# Basic Composite Figures

This page will introduce three powerful techniques for solving composite figure challenges:

- adding in lines and grids;
- finding complementary areas;
- using symmetry.

#### Contents

- Prerequisite Geometry Facts
- Adding in Lines and Grids
- Finding Complementary Areas
- Using Symmetry
- Example 1
- Example 2
- Example 3
- Example 4
- Example 5
- Example 6
- Example 7
- Example 8
- Example 9
- Example 10
- Example 11
- Example 12
- Example 13
- Example 14
- Example 15
- Example 16
- Example 17
- Example 18
- Example 19
- Example 20
- Length and Area - Composite Figures

## Prerequisite Geometry Facts

To successfully solve 2D-composite figures problems, you will frequently need to know

- the properties of triangles, polygons, and circles;
- how to find the perimeter and area of triangles, the perimeter and area of polygons, and the circumference and area of circles;
- how to identify and compare congruent and similar triangles;
- how to find area and perimeter relations in similar polygons.

## Adding in Lines and Grids

## Finding Complementary Areas

## Using Symmetry

## Example 1

Lots of ways to get \(\frac{1}{2}:\)

## Example 2

Solution visuals:

## Example 3

AMC 2007

What is the area of the shaded pinwheel shown in the \(5\times 5\) grid?

## Example 4

AMC 2004

### An annulus is the region between two concentric circles. The concentric circles in the ﬁgure have radii \(\left\lvert\overline{OX}\right\rvert=b\) and \(\left\lvert\overline{OZ}\right\rvert=c,\) respectively. Let \(\overline{XZ}\) be tangent to the smaller circle at \(Z,\) and let \(\overline{OY}\) be the radius of the larger circle that contains \(\overline{OZ}.\) If we let \(a=\left\lvert\overline{XZ}\right\rvert, d=\left\lvert\overline{YZ}\right\rvert\) and \(e=\left\lvert\overline{XY}\right\rvert,\) which of the following is the area of the annulus?

\[ \]

\[\begin{array} &&(A)~\pi a^2 &&(B)~\pi b^2 &&(C)~\pi c^2 &&(D)~\pi d^2 &&(E)~\pi e^2\end{array}\] \[ \]

## Example 5

AMC 2014

A regular hexagon has side length 6. Congruent arcs with radius 3 are drawn with the center at each of the vertices, creating circular sectors as shown. The region inside the hexagon but outside the sectors is shaded as shown. What is the area of the shaded region?

## Example 6

**Find the area of the triangle outlined in black.**

Strategy: Find the area of the colored triangles and subtract it from the total.

We have\[\begin{align} A_{\text{yellow triangle}} &= \frac { 2 \times 5}{2} = 5 \\ A_{\text{red triangle}} &= \frac {2 \times 6}{2} = 6 \\ A_{\text{orange triangle}} &= \frac {3 \times 4}{2} = 6 \\ \\ \text{Total Area} &=5 \times 6 = 30 \\ \\ \Rightarrow \text{Outlined Triangle Area}&= 30 - 6 - 6 - 5 \\ &= 13. \ _\square \end{align} \]

## Example 7

**Find the area of the leaf below.**

Strategy: First split the image in two.

Then realize

1) we want to find the area of the 'leaf'L;

2) the whole square has areaS;

3) a quarter circle has areaQ.Then note that

Q+Q=S+L.Aka, you can think about the area of the two quarter-circles as covering the area of the square plus overlapping twice over the leaf, which lets us calculate the area of a leaf:

L = 2Q - S. \(_\square\)

## Example 8

### What are the area and the perimeter of the above figure?

## Example 9

### If the lengths of the diagonals of the rhombus \( \lozenge ABCD\) are \(\overline{BD}=10\) and \(\overline{AC}=5,\) what is the area of \(\lozenge ABCD?\)

\[ \] There are many ways to solve this problem.

**Strategy 1:** Use the rhombus formula to solve for the area.

We have\[\begin{align} A_{\text{rhombus}} &= \frac{\overline{BD} \times \overline{AC}}{2} \\ &= \frac{ 10 \times 5 }{2} \\ &= 25. \ _\square \end{align} \]

\(\)

**Strategy 2:** Divide it into 2 equivalent triangles.

\( A_{\text{triangle}} = \frac{ b \cdot h}{2}, \) where \(b\) is the base and \(h\) is the height.Plugging in their values, we have \( \frac{ 5 \times 5}{2} .\) Since there are 2 triangles, we multiply it by 2 to obtain

\[ \frac{25}{2} \times 2 = 25. \ _\square \]

## Example 10

### The triangular plot of \(ACD\) lies between Aspen Road, Brown Road and a railroad. Main Street runs east and west, and the railroad runs north and south. The numbers in the diagram indicate distances in miles. The width of the railroad track can be ignored. How many square miles are in the plot of land \(ACD?\)

\[ \]

Strategy: Find the area of \( \triangle{ABC} \) and \( \triangle{ABD}.\)

Since \( \left\lvert\overline{AB}\right\rvert = 3\) and \( \left\lvert\overline{BC}\right\rvert = 3, \) the area of \(\triangle{ABC}\) is \(\frac{3 \times 3}{2} = \frac{9}{2} \) square miles.

