McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
1. Multiplying and Dividing Rational Expressions
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Exercise 40 Page 535

What is the formula for the area of a parallelogram? Use it to find the length of the longer side of each parallelogram.

4x^2+9x-5/2 square meters

Practice makes perfect

Let's consider the given diagram.

We are also given the height and area of parallelogram F and G.

Parallelogram Height Area
F h_F=2x+3 A_F=8x^2+10x-3
G h_G=3x-1 A_G=6x^2+13x-5

To find the area of the right triangle H, we need to find the lengths b_F and b_G.

Recall the formula for the area of a parallelogram. A=bh Here, b is the length of the base of the parallelogram and h is its height. Note that we are given the area and the height of a parallelogram and we want to find the length of its base. This means we want to find b, knowing A and h. A=bh ⇔ b=A/h Using this formula, we can find the lengths b_F and b_G. Let's do it!

Parallelogram F

Substitute the expressions for the height and area of the parallelogram F into the formula.

b_F=A_F/h_F
b_F=8x^2+10x-3/2x+3
â–¼
Simplify
b_F=(4x-1)(2x+3)/2x+3
b_F=(4x-1)(2x+3)/2x+3
b_F=4x-1

We have that b_F=4x-1.

Parallelogram G

Again, let's substitute the expressions for the height and the area of the parallelogram G into the formula.

b_G=A_G/h_G
b_G=6x^2+13x-5/3x-1
â–¼
Simplify
b_G=(2x+5)(3x-1)/3x-1
b_G=(2x+5)(3x-1)/3x-1
b_G=2x+5

We have that b_G=2x+5.

Triangle H

Knowing the lengths b_F and b_G, we can calculate the area of the triangle H. It is a right triangle, so to find its area we have to multiply the lengths of its legs and divide by 2.

Therefore, the formula for the triangle H can be written as follows. A_H=1/2b_Fb_G Finally, let's substitute the values of b_F and b_G and simplify.

A_H=1/2b_F b_G
A_H=1/2(4x-1)(2x+5)
â–¼
Simplify
A_H=1/2(4x(2x+5)-1(2x+5))
A_H=1/2(8x^2+20x-2x-5)
A_H=1/2(8x^2+18x-5)
A_H=4x^2+9x-5/2

The area of the triangle H is 4x^2+9x- 52 square meters.