# 2.10: Perimeter and Area

**At Grade**Created by: CK-12

## Triangles and Parallelograms

**The Importance of Units -** Students will give answers that do not include the proper units, unless it is required by the instructor. When stating an area, square units should be included, and when referring to a length, linear units should be used. Using proper units helps reinforce the basic concepts. With these first simple area problems including the units seems like a small detail, but as the students move to more complex situations combining length, area, and volume, units can be a helpful guide. In physics and chemistry dimensional analysis is an important tool.

**The Power of Labeling -** When doing an exercise where a figure needs to be broken into polynomials with known area formulas, it is important for the student to draw on and label the figure well. Each polygon, so far only parallelograms and triangles, should have their base and height labeled and the individual area should be in the center of each. By solving these exercises in a neat, orderly way student will avoid errors like using the wrong values in the formulas, overlapping polygons, or leaving out some of the total area.

**Subtracting Areas -** Another way of finding the area of a figure that is not a standard polygon is to calculate a larger known area and then subtracting off the areas of polygons that are not included in the target area. This can often result in fewer calculations than adding areas. Different minds work in different ways, and this method might appeal to some students. It is nice to give them as many options as possible so they feel they have the freedom to be creative.

**The Height Must Be Perpendicular to the Base -** Students will frequently take the numbers from a polygon and plug them into the area formula without really thinking about what the numbers represent. In geometry there will frequently be more steps. The students will have to use what they have learned to find the correct base and height and then use those numbers in an area formula. Remind students that they already know how to use a formula; many exercises in this class will require more conceptual work.

**Write Out the Formula -** When using an area formula, it is a good idea to have students first write out the formula they are using, substitute numbers in the next step, and then solve the resulting equation. Writing the formula helps them memorize it and also reduces error when substituting and solving. It is especially important when the area is given and the student is solving for a length measurement in the polygon. Students will be able to do these calculations in their heads for parallelograms, and maybe triangles as well, but it is important to start good habits for the more complex polygons to come.

## Trapezoids, Rhombi, and Kites

**It’s Arts and Crafts Time -** Student have trouble remembering how to derive the area formulas. At this level it is required that they understand the nature of the formulas and why the formulas work so they can modify and apply them in less straightforward situations. An activity where student follow the explanation by illustrating it with shapes that they cut out and manipulate is much more powerful than just listening and taking notes. It will engage the students, keep their attention, and make them remember the lesson longer. Here are examples of how students can do this for each figure in this section.

**Trapezoid -** This procedure models the description in the text. Actually have students physically cut out the figures and do this will deepen understanding. Afterwards, the pictures in the lesson in the text will have much more meaning to them as they will have a more concrete understanding of what is going on.

- Have student use the parallel lines on binder paper to draw a trapezoid. They should draw in the height and label it \begin{align*}h\end{align*}
h . They should also label the two bases \begin{align*}b_1\end{align*}b1 and \begin{align*}b_2\end{align*}b2 . - Now they can trace and cut out a second congruent trapezoid and label it as they did the first.
- The two trapezoids can be arranged into a parallelogram and glued down to another piece of paper.
- Identify the base and height of the parallelogram in terms of the trapezoid variables. Then substitute these expressions into the area formula of a parallelogram to derive the area formula for a trapezoid.
- Remember that two congruent trapezoids were used in the parallelogram, and the formula should only find the area of one trapezoid.

**Kite -** This particular method is different than the description in the text. You can choose to do this either way, but it helps to improve understanding sometimes to show different methods to achieve the same results. Some students may have an easier time grasping the concept one way or the other.

- Have the students draw a kite. They should start by making perpendicular diagonals, one of which is bisecting the other. Then they can connect the vertices to form a kite.
- Now they can draw in the rectangle around the kite.
- Identify the base and height of the parallelogram in terms of \begin{align*}d_1\end{align*}
d1 and \begin{align*}d_2\end{align*}d2 , and then substitute into the parallelogram area formula to derive the kite area formula. - Now have the students cut off the four triangles that are not part of the kite and arrange them over the congruent triangle in the kite to demonstrate that the area of the kite is half the area of the rectangle.

**Rhombus -** You could allow students to decide which way they would rather do this one or just let them come up with their own method. The more they can do this on their own, the more they have gleaned from this activity.

Also, the area of a rhombus can be found using either the kite or parallelogram area formulas. Use this as an opportunity to review subsets and what they mean in terms of applying formulas and theorems.

## Area of Similar Polygons

**Reducing Fractions -** It is helpful in this section to have ratios in reduced form. Once students start square-rooting ratios (in fraction form), however, they may start messing up their reductions. For example, they may start thinking that \begin{align*}\frac{4}{9}\end{align*}

**Adjust the Scale Factor** - It is difficult for students to remember to square and cube the scale factor when writing proportions involving area and volume. Writing and solving a proportion is a skill they know well and have used frequently. Once the process is started, it is hard to remember to add that extra step of checking and adjusting the scale factor in the middle of the process. Here are some ways to reinforce this step in the students’ minds.

