The area of an isosceles triangle with base 8 cm and each equal side is of length 6 cm , is
A$4 \sqrt{5} cm^2$
B$6 \sqrt{5} cm^2$
C$8 \sqrt{5} cm^2$
D$16 \sqrt{5} cm^2$
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C$8 \sqrt{5} cm^2$
(c) $8 \sqrt{5} cm^2$ The area $A$ of an isosceles triangle with base $a$ and each equal side $b$ is given by $A=\frac{a}{4} \sqrt{4 b^2-a^2}$ Here, $a=8 cm$ and $b=6 cm$. $\therefore \quad A=\frac{8}{4} \sqrt{4 \times 6^2-8^2}=2 \sqrt{144-64}=8 \sqrt{5} cm^2$
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