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Pearson Geometry Common Core, 2011

PG

Pearson Geometry Common Core, 2011 View details

1. Areas of Parallelograms and Triangles

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Exercise 41 Page 621

Notice that the area of the figure can be found by evaluating the difference between the area of the square and the area of the triangle.

312.5 ft^2

Practice makes perfect

Let's take a look at the given figure.

The figure is created using two triangles, and we have been given the length of one side on each. To find their areas we will need to know the length of the other sides adjacent to the right angles. While we currently only know that the sum of these lengths is 25, if we add a vertical segment we can create a square.

To find the area of the given figure we will calculate the area of the square first. Then we will subtract the area of the supplement, which is a triangle. Notice that both the base and height of the triangle are equal to 25 feet.

Area of the Square

To find the area of the square, we will substitute s= 25 feet into the formula and evaluate.

A=s^2

s= 25

A=( 25)^2

Calculate power

A=625

The area of the square is 625ft^2.

Area of the Supplement

To find the area of the supplemental triangle, we will substitute b= 25 feet and h=25 feet into the formula for the area of a triangle and simplify.

A=1/2bh

b= 25, h= 25

A=1/2( 25)(25)

▼

Evaluate right-hand side

Multiply

A=1/2(625)

MoveRightFacToNumOne

1/b* a = a/b

A=625/2

Calculate quotient

A=312.5

The area of the triangle is 312.5ft^2.

Area of the Figure

We found that the area of the square is 625 ft^2 and that the area of the triangle is 312.5 ft^2.

To find the area of the original figure, we will subtract the area of the triangle from the area of the square. Area of the Figure 625-312.5=312.5ft^2

Subchapter links

Lesson Check p.619

Practice and Problem-Solving Exercises p.619-621

Standardized Test Prep p.622

Mixed Review p.622