Question

the sides of a triangle measures 13 cm , 14 cm and 15 cm. its area is

Answer 1

by herin's formula
let's say A=13
B=14
C=15
s=A+B+C/2
=42/2
=21

area =√s(s-A)(s-B)(s-c)
=84cm^2

Answer 2

The area of triangle is 84 $cm^{2}$.

Step-by-step explanation:

Here, a = 13 cm, b = 14 m and c = 15 cm

To find, the area of the triangle = ?

∴ Semiperimeter(s) $=\dfrac{a+b+c}{2}$

$=\dfrac{13+14+15}{2}$

$=\dfrac{42}{2}=21cm$

Area of triangle(by Heron's formula)

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

$=\sqrt{21(21-13)(21-14)(21-15)}$

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

= 84 $cm^{2}$

The area of the triangle = 84 $cm^{2}$

Hence, the area of triangle is 84 $cm^{2}$.