Question

sing the box method to calculate the area of a triangle
find the area of the triangle qrs.
area = square units

Explanation:

Step1: Find the length and width of the box

The \(x -\)coordinates of the points are \(-9\), \(2\), and \(6\). The maximum \(x-\)value is \(6\) and the minimum \(x-\)value is \(-9\), so the length of the box in the \(x-\)direction is \(6-(-9)=15\).
The \(y -\)coordinates of the points are \(5\), \(10\), and \(- 10\). The maximum \(y-\)value is \(10\) and the minimum \(y-\)value is \(-10\), so the length of the box in the \(y-\)direction is \(10-(-10)=20\).
The area of the box \(A_{box}=15\times20 = 300\).

Step2: Calculate the areas of the three right - angled triangles outside the triangle \(QRS\)

For the first triangle:
Let's consider the triangle with vertices \(Q(-9,5)\), \(R(6,10)\) and the right - angled corner.
The base \(b_1=15\) (from \(x=-9\) to \(x = 6\)) and the height \(h_1 = 5\) (from \(y = 5\) to \(y = 10\)). The area \(A_1=\frac{1}{2}\times15\times5=\frac{75}{2}=37.5\).
For the second triangle:
Vertices \(R(6,10)\), \(S(2,-10)\) and the right - angled corner.
The base \(b_2 = 4\) (from \(x = 2\) to \(x = 6\)) and the height \(h_2=20\) (from \(y=-10\) to \(y = 10\)). The area \(A_2=\frac{1}{2}\times4\times20 = 40\).
For the third triangle:
Vertices \(Q(-9,5)\), \(S(2,-10)\) and the right - angled corner.
The base \(b_3=11\) (from \(x=-9\) to \(x = 2\)) and the height \(h_3 = 15\) (from \(y=-10\) to \(y = 5\)). The area \(A_3=\frac{1}{2}\times11\times15=\frac{165}{2}=82.5\).

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

Using the formula \(A = A_{box}-A_1 - A_2-A_3\).
\(A=300-(37.5 + 40+82.5)\).
\(A=300 - 160\).
\(A = 140\).

Answer:

\(140\)