# Basic Composite Figures

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

*1) Adding in Lines and Grids*

*2) Finding Complementary Areas*

*3) 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:

## Adding in Lines and Grids

## Finding Complementary Areas

## Using Symmetry

## Example 1

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

## 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.

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

## 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 area.S

3) A quarter circle has area.QThen 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.

This lets us calculate the area of a leaf:. \(_\square\)L = 2Q - S

## 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

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

The figure is completely symmetrical.

The figure is not drawn to scale.

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

## Example 19

## Example 20

## Length and Area - Composite Figures

**Cite as:**Basic Composite Figures.

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