 Measurement problems

Lesson

Sometimes, in mathematical problems we are required to find certain lengths, areas and volumes in terms of a given variable. We will illustrate the idea with two examples:

Example 1:

Suppose we know that the length of a certain rectangle is $6$6 $cm$cm more than twice its width. Can we find its area and perimeter?

On this information alone, we can't. But what we can do is formulate an expression - specifically a polynomial expression - for the perimeter and area in terms of the width of the rectangle.

Suppose we call the width of the rectangle $x$x. Then this means that the length could be expressed as $2x+6$2x+6. The perimeter $P$P, in terms of $x$x, becomes the linear polynomial $P=2\left[x+\left(2x+6\right)\right]=6x+12$P=2[x+(2x+6)]=6x+12 and the area $A$A, in terms of $x$x, becomes the quadratic polynomial $A=x\left(2x+6\right)=2x^2+6x$A=x(2x+6)=2x2+6x.

These are variable quantities, and remain so until further information is provided.

For example, suppose we learn at a later time that the width is $3$3 $cm$cm. Then we immediately know that:

 $P$P $=$= $6x+12$6x+12 $=$= $6\left(3\right)+12$6(3)+12 $=$= $30cm$30cm

and

 $A$A $=$= $2x^2+6x$2x2+6x $=$= $2\left(3\right)^2+6\left(3\right)$2(3)2+6(3) $=$= $36$36 $cm^2$cm2

Alternatively, perhaps we learn that the numerical values of the area and the perimeter are equal. Then, with the variable quantities known we can solve for $x$x in the equation $2x^2+6x=6x+12$2x2+6x=6x+12.

 $2x^2+6x$2x2+6x $=$= $6x+12$6x+12 $2x^2-12$2x2−12 $=$= $0$0 $2x^2$2x2 $=$= $12$12 $x^2$x2 $=$= $6$6 $\therefore$∴     $x$x $=$= $\sqrt{6}$√6, $x>0$x>0
Example 2:

As another example, consider a right-angled triangle with the two shorter sides given as $p^2-1$p21 and $2p$2p.

Can we find an expression for its perimeter and area? By Pythagoras' theorem, we have that the hypotenuse $h$h is determined as:

 $h^2$h2 $=$= $\left(p^2-1\right)^2+\left(2p\right)^2$(p2−1)2+(2p)2 $=$= $p^4-2p^2+1+4p^2$p4−2p2+1+4p2 $=$= $p^4+2p^2+1$p4+2p2+1 $=$= $\left(p^2+1\right)^2$(p2+1)2 $\therefore$∴     $h$h $=$= $p^2+1$p2+1

Thus the perimeter is given by the polynomial:

 $P$P $=$= $\left(p^2+1\right)+\left(p^2-1\right)+2p$(p2+1)+(p2−1)+2p $=$= $2p^2+2p$2p2+2p $=$= $2p\left(p+1\right)$2p(p+1)

The area is given by:

 $A$A $=$= $\frac{1}{2}\left(2p\right)\left(p^2-1\right)$12​(2p)(p2−1) $=$= $\left(p\right)\left(p^2-1\right)$(p)(p2−1) $=$= $\left(p\right)\left(p+1\right)\left(p-1\right)$(p)(p+1)(p−1)

Worked Examples

Question 1

Find the polynomial that represents the area of the triangle. Question 2

Jimmy and a few of his friends are jumping off their local wharf. If the length of the wharf is $\frac{6}{3y+2}$63y+2 metres and Jimmy jumps a distance of $\frac{5}{8y+8}$58y+8 metres, then how far does Jimmy travel from the start of his run up? Give your answer in factorised form. Question 3

Consider the cube shown. 1. Find the polynomial that represents the volume of the cube.

2. Find the volume of the cube if the value of $x$x is $6$6.