The areas of two similar triangles ∆ABC and ∆DEF are 144 cm2 and 81 cm2 respectively. If the longest side of larger ∆ABC be 36 cm, then the longest side of the smaller triangle ∆DEF is
Question:

The areas of two similar triangles ∆ABC and ∆DEF are 144 cm2 and 81 cm2 respectively. If the longest side of larger ∆ABC be 36 cm, then the longest side of the smaller triangle ∆DEF is

(a) 20 cm
(b) 26 cm
(c) 27 cm
(d) 30 cm

Solution:

Given: Areas of two similar triangles $\triangle \mathrm{ABC}$ and $\triangle \mathrm{DEF}$ are $144 \mathrm{~cm}^{2}$ and $81 \mathrm{~cm}^{2}$.

If the longest side of larger ΔABC is 36cm

To find: the longest side of the smaller triangle ΔDEF

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

$\frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DEF})}=\left(\frac{\text { longest side of larger } \Delta \mathrm{ABC}}{\text { longest side of smaller } \Delta \mathrm{DEF}}\right)^{2}$

$\frac{144}{81}=\left(\frac{36}{\text { longest side of smaller } \triangle \mathrm{DEF}}\right)^{2}$

Taking square root on both sides, we get

129=36longest side of smaller ∆DEF

longest side of smaller $\triangle \mathrm{DEF}=\frac{36 \times 9}{12}=27 \mathrm{~cm}$

Hence the correct answer is C

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