Question

# The area of an isosceles triangle having base 2cm and the length of one of the equal sides 4cm, is

A
15cm2
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B
152cm2
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C
215cm2
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D
415cm2
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Solution

## The correct option is A √15cm2Let ABC be an isosceles triangle in which AB = AC = 4cm and BC = 2cm. In right angled ΔADB, AB2=AD2+BD2 [by Pythagoras theorem] ⇒ (4)2=AD2+1 ⇒ AD2=16−1 ⇒ AD2=15 ∴ AD=√15cm [taking positive square root because length is always positive] ∴ Area of ΔABC=12×BC×AD [∵ area of triangle=12(base×height)] =12×2×√15=√15cm2 Alternate Method We know that, Area of an isosceles triangle =a4√4b2−a2 Where, b is the length of equal sides and a is the length of the base. Here, the length of side be b = 4cm and a = 2cm ∴ Area of an isosceles triangle =2√4(4)2−44=√64−42 =√602=2√152=√15cm2

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