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1.10: Perimeter and Area

Difficulty Level: At Grade Created by: CK-12
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Day 1 Day 2 Day 3 Day 4 Day 5
Triangles and Parallelograms Trapezoids, Rhombi, and Kites

More Trapezoids, Rhombi, and Kites

Start Area of Similar Polygons

Finish Area of Similar Polygons

Quiz 1

Start Circumference and Arc Length

Day 6 Day 7 Day 8 Day 9 Day 10

Finish Circumference and Arc Length

Investigation 10-1

Area of Circles and Sectors

Quiz 2

Start Review of Chapter 10

Review of Chapter 10 Chapter 10 Test

Triangles and Parallelograms


This lesson introduces students to the area and perimeter formulas for triangles, parallelograms and rectangles.

Relevant Review

Most of this lesson should be review for students. They have learned about area and perimeter of triangles and rectangles in a previous math class (Math 6, Pre-Algebra, or equivalent).

Notation Note

In this chapter, students need to use square units. If no specific units are given, students can write \begin{align*}\text{units}^2\end{align*}units2 or \begin{align*}u^2\end{align*}u2.

Teaching Strategies

If students are having a hard time with the formulas for area and perimeter of a rectangle, place Example 3 on a piece of graph paper or transparency. Then, students can could the squares for the area and perimeter and you can generate the formula together.

If you count all the squares, there are 36 squares in the area, or square centimeters (red numbers). Counting around the rectangle (blue numbers), we see there are 26 squares. Therefore, the perimeter of this square is 26 cm.

This technique will also work for squares.

An important note, each problem will have some sort of units. Remind students that the shapes might not always be drawn to scale.

Example 5 is a counterexample for the converse of the Congruent Areas Postulate. Therefore, the converse is false. An additional counterexample would to have them draw all the possible rectangles with an area of \begin{align*}20 \ in^2\end{align*}20 in2. Use graph paper so students will see that each rectangle has 20 squares. Possible answers are: \begin{align*}20 \times 1, 10 \times 2\end{align*}20×1,10×2, and \begin{align*}5 \times 4\end{align*}5×4.

The Area Addition Postulate encourages students to separate a figure into smaller shapes. Always divide the larger shape into smaller shapes that students know how to find the area of.

To show students the area of a parallelogram, cut out the picture (or draw a similar picture to cut out) of the parallelogram and then cut the side off and move it over so that the parallelogram is transformed into a rectangle. Explain to students that the line that you cut is the height of the parallelogram, which is not a side of the parallelogram. Then, cut this parallelogram along a diagonal to create a triangle. Here, students will see that the area of a triangle is half the area of a parallelogram. You may need to rotate the halves (triangles) so that they overlap perfectly. This will show the students that the triangles are congruent and each is exactly half of the parallelogram.

Create another set of flashcards for the area formulas in this chapter. These flashcards should be double-sided. The blank side should be a sketch of the figure and its name. The flip side should have the formula for its area and the formula for its perimeter. Students should create flashcards as the chapter progresses.

Trapezoids, Rhombi, and Kites


This lesson further expands upon area formulas to include trapezoids, rhombi, and kites.

Relevant Review

Students might need a quick review of the definitions of trapezoids, rhombi, and kites. Go over their properties (especially that the diagonals of rhombi and kites are perpendicular) and theorems. Students may know the area formula of a trapezoid from a previous math class.

Review the Pythagorean Theorem and special right triangles. There are several examples and review questions that will use these properties. If students do not remember the special right triangle ratios, they can use the Pythagorean Theorem.

Teaching Strategies

Use the same technique discussed in the previous lesson for the area of a parallelogram and triangle. Cut out two congruent trapezoids and demonstrate the explanation at the beginning of the lesson explaining the area of a trapezoid. Going over this with students (rather than just giving them a handout or reading it) will enable them to understand the formula better. These activities are done best on an overhead projector.

