Question

In the given figure, BC is parallel to DE. Area of triangle ABC = 25 cm², Area of trapezium BCED = 24 cm² and DE = 14 cm. Calculate the length of BC. Also, find the area of triangle BCD.

Answer 1

In ΔABC and ΔADE,

As BC || DE, corresponding angles are equal

∠ABC = ∠ADE

∠ACB = ∠AED

ΔABC ~ ΔADE

∴ `(ar(ΔABC))/(ar(ΔADE)) = (BC^2)/(DE^2)`

`25/49 = (BC^2)/14^2`

(ar(∆ADE) = ar(∆ABC) + ar(trapezium BCED))

BC² = 100

BC = 10 cm

In trapezium BCED,

Area = `1/2 ("Sum of parallel sides") xx h`

Given: Area of trapezium BCED

= 24 cm², BC = 10 cm, DE = 14 cm

∴ `24 = 1/2(10 + 14) xx h`

`=> h = 48/((10 + 14))`

`=>h = 48/24`

`=>` h = 2

Area of ΔBCD = `1/2 ` × base × height

= `1/2 xx BC xx h`

= `1/2 xx 10 xx 2`

∴ Area of ∆BCD = 10 cm²