Sometimes it is useful to be able to work backwards from the surface area or volume of an object to find an unknown dimension (for example height, length, width, or radius).

Example

Suppose you are asked to design a crate in the shape of a square prism with a capacity of $1$1 m³ and a height of $0.5$0.5 m. What should the side length of the square base be?

Let $x$x denote the unknown side length of the square base.

Since the crate is a prism its volume is given by the formula

$\text{Volume }=\text{ Area of base }\times\text{ height}$Volume = Area of base × height

We known the volume is $1$1 m³ and the height is $0.5$0.5 m. Since the base is a square with side length $x$x, the area of the base is $A=x^2$A=x2.

Substituting these values into the formula above gives:

$1=x^2\times0.5$1=x2×0.5

Next, divide each side by $0.5$0.5 to isolate the unknown:

$x^2$`x`2	$=$=	$\frac{1}{0.5}$10.5
	$=$=	$2$2

The solutions to this equation are:

$x=\pm\sqrt{2}$x=±√2

Since $x$x is a length it must be a positive quantity, so we take the positive solution:

$x=\sqrt{2}\approx1.41$x=√2≈1.41 m

Therefore the side length of the square base of the crate should be approximately $1.41$1.41 m.

As you can see from the example above, working backwards from surface area or volume typically requires rearranging an area or volume formula. For reference, below is a summary of all such formulas that we have seen so far.

Surface area and volume

Prisms

Surface area	$=$=	sum of area of faces

Volume	$=$=	area of base $\times$× height
	$=$=	$Ah$`Ah`

Pyramids

Surface area	$=$=	sum of areas of the faces

Volume	$=$=	$\frac{1}{3}$13 area of base $\times$× height
	$=$=	$\frac{1}{3}Ah$13`Ah`

Cylinders

Surface area	$=$=	2 $\times$× area of base $+$+ area of rectangle
	$=$=	$2\pi r^2+2\pi rh$2π`r`2+2π`rh`

Volume	$=$=	area of base $\times$× height
	$=$=	$\pi r^2h$π`r`2`h`

Cones

Surface area	$=$=	area of base $+$+ area of sector
	$=$=	$\pi r^2+\pi rs$π`r`2+π`rs`

Volume	$=$=	$\frac{1}{3}$13 area of base $\times$× height
	$=$=	$\frac{1}{3}\pi r^2h$13π`r`2`h`

Spheres

Surface area	$=$=	$4\pi r^2$4π`r`2

Volume	$=$=	$\frac{4}{3}\pi r^3$43π`r`3

Practice questions

Question 1

A ball has a surface area of $50.27$50.27 $mm$ ².

Determine its radius, to the nearest integer.

Question 2

This wild animal house is made out of plywood.

If the nesting box needs to have a volume of $129978$129978 cm³ and a height of $83$83 cm and front width of $54$54 cm, find the depth of the box.

A three-dimensional representation of a birdhouse is shown with dimensions provided on three sides. The birdhouse has a height of $83$83 cm, a width of $54$54 cm, and a depth of $d$`d` cm. The volume of the birdhouse is given as $129978$129978 $cm^3$`cm`3 but not labeled on the image. On the front of the birdhouse, there is an illustration of a bird perched on a protruding wooden ledge below an entrance hole. The birdhouse has a flat roof that overhangs slightly, colored in purple. The main body of the birdhouse is colored in shades of brown. The entrance hole is circular.

Question 3

The volume of the following tent is $4.64$4.64 m³.

Determine the height $h$h of the tent, in cm.

Outcomes

ACMGM019

calculate the volumes of standard three-dimensional objects such as spheres, rectangular prisms, cylinders, cones, pyramids and composites in practical situations; for example, the volume of water contained in a swimming pool

5.08 Finding dimensions from surface area or volume