Question

The perimeter of a triangle $ABC$ is $37\mathrm{cm}$ and the ratio between the lengths of its altitudes be $6:5:4$. Find the lengths of its sides. Let the sides be $x\mathrm{cm},y\mathrm{cm},\mathrm{and}(37-x-y)\mathrm{cm}.$

Answer 1

Step 1. Find the relation between the sides of the triangle.

It is given that the ratio between the lengths of its altitudes be $6:5:4$.

So, let us consider the lengths of altitudes are $6n,5n\mathrm{and}4n$.

Given that, the length of the sides are $x\mathrm{cm},y\mathrm{cm},\mathrm{and}(37-x-y)\mathrm{cm}.$

As we know,

$\mathrm{Area}\mathrm{of}\mathrm{triangle}=\frac{1}{2}\times \mathrm{base}\times \mathrm{altitude}$

$$\begin{gathered} \therefore \frac{1}{2}\times x\times 6n=\frac{1}{2}\times y\times 5n=\frac{1}{2}\times (37-x-y)\times 4n \\ \Rightarrow 6\mathrm{x}=5\mathrm{y}=4(37-\mathrm{x}-\mathrm{y}) \\ \Rightarrow 6\mathrm{x}=5\mathrm{y}=148-4\mathrm{x}-4\mathrm{y} \end{gathered}$$

From the above expression, we can write as,

$$\begin{gathered} \because 6x=5y \\ \therefore 6x-5y=0...\left(1\right) \\ \because 6x=148-4x-4y \\ \therefore 10x+4y=148...\left(2\right) \end{gathered}$$

Step 2. Find the sides of the triangle.

Substitute the value of $x$ from equation $1$ into equation $2$.

$$\begin{gathered} \therefore 10\left(\frac{5}{6}y\right)+4y=148 \\ \Rightarrow \frac{25y+12y}{3}=148 \\ \Rightarrow \frac{37}{3}y=148 \\ \Rightarrow y=148\times \frac{3}{37} \\ \Rightarrow y=4\times 3 \\ \Rightarrow y=12\mathrm{cm} \end{gathered}$$

and $x=\frac{5}{6}\times 12=10\mathrm{cm}$

The third side will be $(37-10-12=15\mathrm{cm})$.

Hence, the lengths of the sides are $10\mathrm{cm},12\mathrm{cm}\mathrm{and}15\mathrm{cm}$.