Question

The sides of a triangle are $7$ cm, $9$ cm and $14$ cm. Its area is

• A. $12\sqrt{5}{\text{cm}}^2$
• B. $12\sqrt{3}{\text{cm}}^2$
• C. $24\sqrt{5}{\text{cm}}^2$
• D. $63{\text{cm}}^2$

Answer 1

The correct option is A $12\sqrt{5}{\text{cm}}^2$

Area of triangle = $\sqrt{S(S-a)(S-b)(S-c)}$

$S=\frac{a+b+c}{2}$

Given $a=7$ cm

$b=9$ cm

$c=14$ cm

$S=\frac{7+9+14}{2}$

$=\frac{30}{2}$

$S=15cm$

$\therefore$ Area of triangle = $\sqrt{S(S-a)(S-b)(S-c)}$

$=\sqrt{15(15-7)(15-9)(15-14)}$

$=\sqrt{\mathrm{15.8.6.1}}$

$=\sqrt{720}$

$=\sqrt{5\times 144}$

$=\sqrt{12\times 12\times 5}$

$=12\sqrt{5}c{m}^2$

$\therefore$ Area of triangle = $12.\sqrt{5}c{m}^2$

So, the answer is A. $12.\sqrt{5}c{m}^2$