The lengths of the three sides of a triangular field are 40 m, 24 m and 32 m respectively. The area of the triangle is
Question:
The lengths of the three sides of a triangular field are 40 m, 24 m and 32 m respectively. The area of the triangle is
(a) 480 m2
(b) 320 m2
(c) 384 m2
(d) 360 m2
Solution:
(c) $384 \mathrm{~m}^{2}$
Let:
$a=40 \mathrm{~m}, b=24 \mathrm{~m}$ and $c=32 \mathrm{~m}$
$s=\frac{a+b+c}{2}=\frac{40+24+32}{2}=48 \mathrm{~m}$
By Heron's formula, we have :
Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{48(48-40)(48-24)(48-32)}$
$=\sqrt{48 \times 8 \times 24 \times 16}$
$=\sqrt{24 \times 2 \times 8 \times 24 \times 8 \times 2}$
$=24 \times 8 \times 2$
$=384 \mathrm{~m}^{2}$