# Find the area of the trapezium ABCD in which AB is parallel to DC, AB=18cm, B=C90°,CD=12 and AD=10cm

2
by MATHO

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by MATHO

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AB II DC

AB = 18 cm

∠B = ∠C = 90°

CD = 12 cm

AD = 10 cm

Now from ∠B we will draw a line which will be parallel to AD and equals to AD

. This line will intersect CD at E.

We have a parallelogram with ED = 12 cm.

CE = 6cm

Now we have an ΔBCE

using Pythagoras theorem,

BE²= BC² + CE²

10 ² = BC² + 6²

100 - 36 = BC ²

64 = BC²

√64 = BC

8 = BC

so we have the height of the trapezium = 8 cm

Now area of trapezium = 1/2 × (base 1 + base2) × h

= 1/2 ×(12 + 18 ) × 8

= 120 cm²