel is 39.8 cm.

14. A well of diameter 160 cm has a stone parapet around it. If the length of the outer edge of the parapet is 516 cm, find the width of the parapet.

Solution:

Given that,

The diameter of the well (d) = 160 cm

So, the radius of the well r = d/2 = 160/2 = 80 cm

The length of the outer edge of the parapet is 516 cm

2πR = 516 cm

R = 516/2π

R = 516/(2 x 3.14)

R = 82.16 cm

Now, the width of the parapet = (Radius of the parapet – Radius of the well)

= 82.16 – 80 = 2.16

Therefore, the width of the parapet is 2.16 cm.

15. Find the area of a circle whose circumference is the same as the perimeter of the square of side 15 cm.

Solution:

Given that,

Square side s = 15 cm

The perimeter of the square P = s²

P = 15² = 15 x 15 = 225

Circumference of the circle = Perimeter of the square

2πr = 225

r = 225/(2π) = 225/(2 x 3.15)

r = 225/6.28 = 35.8 cm

Area of the circle A = πr²

A = 3.14 x 35.8²

A = 3.14 x 35.8 x 35.8 = 4,030.65 cm²

Therefore, the area of the circle is 4,030.65 cm².

16. From a rectangular metal sheet of size 60 by 40, a circular sheet as big as possible is cut. Find the area of the remaining sheet.

Solution:

Given that,

The size of the rectangular metal sheet is 60 by 40.

The diameter of the largest circle = Length of the smallest side of the rectangle

Radius of the circle r = 40/2 = 20

Area of rectangle = l x b = 60 x 40 = 2400 cm²

Area of the circle = πr²

= 3.14 x (20)² = 3.14 x 400 = 1256 cm²

Remaining area = 2400 – 1256

= 1144 cm²

∴ The area of the remaining sheet is 1144 cm².

17. Two circles have areas in the ratio of 16 : 20. Find the ratio of their circumferences.

Solution:

Given that,

The ratio of areas of two circles = 16 : 20

The ratio of circumferences = √(Ratio of area)

= √(16 : 20)

= √(4 : 5)

= 2 : 2.23

Therefore, the ratio of circles circumferences is 2: 2.23.

18. A square metallic frame has a perimeter of 289 cm. It is bent in the shape of a circle. Find the area of the circle, the side length of the square.

Solution:

Given that,

The perimeter of the square = 289 cm

side² = 289

side = √(289)

side = 17

The perimeter of the square = Circumference of a circle

289 = 2πr

r = 289/2π

r = 289/(2 x 3.14)

= 289/6.28 = 46 cm

Area of the circle = πr²

= 3.14 x (46)²

= 6,644.24 cm²

Therefore, the circle area is 6,644.24 cm², side length of the square is 17 cm.

19. From a circular sheet of radius 18 cm, two circles of radii 4.5 cm and a rectangle of length 4 cm and breadth 2 cm are removed; find the area of the remaining sheet.

Solution:

Given that,

The radius of the circle r = 18 cm

Radii of two small circles = 4.5 cm

Length of the rectangle = 4 cm

The breadth of the rectangle = 2 cm

Area of the circle = πr²

= 3.14 x 18²

= 1,017.36 cm²

Area of the smallest circle = 3.14 x 4.5²

= 63.585

As there are 2 small circles so the total area of the circles is 63.64 x 2 = 127.28 square cm.

Area of rectangle = l x w = 4 x 2 = 8 cm²

Total area of cutouts = 127.28 + 8 = 135.28 cm²

Area of sheet left = 1,017.36 – 135.28 = 882.08 cm²

Therefore, area of the remaining sheet is 882.08 cm².

Worksheet on Area of the Path | Area of the Path Worksheet with Solutions

January 19, 2021 by Mounika Kandukuri

Are you looking for guidance on the concept of the Area of the path? Then, get the Worksheet on Area of the Path now. Here, you can get the detailed step by step explanation to solve each and every question. Solve all the questions from the Area of the Path Worksheet and re-check your answers here. By solving these problems, you will learn the concept easily and cover various models.

You can improve problem-solving skills and knowledge by solving the Area of the Path Worksheets frequently and assess your preparation standards. All questions provided here are framed by the subject experts to help the students to have a perfect knowledge of the concept.

1. A rectangular field is of dimensions 30 m × 15 m. Two paths run parallel to the sides of the rectangle through the centre of the field. The width of the longer path is 3 m and that of the shorter path is 2 m.

Find that:

(a) area of the paths

(b) area of the remaining portion of the field.

Solution:

Given that,

The dimensions of the rectangular field is 30 x 15

The width of the longer path is 3 m, the shorter path is 2 m.

The area of the rectangle = length x breadth

= 30 x 15 = 450

Area of the shorter path = 15 x 2 = 30 m

Area of the longer path = 30 x 3 = 90 m

Area of the middle path = 2 x 3 = 6

Area of paths = [Area of the shorter path] + [Area of the londer path] – [Area of the middle common path]

= 30 + 90 – 6

= 120 – 6 = 114

Area of the remaining portion of the field = The area of the rectangle – Area of path

= 450 – 114 = 336

Cost of constructing the roads at the rate of $10 per m²

The total cost of path = 114 x 10 = $1140

∴ The area of path is 114 m², The area of the remaining portion of the field is 336 m², Cost of constructing the roads at the rate of $10 per m² is $1140.

2. Find the area and perimeter of the following path.

Solution:

The perimeter of the path is the sum of all faces.

So, path perimeter = 4(20) + 2(4) + 2(8)

= 80 + 8 + 16

= 104

Area of ABLK = 20 x 4 = 80

Area of EFHG = 20 x 4 = 80

Area of CDIJ = 4 x 8 = 32

Area of the path = Area of ABLK + Area of EFGH + Area of CDIJ

= 80 + 80 + 32

= 192

∴ The area of path is 192 cm², the perimeter of the path is 104 cm.

3. A garden is 90 m long and 75 m board. A path 5 m wide is to be built outside all around it along its border. Find the area of the path?

Solution:

Given that,

The garden length is 90 m, width is 75 m.

Let us take ABCD as the garden.

Then, the area of path = Area of EFGH – Area of ABCD

Area of EFCGH = (90 + 5 + 5) x (75 + 5 + 5)

= 100 x 85 = 8500

Area of ABCD = 90 x 75

= 6750

Area of path = 8500 – 6750 = 1750

∴ The area of the path is 1750 m².

4. A grassy plot is 100 m x 60 m. Two cross paths each 5 m wide are constructed at right angles through the center of the field, such that each path is parallel to one of the sides of the rectangle. Find the total area used as a path.

