The areas of two similar triangles are 81 cm2 and 49 cm2 respectively. Find the ratio of their corresponding heights. What is the ratio of their corresponding medians?

We have,


ΔABC ~ ΔPQR


Area (ΔABC) = 81 cm2


Area (ΔPQR) = 49 cm2


And AD and PS are the altitudes


By area of similar triangle theorem


Area of ΔABC/Area of ΔPQR = AB2 /PQ2


Or, (81)2/(49)2 = AB2/ PQ2


Or, 9/7 = AB/PQ …………………(i) (Taking square root)


In ΔABD and ΔPQS


<B = <Q (ΔABC ~ ΔPQR)


<ADB = < PSQ (Each 90°)


Then, ΔABD ~ ΔPQS (By AA similarity)


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


Compare (1) and (2)


AD/PS = 9/7


So, Ratio of altitudes = 9/7


Hence, ratio of altitudes = Ratio of medians = 9:7


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