Unit 62: Geometric Word Problems

Geometric word problems are exactly like they sound – they describe a shape, ask you to modify it or create an equation.

An example:

A triangle has sides of s, and s + 4.  The third side is twice the length of s.  What is the perimeter, in terms of s

A) \(3s+6s\)
B) \(s+4\)
C) \(s^2+4s\)
D) \(4s+4\)

Strategy:

  1. Draw!
  2. Consider assigning values.
  3. Use the math reference section for formulas.

Back to our initial example:

Draw a picture, labeling our sides.

Don’t worry about drawing a super-accurate triangle.  It doesn’t need to be precise.  It also doesn’t matter the angles are, we just happened to draw a right triangle here.

Perimeter is all the sides added together, so we take \(s+s+4+2s=4s+4\)

Or consider the following:

If a rectangle that has sides of length x and width y is altered by increasing its length by 20 percent and it’s width by p percent.  If these alterations increased the area of the rectangle by 30 percent, what is the value of p ?

A) 8.3
B) 20
C) 30
D) 60.6

Let’s try assigning values. Since we know we’re going to be working with percents, 10 or 100 would be good values.  We’ll use 10 here, but any value will work.  We could use 10 for y, but to make sure we’re keeping everything straight it’s good to use a different number.  Here, we’ll use 12.

The area of original rectangle would be \(10\times12=120\).  We know our new area is increased by 30 percent, so \(120(1.3)=156\).

We know our new length is \(10(1.2)=12\).  So our new rectangle would be \(12\cdot x=156\).  Dividing by 12, we find our new width must be 13.  

Remember our formula for percent change:

$$\frac{new-old}{old}$$

so

$$\frac{13-12}{12}=\frac1{12}=.083=8.3\%$$

Our width would have grown by A).

Alternatively, we could have combined this all into the equation \(v\) and then solved for p.  Either approach is correct and will arrive at the same answer.