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