Question

Type the correct answer in each box. If necessary, round your answer(s) to the nearest hundredth. The vertices of ∆ABC are A(-2, 2), B(6, 2), and C(0, 8). The perimeter of ∆ABC is units, and its area is square units.

Answer 1

Answer:

Perimeter = 22.809 units

Area = 24 square units.

Step-by-step explanation:

The vertices of ΔABC are given by A(-2,2), B(6,2), and C(0,8).

We know, that the length of a line joining two points (x1, y1) and (x2, y2) is given by $\sqrt{(x1-x2)^{2}+(y1-y2)^{2} }$.

Now, length of AB = $\sqrt{(-2-6)^{2}+(2-2)^{2} } =8$ units

length of BC = $\sqrt{(6-0)^{2}+(2-8)^{2} } =6\sqrt{2}=8.485$ units

and the length of CA = $\sqrt{(-2-0)^{2} +(2-8)^{2} } =6.324$ units

Therefore, the perimeter of the triangle is ( 8 + 8.485 + 6.324 ) = 22.809 units. (Answer)

Again the area of the triangle ABC is given by

$\frac{1}{2} |-2(2-8)+6(8-2)+0(2-2)|=24$ square units. (Answer)

Since we know, that area of a triangle having vertices (x1, y1), (x2, y2) and (x3, y3) is given by

$\frac{1}{2} |x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|$.