Question

an isosceles triangle has perimeter 24 cm and each of the equal side is 9 cm find the area of the triangle
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Answer 1

Answer

The area of the triangle is 18√2cm²

$\bf\large\underline{Given}$

• An isosceles triangle has perimeter 24cm and each of the equal side is 9cm

$\bf\large\underline{To \: Find}$

• The area of the triangle

$\bf\large\underline{Solution}$

Let us consider the unequal side of the isosceles triangle be x cm

$\sf\ Perimeter \: of \: triangle = 24 \\\\ \sf\implies 9 + 9 + x = 24 \\\\ \sf\implies 18+x = 24 \\\\ \sf\implies x = 24-18 \\\\ \sf\implies x = 6$

Therefore , the other side of the triangle is 6cm

Let the sides of the triangle be

a = 9cm , b = 9cm and c = 6cm

again ,

➝ s = 24/2 cm

➝ s = 12cm

Now from herons formula we have :

$\sf \longrightarrow \sqrt{s(s-a)(s-b)(s - c) } \\\\ \sf\longrightarrow \sqrt{12(12- 9)(12-9)(12-6)} \\\\ \sf\longrightarrow \sqrt{12\times3\times3\times 6} \\\\ \sf\longrightarrow \sqrt{6\times2\times3^{2}\times6} \\\\ \sf\longrightarrow \sqrt{6^{2}\times3^{2}\times2}\\\\ \sf\longrightarrow 6\times3\sqrt{2} \\\\ \sf\longrightarrow 18\sqrt{2}$

Therefore, the area of the triangle is 18√2 cm²