Each of the equal sides of an isosceles triangle is 2 cm more than its height and the base of the triangle is 12 cm. Find the area of the triangle.
Mensuration
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Each of the equal sides of an isosceles triangle is 2 cm more than its height and the base of the triangle is 12 cm. Find the area of the triangle.

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Let ABC be an isosceles triangle in which AB = AC, AD BC and BC is the base. Given, Each of the equal sides of the isosceles triangle is 2 cm more than its height. Let the height of the triangle be h cm. Equal sides: AB = AC = h + 2 Base: BC = 12 cm In Δ ABD and Δ ACD, AD = AD [Common Side] ∠ADB = ∠ADC [Both equal to 90 ] AB = AC [Δ ABC is an isosceles triangle] ∴ Δ ABD ≅ Δ ACD [By R.H.S. axiom] ∴ BD = CD [C.P.C.T.C.] ∴ BD = CD = = 6 cm. By using the Pythagoras theorem in Δ ABD, ⇒ BD2 + AD2 = AB2 ⇒ 62 + h2 = (h + 2)2 ⇒ 36 + h2 = h2 + 4 + 2 × h × 2 ⇒ 36 = 4 + 4h ⇒ 4h = 32 ⇒ h = ⇒ h = 8 cm. Hence, area of the triangle = 48 cm2.

Mathematics

Each of the equal sides of an isosceles triangle is 2 cm more than its height and the base of the triangle is 12 cm. Find the area of the triangle.

Mensuration

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Mathematics

Each of the equal sides of an isosceles triangle is 2 cm more than its height and the base of the triangle is 12 cm. Find the area of the triangle.

Mensuration

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