Solution:

Area of the longer path (EFGH) = 100 x 5 = 500

Area of the shorter path (PQRS) = 5 x 60 = 300

Area of the centre field (IJKL) = 5 x 5 = 25

Area of the path = (Area of the longer path + Area of the shorter path – Area of the center field)

= 500 + 300 – 25

= 800 – 25 = 775 m².

∴ The total area used as path is 775 m².

5. A floor is 15 m long and 9 m wide. A square carpet of side 5 m is laid on the floor. Find the area of the floor not carpeted?

Solution:

Given that,

The length of the floor = 15 m

The breadth of the floor = 9 m

Side of the square = 5 m

Area of the square = side²

= 5² = 5 x 5

= 25 m²

Area of the floor = length x breadth

= 15 x 9 = 135

The area of floor not carpeted = Area of the floor – Area of carpet

= 135 – 25 = 110 m²

∴ The area of the floor not carpeted is 110 m².

6. Two crossroads, each of width 10 cm, run at right angles through the center of a rectangular garden of length 600 m and breadth 300 m and parallel to its sides.

Find:

(i) The area of the garden excluding crossroads.

(ii) The area of the crossroads

(iii) Convert these areas into hectares and ares.

Solution:

Given that,

(i) Length of the garden excluding crossroads = 600 – 10 = 590 m

Breadth of the garden excluding crossroads = 300 – 10 = 290

The area of the garden excluding crossroads = (length x breadth)

= 590 x 290 = 171100 m²

(ii) Length of the rectangular garden = 600 m

Breadth of the rectangular garden = 300 m

The total area of the rectangular garden = (length x breadth)

= (600 x 300) = 180000 m²

Area of the crossroads = The total area of the rectangular garden – The area of the garden excluding crossroads

= 180000 – 171100

= 8900 m²

(iii) We know that,

1 square metre = 0.0001

The area of the garden excluding crossroads = 171100 x 0.0001 = 17.11 hectares

Area of the crossroads = 8900 x 0.0001 = 0.89 hectares

1 square metre = 0.01 area

The area of the garden excluding crossroads = 171100 x 0.01 = 1711 ares

Area of the crossroads = 8900 x 0.01 = 89 ares

7. A square lawn is surrounded by a path 3 m wide. If the area of the path is 360 m². Find the area of the lawn.

Solution:

Given that,

Area of the path = 360 m²

Let the side of the lawn be x m.

Then, the area of the lawn = x²

Length of outer side = (x + 6) m

Area of outer square = (x + 6)²

= x² + 12x + 36

Area of the path = Area of the outer square – Area of the lower square

360 = x² + 12x + 36 – x²

360 = 12x + 36

12x = 360 – 36

12x = 324

x = 324/12

x = 27

Side of the lawn = 27 m

Area of the lawn = 27 x 27

= 729 m².

∴ Area of the lawn = 729 m²

8. A strip of width 3 m is cut out all round from a sheet of paper with dimensions 50 m × 30 m. Find the area of the strip cut out and the area of the remaining sheet.

Solution:

Given that,

Length of the sheet = 50 m

Breadth of the sheet = 30 m

Area of the sheet = 50 x 30 = 1500 m²

A strip of width 3 cm is cut out all round from a sheet of paper

Let that form a rectangle PQRS

Length of the rectangle PQRS = 50 – 3 – 3 = 50 – 6 = 44 m

Breadth of the rectangle = 30 – 3 – 3 = 30 – 6 = 24 m

Area of the rectangle PQRS = 44 x 24 = 1056 m²

Area of the strip cut out = Area of ABCD – Area of PQRS

= 1500 – 1056

= 444 m²

∴ The area of the strip cut out and the area of the remaining sheet is 444 m².

9. A path 5 m wide runs along inside a rectangular field. The length of the rectangular field is three times the breadth of the field. If the area of the path is 500 m², then find the length and breadth of the field.

Solution:

Let l be the length of the field, breadth of the field is b.

Breadth of the path = 5 m

The area of the path is 500 m².

2 x 5(l – 5) + 2 x 5(b – 5) = 500

10(l – 5) + 10(b – 5) = 500

10(l – 5 + b – 5) = 500l + b – 10 = 50

l + b = 50 + 10

l + b = 60 —- (i)

The length of the rectangular field is three times the breadth of the field

l = 3b —- (ii)

Substitute, l = 3b in equtaion (i)

3b + b = 60

4b = 60

b = 60/4

b = 15 m

Putting b = 15 in equation (i)

l + 15 = 60

l = 60 – 15

l = 45 m

∴ length and breadth of the field is 45 m, 15 m.

10. Four square flowerbeds each of the sides 4 m are dug on a piece of land 15 m long and 8 m wide. Find the area of the remaining portion of the land. Find the cost of leveling this land at a rate of $5 per 100 cm².

Solution:

Length of the land = 15 m

Breadth of the land = 8 m

Area of the land = 15 x 8

= 120 m²

Side of the square flowerbed = 4 m

Area of 1 square bed = side²

= 4² = 4 x 4 = 16

Area of 4 square bed = 16² = 16 x 16

= 256 m²

Area of the remaining portion = 256 – 16

= 240 m²

Cost of levelling 100 cm² = $5

Cost of levelling 0.0001 m² = $5

Cost of levelling of 240 m² = 240 x 50000 = 12000000

Hence, the area of the remaining portion of the land is 240 m² and the cost of leveling is $12000000.

Worksheet on Area and Perimeter of Squares | Questions on Area and Perimeter of a Square

January 18, 2021 by Mounika Kandukuri

Worksheet on Area and Perimeter of Squares contains different questions related to the square that helps the students to prepare for the exam. We have covered each and every model related to the square area, perimeter, and diagonal on this page. You can also get the step by step process to solve the Area and Perimeter Word Problems. Practice all the questions without fail and verify your answers from here.

Practicing questions from the Area and Perimeter of Squares Worksheet makes you feel comfortable at the exam. So that you can attempt all questions with 100% confidence and score better grades.

1. Find the perimeter and area of the following squares whose dimensions are:

(a) 8 m

(b) 6.8 cm

Solution:

(a)

Given that,

Square side length s = 8 m

Square perimeter p = 4s

p = 4 x 8 = 32 m

Square area A = s²

A = 8² = 8 x 8

A = 64 m²

∴ Square area is 64 m², perimeter is 32 m.

(b)

Given that,

Square side length s = 6.8 cm

Square perimeter p = 4s

p = 4 x 6.8

p = 27.2 cm

Square area A = s²

A = 6.8² = 6.8 x 6.8

A = 46.24 cm²

∴ The Square area is 46.24 cm², perimeter is 27.2 cm.

(c)

Given that,

Square side length s = 12 m

Square perimeter p = 4s

p = 4 x 12 = 48 m

Square area A = s²

A = 12²

A = 12 x 12 = 144 m²

∴ The Square area is 144 m², perimeter is 48 m.