- Inform students that this material is frequently used on the SAT and other standardized tests in some of the more difficult problems.
- Play with graph paper. Have students draw similar shape on graph paper. They can estimate the area by counting squares, and then compare the ratio of the areas to the ratio of the side lengths. Creating the shapes on graph paper will give the students a good visual impression of the areas.
- Write out steps, or have the students write out the process they will use to tackle these problems. (1) Write a ratio comparing the two polygons. (2) Identify the type of ratio: linear, area, or volume. (3) Adjust the ratio using powers or roots to get the desired ratio. (4) Write and solve a proportion.
- Mix-up the exercises so that students will have to square the ratio in one problem and not in the next. Keep them on the lookout. Make them analyze the situation instead of falling into a habit.

Example 1: The ratio of the lengths of the sides of two squares is 2:3. What is the ratio of their areas?

Answer: 4:9, The ratio of areas is the ratio of the lengths squared.

Example 2: The area of a small triangle is \begin{align*}15 \ cm^2\end{align*}

Answer: 10 cm

The area ratio is 15:60 or 1:4. The length ratio is then the square root of this, or 1:2. Now set up a proportion to solve for the height of the smaller triangle.

\begin{align*}\frac{1}{2} = \frac{5}{x}\end{align*}

Example 3: The ratio of the lengths of two similar rectangles is 4:5. The larger rectangle has a width of 45 cm. What is the width of the smaller rectangle?

Answer: 36 cm.

In this problem, we did not have to adjust the ratio since we are given a ratio of lengths and a length is what we are trying to find. Just set up the proportion and solve.

\begin{align*}\frac{4}{5} = \frac{x}{45}\end{align*}

Example 4: The ratio of the areas of two regular pentagons is 25:64. What is the ratio of their corresponding sides?

Answer: 5:8

This time students have to square root the area ratio to find the ratio of the lengths of the sides.

## Circumference and Arc Length

**Pi is an Irrational Number -** Many students can give the definition of an irrational number. They know that an irrational number has an infinite decimal that has no pattern, but they have not really internalized what this means. Infinity is a difficult concept. A fun way to help the students develop this concept is to have a pi contest. The students can chose to compete by memorizing digits of pi. They can be given points, possible extra credit, for ever ten digits or so, and the winner gets a pie of their choice. The students can also research records for memorizing digits of pi. The competition can be done on March \begin{align*}14^{th}\end{align*}

**There Are Two Values That Describe an Arc -** The measure of an arc describes how curved the arc is, and the length describes the size of the arc. Whenever possible, have the students give both values with units so that they will remember that there are two different numerical descriptions of an arc. Often student will give the measure of an arc when asked to calculate its length.

**Arc Length Fractions -** Fractions are a difficult concept for many students even when they have come as far as geometry. For many of them putting the arc measure over 360 does not obviously give the part of the circumference included in the arc. It is best to start with easy fractions. Use a semi-circle and show how \begin{align*}\frac{180}{360}\end{align*}

**Exact or Approximate -** When dealing with the circumference of a circle there are often two ways to express the answer. The students can give exact answers, such as \begin{align*}2\pi\end{align*}

## Areas of Circles and Sectors

**Reinforce -** This section on area of a circle and the area of a sector is analogous to the previous section about circumference of a circle and arc length. This gives students another chance to go back over the arguments and logic to better understand, remember, and apply them. Focus on the same key points and methods in this section, and compare it to the previous section. Mix-up exercises so students will see the similarities and learn each more thoroughly.

**Don’t Forget the Units -** Remind students that when they calculate an area the units are squared. When an answer contains the pi symbol, students are more likely to leave off the units. In the answer \begin{align*}7\pi \ cm^2\end{align*}

**Draw a Picture -** When applying geometry to the world around us, it is helpful to draw, label, and work with a picture. Visually organized information is a powerful tool. Remind students to take the time for this step when calculating the areas of the irregular shapes that surround us.

Example 1: What is the area between two concentric circles with radii 5 cm and 12 cm?

(Hint: Don’t subtract the radii.)

Answer: \begin{align*}144\pi - 25\pi = 119 \pi cm^2\end{align*}

Example 2: The area of a sector of a circle with radius 6 cm, is \begin{align*}12\pi \ cm^2\end{align*}

Answer: \begin{align*}12 \pi = \frac{x}{360}* \pi *6^2, x=120^\circ\end{align*}

Example 3: A square with side length \begin{align*}5 \sqrt 2 \ cm\end{align*}

Answer: approx. \begin{align*}28.5 \ cm^2\end{align*}

First find the diagonal of the square using special right triangles: \begin{align*}5\sqrt{2}*\sqrt{2} = 5 * 2 = 10\end{align*}

This makes the radius of the circle 5 cm. Now we can find the area of the circle and subtract the area of the square as shown below.

\begin{align*}&5^2*\pi-\left (5\sqrt{2} \right )^2\\ &25 \pi -50\approx 28.5 cm^2\end{align*}