Exercise 11 Page 435

We start by marking the vertices and the polygon in a coordinate plane.

Now we see that it's a triangle.

We determine one of the line segments by counting squares.

To find the other lengths we use the Distance Formula.

J L = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

Substitute $(- 1, 3)$ & $(2, - 2)$

J L = (- 1 - 2)^{2} + (3 - (- 2))^{2}

▼

Evaluate

Subtract term

J L = (- 3)^{2} + (3 - (- 2))^{2}

$a - (- b) = a + b$

J L = (- 3)^{2} + 5^{2}

Calculate quotient

J L = 9 + 25

Add terms

J L = 34

Use a calculator

J L = 5.83095 \dots

Round to $1$ decimal place(s)

J L \approx 5.8

One of the sides is about

5.8

units. Now we calculate the length of the third.

K L = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

Substitute $(5, 3)$ & $(2, - 2)$

K L = (5 - 2)^{2} + (3 - (- 2))^{2}

▼

Evaluate

Subtract term

K L = 3^{2} + (3 - (- 2))^{2}

$a - (- b) = a + b$

K L = 3^{2} + 5^{2}

Calculate power

K L = 9 + 25

Add terms

K L = 34

Use a calculator

K L = 5.83095 \dots

Round to $1$ decimal place(s)

K L \approx 5.8

The third side is also about

5.8

units.

The perimeter of the triangle is therefore approximately

6 + 5.8 + 5.8 = 17.6 .

The area of a triangle can be found by multiplying the base and height and then dividing by $2 .$

The base is

6

units and the height is

5

units. Therefore, the area is

A = \frac{6 \cdot 5}{2} = 15 .