In Delta ABC,AB = 8 text{ cm} and BC = 10 tex | vedclass

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Question

(A) According to the triangle inequality theorem,the sum of any two sides of a triangle must be greater than the third side.Let the third side be $AC

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(A) According to the triangle inequality theorem,the sum of any two sides of a triangle must be greater than the third side.Let the third side be $AC = x 	ext{ cm}$.Then,$AB + BC > AC \implies 8 + 10 > x \implies x Also,$AB + AC > BC \implies 8 + x > 10 \implies x > 2$.And $BC + AC > AB \implies 10 + x > 8$,which is true for all $x > 0$.So,$2 The perimeter $P = AB + BC + AC = 8 + 10 + x = 18 + x$.Since $x Therefore,the perimeter is less than $36 	ext{ cm}$.

In $\Delta ABC$,$AB = 8 \text{ cm}$ and $BC = 10 \text{ cm}$. Then the perimeter of $\Delta ABC$ is less than how many centimeters?

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