Question

plzz solve it.... ?!​

Answer 1

$\huge\frak{AnswEr :}$

Given Parameters :

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Perimeter = 30 cm

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Length of equal sides = 12cm

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Let us assume that the third side of isosceles triangle be x cm.

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${\underline{\sf \bigstar \: According \: to \: the \: question :}}$

$\implies\sf \:x + 12 + 12 = 30 \\ \\ \\ \implies\sf \:x + 24 = 30 \\ \\ \\ \implies\sf \:x = 30 - 24 \\ \\ \\ \implies \large{ \boxed{\frak{ \purple{ \:x = 6}}}}$

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Now,

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By using Heron's formula :

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$\implies\sf \: s = \dfrac{a + b + c}{2} \\ \\ \\ \implies\sf \: s = \dfrac{Perimeter}{2} \\ \\ \\ \implies\sf \: s = \frac{30}{2} \\ \\ \\ \implies \large{ \boxed{ \frak{\pink{\: s = 15}}}}$

⠀__________________________

We know that,

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$\implies\sf \: Area \: of \: \triangle = \sqrt{s(s - a)(s - b)(s - c)}$

Substituting the values :

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$\implies\sf \: Area \: of \: \triangle = \sqrt{15(15 - a)(15 - b)(15 - c)} \\ \\ \\ \implies\sf \: Area \: of \: \triangle =\sqrt{15(15 - 12)(15 - 12)(15 - 6)} \\ \\ \\ \implies\sf \: Area \: of \: \triangle = \sqrt{15 \times 3 \times 3 \times 9} \\ \\ \\ \implies\sf \: Area \: of \: \triangle = 3 \times 3 \times \sqrt{15} \\ \\ \\ \implies\sf \: Area \: of \: \triangle = 9 \times \sqrt{15} \\ \\ \\ \implies\sf \: Area \: of \: \triangle = 9 \times 3.88 \\ \\ \\ \implies \large{ \boxed{ \frak{\blue{ \: Area \: of \: \triangle = 34.92 \: cm {}^{2} }}}}$

$\therefore \: {\underline{\sf{ Area \: of \: the \: isosceles \: \triangle \: is \: \bf \: 34.92 cm{}^{2}. }}}$

Answer 2

$\huge\frak{AnswEr :}$

Given Parameters :

⠀⠀⠀⠀⠀⠀⠀⠀⠀

Perimeter = 30 cm

⠀⠀⠀⠀⠀⠀⠀⠀⠀

Length of equal sides = 12cm

⠀⠀⠀⠀⠀⠀⠀⠀⠀

Let us assume that the third side of isosceles triangle be x cm.

⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\begin{gathered}{\underline{\sf \bigstar \: According \: to \: the \: question :}} \\ \\ \end{gathered}$

$\begin{gathered}\implies\sf \:x + 12 + 12 = 30 \\ \\ \\ \implies\sf \:x + 24 = 30 \\ \\ \\ \implies\sf \:x = 30 - 24 \\ \\ \\ \implies \large{ \boxed{\frak{ \purple{ \:x = 6}}}} \\ \\ \end{gathered}$

⠀__________________________

Now,

⠀⠀⠀⠀⠀⠀⠀⠀⠀

By using Heron's formula :

⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\begin{gathered}\implies\sf \: s = \dfrac{a + b + c}{2} \\ \\ \\ \implies\sf \: s = \dfrac{Perimeter}{2} \\ \\ \\ \implies\sf \: s = \frac{30}{2} \\ \\ \\ \implies \large{ \boxed{ \frak{\pink{\: s = 15}}}} \\ \\ \end{gathered}$

⠀__________________________

We know that,

⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\begin{gathered}\implies\sf \: Area \: of \: \triangle = \sqrt{s(s - a)(s - b)(s - c)} \\ \\ \\ \end{gathered}$

Substituting the values :

⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\begin{gathered}\implies\sf \: Area \: of \: \triangle = \sqrt{15(15 - a)(15 - b)(15 - c)} \\ \\ \\ \implies\sf \: Area \: of \: \triangle =\sqrt{15(15 - 12)(15 - 12)(15 - 6)} \\ \\ \\ \implies\sf \: Area \: of \: \triangle = \sqrt{15 \times 3 \times 3 \times 9} \\ \\ \\ \implies\sf \: Area \: of \: \triangle = 3 \times 3 \times \sqrt{15} \\ \\ \\ \implies\sf \: Area \: of \: \triangle = 9 \times \sqrt{15} \\ \\ \\ \implies\sf \: Area \: of \: \triangle = 9 \times 3.88 \\ \\ \\ \implies \large{ \boxed{ \frak{\blue{ \: Area \: of \: \triangle = 34.92 \: cm {}^{2} }}}} \\ \\ \end{gathered}$

$\begin{gathered}\therefore \: {\underline{\sf{ Area \: of \: the \: isosceles \: \triangle \: is \: \bf \: 34.92 cm{}^{2}. }}} \\ \end{gathered}$