In Delta ABC,m angle C = 90^circ,AB = 12.5 an | vedclass

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(C) Given that $\Delta ABC$ is a right-angled triangle with $m \angle C = 90^\circ$.According to the Pythagorean theorem,in a right-angled triangle,th

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$3.5$

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(C) Given that $\Delta ABC$ is a right-angled triangle with $m \angle C = 90^\circ$.According to the Pythagorean theorem,in a right-angled triangle,the square of the hypotenuse is equal to the sum of the squares of the other two sides.Here,$AB$ is the hypotenuse,and $AC$ and $BC$ are the other two sides.Therefore,$AB^2 = AC^2 + BC^2$.Substituting the given values: $(12.5)^2 = AC^2 + (12)^2$.$156.25 = AC^2 + 144$.$AC^2 = 156.25 - 144$.$AC^2 = 12.25$.Taking the square root of both sides,$AC = \sqrt{12.25} = 3.5$.

In $\Delta ABC$,$m \angle C = 90^\circ$,$AB = 12.5$ and $BC = 12$. Find $AC$.

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