Sum

The areas of two similar triangles ∆ABC and ∆PQR are 25 cm^{2} and 49 cm^{2} respectively. If QR = 9.8 cm, find BC

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#### Solution

It is being given that ∆ABC ~ ∆PQR, ar (∆ABC) = 25 cm^{2} and ar (∆PQR) = 49 cm^{2} .

We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

`\therefore \text{ }\frac{ar\ (\Delta ABC)}{ar\ (\DeltaPQR)}=\frac{BC^{2}}{QR^{2}}`

`\Rightarrow \frac{25}{49}=\frac{x^{2}}{(9.8)^{2}}`

`\Rightarrowx^{2}=( \frac{25}{49}\times 9.8\times 9.8)`

`\Rightarrow x=\sqrt{\frac{25}{49}\times 9.8\times 9.8}=(5/7xx9.8)=(5xx1.4)=7`

Hence BC = 7 cm.

Concept: Areas of Similar Triangles

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