Exercise 88 Page 425

Practice makes perfect

We are given the following square.

We want to find the side length x. Let's start by recalling that the area of a square A is the side length s squared. A= s^2 In this case, A= 225 and s= x. Let's substitute these values into the formula. 225= x^2 We got an equation that we can solve for x. Let's do it!

225=x^2

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt(225)=sqrt(x^2)

SqrtPowToNumber

sqrt(a^2)=a

sqrt(225)=x

CalcRoot

Calculate root

15=x

RearrangeEqn

Rearrange equation

x=15

We found that x= 15 is a solution to the equation. This means that the missing side length is 15 meters.

We want to find the area of the given square.

Let's do it! Remember, the area of a square is its side length squared. A= 11^2 = 121 The area of the square is 121 square centimeters.

We are given the following rectangle.

We want to find the length w. Let's start by recalling that the area of a rectangle is the product of its length and width. In this case, the area is 100 square meters, the length is w, and the width is 5 meters. Area= Length* Width [0.3em] ↓ [0.3em] 100= w * 5 Let's solve the equation we got for w.

100 = w* 5

DivEqn

.LHS /5.=.RHS /5.

100/5=w* 5/5

SimpQuot

Simplify quotient

100/5=w

CalcQuot

Calculate quotient

20=w

RearrangeEqn

Rearrange equation

w=20

We found that w= 20 is the solution to the equation, which means that the length of the rectangle is 20 meters.

Let's consider the given square.

To find the value of y, let's use the formula for the area of a square. In this case, A= 150 and s= y. A= s^2 ↓ 150= y^2 We got an equation that we can solve for y. Let's do it!

150=y^2

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt(150)=sqrt(y)

SqrtPowToNumber

sqrt(a^2)=a

sqrt(150)=y

CalcRoot

Calculate root

12.247448... =y

RoundDec

Round to 2 decimal place(s)