Measurement problems

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 $p^2-1$p2−1 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

Question 2

Question 3