The circum-radius of the triangle whose sides | vedclass

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(B) First,observe that the sides $5, 12, 13$ satisfy the Pythagorean theorem: $5^2 + 12^2 = 25 + 144 = 169 = 13^2$. Thus,the triangle is a right-angle

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$13/2$

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(B) First,observe that the sides $5, 12, 13$ satisfy the Pythagorean theorem: $5^2 + 12^2 = 25 + 144 = 169 = 13^2$. Thus,the triangle is a right-angled triangle with the hypotenuse $c = 13$. For a right-angled triangle,the circum-radius $R$ is half of the hypotenuse. $R = \frac{	ext{hypotenuse}}{2} = \frac{13}{2} = 6.5$. Alternatively,using the formula $R = \frac{abc}{4\Delta}$: $s = \frac{5+12+13}{2} = 15$. $\Delta = \sqrt{15(15-5)(15-12)(15-13)} = \sqrt{15 	imes 10 	imes 3 	imes 2} = \sqrt{900} = 30$. $R = \frac{13 	imes 12 	imes 5}{4 	imes 30} = \frac{780}{120} = \frac{13}{2}$.

The circum-radius of the triangle whose sides are $13, 12$ and $5$ is

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