The parallel sides of a trapezium …

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The parallel sides of a trapezium are 20 cm and 10 cm and the length of each of the non - parallel sides is 13 cm .Find the area of trapezium.

Posted by Sundram Jha 4 years, 11 months ago

CBSE > Class 08 > Mathematics

1 answers

Yogita Ingle 4 years, 11 months ago

ABCD be the given trapezium in which AB = 20 cm, DC = AE= 10 cm, BC = 13 cm and AD = 13cm.
Through C, draw CE || AD, meeting AB at E.
Draw CF ⊥ AB.
Now, EB = (AB - AE) = (AB - DC)
EB = (20- 10) cm = 10 cm;
CE = AD = 13 cm; AE = DC = 13 cm.
Now, in ∆EBC, we have CE = BC = 13 cm.
It is an isosceles triangle.
Also, CF ⊥ AB
So, F is the midpoint of EB.
Therefore, EF = ¹/₂ × EB = 1/2× 10= 5cm.
Thus, in right-angled ∆CFE, we have CE = 13 cm, EF = 5 cm.
By Pythagoras’ theorem, we have
CF = [√CE² - EF²]
CF = √(13² - 5²)
CF= √169-25= √144 = √12×12
CF= 12cm
Thus, the distance between the parallel sides is 12 cm.
Area of trapezium ABCD = ¹/₂ × (sum of parallel sides) × (distance between them)
Area of trapezium ABCD = ¹/₂ × (20 + 10) × 12 cm²
Area of trapezium ABCD = 1/2×(30)× 12
Area of trapezium ABCD= 30×6 = 180 cm²
Hence, Area of trapezium ABCD= 180 cm²

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CBSE > Class 08 > Mathematics