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Question
The areas of two similar triangles are 100 cm2 and 49 cm2 respectively. If the altitude the bigger triangle is 5 cm, find the corresponding altitude of the other.
Solution
We have, ΔABC ~ ΔPQR
Area(ΔABC) = 100 cm2,
Area (ΔPQR) = 49 cm2
AD = 5 cm
And AD and PS are the altitudes
By area of similar triangle theorem
`("Area"(triangleABC))/("Area"(trianglePQR))="AB"^2/"PQ"^2`
`rArr100/49="AB"^2/"PQ"^2`
`rArr10/7="AB"/"PQ"` ............(i)
In ΔABD and ΔPQS
∠B = ∠Q [ΔABC ~ ΔPQR]
∠ADB = ∠PSQ [Each 90°]
Then, ΔABD ~ ΔPQS [By AA similarity]
`therefore"AB"/"PQ"="AD"/"PS"` .........(ii)[Corresponding parts of similar Δ are proportional]
Compare (i) and (ii)
`"AD"/"PS"=10/7`
`rArr5/"PS"=10/7`
`rArr"PS"=(5xx7)/10=3.5` cm
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