1. ## Similarity..help!

In △ABC, <B=90, AB=12cm, and AC=15cm. D and E are points on AB and AC respectively such that <AED=90 and DE=3cm. Find:
i) Area of △ADE
ii) Area of quadrilateral BCED : Area of △ABC

2. well....i managed to work out i) on my own-

△AED~△ABC->[by A.A. similarity rule]
So, ar△AED:ar△ABC= (DE/BC)^2
=> ar△AED: 1/2BC.AB=(3/9)^2
=> ar△AED= 54*1/9
= 6 sq.cm

So somebody now plz solve the ii) part for me...

3. Originally Posted by snigdha
In △ABC, <B=90, AB=12cm, and AC=15cm. D and E are points on AB and AC respectively such that <AED=90 and DE=3cm. Find:
i) Area of △ADE
ii) Area of quadrilateral BCED : Area of △ABC
$AB=3AD$

$BC=3DE$

Therefore the area of triangle ABC is 3(3)=9 times the area of triangle ADE.

Therefore the quadrilateral DBCE, being triangle ABC with triangle ADE cut off has area 9(6)-6=8(6)

The ratio of the areas is 8:9