Question

Δ ABC ∼ Δ PQR. AD and PS are altitudes from A and P on sides BC and QR respectively. If AD : PS = 4 : 9 , find the ratio of the areas of Δ ABC and Δ PQR.

Answer 1

Given : AD : PS = 4 : 9 and Δ ABC ∼ Δ PQR 

To Find : `("Ar" triangle  "ABC")/("Ar" triangle  "PQR")`

Sol : `("Ar" triangle  "ABC")/("Ar" triangle  "PQR") = ("AB"^2)/("PQ"^2)`   .......(1)

[The ratio of area of two similar triangles is equal to the ratio of square of their corresponding sides]

In Δ BAD and Δ QPS

∠ B = ∠ Q   (Δ ABC ∼ Δ PQR)

∠AOB = ∠PSQ   (90° each) 

Δ BAD ∼ Δ QPS    (AA corollary)

`therefore "AB"/"PQ" = "AD"/"PS"`    ........(2)  (similar sides of similar triangles)

Using (1) and (2)

`("Ar" triangle  "ABC")/("Ar" triangle  "PQR") = "AD"^2/"PS"^2 = (4/9)^2 = 16/81`

Required ratio is 16 : 81.