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Let one of the sides of a triangle be $17 \mathrm{~cm}$ and the sum of all the sides of the triangle be $40 \mathrm{~cm}$. If the sum of two adjacent sides is $35 \mathrm{~cm}$, then the area (in sq. cms) of the triangle is
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Verified Answer
The correct answer is:
$30 \sqrt{2}$
According to given informations,
Let $a=17, a+b+c=40$ and $a+b=35$
So, $b=18$ and $c=5$
$\because$ Area of required triangle is
$\sqrt{s(s-a)(s-b)(s-c)}$
$\left[\right.$ where, $\left.s=\frac{a+b+c}{2}=20\right]$
$=\sqrt{20(3)(2)(15)}=30 \sqrt{2}$
Hence, option (c) is correct.
Let $a=17, a+b+c=40$ and $a+b=35$
So, $b=18$ and $c=5$
$\because$ Area of required triangle is
$\sqrt{s(s-a)(s-b)(s-c)}$
$\left[\right.$ where, $\left.s=\frac{a+b+c}{2}=20\right]$
$=\sqrt{20(3)(2)(15)}=30 \sqrt{2}$
Hence, option (c) is correct.
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