Similarly, since \( \left\lvert\overline{AB}\right\rvert = 3\) and \( \left\lvert \overline{BD}\right\rvert = \left\lvert\overline{BC}\right\rvert + \left\lvert\overline{CD} \right\rvert = 3 + 3=6, \) the area of \(\triangle{ABD}\) is \(\frac{3 \times 6}{2} = 9 \) square miles.

Then the area of \(\triangle{ACD}\) (in square miles) is

\[(\text{Area of }\triangle{ABD})-(\text{Area of }\triangle{ABC})=9 - \frac{9}{2} = \frac{9}{2}. \ _\square\]

## Example 11

### What is the area of the shaded region in the above diagram, given that the two smaller circles are identical?

## Example 12

### John owns a grassy plot of land, of which he built a house on top. What are the area and the perimeter of the grassy patch of land?

**Solution**:

The horizontal length of the house is \( 11 - 2 - 2 = 7 \). The vertical length of the house is \(5\). Thus, the area of the house is \( 7 \times 5 \). Now, observe that the area of the grass patch is equal to the area of the original rectangle, minus the area of the house. Then the area of the grassy path of land is \( 11 \times 9 - 7 \times 5 = 99 - 35 = 64 \).

The perimeter of the grass patch is \( 2 + 5 + 7 + 5 + 2 + 9 + 11 +9 = 50. \ _\square \)

## Example 13

### Find the area of the figure below.

**Solution**:

By joining some points in the figure, we can break it down into elementary components.

We now have broken the composite figure into two rectangles and two triangles. We then individually find the area of each figure and sum it up.

Thus we get the area to be \(15 + 2 + 6 + 1.5 = 24.5 \) square units. \(_\square\)

## Example 14

### The square below is inscribed in the circle. If the area of the circle is \(36,\) what are the area and the perimeter of the shaded region?

\[ \]

## Example 15

## Given that the small square and the equilateral triangle in the figure below have equal area and their centroids coincide, find the area of the unshaded region in terms of the side length of the square.

\[ \]

\[ \]

## Example 16

The diagram above shows a Reuleaux septagon drawn from a regular septagon with circumradius 1, where arc \(A_1 A_2 \) is drawn from center \(A_5 \), arc \(A_2 A_3 \) is drawn from center \(A_6\), arc \(A_3 A_4 \) is drawn from center \(A_7\), and so on.

Find the area of the Reuleaux septagon to 5 decimal places.

**Bonus**: Generalize this for all Reuleaux \(n\)-gon where \(n\) is an odd number.

Inspiration

###### You may want to read up Reuleaux triangle first.

###### Image Credit: Wikimedia Reuleaux polygons by LEMeZza

## Example 17

## Example 18

Mandy tried to draw a rose for Mother's Day using many circles. He started out with a large circle of radius 7cm, and then drew 4 smaller circles in a symmetrical way. He then colored in parts of the flower in yellow. What is the area \((\)in \(\text{cm}^2)\) of the yellow part?

The figure is completely symmetrical.

The figure is not drawn to scale.

Take \( \pi = \frac{ 22}{7} \).

## Example 19

## Example 20

While Mai and Joey were preparing for the party, Tea Gardner and Yugi Muto, both lovers, hung out at the church, where they stared at a stain-glassed Reuleaux pentagon window, filled with pink and red colors. The conversation started:

**Tea**: Wow! What a beautiful rose! Isn't it beautiful, Yugi?

**Yugi**: Yes, it is!

**Tea**: I have a math problem for you to try out. Suppose we start with five unit circles, where each center is intersected by two circular arcs. Then, this makes a Reuleaux pentagon.

**Yugi**: That is easy to remember...

**Tea**: But there's more! Draw another Reuleaux pentagon inside, where each vertex intersects the midpoint of the circular arcs. Repeat this infinitely, so we have much like the stain-glassed one. Here is the following:

*[After minutes of perfect-sketching]*

Here is the rose with alternating red and white petals. Your goal is to compute the area of all red petals.

**Yugi**: Touche! What a long geometry problem you asked here! Is there a specific formula to generalize the area of the red petals after finite number of iterations?

**Tea**: If I give that away, then you will know the answer to this fun problem! Solve this correctly, and we will enjoy a nice party! :)

What is the area of all red petals in the second diagram, where the diagonal of the largest Reuleaux pentagon is 1? If your area is \(A\), input \(\left\lfloor A \cdot 10^3 \right\rfloor\) as your answer.

**Preliminaries**

A Reuleaux polygon is a curvilinear polygon formed by an odd number of circular arcs. Like a Reuleaux triangle, the apex points of a Reuleaux polygon are centers of identical circles.

For the main problem, determine the area of a Reuleaux pentagon.

**Bonus:** Generalize this for any odd number of vertices and circular arcs. You should notice that as the number of edges approaches \(\infty\), a Reuleaux polygon becomes close to a circle of area \(\frac{\pi}{4}r^2\), where \(r\) is the radius of the large circle.

*This is the end of the fourth chapter of the story. Check chapter directory if you are interested:*

First - Second - Third - Fourth - Fifth

## Length and Area - Composite Figures

**Cite as:**Basic Composite Figures.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/basic-composite-figures/