Question

The sides of a triangle are $4,5$ and $6$ cm. The area of the triangle is equal to

• A. $\frac{15}{4}c{m}^2$
• B. $\frac{15}{4}\sqrt{7}c{m}^2$
• C. $\frac{4}{15}\sqrt{7}c{m}^2$
• D. None of these

Answer 1

Answer: (B)

$\frac{15}{4}\sqrt{7}c{m}^2$

Given that, a = 4 cm, b = 5 cm, and c = 6 cm
We know $s=\frac{a+b+c}{2}=\frac{4+5+6}{2}=\frac{15}{2}$
$\therefore$ Area of the triangle $=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{\left(\frac{15}{2}\right)(\frac{15}{2}-4)(\frac{15}{2}-5)(\frac{15}{2}-6)}$
$=\sqrt{\frac{15}{2}\times \frac{7}{2}\times \frac{5}{2}\times \frac{3}{2}}=\frac{15}{4}\sqrt{7}c{m}^2$