Question

the parallel sides of the trapezium are 10cm and 20am.Its non parallel sides are both equal each being 13cm.Find the area of the trapezium

Answer 1

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

Answer 2

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