2. Each side of a square is 2.6 cm. Find its perimeter, area?

Solution:

Given that,

Square side = 2.6 cm

Square perimeter = 4 x side length

P = 4 x 2.6

P = 10.4 cm

Square area = side²

= 2.6² = 2.6 x 2.6

= 6.76 cm²

∴ Square perimeter is 10.4 cm, area is 6.76 cm².

3. Find the perimeter of a square whose area is 120 m².

Solution:

Given that,

Square area A = 120 m²

Area A = Side²

120 = Side²

Side = √120

Side = 10.95

Square Perimeter P = 4side

P = 4 x 10.95

P = 43.817 m

∴ The perimeter of the square is 43.817 m.

4. Find the area of the square field whose perimeter is 240 m.

Solution:

The perimeter of the square p = 240 m

p = 4 x side

side = p / 4

side = 240 / 4

side = 60

Area of the square = side²

A = 60²

A = 60 x 60 = 3600

∴ Area of the square is 3600 m².

5. A rope of length of 104 m is used to fence a square garden. What is the length of the side of the garden?

Solution:

Given that,

The perimeter of the garden P = 104 m

We know that perimeter of a square = 4 × length of a side

So, 4 × length of a side = 104

The length of a side = 104/4

Side length = 26 m

∴ The length of the side of the garden is 26 m.

6. Lila has 16 square stamps of side 4 cm each. She glues them onto an envelope to form a bigger square. What area of the envelope does the bigger square cover?

Solution:

16 square-shaped stamps can be arranged as 4 in each row. So it forms 4 rows and 4 columns.

Side of the formed square s = 4 + 4 + 4 + 4

s = 16 cm

Area of the formed square A = 16 x 16

= 256

∴ The area of the bigger square is 256 cm².

7. If the diagonal length of a square is 7 cm, find the square area, perimeter?

Solution:

Given that,

The diagonal length of a square d = 7 cm

We know that, when you draw a diagonal in the square, it forms a right-angled triangle.

By using the Pythagorean theorem,

side² + side² = diagonal²

2side² = 7²

2side² = 49

side² = 49/2

side² = 24.5

side = √24.5

= 4.94 cm

The square perimeter p = 4 x side

p = 4 x 4.94

= 19.79 cm

The square area A = side²

A = 4.94² = 4.94 x 4.94

A = 24.4034 cm²

∴ The square area is 24.4034 cm², side length is 4.94 cm, and perimeter is 19.79 cm.

8. The diagonals of two squares are in the ratio 2:5. Find the ratio of their areas.

Solution:

Let us take the diagonals of two squares as 2x, 5x

Area of the square formula when diagonal is given,

A = (1/2) x d²

Area of the first square = (1/2) x (2x)²

= (1/2)(4x²)

= 2x²

Area of the second square = (1/2) x (5x)²

= (1/2) x (25x²)

= 12.5x²

The ratio of their areas = 2x² : 12.5x²

= 4 : 25

So, the ratio of the two squares is 4 : 25.

9. The areas of a square and rectangle are equal. If the side of the square is 15 cm and the breadth of the rectangle 10 cm, find the length of the rectangle and its perimeter.

Solution:

Given that,

Side of the square s = 15 cm

The breadth of the rectangle b = 10 cm

The areas of a square and rectangle are equal

Area of square = Area of Rectangle

s² = l x b

15² = l x 10

225 = l x 10

l = 225/10

l = 22.5

The perimeter of the rectangle p = 2(l + b)

= 2(22.5 + 10)

= 2(32.5) = 65

The perimeter of a square = 4 x side

= 4 x 15 = 60 cm

∴ The square, rectangle area is 225 cm², square perimeter is 60 cm, rectangle perimeter is 65 cm.

10. A wire is in the shape of a rectangle whose width is 12 cm is bent to form a square of side 17 cm. Find the rectangle length and also find which shape encloses more area.

Solution:

Given that,

Rectangle width w = 12 cm

Square side = 17 cm

The perimeter of a rectangle = Perimeter of a square

2(l + w) = 4side

2(l + 12) = 4 x 17

2l + 24 = 68

2l = 68 – 24

2l = 44

l = 44/2

l = 22

Area of square = side²

= 17² = 17 x 17

= 289

Area of the rectangle = l x b

= 22 x 12 = 264

∴ Rectangle length is 22 cm, and square has more area.

11. The area of a square field is 49 hectares. Find the cost of fencing the field with a wire at the rate of $5 per m.

Solution:

Given that,

Area of the square field = 49 hectares

1 hectare = 10000 sq.m.

So, 49 hectares = 49 x 10,000 = 4,90,000

So, the area of the square field = 4,90,000

Square area = side²

4,90,000 = side²

side = √(4,90,000)

side = 700

The perimeter of the square = 4 x side

P = 4 x 700 = 2800

Cost of fencing 1 m = $5

Cost of fencing 2800 m = 2800 x 5 = 14000

Hence The cost of fencing the square field is $14000.

12. How many square tiles of side 6 cm will be needed to fit in a square floor of a bathroom of side 600 cm. Find the cost of tiling at the rate of $65 per tile.

Solution:

Given that,

Side length of a tile = 6 cm

Side length of bathroom = 600 cm

Area of the square = 600 x 600 = 360000

Area of the tile = 6 x 6 = 36

Number of tiles = Area of the square / Area of the tile

= 360000 / 36 = 10000

Cost of 1 tile = $65

Cost of tiling = 65 x 10000 = 650000

∴ Number of square tiles required is 10000, cost of tiles is $650000.

13. If it costs $420 to fence a square field at the rate of $4 per m, find the length of the side and the area of the field.

Solution:

Given that,

The total cost of fencing = $420

The cost of fencing per m = $4

So, the Perimeter of the square field = 420/4

P = 105

Length of square = 105 / 4

Side = 26.25

Area of the square A = side²

A = (26.25)²

A = 26.25 x 26.25 = 689.0625

∴ The area of the field is 689.0625 m², side length is 26.25 m.

Worksheet on Area and Perimeter of Rectangles | Rectangle Area and Perimeter Worksheets with Answers

January 18, 2021 by Mounika Kandukuri

Refer to Worksheet on Area and Perimeter of Rectangles while preparing for the exam. You can perform well by practicing various questions from Rectangle Area and Perimeter Worksheets. Multiple models of questions along with the detailed solutions are given here. This Area and Perimeter of Rectangles Worksheet are designed as per the latest syllabus. So, students can learn the rectangles topic and score well in the examinations. Check out all the questions and rectangle area, rectangle perimeter formulas in the following sections of this page.

