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Question
The areas of two similar triangles are 25 cm2 and 36 cm2 respectively. If the altitude of the first triangle is 2.4 cm, find the corresponding altitude of the other.
Solution
We have,
ΔABC ~ ΔPQR
Area (ΔABC) = 25 cm2
Area (ΔPQR) = 36 cm2
AD = 2.4 cm
And AD and PS are the altitudes
To find: PS
Proof: Since, ΔABC ~ ΔPQR
Then, by area of similar triangle theorem
`("Area"(triangleABC))/("Area"(trianglePQR))="AB"^2/"PQ"^2`
`rArr25/36="AB"^2/"PQ"^2`
`rArr5/6="AB"/"PQ"` .........(i)
In ΔABD and ΔPQS
∠B = ∠Q [Δ ABC ~ ΔPQR]
∠ADB ~ ∠PSQ [Each 90°]
Then, ΔABD ~ ΔPQS [By AA similarity]
`therefore"AB"/"PS"="AD"/"PS"` ......(ii) [Corresponding parts of similar Δ are proportional]
Compare (i) and (ii)
`"AD"/"PS"=5/6`
`rArr2.4/"PS"=5/6`
`"PS"=(2.4xx6)/5=2.88` cm
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