Question

question 7 (multiple choice worth 4 points)
(04.03r mc)

calculate the area of triangle abc with altitude cd, given a (-7, -1), b (-1, 5), c (0, 0), and d (-3, 3).

9 square units

18 square units

18.5 square units

21 square units

Explanation:

Step1: Calculate the length of base \(AB\)

Use the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
For \(A(-7,-1)\) and \(B(-1,5)\), \(x_1=-7,y_1 = - 1,x_2=-1,y_2 = 5\).
\(AB=\sqrt{(-1+7)^2+(5 + 1)^2}=\sqrt{6^2+6^2}=\sqrt{36 + 36}=\sqrt{72}=6\sqrt{2}\)

Step2: Calculate the length of altitude \(CD\)

For \(C(0,0)\) and \(D(-3,3)\), \(x_1 = 0,y_1=0,x_2=-3,y_2 = 3\)
\(CD=\sqrt{(-3-0)^2+(3 - 0)^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}\)

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

The area of a triangle formula is \(A=\frac{1}{2}\times base\times height\)
Substitute \(base = AB = 6\sqrt{2}\) and \(height=CD = 3\sqrt{2}\)
\(A=\frac{1}{2}\times6\sqrt{2}\times3\sqrt{2}\)

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Answer:

18 square units