Question

The length and the breadth of a rectangle are 6 cm and 4 cm respectively. Find the height of a triangle whose base is 6 cm and whose area is 3 times that of the rectangle.
\[\begin{align}
  & A.\text{ 24 cm} \\
 & \text{B}\text{. 20 cm} \\
 & \text{C}\text{. 26 cm} \\
 & \text{D}\text{. 12 cm} \\
\end{align}\]

Answer 1

Hint: In the given question the length (6 cm) and breadth (4cm) of a rectangle is given. Also, in question we have given the base length (6 cm) of a triangle and we have to find the height of the triangle. So, in order to calculate the height, we have also given the relation between the area of the rectangle and area of the triangle. As we know that area of rectangle is given by $A=LB$where $\text{L and B}$are the length and breadth of rectangle respectively, and area of triangle is given by ${A}'=\dfrac{1}{2}bh$ where $b\text{ and }h$are the length of base (6cm) and length of height of triangle. It is given from question ${A}'=3A$. So, we have to first calculate the area of the rectangle as well as the triangle and by equating both we can find the height of the triangle.

Complete step by step answer:

From question we have
Length of rectangle $L=6\text{ cm}$
Breadth of rectangle $B=4\text{ cm}$
As we know that,
Area of rectangle is given by
$A=LB$
So, area of rectangle can be found as
$\begin{align}
  & A=\left( 6cm \right)\left( 4cm \right) \\
 & A=24c{{m}^{2}}-------(1) \\
\end{align}$
Also, we have
Base length of triangle
$b=6cm$
Suppose the height of the triangle is $h$.
As we know that area of triangle
$\begin{align}
  & {A}'=\dfrac{1}{2}bh \\
 & {A}'=\dfrac{1}{2}(6cm)h------(2) \\
\end{align}$
Now it is given ${A}'=3A$, so we can write
$\begin{align}
  & \dfrac{1}{2}6cm(h)=\left( 3 \right)24c{{m}^{2}} \\
 & \Rightarrow h=\dfrac{\left( 3 \right)\left( 24 \right)\left( 2 \right)c{{m}^{2}}}{6c{{m}^{2}}} \\
 & \Rightarrow h=24cm \\
\end{align}$
Hence, the height of the triangle is 24 cm.

So, the correct answer is “Option A”.

Note:
Whenever we have to find such a type of question, we must see which constraint is applicable, here area is equal. Also, while we have to write an answer, we must take care of the unit. For one-dimension figure we have length and its unit is cm, for two-dimension figure we have area and its unit is $c{{m}^{2}}$