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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.
Solution
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.
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