1. Find the perimeter and area of the following rectangles whose dimensions are:

(i) length = 11 cm, breadth = 5 cm

(ii) length = 6.7 cm, breadth = 5.1 cm

(iii) length = 14 m, breadth = 6 m

(iv) length = 8 feet, breadth = 3 feet

Solution:

(i)

Given that,

length = 11 cm, breadth = 5 cm

Rectangle Area = length x breadth

= 11 x 5 = 55 cm²

Rectangle Perimeter = 2(length + breadth)

= 2(11 + 5) = 16 x 2 cm

= 32 cm

∴ The Rectangle area is 55 cm², perimeter is 32 cm.

(ii)

Given that,

length = 6.7 cm, breadth = 5.1 cm

Rectangle Area = length x breadth

= 6.7 x 5.1 = 34.17 cm²

Rectangle Perimeter = 2(length + breadth)

= 2(6.7 + 5.1) = 2 x 11.8 cm

= 23.6

∴ The Rectangle area is 34.17 cm², perimeter is 23.6 cm.

(iii)

Given that,

length = 14 m, breadth = 6 m

Rectangle Area = length x breadth

= 14 x 6 = 84 m²

Rectangle Perimeter = 2(length + breadth)

= 2(14 + 6) = 20 x 2 = 40 m

∴ The Rectangle area is 84 m², perimeter is 40 m.

(iv)

Given that,

length = 8 feet, breadth = 3 feet

Rectangle Area = length x breadth

= 8 x 3 = 24 sq feet

Rectangle Perimeter = 2(length + breadth)

= 2(8 + 3) = 11 x 2 = 22 feet

∴ The Rectangle area is 24 sq feet, perimeter is 22 feet.

2. The area of a rectangle is 92 m², its length is 8 m. Find the rectangle breadth and perimeter?

Solution:

Given that,

Rectangle area = 92 m²

Rectangle length = 8 m

The rectangle area formula is

Area = length x breadth

So, breadth = area / length

Breadth = 92 / 8

= 11.5 m

Rectangle Perimeter = 2(Length + breadth)

= 2(8 + 11.5) = 2(19.5) = 39 m

∴ The Rectangle breadth is 11.5 m, perimeter is 39 m.

3. If the rectangle perimeter is 28 cm, its width is 18 cm. Find the rectangle length and area?

Solution:

Given that,

Rectangle Perimeter p = 28 cm

Width w = 18 cm

Rectangle perimeter p = 2(l + w)

28 = 2l + 36

2l = 28 – 36

2l = 8

l = 8/2

l = 4 cm

Rectangle area A = (l x w)

= 18 x 4 = 72 cm²

∴ The Rectangle length is 4 cm, area is 72 cm².

4. Find the cost of tiling a rectangular plot of land 250 m long and 500 m wide at the rate of $8 per hundred square m?

Solution:

Given that,

The rectangular plot length = 250 m

Rectangular plot width = 500 m

Area of rectangular polt = length x width

= 250 x 500 = 125000 sq. m

Cost of tiling = $8 per 100 sq. m = $8/100 per 1 sq. m

Cost of tiling of rectangular polt of 125000 sq. m = (8/100) x 125000 = 10,000

∴ The cost of tiling of the rectangular plot is Rs. 10,000/-.

5. A room is 4 feet long and 6 feet wide. How many square feet of carpet is needed to cover the floor of the room?

Solution:

Given that,

Rectangle length l = 4 feet

Rectangle width w = 6 feet

Area of the rectangle A = l x w

A = 4 x 6

A = 24 sq feet

So, 24 sq feet of carpet is needed to cover the room floor.

6. A table-top measures 5 m by 3 m 50 cm. What is the area and perimeter of the table?

Solution:

Given that,

Table length l = 5 m

width w = 3 m 50 cm

= 3 + 50 x (1/100)

= 3 + 1/2

= 7/2 = 3.5 m

Table Perimeter p = 2(l + w)

= 2(5 + 3.5) = 2(8.5) m

= 17 m

Table area A = l x w

A = 5 x 3.5

= 17.5 m²

∴ The table area is 17.5 m², the perimeter is 8.5 m.

7. A floor is 25 m long and 14 m wide. A square carpet of sides 8 m is laid on the floor. Find the area of the floor that is not carpeted and floor perimeter?

Solution:

Given that,

Rectangular floor-length l = 25 m

Rectangular floor breadth b = 14 m

Square carpet side s = 8 m

Rectangular floor area A = l x b

A = 25 x 14

A = 350 m²

Area of the square carpet a = s x s

a = 8 x 8

a = 64 m²

Area of the floor that is not carpeted = Rectangular floor area – Area of the square carpet

= A – a = 350 – 64

= 286 m²

Rectangular flooe perimeter P = 2(l + b)

P = 2(25 + 14)

P = 2(39) = 78 m

∴ Area of the floor that is not carpeted is 286 m², rectangular floor perimeter is 39 m.

8. How many tiles whose length and breadth are 15 cm and 6 cm respectively are needed to cover a rectangular region whose length and breadth are 510 cm and 135 cm?

Solution:

Given that,

Length of the tile l = 15 cm

Breadth of the tile b = 6 cm

Rectangular region length = 510

Rectangular region breadth = 135

Area of the tiles = l x b

= 15 x 6 = 90 cm²

Area of the plot = 510 x 135

= 68,850 cm²

Number of tiles required = Area of plot / Area of tiles

= 68850 / 90

= 765

Therefore, the required number of tiles are 765.

9. How many rectangles can be drawn with 22 cm as a perimeter? Also, find the dimensions of the rectangle whose area will be maximum?

Solution:

Given that,

The perimeter of the rectangle = 22 cm

2(l + w) = 22

(l + w) = 22/2

= 11

Possible dimensions of the rectangle are (1, 10), (2, 9), (3, 8), (4, 7), (5, 6), (6, 5), (7, 4), (10, 1), (9, 2), (8, 3), (7, 4).

∴ The dimensions of the rectangle whose area is maximum is (6, 5) or (5, 6) and 11 rectangles can be drawn.

10. The perimeter of a rectangular pool is 140 m, and its length is 60 m. Find the pool width and area?

Solution:

The rectangular pool perimeter p = 140 m

Rectangular pool length l = 60 m

Rectangular pool width w = ?

Rectangular pool perimeter p = 2(l + w)

140 = 2(60 + w)

140 = 120 + 2w

2w = 140 – 120

2w = 20

w = 20/2

w = 10

Rectangular pool area A = l x w

A = 60 x 10

A = 600

∴ The rectangular pool width is 10 m, area is 600 m².

11. The length of a rectangular wooden board is four times its width. If the width of the board is 155 cm, find the cost of framing it at the rate of $5 for 20 cm?

