# The sides of a triangle are 56 cm,

Question:

The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then, the area of the triangle is

(a) 1322 cm2

(b) 1311 cm2

(c) 1344 cm2

(d) 1392 cm2

Thinking Process

(i) First, determine the semi-perimeter of a triangle by using the formula, s = (a + b + c)/2

(ii) Further, determine the area of triangle by using the formula, area of triangle (Fieron’s formula) =

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

Solution:

(c) Since, the three sides of a triangle are $a=56 \mathrm{~cm}, b=60 \mathrm{~cm}$ and $c=52 \mathrm{~cm}$.

Then, semi-perimeter of a triangle,

$s=\frac{a+b+c}{2}=\frac{56+60+52}{2}=\frac{168}{2}=84 \mathrm{~cm}$

Area of a triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ [by Heron's formula]

$=\sqrt{84(84-56)(84-60)(84-52)}$

$=\sqrt{84 \times 28 \times 24 \times 32}$

$=\sqrt{4 \times 7 \times 3 \times 4 \times 7 \times 4 \times 2 \times 3 \times 4 \times 4 \times 2}$

$=\sqrt{(4)^{6} \times(7)^{2} \times(3)^{2}}$

$=(4)^{3} \times 7 \times 3=1344 \mathrm{~cm}^{2}$

Hence, the area of triangle is $1344 \mathrm{~cm}^{2}$.