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Question
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,
AB2 = AD2 + BD2 (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` cm2
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