Question

calculate the area of triangle abc with altitude bd, given a(-6,0), b(0,0), c(0,6), and d(-3,3).
21 square units
18.5 square units
9 square units
18 square units

Explanation:

Step1: Calculate the base length

The base \(AC\) can be calculated using the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). For \(A(-6,0)\) and \(C(0,6)\), \(AC=\sqrt{(0 + 6)^2+(6 - 0)^2}=\sqrt{36 + 36}=\sqrt{72}=6\sqrt{2}\). Another way: since \(A(-6,0)\), \(C(0,6)\), the length of \(AC\) using the formula for the distance between two points \(d=\sqrt{(x_2-x_1)^2+(y_2 - y_1)^2}\), or we can also note that in a right - angled coordinate system, the base \(AC\) (if we consider the triangle with vertices \(A\), \(B\), \(C\) where \(AB\) is along the \(x\) - axis from \(x=-6\) to \(x = 0\) and \(BC\) is along the \(y\) - axis from \(y = 0\) to \(y=6\)) has length \(l=\sqrt{(0+6)^2+(6 - 0)^2}\), but more simply, if we use the formula for the area of a triangle \(S=\frac{1}{2}\times base\times height\). The base \(AC\): \(|AC|=\sqrt{(0 + 6)^2+(6-0)^2}=6\sqrt{2}\), but we can also use the formula \(S=\frac{1}{2}\times|x_A - x_C|\times|y_C - y_B|\) (since \(B(0,0)\)). The length of \(AC\): \(|AC|=\sqrt{(0+6)^2+(6 - 0)^2}=6\sqrt{2}\), but the formula for the area of \(\triangle ABC\) with \(A(-6,0)\), \(B(0,0)\), \(C(0,6)\) is \(S=\frac{1}{2}\times|AB|\times|BC|\). \(|AB|=6\) (since \(A(-6,0)\) and \(B(0,0)\)), \(|BC| = 6\) (since \(B(0,0)\) and \(C(0,6)\)).

Step2: Apply the area formula

The formula for the area of a triangle is \(S=\frac{1}{2}\times base\times height\). Here, the base \(AB\) (distance between \(A(-6,0)\) and \(B(0,0)\)) is \(|AB|=\sqrt{(0 + 6)^2+(0-0)^2}=6\), and the height \(BC\) (distance between \(B(0,0)\) and \(C(0,6)\)) is \(|BC|=\sqrt{(0 - 0)^2+(6 - 0)^2}=6\). Using the formula \(S=\frac{1}{2}\times base\times height\), where base \(=|AB| = 6\) and height \(=|BC|=6\).

$$S=\frac{1}{2}\times6\times6 = 18$$

Answer:

18 square units