Question

ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that :

1. ΔADE ∼ ΔACB.
2. If AC = 13 cm, BC = 5 cm and AE = 4 cm. Find DE and AD.
3. Find, area of ΔADE : area of quadrilateral BCED.

Answer 1

i. Consider ΔADE and ΔACB

∠A = ∠A  ...[Common]

m∠A = m∠E = 90°

Thus, by Angle-Angle similarity, triangles ΔACB ∼ ΔADE

ii. Since ΔADE ∼ ΔACB, their sides are proportional

`=> (AE)/(AB) = (AD)/(AC) = (DE)/(BC)`  ...(1)

In ΔABC, by Pythagoras Theorem, we have

AB² + BC² = AC²

`=>` AB² + 5² = 13²

`=>` AB = 12 cm

From equation 1 we have

`4/12 = (AD)/13 = (DE)/5`

`=> 1/3 = (AD)/13`

`=> AD = 13/3 cm`

Also, `4/12 = (DE)/5`

`=> DE  = 20/12 = 5/3 cm`

iii. We need to find the area of ΔADE and quadrilateral BCED

Area of ΔADE = `1/2 xx AE xx DE`

= `1/2  xx 4 xx 5/3`

= `10/3 cm^3`

Area of quadrilateral BCED = Area of ΔABC – Area of ΔADE

= `1/2 xx BC xx AB - 10/3`

= `1/2 xx 5 xx 12 - 10/3`

= `30 - 10/3`

= `80/3 cm^2`

Thus, ratio of areas of ADE to quadrilateral BCED = `(10/3)/(80/3) = 1/8`