An isosceles right triangle has area $112.5$ sq. cm. The length of its hypotenuse (in cm) is

15

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15 \sqrt{2}

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16

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16 \sqrt{2}

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Solution

The correct option is B $15 \sqrt{2}$
Let equal and pendicular side of triangle be $a$
$\frac{1}{2} a^{2} = 112.5 \Rightarrow a^{2} = 225.0 \Rightarrow a = 15$
$\Rightarrow H y p . = \sqrt{15^{2} + 15^{2}} = 15 \sqrt{2} c m$ .

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An isosceles right triangle has area 112.5 sq. cm. The length of its hypotenuse (in cm) is

The correct option is B 15√2Let equal and pendicular side of triangle be a12a2=112.5⇒a2=225.0⇒a=15⇒Hyp.=√152+152=15√2cm.

An isosceles right triangle has area $112.5$ sq. cm. The length of its hypotenuse (in cm) is

The correct option is B $15 \sqrt{2}$
Let equal and pendicular side of triangle be $a$
$\frac{1}{2} a^{2} = 112.5 \Rightarrow a^{2} = 225.0 \Rightarrow a = 15$
$\Rightarrow H y p . = \sqrt{15^{2} + 15^{2}} = 15 \sqrt{2} c m$ .