If $△ A B C$ is similar to $△ D E F$ such that $B C = 3$ cm, $E F = 4$ cm and area of $△ A B C = 54 {cm}^{2} .$ Find the area of $△ D E F .$ (in cm $^{2}$ )

54

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36

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72

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96

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Solution

The correct option is C $96$
Since the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides,
Therefore, $\frac{a r (△ A B C)}{a r (△ D E F)} = \frac{B C^{2}}{E F^{2}}$
$\Rightarrow$ $\frac{54}{a r (△ D E F)} = \frac{3^{2}}{4^{2}}$
Thus $a r (△ D E F) = \frac{54 \times 16}{9} = 96 {cm}^{2}$

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If △ABC is similar to △DEF such that BC=3 cm, EF=4 cm and area of △ABC=54cm2. Find the area of △DEF. (in cm2)

The correct option is C 96Since the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides,Therefore, ar(△ABC)ar(△DEF)=BC2EF2 ⇒ 54ar(△DEF)=3242 Thus ar(△DEF)=54×169=96cm2

If $△ A B C$ is similar to $△ D E F$ such that $B C = 3$ cm, $E F = 4$ cm and area of $△ A B C = 54 {cm}^{2} .$ Find the area of $△ D E F .$ (in cm $^{2}$ )