Solution:

Given that,

Width of a rectangular wooden board w = 155 cm

The cost of framing = $5 for 20 cm

The length of a rectangular wooden board l = 4w

l = 4 x 155

l = 620

Wooden board perimeter P = 2(l + w)

P = 2(620 + 155)

= 2(775) = 1550

The cost of framing = $5/20 for 1 cm

The cost of wooden board framing = 1550 x (5/20)

= 7750/20

= 387.5

∴ The cost of framing is $387.5.

Worksheet on Problems on Simultaneous Linear Equations | Simultaneous Equations Problems with Solutions

January 16, 2021 by Mounika Kandukuri

Students who want to get complete knowledge on Simultaneous Linear Equations Word problems can check this Worksheet on Problems on Simultaneous Linear Equations. Our System of Linear Equations Word Problems Worksheet is helpful to improve your preparation levels. It contains a number of examples of Simultaneous Linear Equations. So, practice all the questions from the Simultaneous Linear Equations Worksheet and develop your skills. Also, have a look at Worksheet on Simultaneous Linear Equations and prepare well for the exam.

1. The sum of the digits of a two-digit number is 7. When the digits are reversed, the number is increased by 27. Find the number.

Solution:

Let the two-digit number be xy.

According to the data provided in the question,

The sum of the digits of a two-digit number is 7.

x + y = 7 —– (i)

If the numbers are reversed, then the number is increased by 27.

The sm of two digits of the number as can be written as 10x + y

If the numbers are reversed, then the number can be written as 10y + x

10y + x = 10x + y + 27

10x + y + 27 – x – 10y = 0

9x – 9y = -27

9(x – y) = -27

x – y = -3 —– (ii)

Adding equations (i) & (ii)

x + y + x – y = 7 + (-3)

2x = 7 – 3

2x = 4

x = 4/2

x = 2

Substituting x = 2 in equation (i)

2 + y = 7

y = 7 – 2

y = 5

Therefore, the required 2 digit number is 25.

2. Mahesh bought 13 bushes and 4 trees from the nursery for the first time and its total cost was $487. For the second time, he bought 6 bushes and 2 trees and totaled $232. The bills do not list the per-item price. What were the costs of one bush and of one tree?

Solution:

Let the cost of each bush be x, one tree is y.

As per the question,

The first condition is he bought 13 bushes and 4 trees from the nursery for the first time and its total cost was $487

13x + 4y = 487 —- (i)

The second condition is he bought 6 bushes and 2 trees and totaled $232

6x + 2y = 232 —– (ii)

Multiply the second equation by 2.

2(6x + 2y = 232)

12x + 4y = 464 —– (iii)

Subtract equation (iii) from equation (i)

12x + 4y – (13x + 4y) = 464 – 487

12x + 4y – 13x – 4y = -23

-x = -23

x = 23

Putting x = 23 in equation (i)

13(23) + 4y = 487

299 + 4y = 487

4y = 487 – 299

4y = 188

y = 188/4

y = 47

So, costs of one bush is $23, and of one tree is $47.

3. The first number is six times the second number. The difference between the numbers is 60. Find those two numbers?

Solution:

Let the first number be x, the second number be y.

As per the given data,

The first condition is the first number is six times the second number.

x = 6y —- (i)

The second condition is the difference between the numbers is 60.

x – y = 60 —– (ii)

Put equation (i) in equation (ii)

6y – y = 60

5y = 60

y = 60/6

y = 10

Substituting y = 10 in equation (i)

x = 6 x 10

x = 60

Therefore, the two numbers are 60, 10.

4. One number is three times the other number. The sum of two numbers is 24. Find the two numbers?

Solution:

Assume that two numbers are x, y.

Given that,

One number is three times the other number.

x = 3y —– (i)

The sum of two numbers is 24.

x + y = 24 —– (ii)

Substituting x = 3y in equation (ii)

3y + y = 24

4y = 24

y = 24/4

y = 6

Putting y = 6 in equation (ii)

x + 6 = 24

x = 24 – 6

x = 18

so, the two numbers are 18, 6.

5. It takes 3 hours for a boat to travel 27 miles upstream. The same boat can travel 30 miles downstream in 2 hours. Find the speeds of the boat and the current?

Solution:

Let x be the speed of the boat (without current), let y be the speed of the current.

According to the given problem,

The boat can travel 27 miles upstream in 3 hours

Upstream speed = (Upstream distance)/(Upstream time)

Upstream speed = 27/3 = 9 miles/hour

The boat can travel 30 miles downstream in 2 hours.

Downstream speed = (Downstream distance) / (Downstream time)

Downstream speed = 30/2 = 15 miles/hour

The linear equation for upstream speed is x – y = 9 —— (i)

downstream speed is x + y = 15 ——- (ii)

Adding equations (i) and (ii)

x – y + x + y = 9 + 15

2x = 24

x = 24/2
x = 12

Substituting x = 12 in equation (ii)

12 + y = 15

y = 15 – 12

y = 3

Therefore, the speed of boat is 12 miles/hour, current is 3 miles/hour.

6. The sum of two numbers is 28 and their difference is 4. Find the two numbers?

solution

Let us take the two numbers as x, y.

As per the data given in the question,

The sum of two numbers is 28.

x + y = 28 —— (i)

The difference between the two numbers is 4.

x – y = 4 ——- (ii)

Adding equations (i) and (ii)

x + y + x – y = 28 + 4

2x = 32

x = 32/2

x = 16

Substitute x = 16 in equation (i)

16 + y = 28

y = 28 – 16

y = 12

So, the two numbers are 16, 12.

7. The sum of two numbers is 14. The difference in their squares is 28. Find the numbers?

Solution:

Assume that the two numbers are x and y.

According to the first condition in the question,

The sum of two numbers is 14.

x + y = 14 —— (i)

y = 14 – x

According to the second condition in the question,

x² – y² = 28 —- (ii)

Substitute y = 14 – x in equation (ii)

x² – (14 – x)² = 28

x² – (14² + x² – 28x) = 28

x² – 196 – x² + 28x = 28

28x – 196 = 28

28x = 28 + 196

28x = 224

x = 224/28

x = 4.6

Substitute x = 4.6 in equation (i)

4.6 + y = 14

y = 14 – 4.6

y = 9.4

Therefore, the two numbers are 4.6, 9.4.

8. The admission fee at a small fair is $1.50 for children and $4.00 for adults. On a certain day, 2200 people enter the fair and $5050 is collected. How many children and how many adults attended?

Solution:

Let us take the number of children who attended the fair as x, the number of adults who attended the fair as y.

The total number of people who attended the fair is 2200.

x + y = 2200 —- (i)

x = 2200 – y

The admission fee at a small fair is $1.50 for children and $4.00 for adults and the total amount collected is $5050.

