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The perimeter of an isosceles triangle is $32 \operatorname{cm}$. If each equal side is $\frac{5}{6}$ th of the base, find the area of the triangle.
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Sol. Let each equal side of triangle be $x$ and base be $y$
ATQ, $x + x + y = 32$
$2x + y = 32$
Also, $x = \frac{5}{6} y$
On solving these equations, we get $x = 10$ and $y = 12$
$\therefore$ sides of the triangle are $10 \operatorname{cm}, 10 \operatorname{cm}, 12 \operatorname{cm}$
Semi - perimeter of the triangle $= 16 \operatorname{cm}$
Area of the triangle $= \sqrt{16 \times (16-10) \times (16-10) \times (16-12)}$
$= \sqrt{16 \times 6 \times 6 \times 4}$
$= 48 \operatorname{cm}^2$
ATQ, $x + x + y = 32$
$2x + y = 32$
Also, $x = \frac{5}{6} y$
On solving these equations, we get $x = 10$ and $y = 12$
$\therefore$ sides of the triangle are $10 \operatorname{cm}, 10 \operatorname{cm}, 12 \operatorname{cm}$
Semi - perimeter of the triangle $= 16 \operatorname{cm}$
Area of the triangle $= \sqrt{16 \times (16-10) \times (16-10) \times (16-12)}$
$= \sqrt{16 \times 6 \times 6 \times 4}$
$= 48 \operatorname{cm}^2$