Quadrilateral area formulas

Here are the five area formulas for the seven special quadrilaterals. There are only five formulas because some of them do double duty — for example, you can calculate the area of a rhombus with the kite formula.

Quadrilateral area formulas:

• 
• (because a rhombus is a type of parallelogram, you can use this formula for a rhombus)
• , or (a rhombus is also a type of kite, so you can use this formula for a rhombus as well)
• , or (this second formula works because a square is a type of kite)

• 

  Note: The median of a trapezoid is the segment that connects the midpoints of the legs. Its length equals the average of the lengths of the bases.

Why the formulas work

The area formulas for the parallelogram, kite, and trapezoid are based on the area of a rectangle. The following figures show you how each of these three quadrilaterals relates to a rectangle, and the following list gives you the details:

• Parallelogram: In Figure 7-1, if you cut off the little triangle on the left and fill it in on the right, the parallelogram becomes a rectangle (and the area obviously hasn’t changed). This rectangle has the same base and height as the original parallelogram. The area of the rectangle is , so that formula gives you the area of the parallelogram as well. If you don’t believe me (even though you should by now) you can try this yourself by cutting out a paper parallelogram and snipping off the triangle, as shown in Figure 7-1.
• Kite: Figure 7-2 shows that the kite has half the area of the rectangle drawn around it (this follows from the fact that , , and so on). You can see that the length and width of the large rectangle are the same as the lengths of the diagonals of the kite. The area of the rectangle thus equals , and because the kite has half that area, its area is .
• Trapezoid: If you cut off the two triangles and move them as I show you in Figure 7-3, the trapezoid becomes a rectangle. This rectangle has the same height as the trapezoid, and its base equals the median (m) of the trapezoid. Thus, the area of the rectangle (and therefore the trapezoid as well) equals .

Trying a few area problems

The key for many quadrilateral area problems is to draw altitudes and other perpendicular segments on the diagram. Doing so creates one or more right triangles, which allows you to use the Pythagorean Theorem or your knowledge of special right triangles, such as the and triangles.

Locating special right triangles in parallelograms

Find the area of parallelogram ABCD in Figure 7-4.

When you see a angle in a problem, a triangle is likely lurking somewhere in the problem. (Of course, a or angle is a dead giveaway of a triangle.) And if you see a angle, a triangle is likely lurking.

To get started, draw in the height of the parallelogram straight down from B to base to form a right triangle, as shown in Figure 7-5.

Consecutive angles in a parallelogram are supplementary. Angle ABC is , so is , and is thus a triangle. Now, if you know the ratio of the lengths of the sides in a triangle, , the rest is a snap. (the 2x side) equals and is thus 6. Then (the x side) is half of that, or 3; (the side) is therefore . Here’s the finish with the area formula:

Using triangles and ratios in a rhombus problem

Now for a rhombus problem: Find the area of rhombus RHOM given that is 6 and that the ratio of to is 4 : 1 (see Figure 7-6).

This one’s a bit tricky. You might feel that you’re not given enough information to solve it or that you just don’t know how to begin. If you ever feel this way when you’re in the middle of a problem, I have a great tip for you.

If you get stuck when doing a geometry problem — or any kind of math problem, for that matter — do something, anything! Begin anywhere you can: Use the given information or any ideas you have (try simple ideas before more-advanced ones) and write something down. Maybe draw a diagram if you don’t have one. Put something down on paper. One idea may trigger another, and before you know it, you’ve solved the problem. This tip is surprisingly effective.

Because the ratio of to is 4 : 1, you can give a length of 4x and a length of x. is thus , or 5x, and so is , because all sides of a rhombus are congruent. Now you have a right triangle with legs of 4x and 6 and a hypotenuse of 5x, so you can use the Pythagorean Theorem:

Because side lengths must be positive, you reject the answer . The length of the base, , is thus 5(2), or 10. (Triangle RBM is your old, familiar friend, a 3-4-5 triangle blown up by a factor of 2.) Now use the parallelogram-rhombus area formula:

Drawing in diagonals to find a kite’s area

What’s the area of kite KITE in Figure 7-7?

Draw in diagonals if necessary. For kite and rhombus area problems (and sometimes other quadrilateral problems), the diagonals are almost always necessary for the solution (because they form right triangles). You may have to add them to the figure.

So draw in and . Use “X” for the point where the diagonals cross. Use some scratch paper, or draw on Figure 7-7; no one will know.

Triangle KIT is a right triangle with congruent legs, so it’s a triangle with sides in the ratio of . The length of the hypotenuse, , thus equals one of the legs times ; that’s , or 16. is half of that, or 8.

