Question

an isosceles triangle has a perimeter of 30 cm and each of the equal side is 12 cm. find the area of triangle.

Answer 1

Hope u like my process
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Though there are many formulae..

I prefer this one..

= >$area \: \: of \: \: triangle = \sqrt{s(s - a)(s - b)(s - c)}$
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Where

=> s = semi perimeter

=> a, b, c= three sides of triangle.

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=> Perimeter of triangle = 30 cm

=> semi perimeter = 30/2 = 15 cm

=> length of 2 sides (a, b) = 12 cm

=> 3rd side (c) = 30 - 12 - 12 = 6 cm
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Now..

=> Area of triangle

$=\sqrt{15(15 - 12)(15 - 12)(15 - 6)} \\ \\ = (15 - 12) \sqrt{15 (15 - 6)} \\ \\ = 3 \sqrt{15 \times 9} \\ \\ = 3 \times 3 \sqrt{15} = \bf \underline{9 \sqrt{15} \: \: cm {}^{2} }$
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Hope this is ur required answer

Proud to help you

Answer 2

✍✍HERE IS YOUR ANSWER..⬇⬇

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$\underline{SOLUTION}$ ➡

$Let, \\ \\ Base \: \: of \: \: isosceles \: \: triangle \: = b \\ \\ Length \: \: of \: \: each \: \: of \: \: each \: \: equal \: \: sides \: = a \\ \\ So, \: \:Perimeter \: of \: the \: isosceles \: triangle \: 2s = a + a + b \\ \\ = 2a + b$

$Given ,\: \\ \\Length \: \: of \: \: each \: \: equal \: \: sides \: a=12 \: \: cm \\ \\ Perimeter \: \: of \: \: the \: triangle = 30 \: \: cm \\ \\ = > 2 a+ b = 30 \\ \\ = > 2 \times 12 + b = 30 \\ \\ = > 24 + b = 30 \\ \\ = > b= 30 - 24 \\ \\ = > b=6 \\ \\ base \: o f\: the \: \: triangle = 6 \: \: cm$

$We \: \: know \: \: that, \\ \\ Area \: of \: isosceles \: triangle = \frac{b}{4} \sqrt{4a {}^{2} - b {}^{2} } \\ \\ = \frac{6}{4} \sqrt{4(12) {}^{2} - (6) {}^{2} } \\ \\ = \frac{3}{2} \sqrt{4 \times 144 - 36} \\ \\ = \frac{3}{2} \sqrt{576 - 36} \\ \\ = \frac{3}{2 } \sqrt{540} =\frac{3}{2}\sqrt{9×4×15}\\ \\=\frac{3}{2}×3×2\sqrt{15}\\ \\ = 9\sqrt{15}$

•°• AREA OF THE ISOSCELES TRIANGLE = $9\sqrt{15} \: \: \: sq.cm$


$\underline{ANSWER}$ ➡ $\boxed{9\sqrt{15} \: \: \: sq.cm}$
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