Question:
The base of an isosceles triangle is 6 cm and each of its equal sides is 5 cm. The height of the triangle is
(a) 8 cm
(b) $\sqrt{30} \mathrm{~cm}$
(c) $4 \mathrm{~cm}$
(d) $\sqrt{11} \mathrm{~cm}$
Solution:
(c) 4 cm
Height of isosceles triangle $=\frac{1}{2} \sqrt{4 a^{2}-b^{2}}$
$=\frac{1}{2} \sqrt{4(5)^{2}-6^{2}} \quad(a=5 \mathrm{~cm}$ and $b=6 \mathrm{~cm})$
$=\frac{1}{2} \times \sqrt{100-36}$
$=\frac{1}{2} \times \sqrt{64}$
$=\frac{1}{2} \times 8$
$=4 \mathrm{~cm}$
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