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Find the area of an isosceles triangle whose equal sides are 12 cm each and the perimeter is 30 cm.

Mensuration

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Answer

Consider △ABC as the isosceles triangle.

Find the area of an isosceles triangle whose equal sides are 12 cm each and the perimeter is 30 cm. Mensuration, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Here, AB = AC = 12 cm.

Perimeter = 30 cm

⇒ AB + AC + BC = 30

⇒ 12 + 12 + BC = 30

⇒ BC = 30 - 24 = 6 cm.

We know that,

Semi-perimeter (s) = Perimeter2=302\dfrac{\text{Perimeter}}{2} = \dfrac{30}{2} = 15 cm.

Area of an isosceles triangle = 14b4a2b2\dfrac{1}{4}b\sqrt{4a^2 - b^2}, where a is length of equal sides and b is the length of other side.

Substituting values we get,

A=14×6×4(12)262=14×6×4×14436=14×6×540=14×6×23.24=139.446=34.86 cm2.A = \dfrac{1}{4} \times 6 \times \sqrt{4(12)^2 - 6^2} \\[1em] = \dfrac{1}{4} \times 6 \times \sqrt{4 \times 144 - 36} \\[1em] = \dfrac{1}{4} \times 6 \times \sqrt{540} \\[1em] = \dfrac{1}{4} \times 6 \times 23.24 \\[1em] = \dfrac{139.44}{6} \\[1em] = 34.86 \text{ cm}^2.

Hence, area of isosceles triangle = 34.86 cm2.

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