Geometry (B): READ: Area of Rhombi and Kites Reading

Rhombi and Kites

Learning Objectives

Understand the relationships between the areas of two categories of quadrilaterals: basic quadrilaterals (rectangles and parallelograms), and special quadrilaterals (trapezoids, rhombi, and kites).
Derive area formulas for rhombi, and kites.
Apply the area formulas for these special quadrilaterals.

Introduction

We’ll use the area formulas for basic shapes to work up to the formulas for special quadrilaterals. It’s an easy job to convert a trapezoid to a parallelogram. It’s also easy to take apart a rhombus or kite and rebuild it as a rectangle. Once we do this, we can derive new formulas from the old ones.

We’ll also need to review basic facts about the rhombus and kite.

Area of a Rhombus or Kite

First let’s start with a review of some of the properties of rhombi and kites.

	Kite	Rhombus
Congruent sides	$2$ Pairs	All $4$
Opposite angles congruent	$1$ Pair yes. $1$ Pair maybe	Both pairs yes
Perpendicular diagonals	Yes	Yes
Diagonals bisected	$1$ Yes. $1$ maybe	Both yes

Now you’re ready to develop area formulas. We’ll follow the command: “Frame it in a rectangle.” Here’s how you can frame a rhombus in a rectangle.

Notice that:

The base and height of the rectangle are the same as the lengths of the two diagonals of the rhombus.
The rectangle is divided into $8$ congruent triangles; $4$ of the triangles fill the rhombus, so the area of the rhombus is one-half the area of the rectangle.

Area of a Rhombus with Diagonals $d_1$ and $d_2$

$A=\frac{1}{2}{d_1d_2}=\frac{d_1d_2}{2}$

We can go right ahead with the kite. We’ll follow the same command again: “Frame it in a rectangle.” Here’s how you can frame a kite in a rectangle.

Notice that:

The base and height of the rectangle are the same as the lengths of the two diagonals of the kite.
The rectangle is divided into $8$ triangles; $4$ of the triangles fill the kite. For every triangle inside the kite, there is a congruent triangle outside the kite so the area of the kite is one-half the area of the rectangle.

Area of a Kite with Diagonals $d_1$ and $d_2$

$A = \frac{1}{2}{d_1d_2} = \frac{d_1d_2}{2}$

Lesson Summary

We see the principle of “no need to reinvent the wheel” in developing the area formulas in this section. If we wanted to find the area of a trapezoid, we saw how the formula for a parallelogram gave us what we needed. In the same way, the formula for a rectangle was easy to modify to give us a formula for rhombi and kites. One of the striking results is that the same formula works for both rhombi and kites.

Points to Consider

You’ll use area concepts and formulas later in this course, as well as in real life.

Surface area of solid figures: the amount of outside surface.
Geometric probability: chances of throwing a dart and landing in a given part of a figure.
Carpet for floors, paint for walls, fertilizer for a lawn, and more: areas needed.

Tech Note - Geometry Software

You saw earlier that the area of a rhombus or kite depends on the lengths of the diagonals.

$A=\frac{1}{2}{d_1d_2}=\frac{d_1d_2}{2}$

This means that all rhombi and kites with the same diagonal lengths have the same area.

Try using geometry software to experiment as follows.

Construct two perpendicular segments.
Adjust the segments so that one or both of the segments are bisected.
Draw a quadrilateral that the segments are the diagonals of. In other words, draw a quadrilateral for which the endpoints of the segments are the vertices.
Repeat with the same perpendicular, bisected segments, but making a different rhombus or kite. Repeat for several different rhombi and kites.
Regardless of the specific shape of the rhombus or kite, the areas are all the same.

The same activity can be done on a geoboard. Place two perpendicular rubber bands so that one or both are bisected. Then place another rubber band to form a quadrilateral with its vertices at the endpoints of the two segments. A number of different rhombi and kites can be made with the same fixed diagonals, and therefore the same area.

The questions are for your own review. The answers are listed below to help you check your work and understanding.

Review Questions

Quadrilateral $ABCD$ has vertices $A(-2, 0), B(0, 2), C(4, 2),$ and $D(0, -2)$ in a coordinate plane.

Show that $ABCD$ is a trapezoid.
What is the area of $ABCD$ ?
Prove that the area of a trapezoid is equal to the area of a rectangle with height the same as the height of the trapezoid and base equal to the length of the median of the trapezoid.
Show that the trapezoid formula can be used to find the area of a parallelogram.
Sasha drew this plan for a wood inlay he is making.

$10$ is the length of the slanted side. $16$ is the length of the horizontal line segment. Each shaded section is a rhombus.

The shaded sections are rhombi. Based on the drawing, what is the total area of the shaded sections?
Plot on a coordinate plane.
- The points are the vertices of a rhombus.
- The area of the rhombus is $24 \;\mathrm{square\ units}$ .
Tyra designed the logo for a new company. She used three congruent kites.

What is the area of the entire logo?
In the figure below:
- $ABCD$ is a square
- $AP = PB = BQ$
- $DC = 20\;\mathrm{feet}$
What is the area of $PBQC$ ?

In the figure below:
- $ABCD$ is a square
- $AP = 20\;\mathrm{feet}$
- $PB = BQ = 10\;\mathrm{feet}$
What is the area of $PBQC$ ?
The area of $PBQD$ is what fractional part of the area of $ABCD$ ?

Review Answers

Slope of $\overline{A B} = 1$ , slope of $\overline{D C} = 1$

$\overline{AB}\parallel\overline{DC}$ are parallel.

$AB&=\sqrt{2^2+2^2}=\sqrt{8}=2\sqrt{2}\\ DC&=\sqrt{4^2+4^2}=\sqrt{32}=4\sqrt{2}\\ AD&=\sqrt{2^2+2^2}=\sqrt{8}=2\sqrt{2}$

slope of $\overline{A D} = -1$

$\overline{A B}$ and $\overline{D C}$ are the bases, $\overline{A D}$ is an altitude.

$A = \frac{(b_1 + b_2)h} {2} = \frac{(2\sqrt{2} + 4\sqrt{2})2\sqrt{2}} {2} = \frac{6\sqrt{2}(2\sqrt{2})} {2} = 12$

.
For a parallelogram, $b_1$ = $b_2 = b$ (the “bases” are two of the parallel sides), so by the trapezoid formula the area is
$\frac{(b_1 + b_2)h} {2} = \frac{(b + b)h} {2} = \frac{2bh} {2} = bh.$
Length of long diagonal of one rhombus is $16$ . Length of other diagonal is $12$ (each rhombus is made of $4 6-8-10$ right triangles).

Total area is $2\left [ \frac{d_1d_2}{2} \right ]=d_1d_2=16\times12=192.$

Many rhombi work, as long as the product of the lengths of the diagonals is $48$ .
$90\;\mathrm{cm}^2$
$200\;\mathrm{square\ feet}$
$300\;\mathrm{square\ feet}$
$\frac{1}{3}$

Last modified: Tuesday, June 29, 2010, 11:47 AM