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

$\;\;\underline{\textbf{\textsf{ Given:-}}}$

• Perimeter of triangle is = 24cm

( Where each of the equal side is 9cm)

$\;\;\underline{\textbf{\textsf{ To Find :-}}}$

• The area of the triangle

$\;\;\underline{\textbf{\textsf{ Solution :-}}}$

Let the unequal side of the isosceles triangle be x cm.

Given that,

Perimeter of triangle is 24 cm.

Let the sides of the triangle be

a = 9cm , b = 9cm and c or x cm.

Then,

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

Hence,

$\underline{\textsf{ The other side of the triangle is \textbf{6cm}}}.$

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Now, find the semi-perimeter of the triangle.

We know that,

➝s = (a+b+c)/2

➝ s = 24/2 cm

➝ s = 12cm

$\underline{\:\textsf{ Using Heron's formula:}}$

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

$\;\;\underline{\textbf{\textsf{ Hence-}}}$

$\underline{\textsf{The area of the triangle is \textbf{18√2 cm² }}}.$

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Answer 2

Answer:

$18\sqrt{2$ $cm^{2}$

Step-by-step explanation:

We know that an isosceles triangle has 2 sides equal and the other side inequivalent. Hence, each of the equal side = 9 cm.

9 * 2 = 18 cm ; Perimeter = 24 cm (given).

Other side = 24 cm  - 18 cm = 6 cm.

Area (by heron's formula) ;

Heron's formula : Area = $\sqrt{s(s-a)(s-b)(s-c)$, where :

s = semi perimeter (perimeter/ 2)

a, b, c = sides.

We get $\sqrt{12(12 - 6)(12 - 9)(12 - 9)$

This results in $\sqrt{12 * 6 * 3 * 3)$

And the answer is : $18\sqrt{2$ $cm^{2}$