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 $=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{48(48-40)(48-24)(48-32)}$

$=\sqrt{48 \times 8 \times 24 \times 16}$

$=\sqrt{24 \times 2 \times 8 \times 24 \times 8 \times 2}$

$=24 \times 8 \times 2$

$=384 \mathrm{~m}^{2}$

 

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