If AB = 12 , cm,BC = 16 , cm and AB is perpen | vedclass

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(A) $AB$ is perpendicular to $BC$,therefore $\Delta ABC$ is a right-angled triangle with $\angle B = 90^{\circ}$.Since the triangle is right-angled,th

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$10$

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(A) $AB$ is perpendicular to $BC$,therefore $\Delta ABC$ is a right-angled triangle with $\angle B = 90^{\circ}$.Since the triangle is right-angled,the hypotenuse $AC$ is the diameter of the circle passing through points $A, B$ and $C$.Using the Pythagorean theorem in $\Delta ABC$:$AC = \sqrt{AB^2 + BC^2}$$AC = \sqrt{12^2 + 16^2}$$AC = \sqrt{144 + 256}$$AC = \sqrt{400} = 20 \, cm$Since the diameter $AC = 20 \, cm$,the radius $r = \frac{AC}{2} = \frac{20}{2} = 10 \, cm$.

If $AB = 12 \, cm$,$BC = 16 \, cm$ and $AB$ is perpendicular to $BC$,then the radius of the circle passing through the points $A, B$ and $C$ is (in $cm$)

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