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Question
In ΔABC, AB = 8cm, AC = 10cm and ∠B = 90°. P and Q are the points on the sides AB and AC respectively such that PQ = 3cm ad ∠PQA = 90. Find: The area of ΔAQP.
Solution
In ΔAQP and ΔABC
∠A = ∠A
∠PQA = ∠ABC ...(right angles)
Therefore, ΔAQP ∼ ΔABC
By Pythagoras theorem,
BC2 = AC2 - AB2
⇒ BC2 = 102 - 82
⇒ BC2 = 100 - 64
⇒ BC2 = 36
⇒ BC = 6cm
Area (ΔABC) `(1)/(2) xx "AB" xx "BC"`
Area (ΔABC) `(1)/(2) xx 8 xx 6`
Area (ΔABC) = 24cm2
Since ΔAQP ∼ ΔABC
`("Area"(Δ"AQP"))/("Area"(Δ"ABC")) = "PQ"^2/"BC"^2`
`("Area"(Δ"AQP"))/("Area"(Δ"ABC")) = 3^2/6^2`
⇒ Area(ΔAQP) = `(9 xx 24)/(36)`
⇒ Area(ΔAQP) = 6cm2 .
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