Question.
Sides of a triangle are in the ratio of 12: 17: 25 and its perimeter is 540 cm. Find its area.
Solution:
Let the common ratio between the sides of the given triangle be x.
Therefore, the side of the triangle will be $12 x, 17 x$, and $25 x$.
Perimeter of this triangle $=540 \mathrm{~cm}$
$12 x+17 x+25 x=540 \mathrm{~cm}$
$54 x=540 \mathrm{~cm}$
$x=10 \mathrm{~cm}$
Sides of the triangle will be $120 \mathrm{~cm}, 170 \mathrm{~cm}$, and $250 \mathrm{~cm}$.
$s=\frac{\text { Perimeter of triangle }}{2}=\frac{540 \mathrm{~cm}}{2}=270 \mathrm{~cm}$
By Heron’s formula,
Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$
$=[\sqrt{270(270-120)(270-170)(270-250)}] \mathrm{cm}^{2}$
$=[\sqrt{270 \times 150 \times 100 \times 20}] \mathrm{cm}^{2}$
$=9000 \mathrm{~cm}^{2}$
Therefore, the area of this triangle is $9000 \mathrm{~cm}^{2}$.
Let the common ratio between the sides of the given triangle be x.
Therefore, the side of the triangle will be $12 x, 17 x$, and $25 x$.
Perimeter of this triangle $=540 \mathrm{~cm}$
$12 x+17 x+25 x=540 \mathrm{~cm}$
$54 x=540 \mathrm{~cm}$
$x=10 \mathrm{~cm}$
Sides of the triangle will be $120 \mathrm{~cm}, 170 \mathrm{~cm}$, and $250 \mathrm{~cm}$.
$s=\frac{\text { Perimeter of triangle }}{2}=\frac{540 \mathrm{~cm}}{2}=270 \mathrm{~cm}$
By Heron’s formula,
Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$
$=[\sqrt{270(270-120)(270-170)(270-250)}] \mathrm{cm}^{2}$
$=[\sqrt{270 \times 150 \times 100 \times 20}] \mathrm{cm}^{2}$
$=9000 \mathrm{~cm}^{2}$
Therefore, the area of this triangle is $9000 \mathrm{~cm}^{2}$.