Exercise 16 Page 30

Talia wants to pack a square-framed poster into a cube-shaped box.

Side Length of the Poster

We know the square-framed poster has an area of 9 square feet. We will begin by finding the length of each side of the poster. Recall that a square has four equal sides. Let's call the side length s.

When we find the area of a square, we multiply its side measurements two times. A = s * s ⇒ A= s^2We know that the area A is equal to 9 square feet. Therefore, we can substitute this value into the equation for the area.

A=s^2

Substitute

A= 9

9=s^2

RearrangeEqn

Rearrange equation

s^2 = 9

We need to find a number that, when squared, is equal to 9.

s^2 = 9

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt(s^2) = sqrt(9)

SplitIntoFactors

Split into factors

sqrt(s^2) = sqrt(3 * 3)

ProdToPowTwoFac

a* a=a^2

sqrt(s^2) = sqrt(3^2)

SqrtPowToNumber

sqrt(a^2)=a

s= 3

We know that the length of the side of the poster is 3 feet.

Side Length of the Box

We also know that the volume of the box is 30 cubic feet. We want to calculate also the length of each side of the box.

Three side measurements of a cube — the length, the width, and the height — all happen to be the same and we called the side length l. We can find the volume of a cube by multiplying its side length three times. V= l* l* l Recall that multiplying a number by itself three times is equal to raising the number to the power of 3. V= l* l* l ⇒ V= l^3 We know that the volume of a cube-shaped box is 30 cubic foot. Therefore, we can substitute it into the equation for the volume.

V=l^3

Substitute

V= 30

30=l^3

RearrangeEqn

Rearrange equation

l^3 = 30

RadicalEqn

sqrt(LHS)=sqrt(RHS)

sqrt(l^3)=sqrt(30)

RootPowToNumber

sqrt(a^3)=a

l = sqrt(30)

Comparison

Now, we need to compare the length side of the poster with the box side length. If the side length of the poster appears to be smaller than side of the box, then the poster will lie flat. Therefore, we need to compare the hole edge length with the block edge length. c|c Poster Side Length& Box Side Length 3 & sqrt(30) We want to find out whether 3 is greater than, less than, or equal to sqrt(30). To do it, we will estimate sqrt(30) to the whole number. Let's narrow down our estimate by looking at nearby perfect cubes. The two nearest perfect cubes are 27 and 64.

27<30<64

sqrt(LHS)

sqrt(27)< sqrt(30)

SplitIntoFactors

Split into factors

sqrt(3 * 3 * 3)< sqrt(30)

ProdToPowThreeFac

a* a* a=a^3

sqrt(3^3)< sqrt(30)

RootPowToNumber

sqrt(a^3)=a

3< sqrt(30)<4

Therefore, we see that the side length of the poster of 3 feet is smaller than side of the box sqrt(30). The poster will lie flat in the box.