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**Hint:**We will suppose the sides of the triangle to be $11x,19x{\text{ and }}24x$and then by definition of perimeter, we can calculate the value of x since perimeter of a triangle is equal to the sum of the sides of the triangle. Then, the area of the triangle can be calculated by Heron’s formula: $Area = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} $ where s is the semi – perimeter of the triangle and a, b, c are the sides of the triangle.

**Complete step-by-step answer:**We are given the ratio of sides of a triangle as $11:19:24$.

The perimeter of the triangle is given as $540cm$.

We are required to calculate the area of the triangle.

Let the sides of the triangle be $11x,19x{\text{ and }}24x$.

By definition of the perimeter of the triangle, the perimeter of the triangle is defined as the total of its boundary i.e., the sum of all three sides.

Therefore, the perimeter of the triangle can be written as: ${\text{Perimeter = }}11x + 19x + 24x$

$

\Rightarrow 540 = 11x + 19x + 24x \\

\Rightarrow 540 = 54x \\

\Rightarrow x = 10 \\

$

Since $x = 10$, the sides of the triangle will be $11x = 11\left( {10} \right) = 110,{\text{ }}19x = 19\left( {10} \right) = 190{\text{ and }}24x = 24\left( {10} \right) = 240$

Figure of the triangle, with sides a, b and c, will be:

Using the Heron’s formula to calculate the value of the area of the triangle ABC given by: $Area = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} $ where s is the semi – perimeter of the triangle and a, b, c are the sides of the triangle.

The semi – perimeter of the triangle is half of the perimeter of the triangle i.e., here $s = \dfrac{{540}}{2} = 270cm$ and a = 110, b = 190 and c = 240 cm.

$

\Rightarrow area\left( {ABC} \right) = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \\

\Rightarrow area\left( {ABC} \right) = \sqrt {270\left( {270 - 110} \right)\left( {270 - 190} \right)\left( {270 - 240} \right)} \\

\Rightarrow area\left( {ABC} \right) = \sqrt {270\left( {160} \right)\left( {80} \right)\left( {30} \right)} = \sqrt {103680000} \\

\Rightarrow area\left( {ABC} \right) = 7200\sqrt 2 c{m^2} \\

$

**Therefore, the area of the triangle whose sides are in the ratio $11:19:24$ is $7200\sqrt 2 c{m^2}$.**

**Note:**In this question, you may get confused while calculating the value of sides using the perimeter of the triangle by assuming the sides to be a multiple of x. You can also simplify further the calculation of the area of the triangle by putting the value of $\sqrt 2 $ in the area of the triangle obtained to be$7200\sqrt 2 c{m^2}$.

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