Exercise 27 Page 34

We are given an expression for the side lengths of an isosceles triangle. We want to find the perimeter of two possibles triangles. To do so, we will consider that an isosceles triangle is a triangle that has two congruent sides. This means that two side lengths must be equal. We will use this information to write an equation that helps us find the perimeter of each possible triangle.

First Triangle

We will begin by considering a triangle where the side that is 4x+5 is equal to the side that is 2x+7. 4x+5=2x+7 Before finding the perimeter of this triangle, we need to find the value of x. To do so, we will use inverse operations to isolate x on one side of the equation. Let's do it!

4x+5=2x+7

SubEqn

LHS-5=RHS-5

4x+5-5=2x+7-5

▼

Solve for x

SubEqn

LHS-2x=RHS-2x

4x+5-5-2x=2x+7-5-2x

SubTerms

Subtract terms

2x=2

DivEqn

.LHS /2.=.RHS /2.

2x/2=2/2

QuotOne

a/a=1

2x/2=1

CrossCommonFac

Cross out common factors

2x/2=1

SimpQuot

Simplify quotient

x=1

Recall that the perimeter is the sum of all the side lengths. P=(3x+1)+(4x+5)+(2x+7) Now, we will substitute x= 1 into this equation and solve it to find the perimeter.

P=(3x+1)+(4x+5)+(2x+7)

RemovePar

Remove parentheses

P=3x+1+4x+5+2x+7

Substitute

x= 1

P=3( 1)+1+4( 1)+5+2( 1)+7

Multiply

P=3+1+4+5+2+7

AddTerms

Add terms

P=22

Therefore, the perimeter of the first possible triangle is 22 inches.

Second Triangle

Now, we will consider that the side that is 3x+1 is equal to the side that is 2x+7. We can equate both expressions and solve the equation to obtain the value of x. Let's do it!

3x+1=2x+7

SubEqn

LHS-1=RHS-1

3x+1-1=2x+7-1

SubEqn

LHS-2x=RHS-2x

3x+1-1-2x=2x+7-1-2x

SubTerms

Subtract terms

x=6

Because we know that the perimeter is the sum of all the side lengths, we can find it by substituting x= 6 into the perimeter formula.

P=(3x+1)+(4x+5)+(2x+7)