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Question
Find the area of an isosceles triangle whose base is 16 cm and the length of each of the equal sides is 10 cm.
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Solution
In isosceles Δ ABC
Base BC = 16 cm
and AB = AC = 10 cm
Let AD ⊥ BC and BD = `1/2"BC"=16/2`
∴ BD = 8 cm
In right Δ ABD
AB2 = AD2 + BD2 ..............(Pythagoras Theorem)
(10)2 = AD2 + (8)2
100 = AD2 + 64
100 − 64 = AD2
36 = AD2
AD = `sqrt36=sqrt(6xx6)`
∴ AD = 6 cm
Now, the area of triangle =`("Base"xx"Altitude")/2`
= `(16xx6)/2` = 48 cm2
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