Question

using the box method to calculate the area of a triangle
find the area of the triangle qrs
area = square units
q(-9, 5)
r(6, 10)
s(2, -10)

Explanation:

Step1: Enclose the triangle in a rectangle

The rectangle has vertices with \(x - \)coordinates from \(-9\) to \(6\) and \(y - \)coordinates from \(- 10\) to \(10\). The length of the rectangle along the \(x\) - axis is \(6-(-9)=15\) and along the \(y\) - axis is \(10 - (-10)=20\). The area of the rectangle \(A_{r}=15\times20 = 300\).

Step2: Calculate the areas of the three right - angled triangles outside the given triangle

For the first right - angled triangle with vertices \(Q(-9,5)\), \(R(6,10)\) and the point \((6,5)\):
The base is \(6-(-9)=15\) and the height is \(10 - 5=5\). Its area \(A_1=\frac{1}{2}\times15\times5=\frac{75}{2}\).
For the second right - angled triangle with vertices \(R(6,10)\), \(S(2,-10)\) and the point \((6,-10)\):
The base is \(6 - 2=4\) and the height is \(10-(-10)=20\). Its area \(A_2=\frac{1}{2}\times4\times20 = 40\).
For the third right - angled triangle with vertices \(Q(-9,5)\), \(S(2,-10)\) and the point \((-9,-10)\):
The base is \(2-(-9)=11\) and the height is \(5-(-10)=15\). Its area \(A_3=\frac{1}{2}\times11\times15=\frac{165}{2}\).

Step3: Calculate the area of triangle \(QRS\)

The area of triangle \(QRS\) is \(A = A_{r}-A_1 - A_2 - A_3\).
\[

$$\begin{align*} A&=300-\frac{75}{2}-40-\frac{165}{2}\\ &=300-( \frac{75 + 165}{2})-40\\ &=300 - \frac{240}{2}-40\\ &=300 - 120-40\\ &=140 \end{align*}$$

\]

Answer:

140