Question

One side of a right angled triangular scarf is $80cm$ and its longest side is $1m$. Find its cost at the rate of Rs $250$ per ${{m}^{2}}$.

Answer 1

Hint: Since the triangle is right angled, the longest side will be hypotenuse, and the other will be the perpendicular. We can label the triangle as ABC so that $AB=80cm$ and $AC=1m$. The base $BC$ can be found by using the Pythagoras theorem which is given by $A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}$. Then using the formula for the area of a triangle given by $A=\dfrac{1}{2}\times \text{base}\times \text{height}$ we can ca;culate the area. Finally, using the rate value which is Rs $250$ per ${{m}^{2}}$, we can calculate the required cost.

Complete step-by-step answer:
According to the question, one side of the right angled triangular scarf is equal to $80cm$ and the longest side is equal to $1m$. Since the triangle is right angled, the three sides will be base, perpendicular and the hypotenuse. We know that the hypotenuse is the longest side. So the length of the hypotenuse is equal to $1m$, and that of the perpendicular is equal to $80cm$. Using this information, we can draw the right angled triangle as shown below.

From the above figure, we have
\[\begin{align}
  & \Rightarrow AC=1m......\left( i \right) \\
 & \Rightarrow AB=80cm \\
\end{align}\]
We know that $1cm=0.01m$. Therefore, we have
$\begin{align}
  & \Rightarrow AB=80\times 0.01m \\
 & \Rightarrow AB=0.8m.........\left( ii \right) \\
\end{align}$
Now, applying the Pythagoras theorem in the triangle ABC, we can write
$\Rightarrow A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}$
Subtracting $A{{B}^{2}}$ from both the sides, we get
\[\begin{align}
  & \Rightarrow A{{C}^{2}}-A{{B}^{2}}=A{{B}^{2}}+B{{C}^{2}}-A{{B}^{2}} \\
 & \Rightarrow A{{C}^{2}}-A{{B}^{2}}=B{{C}^{2}} \\
 & \Rightarrow B{{C}^{2}}=A{{C}^{2}}-A{{B}^{2}} \\
\end{align}\]
Substituting the equations (i) and (ii) in the above equation, we get
$\begin{align}
  & \Rightarrow B{{C}^{2}}={{1}^{2}}-{{0.8}^{2}} \\
 & \Rightarrow B{{C}^{2}}=1-0.64 \\
 & \Rightarrow B{{C}^{2}}=0.36 \\
\end{align}$
Taking square root both the sides, we get
$\begin{align}
  & \Rightarrow \sqrt{B{{C}^{2}}}=\sqrt{0.36} \\
 & \Rightarrow BC=0.6m \\
\end{align}$
Now, we know that the area of a triangle is given by
$\Rightarrow A=\dfrac{1}{2}\times \text{base}\times \text{height}$
Therefore, the area of the triangle ABC will be given by
$\begin{align}
  & \Rightarrow A=\dfrac{1}{2}\times BC\times AB \\
 & \Rightarrow A=\dfrac{1}{2}\times 0.6\times 0.8 \\
\end{align}$
On solving we get
$\Rightarrow A=0.24{{m}^{2}}$
Now, the rate is given to be equal to Rs $250$ per ${{m}^{2}}$. Therefore, the cost of $0.24{{m}^{2}}$ is
$\begin{align}
  & \Rightarrow \text{Rs }250\times 0.24 \\
 & \Rightarrow \text{Rs 60} \\
\end{align}$
Hence, the required cost is equal to Rs $\text{60}$.

Note: We can also use the Heron’s formula for calculating the area of the triangular scarf. Do not forget to convert the values of the sides in the same units. Since the cost was given per metre squares, we obtained the sides in the units of metre.