1.5x + 4y = 5050 —— (ii)

Putting x = 2200 – y in the second equation.

1.5(2200 – y) + 4y = 5050

3300 – 1.5y + 4y = 5050

3300 + 2.5y = 5050

2.5y = 5050 – 3300

2.5y = 1750

y = 1750/2.5

y = 700

Substitute y = 700 in equation (i)

x + 700 = 2200

x = 2200 – 700

x = 1500

So, the number of children attended the small fair is 1500, the number of adults attended the fair is 700.

9. Emma has 23 notes of £20 and £5 in her handbag. The amount of money she has in the bag is £340.00. Find the number of notes of each type.

Solution:

Let us take the number of £20 notes as x, £5 notes as y.

The total number of notes in the handbag is 23.

x + y = 23 —- (i)

As per the second condition in the question,

20x + 5y = 340 —– (ii)

Multiply the equation (i) by 5.

5(x + y) = 23 x 5

5x + 5y = 115 —— (iii)

Subtracting equation (i) from equation (ii)

5x + 5y – (20x + 5y) = 115 – 340

5x + 5y – 20x – 5y = -225

-15x = -225

x = 225/15

x = 15

Putting x = 15 in equation (i)

15 + y = 23

y = 23 – 15

y = 8

So, Emma has 15 notes of £20, 8 notes of £5 in her handbag.

10. A family goes to the cinema. 4 adults and 2 child tickets cost $47, 1 adult and 3 child tickets cost $25.50. Calculate the costs of each adult and child ticket.

Solution:

Let us take the cost of the ticket for adults as x, cost of the ticket for the child as y.

According to the question,

4 adults and 2 child tickets cost $47, 1 adult and 3 child tickets cost $25.50.

4x + 2y = 47 —— (i)

x + 3y = 25.5 —– (ii)

x = 25.5 – 3y

Putting x = 25.5 – 3y in equation (i)

4(25.5 – 3y) + 2y = 47

102 – 12y + 2y = 47

102 – 10y = 47

102 – 47 = 10y

10y = 55

y = 55/10

y = 5.5

Substitute y = 5.5 in equation (ii)

x + 3(5.5) = 25.5

x + 16.5 = 25.5

x = 25.5 – 16.5

x = 9

Therefore, the ticket cost for adults is $9, the ticket cost for child is $5.5.

11. The area of a rectangle gets reduced by 10 square units if its length is reduced by 4 units and breadth is increase by 2 units. If we increased the length by 3 units and breadth by 4 units, the area is increased by 96 square units. Find the length and breadth of the rectangle.

Solution:

Let the length and breadth of the rectangle is l and b respectively.

The area of a rectangle is lb.

The area of a rectangle gets reduced by 10 square units if its length is reduced by 4 units and breadth is increased by 2 units.

(l – 4)(b + 2) = (lb – 10) sq units

lb + 2l – 4b – 8 = lb – 10

lb – lb – 2l + 4b = 10 – 8

4b – 2l = 2 —- (i)

If we increased the length by 3 units and breadth by 4 units, the area is increased by 96 square units.

(l + 3)(b + 4) = (lb + 96)

lb + 3b + 4l + 12 = lb + 96

lb + 3b + 4l – lb = 96 – 12

3b + 4l = 84 —— (ii)

Multiplying equation (i) by 2

4b – 2l = 2 x 2

8b – 4l = 4 —- (iii)

Adding equation (iii) in equation (ii)

8b – 4l + 3b + 4l = 84 + 4

11b = 88

b = 88/11

b = 8

Substituting b = 11 in equation (i)

4(11) – 2l = 2

44 – 2 = 2l

42 = 2l

l = 42/2

l = 21

So, the length and breadth of the rectangle is 21 units, 8 units.

12. If 5 is added to the numerator and 9 is subtracted from the denominator it becomes 21/12 and if 3, 8 is subtracted from the numerator and denominator it becomes 4/5. Find fractions.

Solution:

Let the numerator of the fraction be x, the denominator of the fraction be y.

As stated in the question,

If 5 is added to the numerator and 9 is subtracted from the denominator it becomes 21/12.

(x + 5)/(y – 9) = 21/12

Cross multiply the fractions.

12(x + 5) = 21(y – 9)

12x + 60 = 21y – 189

12x – 21y + 189 + 60 = 0

12x – 21y + 249 = 0 —- (ii)

If 3, 8 is subtracted from the numerator and denominator it becomes 4/5.

(x – 3)/(y – 8) = 4/5

Cross multiply the fractions.

5(x – 3) = 4(y – 8)

5x – 15 = 4y – 32

5x – 4y = -32 + 15

5x – 4y = -17

5x – 4y + 17 = 0

5x = 4y – 17

x = (4y – 17)/5 — (ii)

substitute equation (ii) value in equation (i)

12[(4y – 17)/5] – 21y + 249 = 0

[(48y – 204)/5] – 21y + 249 = 0

48y – 204 – 105y + 1245 = 0

-57y + 1041 = 0

57y = 1041

y = 1041/57

y = 18.26

Putting y = 18.26 in equation (ii)

x = (4(18.26) – 17)/5

x = (73.052 – 17)/5

x = 56.0526/5

x = 11.210

Therefore, the required fraction is 1121/1826.

13. Chole and Tino have a combined age of 48. Three years ago Chole was double the age Tino is now. Find the present ages of Chole and Tino.

Solution:

Let us take the age of chole and Tino as x, y.

As per the data given in the question,

The combined age of Chole and Tino is 48.

x + y = 48 —- (i)

Three years ago Chole was double the age Tino is now

x – 2y = 3

x = 3 + 2y —– (ii)

Putting x value in equation (i)

3 + 2y + y = 48

3 + 3y = 48

3y = 48 – 3

3y = 45

y = 45/3

y = 15

Substituting y = 15 in equation (ii)

x = 3 + 2(15)

x = 3 + 30

x = 33

So, the present age of Chole is 33 years, Tino is 15 years.

Worksheet on Simultaneous Linear Equations | Word Problems on System of Linear Equations

January 16, 2021 by Mounika Kandukuri

Worksheet on Simultaneous Linear Equations is available here. Students can get various methods to solve the system of linear equations from this page. One can learn how to solve two simultaneous equations in two variables using comparison, substitution, elimination, and cross-multiplication methods. You can even find different types of problems on linear equations in two variables. So, practice as many questions as possible for a better understanding of the concept.

This System of Linear Equations in Two Variables Worksheet page includes the questions with detailed step by step solution. Therefore, have a look at these problems and practice them to score good marks in the examination easily.

