In Delta ABC,m angle B = 90^{circ}. If a = 9 | vedclass

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(D) In $\Delta ABC$,since $m \angle B = 90^{\circ}$,the triangle is a right-angled triangle.According to the Pythagorean theorem,the square of the hyp

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$15$

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(D) In $\Delta ABC$,since $m \angle B = 90^{\circ}$,the triangle is a right-angled triangle.According to the Pythagorean theorem,the square of the hypotenuse is equal to the sum of the squares of the other two sides.Here,$b$ is the hypotenuse (side opposite to $\angle B$).Therefore,$b^{2} = a^{2} + c^{2}$.Substituting the given values: $b^{2} = 9^{2} + 12^{2}$.$b^{2} = 81 + 144 = 225$.Taking the square root on both sides,$b = \sqrt{225} = 15$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$. If $a = 9$ and $c = 12$,then $b = \dots$

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