Conveniently, the area formula of the rhombus and kite are the same. Again, you can cut out a rhombus and kite, then cut them on the diagonals and piece each together to form a rectangle. Generate the formula with students. Another way to write the formula of a rhombus is to say that it has 4 congruent triangles, with area \begin{align*}\frac{1}{2} \left (\frac{1}{2} d_1 \right ) \left (\frac{1}{2} d_2 \right ) = \frac{1}{8} d_1d_2\end{align*}12(12d1)(12d2)=18d1d2. Multiplying this by 4, we get \begin{align*}\frac{4}{8}d_1d_2 = \frac{1}{2}d_1d_2\end{align*}48d1d2=12d1d2. This process is not as easily done with a kite because one of the diagonals is not bisected.

Additional Example: Find two different rhombi that have an area of \begin{align*}48 \ \text{units}^2\end{align*}48 units2.

Solution: The diagonals are used to find the area, so when solving this problem, we are going to be finding the diagonals’ lengths. \begin{align*}\frac{1}{2} d_1d_2 = 48\end{align*}12d1d2=48, so \begin{align*}d_1d_2 = 96\end{align*}d1d2=96. This means that the product of the diagonals is double the area.

The diagonals can be: 1 and 96, 2 and 48, 3 and 32, 4 and 24, 6 and 16, 8 and 12.

As an extension, you can students draw the rhombi. The diagonals bisect each other, so have the diagonals cut each other in half and then connect the endpoints of the diagonals to form the rhombus. Three examples are below.

Areas of Similar Polygons


Students will learn about the relationship between the scale factor of similar polygons and their areas. Students should also be able to apply area ratios to solving problems.

Relevant Review

Review the properties of similar shapes, primarily triangles and quadrilaterals, from Chapter 7. Remind students that the perimeter, sides, diagonals, etc. have the same ratio as the scale factor. The Review Queue reviews similar squares. As an additional question, ask students to find the perimeter of both squares and then reduce the ratio (smaller square = 40, larger square = 100, ratio is 2:5, the same as the ratio of the sides). Ask students why they think the ratio of the sides is the same as the ratio of the perimeters.

Teaching Strategies

Examples 1 and 2 lead students towards the Area of Similar Polygons Theorem. As an additional example (before introducing the Area of Similar Polygons Theorem), ask students to find the area of two more similar shapes. Having students repeat problems like Example 2, they should see a pattern and arrive at the theorem on their own.

Additional Example: Two similar triangles are below. Find their areas and the ratio of the areas. How does the ratio of the areas relate to the scale factor?

Solution: Each half of the isosceles triangles are 3-4-5 triangles. The smaller triangle has a height of 3 and the larger triangle has a height of 9 (because 12 is 3 time 4, so this triangle is three time larger than the smaller triangle). The areas are: \begin{align*}A_{larger \ \Delta} = \frac{1}{2} \cdot 24 \cdot 9 = 108\end{align*}Alarger Δ=12249=108 and \begin{align*}A_{smaller \ \Delta} = \frac{1}{2} \cdot 8 \cdot 3 = 12\end{align*}Asmaller Δ=1283=12. The ratio of the area is \begin{align*}\frac{12}{108} = \frac{1}{9}\end{align*}12108=19. The ratios of the scale factor and areas relate by squaring the scale factor, \begin{align*}\frac{1}{9} = \left (\frac{1}{3} \right )^2\end{align*}19=(13)2.

Circumference and Arc Length


The purpose of this lesson is to review the circumference formula and then derive a formula for arc length.

Relevant Review

The Review Queue is a necessary review of circles. Students need to be able to apply central angles, find intercepted arcs and inscribed angles. They also need to know that there are \begin{align*}360^\circ\end{align*}360 in a circle.

Teaching Strategies

Students may already know the formula for circumference, but probably do not remember where \begin{align*}\pi\end{align*}π comes from. Investigation 10-1 is a useful activity so that students can see how \begin{align*}\pi\end{align*}π was developed and why it is necessary to find the circumference and area of circles. You can decide to make this investigation teacher-led or allow students to work in pairs or groups. From this investigation, we see that the circumference is dependent upon \begin{align*}\pi\end{align*}π.

