Question

B.) The perimeter of an equilateral triangle measures 63cm. Find the length of each side.​

Answer 1

Given: The area of an isosceles triangle whose Perimeter is 16 cm. & The Base of the isosceles triangle is 6 cm.

Need to find: Area of the triangle?

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» Let's say, the equal sides of the isosceles triangle be x cm.

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$\underline{\bf{\dag} \:\mathfrak{As\;we\;know\: that\: :}}$⠀⠀⠀⠀

Perimeter of the triangle is sum of all sides. & Perimeter is Given that is 16 cm.

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$:\implies\bf \Big\{Perimeter = x + x + c\Big\}\\\\\\:\implies\sf 2x + 6 = 16\\\\\\:\implies\sf 2x = 16 - 6\\\\\\:\implies\sf 2x = 10\\\\\\:\implies\sf x = \cancel\dfrac{10}{2}\\\\\\:\implies{\pmb{\boxed{\frak{x = 5}}}}$

∴ Hence, the sides of the isosceles triangle are 5 cm, 5 cm & 6 cm respectively.

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✇ & If the perimeter of the Given Isosceles triangle is 16 cm then the semi – perimeter of the triangle would be 8 cm. i.e ( s ) = 8 cm.

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• C a l c u l a t i n G⠀A r e a :

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

$\frak{Sides}\begin{cases}\sf{\quad a =\bf{5\;cm}}\\\sf{\quad b =\bf{5\;cm}}\\\sf{\quad c=\bf{6\;cm}}\\\sf{\quad s = \bf{8\:cm}}\end{cases}$

$\dashrightarrow\sf Area_{\:(triangle)} = \sqrt{8\Big(8 - 5\Big)\Big(8 - 5\Big)\Big(8 - 6\Big)}\\\\\\\dashrightarrow\sf Area_{\;(triangle)} = \sqrt{8 \times 3 \times 3 \times 2}\\\\\\\dashrightarrow\sf Area_{\:(triangle)} = \sqrt{144}\\\\\\\dashrightarrow{\pmb{\boxed{\frak{Area_{\:(triangle)} = 12\;cm^2}}}}\;\bigstar$

$\therefore{\underline{\sf{Hence,\;the\;area\;of\; isosceles\; triangle\;is\;{\pmb{\sf{12\;cm^2}}}.}}}$⠀⠀⠀⠀⠀⠀