Question

An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.​

Answer 1

Answer:

Given: An isosceles triangle has Perimeter 30 cm. & each of the equal sides of triangle is 12 cm.

Need to find: Area of the triangle?

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So, we will use Heron's Formula to find the Area of Triangle (let's say ∆ABC).

✇ If the perimeter of given isosceles triangle is 30 cm then the semi perimeter of the isosceles triangle would be 15 cm. i.e ( s ) = 15 cm.

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( I ) Finding third Side :

» As We know that, Perimeter of triangle is equal to sum of all sides of triangle. & The perimeter is Given that is 30 cm. Therefore:

$:\implies\bf Perimeter_{\:(triangle)} = \Big\{a + b + c\Big\}\\\\\\:\implies\sf 30 = 12 + 12 + c\\\\\\:\implies\sf 30 = 24 + c\\\\\\:\implies\sf 30 - 24 = c\\\\\\:\implies\sf 6 = c.$

∴ Hence, third side of the triangle is 6 cm.

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( II ) Area of Triangle :

$\star\:\underline{\boxed{\pmb{\sf{Area_{\;(triangle)} = \sqrt{s\Big(s - a\Big)\Big(s - b\Big)\Big(s - c\Big)}}}}}$

$\frak{Sides}\begin{cases}\sf{\quad a =\bf{12\;cm.}}\\\sf{\quad b =\bf{12\;cm.}}\\\sf{\quad c=\bf{6\;cm.}}\\\sf{\quad s = \bf{15\:cm.}}\end{cases}$

$:\implies\sf Area_{\;(triangle)} = \sqrt{15\Big(15 - 12\Big)\Big(15 - 12\Big)\Big(15 - 6\Big)}\\\\\\:\implies\sf Area_{\;(triangle)} = \sqrt{15 \times 3 \times 3 \times 9} \\\\\\:\implies\sf Area_{\:(triangle)} = \sqrt{15 \times 9 \times 9}\\\\\\:\implies{\pmb{\boxed{\frak{Area_{\;(triangle)} = 9\sqrt{15}\;cm^2}}}}\;\bigstar$

$\therefore{\underline{\sf{Hence,\;the\;area\;of\; isosceles\; triangle\;is\;{\pmb{\sf{9\sqrt{15}\;cm^2}}}.}}}$

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Answer 2

Answer:

Step-by-step explanation:

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