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.