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.

We have,

ΔABC ~ Δ PQR


Area (ΔABC) = 25 cm2


Area (PQR) = 36 cm2


And 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 of ΔABC/Area of ΔPQR = AB2 /PQ2


25/36 = AB2/PQ2


5/6 = AB/PQ………………..(i)


In ΔABD and Δ PQS


<B = <Q (ΔABC ~ ΔPQR)


<ADB = <PSQ (Each 90°)


Then, ΔABD ~ Δ PQS (By AA similarity)


So, AB/PS = AD/PS…………(ii) (Corresponding parts of similar Δ are proportional )


Compare (i) and (ii)


AD/PS = 5/6


2.4/PS = 5/6


PS = 2.4 x 6/5


PS = 2.88cm


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