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Big Ideas Math Integrated I, 2016

BI

Big Ideas Math Integrated I, 2016 View details

Chapter Test

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Exercise 11 Page 435

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)

SubstitutePoints

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

JL = sqrt(( -1- 2)^2 + ( 3-( -2))^2)

▼

Evaluate

Subtract term

JL=sqrt((-3)^2+(3-(-2))^2)

a-(- b)=a+b

JL=sqrt((-3)^2+5^2)

Calculate quotient

JL=sqrt(9+25)

Add terms

JL=sqrt(34)

Use a calculator

JL=5.83095...

Round to 1 decimal place(s)

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)

SubstitutePoints

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

KL = sqrt(( 5- 2)^2 + ( 3-( -2))^2)

▼

Evaluate

Subtract term

KL=sqrt(3^2+(3-(-2))^2)

a-(- b)=a+b

KL=sqrt(3^2+5^2)

Calculate power

KL=sqrt(9+25)

Add terms

KL=sqrt(34)

Use a calculator

KL=5.83095...

Round to 1 decimal place(s)

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.

Subchapter links

Exercises p.435