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Question
ΔABC ~ ΔDEF and their areas are respectively `100cm^2` and `49cm2`. If the altitude of ΔABC is 5cm, find the corresponding altitude of ΔDEF.
Solution
It is given that ΔABC ~ ΔDEF.
Therefore, the ration of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
Also, the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.
Let the altitude of ΔABC be AP, drawn from A to BC to meet BC at P and the altitude of ΔDEF be DQ, drawn from D to meet EF at Q.
`(ar(Δ ABC))/(ar(ΔDEF))=(AP^2)/(DQ^2)`
⇒` 100/49=5^2/(DQ^2)`
⇒ `100/49=25/(DQ^2)`
⇒ `DQ^2=(49xx25)/100`
⇒`DQ=sqrt((49xx25)/100)`
⟹ 𝐷𝑄=3.5 𝑐𝑚
Hence, the altitude of ΔDEF is 3.5 cm
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