1. Use the comparison method to solve the following simultaneous linear equations:

(i) x + y = 1, x – 2y = 5

(ii) 6x + 7y = 5, 2x – 3y – 8 = 0

(iii) 9x – 6y = 12, 4x + 6y = 14

(iv) x – y = -1, 2y + 3x = 12

Solution:

(i) x + y = 1, x – 2y = 5

Given pair of the system of linear equations are

x + y = 1 ——– (1)

x – 2y = 5 ——- (2)

Express x in terms of y

From equation (1) x + y = 1, we get

x = 1 – y

From equation (2), we get

x = 5 + 2y

Equate the values of x obtained from both equations.

1 – y = 5 + 2y

2y + y = 1 – 5

3y = -4

y = -4/5

Substitute y = -4/5 in equation (1)

x – 4/5 = 1

x = 1 + 4/5

x = (5 + 4)/5 = 9/5

Therefore, the required solution is x = 9/5, y = -4/5.

(ii) 6x + 7y = 5, 2x – 3y – 8 = 0

Given simultaneous linear equations are

6x + 7y = 5 ——- (1)

2x – 3y – 8 = 0 ——– (2)

Express x in terms of y

From equation (1) 6x + 7y = 5, we get

6x = 5 – 7y

x = (5 – 7y)/6

From equation (2) 2x – 3y – 8 = 0, we get

2x = 3y + 8

x = (3y + 8)/2

Equate the values of x obtained from both equations.

(5 – 7y)/6 = (3y + 8)/2

2(5 – 7y) = 6(3y + 8)

10 – 14y = 18y + 48

10 – 48 = 18y + 14y

32y = -38

y = -38/32 = -19/16

Putting y = -19/16 in equation (2)

2x – 3(-19/16) – 8 = 0

2x + 57/16 – 8 = 0

2x + (57 – 128)/16 = 0

2x -71/16 = 0

2x = 71/16

x = 71/32

Therefore, the required solution set is x = 71/32, y = -19/16.

(iii) 9x – 6y = 12, 4x + 6y = 14

Given the system of linear equations are

9x – 6y = 12 —— (i)

4x + 6y = 14 —— (ii)

Express y in terms of x

From equation (i), we get

9x – 12 = 6y

y = (9x – 12)/6 —— (iii)

From equation (ii), we get

6y = (14 – 4x)

y = (14 – 4x)/6 ——- (iv)

Equation equation (iii) and equation (iv)

(9x – 12)/6 = (14 – 4x)/6

Cross multiply the fractions

6(9x – 12) = 6(14 – 4x)

9x – 12 = 14 – 4x

9x + 4x = 14 + 12

13x = 26

x = 26/13

x = 2

Putting x = 2 in equation (ii)

4(2) + 6y = 14

8 + 6y = 14

6y = 14 – 8

6y = 6

y = 6/6

y = 1

Therefore, the required solution set is x = 2, y = 1.

(iv) x – y = -1, 2y + 3x = 12

Given pair of linear equations are

x – y = -1 —— (i)

2y + 3x = 12 —— (ii)

Express y in terms of x

From equation (i), we get

x + 1 = y

From equation (ii), we get

2y = (12 – 3x)

y = (12 – 3x)/2

Equate the values of x obtained from both equations.

x + 1 = (12 – 3x)/2

2(x + 1) = (12 – 3x)

2x + 2 = 12 – 3x

2x + 3x = 12 – 2

5x = 10

x = 10/5

x = 2

Substituting x = 2 in equation (ii)

2y + 3(2) = 12

2y + 6 = 12

2y = 12 – 6

2y = 6

y = 6/2

y = 3

Therefore, the required solution is x = 2, y = 3.

2. Solve the following simultaneous equations by using the Elimination Method:

(i) 2x – y = 5, x + 3y – 9 = 0

(ii) 2x – 3y = 1, 3x – 4y = 1

(iii) (2x/3) + (y/2) = -1, (-x/3) + y = 3

Solution:

(i) 2x – y = 5, x + 3y – 9 = 0

Given the system of linear equations are

2x – y = 5 —– (i)

x + 3y = 9 —— (ii)

Multiply the equation (i) by 3.

3(2x – y) = 3(5)

6x – 3y = 15 —– (iii)

Add equation (ii) and equation (iii)

x + 3y + 6x – 3y = 9 + 15

7x = 24

x = 24/7

Substitute x = 24/7 in equation (ii)

24/7 + 3y = 9

3y = 9 – 24/7

3y = (63 – 24)/7

y = 39/21

Therefore, the required solution is x = 24/7, y = 39/21.

(ii) 2x – 3y = 1, 3x – 4y = 1

Given the system of linear equations are

2x – 3y = 1 —- (i)

3x – 4y = 1 —- (ii)

Multiply equation (i) by 3

3(2x – 3y) = 3(1)

6x – 9y = 3 —— (iii)

Multiply equation (ii) by 2

2(3x – 4y) = 2(1)

6x – 8y = 2 —– (iv)

Subtract equation (iii) from equation (iv)

(6x – 9y) – (6x – 8y) = 3 – 2

6x – 9y – 6x + 8y = 1

-y = 1

y = -1

Putting y = -1 in equation (iv)

6x – 8(-1) = 2

6x + 8 = 2

6x = 2 – 8

6x = -6

x = -1

Therefore, the required solution set is x = -1, y = -1.

(iii) (2x/3) + (y/2) = -1, (-x/3) + y = 3

Given pair of linear equations are

(2x/3) + (y/2) = -1

(4x + 3y)/6 = -1

(4x + 3y) = -6 —– (i)

(-x/3) + y = 3

(-x + 3y)/3 = 3

-x + 3y = 9 —– (ii)

Subtract equation (i) from equation (ii)

-x + 3y – (4x + 3y) = 9 – (-6)

-x + 3y – 4x – 3y = 9 + 6

-5x = 15

x = -15/5

x = -3

Put x = -3 in equation (ii)

-(-3) + 3y = 9

3 + 3y = 9

3y = 9 – 3

3y = 6

y = 6/3

y = 2

Therefore, the required solution is x = -3, y = 2.

3. Solve each other pair of the equation given below using the Substitution Method:

(i) x + y = 12, x – y = 2

(ii) 2x – 5y = 9, 4x – y = 9

(iii) x – y = 5, 2x – y = 11

Solution:

(i) x + y = 12, x – y = 2

Given pair of simultaneous linear equations are

x + y = 12 —— (i)

x – y = 2 —— (ii)

From equation (ii)

x = y + 2

Substitute the obtained x value in equation (i)

(y + 2) + y = 12

2y + 2 = 12

2y = 12 – 2

2y = 10

y = 10/2

y = 5

Putting y = 5 in equation (ii)

x – 5 = 2

x = 2 + 5

x = 7

Therefore, the required solution is x = 7, y = 5.

