Question

The sides of a triangle are in the ratio 3 : 5 : 7 and its perimeter is 75 m. The area of the triangle is

• A. $\frac{375}{4}{m}^2$
  
• B. $\frac{125}{4}\sqrt{3}{m}^2$
  
• C. $\frac{375}{4}\sqrt{3}{m}^2$
  
• D. $\frac{300}{7}\sqrt{3}{m}^2$

Answer 1

The correct option is C $\frac{375}{4}\sqrt{3}{m}^2$


Let the sides of the triangle be a, b and c.
Given that the sides are in the ratio 3 : 5 : 7.
Let a = 3k, b = 5k, c = 7k, where k is any natural number.
Now, perimeter of the triangle is 75 m.
$\Rightarrow a+b+c=75$
$\Rightarrow 3k+5k+7k=75$
$\Rightarrow 15k=75$
$\Rightarrow k=5$
$\therefore a=15m,b=25m,c=35m$
$\therefore$ Semi-perimeter, (s) = $\frac{a+b+c}{2}$
$=\frac{75}{2}$
$\therefore$ Area of a triangle = $\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{\frac{75}{2}(\frac{75}{2}-15)(\frac{75}{2}-25)(\frac{75}{2}-35)}$
$=\sqrt{\frac{75}{2}\times \frac{45}{2}\times \frac{25}{2}\times \frac{5}{2}}$
$=\frac{375}{4}\sqrt{3}{m}^2$
Hence, the correct answer is option (c).