Problems Involving Areas

Review of Formulas

Let us review the main formulas for areas:

Parallelogram, Rectangle, Square, Rhombus:

The area is the length of the base times the height:

Triangle:

A triangle is half a parallelogram, so its area is half its base times its height;

Trapezoid:

The area of a trapezoid is the length of the midsegment times the height:

Circle:

The area of a circle is pi times the square of its radius:

EXAMPLE 1:  A rectangular picture is 8 inches by 10 inches.  A framer wants to place a boarder around the picture so the area of the boarder is equal to the area of the picture, and the spaces between the edges of the picture and the boarder are the same.  What dimensions should the matt have?

SOLUTION:

Let the common width of the boarder be x.  Then the width of the matt is 2x + 8 and its length is 2x + 10:

The area of the boarder is the area of the matt minus the area of the picture.  Since this is to equal the area of the picture, we can write the following equation:

(2x + 8)(2x + 10) – 8 . 10 = 8 . 10

(2x + 8)(2x + 10) – 80  = 80

Simplifying and subtracting 80 from both sides, we have the following quadratic equation:

4x2 + 36x – 80 = 0

We can simplify further by dividing by 4:

x2 + 9x – 20 = 0

The left side does not factor, so we will use the quadratic formula, but only for the case where x is positive:

Using a calculator, we find that x is approximately 1.844, so 2x is approximately 3.7.

The width is 2x + 8 = 11.7 inches, and the length is 2x + 10 = 13.7 inches.

EXAMPLE 2:  If the area of the blue region is 8x square inches and the width of the rectangle is x inches, then what is the radius of each circle?

SOLUTION:  The radius of each circle is x/2 so the combined area of the two circles is

The area of the whole rectangle is x(2x) = 2x2

Therefore the area of the blue region is the difference of these two, so

We can factor x2 on the left:

Since x isn't 0, we can divide both sides by x to get

Therefore,

The radius of each circle is half this, or 9.32 inches.

EXAMPLE 3:  If a dart is thrown and lands inside rectangle ABCD, what is the probability that it lands inside one of the circles?

SOLUTION:

The probability is the ratio of the area of the circles to the area of the rectangle (which is 1800 square units).  The two circles touch at the center of the rectangle.  Let the radius of each circle be x.  Then we have:

We can apply the Pythagorean Theorem to the right triangle:

x2 = (25 – x)2 + (18 – x)2

x2 = 625 – 50x + x2 + 324 – 36x + x2

0 = x2 – 86x + 949

Now 949 factors as 13 . 73, and since 12 + 73 = 86, we can factor the right side:

0 = (x – 13)(x – 73)

So x = 13 or x = 73.  But x cannot be 73 since that is more than the length of a side of the rectangle.  Therefore x is 13, and the ratio of the areas of the circles to the area of the rectangle gives us the probability: