Question

In the following figure, DE || BC, then:

1. If DE = 4 cm, BC = 8 cm, A(ΔADE) = 25 cm², find A(ΔABC).
2. If DE : BC = 3 : 5, then find A(ΔADE) : A(`square`DBCE).

Answer 1

In ΔABC, DE || BC, DE = 4 cm, BC = 8 cm and A(△ADE) = 25 cm²

i. In ΔADE and ΔABC

∠ABC ≅ ∠ADE  ...(Common angle of triangle)

∠BAC ≅ ∠DAE  ...(Corresponding angles)

∴ ΔABC ∼ ΔADE  ...(By AA similarity criteria)

∴ By area of similarity theorem,

`(A(triangle ADE))/(A(triangle ABC)) = (DE^2)/(BC^2)`

`25/(A(ΔABC)) = 4^2/8^2`

`25/(A(ΔABC)) = 16/64`

(ΔABC) × 16 = 25 × 64

A(ΔABC) = `(25 xx 64)/16`

A(ΔABC) = 25 × 4

A(ΔABC) = 100 cm²

ii. In ΔABC and ΔADE,

∠BAC ≅ ∠DAE   .....(Common angle)

∠ADE ≅ ∠ABC  .....(Corresponding angles of parallel lines)

∠AED ≅ ∠ACB  .....(Corresponding angles of parallel lines)

∴ ΔАВС ∼ ΔADE   ....(By AAA test for similarity of triangles)

Now, by using areas of similar triangle theorem,

`(A(triangle ADE))/(A(triangle ABC)) = (DE^2)/(BC^2)`

`(A(triangle ADE))/(A(triangle ABC)) = 3^2/5^2`

`(A(triangle ADE))/(A(triangle ABC)) = 9/25`

∴ `(A(triangle ADE))/(A(triangle ADE) + A(squareDBCE)) = 9/25`

25 × A(ΔADE) = 9[A(ΔADE) + A(`square`DBCE)]

25A(ΔADE) = 9A(ΔADE) + 9A(`square`DBCE)

25A(ΔADE) − 9A(ΔADE) = 9A(`square`DBCE)

16A(∆ADE) = 9A(`square`DBCE)

`(A(∆ADE))/(A(squareDBCE)) = 9/16`

∴ A(△ADE) : A(`square`DBCE) = 9 : 16