Triangle KIX is another triangle (), the kite’s main diagonal, bisects opposite angles KIT and KET, and half of is ; therefore, , like , is 8. You’ve got another right triangle, , with a side of 8 and a hypotenuse of 17. I hope that rings a bell! You’re looking at an 8-15-17 triangle, so without any work, you see that is 15. (No bells? No worries. You can get with the Pythagorean Theorem instead.) Add to and you get for diagonal .

Now that you know the diagonal lengths, you have what you need to finish. The length of diagonal is 16, and diagonal is 23. Plug these numbers into the kite area formula for your final answer:

The polygon area formulas

A regular polygon is equilateral (it has equal sides) and equiangular (it has equal angles). To find the area of a regular polygon, you use an apothem — a segment that joins the polygon’s center to the midpoint of any side and that is perpendicular to that side. ( in upcoming Figure 7-8 is an apothem.)

Area of a regular polygon: Use the following formula to find the area of a regular polygon.

Note: This formula is usually written as , but if I do say so myself, the way I’ve written it, , is better. I like this way of writing it because the formula is based on the triangle area formula, : The polygon’s perimeter (p) is related to the triangle’s base (b), and the apothem (a) is related to the height (h).

An equilateral triangle is the regular polygon with the fewest possible number of sides. To figure its area, you can use the regular polygon formula; however, it also has its own area formula.

Area of an equilateral triangle: Here’s the area formula for an equilateral triangle.

(where s is the length of each of the triangle’s sides)

Tackling an area problem

Don’t tell me about your problems; I’ve got problems of my own — and here’s one of them.

What’s the area of a regular hexagon with an apothem of ?

For hexagons, use and equilateral triangles. A regular hexagon can be cut into six equilateral triangles, and an equilateral triangle can be divided into two triangles. So if you’re doing a hexagon problem, you may want to cut up the figure and use equilateral triangles or triangles to help you find the apothem, perimeter, or area.

First, sketch the hexagon with its three diagonals, creating six equilateral triangles. Then draw in an apothem, which goes from the center to the midpoint of a side. Figure 7-8 shows hexagon EXAGON.

Note that the apothem divides into two triangles (halves of an equilateral triangle). The apothem is the long leg (the side) of a triangle, so

is the short leg (the x side), so its length is 10. is twice as long, so it’s 20. And the perimeter is six times that, or 120.

Now you can finish with either the regular polygon formula or the equilateral triangle formula (multiplied by 6). They’re equally easy. Take your pick. Here’s what it looks like with the regular polygon formula:

Interior and exterior angles

You use two kinds of angles when working with polygons (see Figure 7-9):

• Interior angle: An interior angle of a polygon is an angle inside the polygon at one of its vertices. Angle Q is an interior angle of quadrilateral QUAD.
• Exterior angle: An exterior angle of a polygon is an angle outside the polygon formed by one of its sides and the extension of an adjacent side. Angle ADZ, , and are exterior angles of QUAD; vertical angle is not an exterior angle of QUAD.

Interior and exterior angle formulas:

• The sum of the measures of the interior angles of a polygon with n sides is .
• The measure of each interior angle of an equiangular n-gon is (the supplement of an exterior angle).
• If you count one exterior angle at each vertex, the sum of the measures of the exterior angles of a polygon is .
• The measure of each exterior angle of an equiangular n-gon is .

A polygon angle problem

You can practice the interior and exterior angle formulas in the following three-part problem: Given a regular dodecagon (12 sides),

1. Find the sum of the measures of its interior angles.

   Just plug the number of sides (12) into the formula for the sum of the interior angles of a polygon:

   

2. Find the measure of a single interior angle.

   This polygon has 12 sides, so it has 12 angles; and because you’re dealing with a regular polygon, all its angles are congruent. So to find the measure of a single angle, just divide your answer from the first part of the problem by 12. (Note that this is basically the same as using the first formula for a single interior angle.)

   

3. Find the measure of a single exterior angle with the exterior angle formula; then check that its supplement, an interior angle, equals the answer you got from part 2 of the problem.

   First, plug 12 into the oh-so-simple exterior angle formula:

   

   Now take the supplement of your answer to find the measure of a single interior angle, and check that it’s the same as your answer from part 2:

   

   It checks. (And note that this final computation is basically the same thing as using the second formula for a single interior angle.)

Criss-crossing with diagonals

Number of diagonals in an n-gon: The number of diagonals that you can draw in an n-gon is .

Here’s one last problem for you: If a polygon has 90 diagonals, how many sides does it have?

You know what the formula for the number of diagonals in a polygon is, and you know that the polygon has 90 diagonals, so plug 90 in for the answer and solve for n:

Thus, n equals 15 or . But because a polygon can’t have a negative number of sides, n must be 15. So you have a 15-sided polygon (a pentadecagon, in case you’re curious).