When introducing arc length, first have students find the circumference of a circle with radius of \begin{align*}6 (12 \pi)\end{align*}6(12π). Then, see what the length of the arc of a semicircle \begin{align*}(6 \pi)\end{align*}(6π). Students should make the connection that the arc length of the semicircle will be half of the circumference. Then ask students what the arc length of half of the semicircle is \begin{align*}(3 \pi)\end{align*}(3π). Ask what the corresponding angle measure for this arc length would be \begin{align*}(90^\circ)\end{align*}(90). See if students can reduce \begin{align*}\frac{90}{360}\end{align*}90360 and if they make the correlation that the measure of this arc is a quarter of the total circumference, just like \begin{align*}90^\circ\end{align*}90 is a quarter of \begin{align*}360^\circ\end{align*}360. Using this same circle, see if students can find the arc length of a \begin{align*}30^\circ\end{align*}30 portion of the circle \begin{align*}\left (\frac{3 \pi}{3} = \pi \right )\end{align*}(3π3=π). Then, as students what portion of the total circumference \begin{align*}\pi\end{align*}π is. \begin{align*}\pi\end{align*}π is \begin{align*}\frac{1}{12}\end{align*}112 of \begin{align*}12 \pi\end{align*}12π, just like \begin{align*}30^\circ\end{align*}30 is \begin{align*}\frac{1}{12}\end{align*}112 of \begin{align*}360^\circ\end{align*}360. This should lead students towards the Arc Length Formula.

Students may wonder why it is necessary to leave answer in exact value, in terms of \begin{align*}\pi\end{align*}π, instead of approximate (multiplying by 3.14). This is usually a teacher preference. By using the approximate value for \begin{align*}\pi\end{align*}π, the answer automatically has a rounding error. Rounding the decimal too short will cause a much larger error than using the decimal to the hundred-thousandths place. Whatever your preference, be sure to explain both methods to your students. The review questions request that answers be left in terms of \begin{align*}\pi\end{align*}, but this can be easily changed, depending on your decision.

Area of Circles and Sectors


This lesson reviews the formula for the area of a circle and introduces the formula for the area of a sector and segment of a circle.

Teaching Strategies

If you have access to an LCD display or a computer lab, show students the animation of the area of a circle formula (link is in the FlexBook).

The formula for the area of a sector is very similar to the formula for arc length. Ask students to compare the two formulas. Stress to students that the angle fraction in the sector formula is the same as it is for the arc length formula. Therefore, students do not need to memorize a new formula; they just need to remember the angle fraction for both.

To find the area of the shaded regions (like Example 8), students will need to add or subtract areas of circles, triangles, rectangles, or squares in order to find the correct area. Encourage students to identify the shapes in these types of problems before they begin to solve it. At the end of this lesson, quickly go over problems 23-25, so that students know how to solve the problems that evening. Remind students to use the examples in the lesson to help them with homework problems.

Finding the area of a segment can be quite challenging for students. This text keeps the angles fairly simple, using special right triangle ratios. Depending on your level of student, you may decide to omit this portion of this lesson. If so, skip Example 9 and review questions 26-31.

Additional Example: Find the area of the blue shaded region below.

Solution: The triangle that is inscribed in the circle is a 45-45-90 triangle and its hypotenuse is on the diameter of the circle. Therefore, the hypotenuse is \begin{align*} 24 \sqrt{2}\end{align*} and the radius is \begin{align*}12 \sqrt{2}\end{align*}. The area of the shaded region is the area of the circle minus the area of the triangle.

\begin{align*}A_{\bigodot} &= \pi (12\sqrt{2})^2 = \pi \cdot 144 \cdot 2 = 288 \pi\\ A_{\Delta} &= \frac{1}{2} \cdot 24 \cdot 24 = 288\end{align*}

The area of the shaded region is \begin{align*}288 \pi - 288 \approx 616.78 \ \text{units}^2\end{align*}

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