Question

If the area of the triangle shown is 30 square units. What is the value of y?

\[\begin{align}
  & \text{A}.\text{ 5} \\
 & \text{B}.\text{ 6} \\
 & \text{C}.\text{ 8} \\
 & \text{D}.\text{ 12} \\
\end{align}\]

Answer 1

b> To solve this question, we will first observe the figure closely and see that what type of triangle is formed using axis. After observing that, it is a right angled triangle, we will use formula of area of triangle to solve this, which is given by \[A=\dfrac{1}{2}\times \text{base}\times \text{height}\] to get the base and height, we will use the distance formula given by \[\text{PQ=}\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}\], where, \[\text{P=}\left( {{x}_{1}},{{y}_{1}} \right)\text{ and Q=}\left( {{x}_{2}},{{y}_{2}} \right)\]

Complete step-by-step solution:
Given that, the area of the triangle is 30 square units. Observing the figure, we get that, two of the sides of the triangle are formed at the x and y-axis.
We know that x-axis and y-axis are perpendicular to each other.
Therefore, these two sides of the given triangle are perpendicular to each other, hence we have an angle between them is ${{90}^{\circ }}$ So, we get that, the given triangle is a right-angled triangle.
The area of a right-angled triangle is given by,
\[A=\dfrac{1}{2}\times \text{base}\times \text{height}\]
We have figured as:

Let us denote the vertices of the triangle as OAB.
Then, the base of the triangle is OB and the height is OA. We will use the distance formula to calculate distance OB and OA.
Distance formula is given as:
\[\text{PQ=}\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}\]
Where, \[\text{P=}\left( {{x}_{1}},{{y}_{1}} \right)\text{ and Q=}\left( {{x}_{2}},{{y}_{2}} \right)\]
Here, O = (0, 0)
A = (0, y)
And B = (5, 0)
Then, \[\begin{align}
  & \text{OB=}\sqrt{{{\left( 5-0 \right)}^{2}}+{{\left( 0-0 \right)}^{2}}} \\
 & \text{OB=}\sqrt{25} \\
 & \text{OB=5} \\
\end{align}\]
And similarly using distance formula,
\[\begin{align}
  & \text{OA=}\sqrt{{{\left( 0-y \right)}^{2}}+{{\left( 0-0 \right)}^{2}}} \\
 & \text{OA=}\sqrt{{{\left( y \right)}^{2}}} \\
 & \text{OA=y} \\
\end{align}\]
We have \[\text{Area of }\Delta \text{AOB}=\dfrac{1}{2}\times \text{base}\times \text{height}\]
\[\text{Area A=}\dfrac{1}{2}\times OB\times OA\]
Substituting the values of OB and OA we get, and area A = 30 square unit.
\[\begin{align}
  & \text{Area A=30=}\dfrac{1}{2}\times 5\times y \\
 & \Rightarrow 30=\dfrac{1}{2}\times 5y \\
\end{align}\]
Multiplying 2 on both sides, we get:
\[\Rightarrow 60=5y\]
Dividing by 5, we get:
\[\Rightarrow 12=y\]
Therefore, the value of y = 12 which is option D.

Note: The key point in this question is using area of triangle as \[A=\dfrac{1}{2}\times \text{base}\times \text{height}\] Here, the triangle is a right-angled triangle. Therefore we can apply this formula of area of the triangle. If it was a normal triangle other than the right-angled triangle then we cannot apply this formula. While applying the distance formula, even if students interchange the coordinates by mistake, they will not go wrong as there is a square and the result won't be altered.