Big Ideas Math Geometry, 2014
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Big Ideas Math Geometry, 2014 View details
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Exercise 11 Page 59

Draw the vertices and polygon in a coordinate plane.

Perimeter: ≈ 17.6
Area: 15

Practice makes perfect

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

Now we see that it's a triangle.

Perimeter

We determine one of the line segments by counting squares.
To find the other lengths we use the Distance Formula.
JL = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
JL = sqrt(( -1- 2)^2 + ( 3-( -2))^2)
Evaluate
JL=sqrt((-3)^2+(3-(-2))^2)
JL=sqrt((-3)^2+5^2)
JL=sqrt(9+25)
JL=sqrt(34)
JL=5.83095...
JL≈5.8
One of the sides is about 5.8 units. Now we calculate the length of the third.
KL = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
KL = sqrt(( 5- 2)^2 + ( 3-( -2))^2)
Evaluate
KL=sqrt(3^2+(3-(-2))^2)
KL=sqrt(3^2+5^2)
KL=sqrt(9+25)
KL=sqrt(34)
KL=5.83095...
KL≈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.

Area

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=6*5/2=15.