Question

What is the number of square units in the area of the triangle whose vertices are points A(2,0), B(6,0), and C(8,5)?A. 15 units²B. 30 units²C. 10 units²D. 6 units²

Answer 1

Given the triangle;

a=|BC|;

$\begin{gathered} a=\sqrt[]{(8-6)^2+(5-0)^2} \\ a=\sqrt[]{29} \\ a=5.39units \end{gathered}$

b=|AC|;

$\begin{gathered} b=\sqrt[]{(8-2)^2+(5-0)^2} \\ b=\sqrt[]{61} \\ b=7.81\text{units} \end{gathered}$

c=|AB|;

$\begin{gathered} c=\sqrt[]{(6-2)^2} \\ c=\sqrt[]{16} \\ c=4\text{units} \end{gathered}$

Thus, using Heron's formula;

$\begin{gathered} A=\sqrt[]{s(s-a)(s-b)(s-c)} \\ \text{Where s= }\frac{a+b+c}{2} \\ s=\frac{5.39+7.81+4}{2} \\ s=8.6 \end{gathered}$

Then, we have;

$\begin{gathered} A=\sqrt[]{8.6(8.6-5.39)(8.6-7.81)(8.6-4)} \\ A=\sqrt[]{8.6(3.21)(0.79)(4.6)} \\ A=\sqrt[]{100.32} \\ A=10.32 \\ A\cong10\text{ square units} \end{gathered}$

CORRECT OPTION: C