Question

find the area of an isosceles triangle whose sides are 10 cm and base is 12cm​

Answer 1

Given :

• Equal Sides = 10 cm
• Base = 12 cm

To Find :

• Area of the Triangle = ?

$\\ \qquad{\rule{200pt}{2pt}}$

SolutioN :

$\dag$ Formula Used :

• ${\underline{\boxed{\pmb{\sf{ Area{\small_{(Triangle)}} = \sqrt{s \bigg( s - a \bigg) \bigg( s - b \bigg) \bigg( s - c \bigg) } }}}}}$

Where :

• s = Semi - Perimeter
• a = Side 1
• b = Side 2
• c = Side 3

$\dag$ Calculating the Semi - Perimeter :

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { S = \dfrac{ a + b + c }{2} } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { S = \dfrac{ 10 + 10 + 12 }{2} } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { S = \dfrac{ 32 }{2} } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { S = \cancel\dfrac{ 32 }{2} } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; {\underline{\boxed{\pmb{\sf{ Semi - Perimeter = 16 \; cm }}}}} \; {\red{\pmb{\bigstar}}} \\ \\ \\ \end{gathered}$

$\dag$ Calculating the Area :

$\begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Area = \sqrt{ \bigg( s - a \bigg) \bigg( s - b \bigg) \bigg( s - c \bigg) } } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Area = \sqrt{ 16 \bigg( 16 - 10 \bigg) \bigg( 16 - 10 \bigg) \bigg( 16 - 12 \bigg) } } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Area = \sqrt{ 16 \times 6 \times 6 \times 4 } } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Area = \sqrt{ \underline{4 \times 4} \times \underline{6 \times 6} \times 4 } } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Area = 4 \times 6 \times \sqrt{4} } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Area = 24 \times \sqrt{4} } \qquad \; \; \bigg\lgroup {\pink{\sf{ Value \; of \; \sqrt{4} = 2 }}} \bigg\rgroup \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Area = 24 \times 2 } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; {\underline{\boxed{\pmb{\sf{ Area = 48 \; {cm}^{2} }}}}} \; {\purple{\pmb{\bigstar}}} \\ \\ \\ \end{gathered}$

$\therefore \;$ Area of the Triangle is 48 cm² .

$\\ \qquad{\rule{200pt}{2pt}}$

:)

Answer 2

$\small \dag \sf \blue{ \underline{ \red{question}}}$

find the area of an isosceles triangle whose sides are 10 cm and base is 12cm.

$\small \dag \sf \blue{ \underline{ \red{Answer :-}}}$

GIVEN :

• EQUAL SIDES = 10 cm
• BASE = 12 cm

➻$\sf\purple{\underline{FORMULA}}$

$\sqrt{s(s - a)(s - b)(s - c)}$

➻$\sf\purple{\underline{<strong>S</strong><strong>E</strong><strong>M</strong><strong>I</strong><strong> </strong>PERIMETER = A+B+C/2}}$

⟶ 10+10+12/2

⟶ 32/2

⟶ 16

$\underline{\rule{220pt}{3pt}}$

➻$\large\bigstar\mathtt{AREA}$

(s - a) = 16 - 10 = 6

(s - b) = 16 - 10 = 6

(s - c) = 16 - 12 = 4

$\sqrt{16 \: \times 6 \times 6 \times 4} \\ 2\sqrt{4 \times 4 \times 6 \times 6} \\ 2 \times 4 \times 6 \\8 \times 6 \\ 48$

$\sf\mathfrak \purple{\underline{Final\:Answer:-}}$

Area of the Triangle = $\fcolorbox{red}{blue}{ 48 cm }$

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Hope it helps you