Question

The area of the triangle with vertices $(0, 0)$, $(6, 0)$, and $(0, 5)$ is:

• A. $11$ Square units
• B. $12$ Square units
• C. $13$ Square units
• D. $15$ Square units

Hint:

For a triangle with one vertex at the origin $(0,0)$ and the other two vertices on the axes (e.g., $(a,0)$ and $(0,b)$), the area is simply $\frac{1}{2} \times |a| \times |b|$. This is a special case of a right-angled triangle.

Answer 1

The Correct Option is D

Step 1: Identify the type of triangle. 
The given vertices are $(0, 0)$, $(6, 0)$, and $(0, 5)$.
The vertex $(0, 0)$ is the origin.
The vertex $(6, 0)$ lies on the x-axis.
The vertex $(0, 5)$ lies on the y-axis.
Since two sides of the triangle lie along the x and y axes, this is a right-angled triangle.
Step 2: Determine the base and height of the triangle.
The length of the base can be calculated as the distance between $(0, 0)$ and $(6, 0)$ along the x-axis.
Base ($b$) = $6 - 0 = 6$ units.
The height of the triangle can be calculated as the distance between $(0, 0)$ and $(0, 5)$ along the y-axis.
Height ($h$) = $5 - 0 = 5$ units.
Step 3: Calculate the area of the triangle.
The formula for the area of a triangle is $\frac{1}{2} \times \text{base} \times \text{height}$.
Area = $\frac{1}{2} \times b \times h$
Area = $\frac{1}{2} \times 6 \times 5$
Area = $\frac{1}{2} \times 30$
Area = $15$ Square units.
Alternatively, using the determinant formula for the area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$:
Area = $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$ Let $(x_1, y_1) = (0, 0)$, $(x_2, y_2) = (6, 0)$, and $(x_3, y_3) = (0, 5)$.
Area = $\frac{1}{2} |0(0 - 5) + 6(5 - 0) + 0(0 - 0)|$
Area = $\frac{1}{2} |0 + 6(5) + 0|$
Area = $\frac{1}{2} |30|$
Area = $15$ Square units.
Step 4: Compare with the given options.
The calculated area is $15$ Square units, which matches option (4). (4) 15 Square units