The area of an isosceles triangle having base | vedclass

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Question

(A) The sides of the isosceles triangle are $a = 4 \, cm$,$b = 4 \, cm$,and $c = 2 \, cm$.The semi-perimeter $s$ is calculated as:$s = \frac{a + b + c

Vedclass · Accepted Answer

$\sqrt{15} \, cm^2$

Vedclass · Answer

(A) The sides of the isosceles triangle are $a = 4 \, cm$,$b = 4 \, cm$,and $c = 2 \, cm$.The semi-perimeter $s$ is calculated as:$s = \frac{a + b + c}{2} = \frac{4 + 4 + 2}{2} = \frac{10}{2} = 5 \, cm$.Using Heron's formula,the area of the triangle is:$	ext{Area} = \sqrt{s(s - a)(s - b)(s - c)}$Substituting the values:$	ext{Area} = \sqrt{5(5 - 4)(5 - 4)(5 - 2)}$$	ext{Area} = \sqrt{5 	imes 1 	imes 1 	imes 3}$$	ext{Area} = \sqrt{15} \, cm^2$.

The area of an isosceles triangle having base $2 \, cm$ and the length of one of the equal sides $4 \, cm$ is

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