Question

altitude down to the base of an isosceles triangleis 8 cm and the perimeter is 28 cm find the area of a triangle ​

Answer 1

$\huge\mathbb{SOLUTION:-}$

$\bold{Given:-}\begin{cases}\sf{Base\:of\: isosceles\:\triangle}\\ \sf{perimeter=28cm}\\ \sf{base=8cm}\end{cases}$

• Let the equal sides of isosceles triangle be x

$\bf\underline{\underline{\red{According\:to\: Question:-}}}$

$\sf perimeter\: of\: \triangle =sum\:of\:all\:sides\:of\:\triangle\\ \\ \sf\mapsto 28=x+x+8\\ \\ \sf\mapsto 28-8=2x\\ \\ \sf\mapsto 20=2x\\ \\ \sf\mapsto \cancel{\frac{20}{2}}=x\\ \\ \sf\implies x=10cm$

• The equal sides of isosceles triangle is 10 cm
• Now we have the three sides of triangle which is 10cm ,10cm and 8cm

• we have to find the area of∆

• By herons formula of triangle we find the area of triangle

$\boxed{\sf{Area\:of\: \triangle=\sqrt{s(s-a)(s-b)(s-c)}}}$

• Where s is the semi perimeter

$\sf\mapsto s=\frac{a+b+c}{2}\\ \\ \sf\mapsto s=\frac{8+10+10}{2}\\ \\ \sf\mapsto s=\cancel{\frac{28}{2}}=14\\ \\ \sf\mapsto semi\: perimeter=14cm$

• Now the area of ∆

$\bold{we\:have:-}\begin{cases}\sf{Sides\:of\: isosceles\:\triangle}\\ \sf{10cm\:10cm\:8cm}\end{cases}$

$\sf\longmapsto Area=\sqrt{14(14-10)(14-10)(14-8)}\\ \\ \sf\longmapsto Area\:of\:triangle=\sqrt{14\times 4\times 4\times 6}\\ \\ \sf\longmapsto Area\:of\: triangle \sqrt{7\times 2\times 4\times 4\times 3\times 2}\\ \\ \sf\mapsto Area\:of\: triangle=8\sqrt{21}cm{}^{2}$

• The area of isosceles triangle is
• $\sf\implies 8\sqrt{21}cm{}^{2}$

$\boxed{\boxed{\sf{\purple{Area\:of\: triangle=8\sqrt{21}cm{}^{2}}}}}$

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