Let one of the sides of a triangle be 17 text | vedclass

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(C) Let the sides of the triangle be $a, b, c$. Given $a = 17 	ext{ cm}$ and the perimeter $P = a + b + c = 40 	ext{ cm}$.Given the sum of two sides

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(C) Let the sides of the triangle be $a, b, c$. Given $a = 17 	ext{ cm}$ and the perimeter $P = a + b + c = 40 	ext{ cm}$.Given the sum of two sides is $a + b = 35 	ext{ cm}$.Substituting $a = 17$, we get $17 + b = 35$, so $b = 18 	ext{ cm}$.Since $a + b + c = 40$, we have $17 + 18 + c = 40$, which gives $c = 5 	ext{ cm}$.The semi-perimeter $s = \frac{40}{2} = 20 	ext{ cm}$.Using Heron's formula, the area $A = \sqrt{s(s-a)(s-b)(s-c)}$.$A = \sqrt{20(20-17)(20-18)(20-5)} = \sqrt{20 	imes 3 	imes 2 	imes 15}$.$A = \sqrt{1800} = \sqrt{900 	imes 2} = 30 \sqrt{2} 	ext{ cm}^2$.Thus, the correct option is $C$.

Let one of the sides of a triangle be $17 \text{ cm}$ and the sum of all the sides of the triangle be $40 \text{ cm}$. If the sum of two adjacent sides is $35 \text{ cm}$, then the area (in $\text{cm}^2$) of the triangle is (in $\sqrt{2}$)

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