In a triangle ABC,AB + BC = 12 text{ cm},BC + | vedclass

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(B) Given:$AB + BC = 12 	ext{ cm}$$BC + CA = 14 	ext{ cm}$$CA + AB = 18 	ext{ cm}$Adding these three equations:$(AB + BC) + (BC + CA) + (CA + AB) =

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$\frac{7}{2}$

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(B) Given:$AB + BC = 12 	ext{ cm}$$BC + CA = 14 	ext{ cm}$$CA + AB = 18 	ext{ cm}$Adding these three equations:$(AB + BC) + (BC + CA) + (CA + AB) = 12 + 14 + 18$$2(AB + BC + CA) = 44$$AB + BC + CA = 22 	ext{ cm}$The perimeter of the triangle is $22 	ext{ cm}$.Let $r$ be the radius of the circle. The perimeter (circumference) of the circle is $2 \pi r$.According to the problem,the perimeter of the circle is equal to the perimeter of the triangle:$2 \pi r = 22$$2 	imes \frac{22}{7} 	imes r = 22$$r = \frac{22 	imes 7}{2 	imes 22}$$r = \frac{7}{2} 	ext{ cm}$

In a triangle $ABC$,$AB + BC = 12 \text{ cm}$,$BC + CA = 14 \text{ cm}$,and $CA + AB = 18 \text{ cm}$. Find the radius of the circle (in $\text{cm}$) which has the same perimeter as the triangle.

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