Sum

One of the equal sides of an isosceles triangle is 13 cm and its perimeter is 50 cm. Find the area of the triangle.

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#### Solution

In isosceles ∆ABC

AB = AC = 13 cm But perimeter = 50 cm

∴ BC = 50 - (13 + 13) cm

= 50 - 26 = 24 cm

AD ⊥ BC

∴ AD = DC = `24/2 = 12` cm.

In right ΔABD,

AB^{2} = AD^{2} + BD^{2} (Pythagoras Therorem)

`(13)^2 = "AD"^2 + (12)^2`

⇒ 169 = `"AD"^2 + 144`

⇒ `"AD"^2 = 169 - 144`

= 25 = (5)^{2}

∴ AD = 5 cm.

Now area of ΔABC = `1/2 "Base" xx "Altitude"`

= `1/2 xx "BC" xx "AD"`

= `1/2 xx 24 xx 5 = 60` cm^{2}

Concept: Perimeter of Triangles

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