(ii) 2x – 5y = 9, 4x – y = 9

Given pair of linear equations are

2x – 5y = 9 —– (i)

4x – y = 9 —– (ii)

From equation (ii), we can write

4x – 9 = y

Substituting the obtained y value in equation (i)

2x – 5(4x – 9) = 9

2x – 20x + 45 = 9

-18x = 9 – 45

-18x = -36

x = 36/18

x = 2

Putting x = 2 in equation (ii)

4(2) – y = 9

8 – 9 = y

y = -1

Therefore, the required solution is x = 2, y = -1.

(iii) x – y = 5, 2x – y = 11

Given simultaneous linear equations are

x – y = 5 —— (i)

2x – y = 11 —— (ii)

From equation (i), we can write as

x = 5 + y

Substituting the new x value in equation (ii)

2(5 + y) – y = 11

10 + 2y – y = 11

10 + y = 11

y = 11 – 10

y = 1

Substitute y = 1 in equation (i)

x – 1 = 5

x = 5 + 1

x = 6

Therefore, the solution is x = 6, y = 1.

4. Solve the below-mentioned simultaneous linear equations by the method of cross-multiplication:

(i) 2x + y = 4, x + y = 6

(ii) 4x – 3y = 7, 3x – y = 4

(iii) 3x + y = 8, 4x + 3y = 14

Solution:

(i) 2x + y = 4, x + y = 6

The transposition of the given simultaneous linear equations are

2x + y – 4 = 0 ——– (i)

x + y – 6 = 0 ———- (ii)

Multiply equation (i) by 1 and equation (ii) by 1, we get

1(2x + y – 4) = 1 x 0

2x + y – 4 = 0 ——- (iii)

1(x + y – 6) = 1x 0

x + y – 6 = 0 ——- (iv)

Subtract equation (iv) from equation (iii)

[2x + y – 4 = 0] – [x + y – 6 = 0]

2x + y – 4 – x – y + 6 = 0

x + 2 = 0

x = -2

Substitute x = -2 in equation (ii)

-2 + y – 6 = 0

y – 8 = 0

y = 8

Therefore, the required solution set is x = -2, y = 8.

(ii) 4x – 3y = 7, 3x – y = 4

The transposition of the given system of linear equations are

4x – 3y – 7 = 0 —— (i)

3x – y – 4 = 0 ——– (ii)

Multiply the equation (i) by -1 and equation (ii) by -3, we get

-1(4x – 3y – 7) = -1 x 0

-4x + 3y + 7 = 0 ——— (iii)

-3(3x – y – 4) = -3 x 0

-9x + 3y + 12 = 0 ——- (iv)

Subtract equation (iv) from equation (iii)

[-4x + 3y + 7] – [-9x + 3y + 12] = 0

-4x + 3y + 7 + 9x – 3y – 12 = 0

5x – 5 = 0

5x = 5

x = 5/5

x = 1

Putting x = 1 in equation (i)

4(1) – 3y – 7 = 0

4 – 3y – 7 = 0

-3y – 3 = 0

-3y = 3

y = -3/3

y = -1

Therefore, the required solution set is x = 1, y = -1.

(iii) 3x + y = 8, 4x + 3y = 14

The transposition of the given linear equations are

3x + y – 8 = 0 —– (i)

4x + 3y – 14 = 0 —— (ii)

Multiplying the equation (i) by 3, we get

3(3x + y – 8) = 0 x 3

9x + 3y – 24 = 0 —– (iii)

Multiplying the equation (ii) by 1, we get

1(4x + 3y – 14) = 1 x 0

4x + 3y – 14 = 0 —— (iv)

Subtracting equation (iv) from equation (iii)

[9x + 3y – 24 = 0] – [4x + 3y – 14 = 0]

(9x + 3y – 24) – (4x + 3y – 14) = 0

9x + 3y – 24 – 4x – 3y + 14 = 0

5x – 10 = 0

5x = 10

x = 10/5

x = 2

Putting x = 2 in equation (ii)

4(2) + 3y – 14 = 0

8 + 3y – 14 = 0

3y – 6 = 0

3y = 6

y = 6/3

y = 2

Therefore, the required solution set is x = 2, y = 2.

5. Solve the following simultaneous equations:

(i) 4/(x – 3) + 6/(y – 4) = 5, 5/(x – 3) – 3/(y – 4) = 1

(ii) (y/6) – (x/15) = 4, (y/3) – (x/12) = 19/4

Solution:

(i) 4/(x – 3) + 6/(y – 4) = 5, 5/(x – 3) – 3/(y – 4) = 1

Given simultaneous linear equations are

4/(x – 3) + 6/(y – 4) = 5 —– (i)

5/(x – 3) – 3/(y – 4) = 1

5/(x – 3) – 1 = 3/(y – 4) —— (ii)

Putting equation (ii) in equation (i)

4/(x – 3) + 2(5/(x – 3) – 1) = 5

4/(x – 3) + 10/(x – 3) – 2 = 5

14/(x – 3) = 5 + 2

14/(x – 3) = 7

2/(x – 3) = 1

2 = (x – 3)

x = 2 + 3

x = 5

Substituting x = 5 in equation (i)

4/(5 – 3) + 6/(y – 4) = 5

4/2 + 6/(y – 4) = 5

2 + 6/(y – 4) = 5

6/(y – 4) = 5 – 2

6/(y – 4) = 3

2/(y – 4) = 1

2 = y – 4

y = 2 + 4

y = 6

Therefore, the required solution set is x = 5, y = 6

(ii) (y/6) – (x/15) = 4, (y/3) – (x/12) = 19/4

Given linear equations are

(y/6) – (x/15) = 4

(15y – 6x) /90 = 4

15y – 6x = 90 x 4

15y – 6x = 360 ——- (i)

(y/3) – (x/12) = 19/4

(12y – 3x)/36 = 19/4

4(12y – 3x) = 19 x 36

(12y – 3x) = 171 —- (ii)

Express y in terms of x

12y = 171 + 3x

y = (171 + 3x) / 12

Substitute the obtained y value in equation (i)

15[(171 + 3x) / 12] – 6x = 360

3(5[(171 + 3x) / 12] – 2x) = 360

5[(171 + 3x) / 12] – 2x = 120

(855 + 15x)/12 – 2x = 120

855 + 15x – 24x = 120 x 12

855 – 9x = 1440

9x = 855 – 1440

9x = -585

x = -585/9

x = -65

Putting the value of x in equation (ii)

(12y – 3(-65)) = 171

12y + 195 = 171

12y = 171 – 195

12y = -24

y = -24/12

y = -2

Therefore, the required solution is x = -65, y = -2.