Question

Q11, If an isosceles triangle having perimeter 30cm and each of equal sides is 12cm, then find the area of triangle


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Answer 1

GIVEN :-

• Perimeter of isosceles triangle = 30 cm.
• Equal side = 12 cm.

TO FIND :-

• The area of isosceles triangle.

SOLUTION :-

$\\ : \implies \displaystyle \sf \: Perimeter \: of \: \triangle = a + b + c$

• a = 12 cm.
• b = 12 cm.
• c = ?

$\\ : \implies \displaystyle \sf \:30 = 12 + 12 + c$

$\\ : \implies \displaystyle \sf \:c = 30 - 24$

$\\ : \implies \underline{ \boxed{\displaystyle \sf \:c = 6 \: cm}}$

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$\\ \dashrightarrow\displaystyle \sf Semi - Perimeter \: of \: \triangle = \frac{a + b + c}{2}$

$\dashrightarrow\displaystyle \sf Semi - Perimeter \: of \: \triangle = \frac{12 + 12 + 6}{2}$

$\dashrightarrow\displaystyle \sf Semi - Perimeter \: of \: \triangle = \frac{30}{2}$

$\dashrightarrow \underline{ \boxed{\displaystyle \sf Semi - Perimeter \: of \: \triangle = 15 \: cm}}$

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

• s = semi - perimeter
• a = side
• b = side
• c = side

$\\ \longmapsto\displaystyle \sf Area \: of \: \triangle = \sqrt{15(15 - 12)(15 - 12)(15 - 6)}$

$\longmapsto\displaystyle \sf Area \: of \: \triangle = \sqrt{15 \times 3 \times 3 \times 9}$

$\longmapsto\displaystyle \sf Area \: of \: \triangle = 3 \sqrt{135} \: cm ^{2}$

$\longmapsto\underline{ \boxed{\displaystyle \sf Area \: of \: \triangle = 9 \sqrt{15} \: cm ^{2} }}$

Answer 2

Please refer to the attachment