Mid-points of a triangle containing sides 12 | vedclass

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Question

(C) The semi-perimeter $(s)$ of the given triangle with sides $a = 12 \, cm$,$b = 18 \, cm$,and $c = 26 \, cm$ is calculated as:$s = \frac{12 + 18 + 2

Vedclass · Accepted Answer

$4 \sqrt{35} \, cm^2$

Vedclass · Answer

(C) The semi-perimeter $(s)$ of the given triangle with sides $a = 12 \, cm$,$b = 18 \, cm$,and $c = 26 \, cm$ is calculated as:$s = \frac{12 + 18 + 26}{2} = \frac{56}{2} = 28 \, cm$.Using Heron's formula,the area of the triangle is:Area $= \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{28(28-12)(28-18)(28-26)}$Area $= \sqrt{28 	imes 16 	imes 10 	imes 2} = \sqrt{8960} = \sqrt{256 	imes 35} = 16 \sqrt{35} \, cm^2$.The triangle formed by joining the mid-points of the sides of a triangle has an area equal to one-fourth of the area of the original triangle.Area of the new triangle $= \frac{1}{4} 	imes 16 \sqrt{35} = 4 \sqrt{35} \, cm^2$.

Mid-points of a triangle containing sides $12 \, cm$,$18 \, cm$,and $26 \, cm$ are joined to each other. Find the area of the triangle formed by joining the mid-points.

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