# Lesson A circle inscribed in a rhombus

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## A circle inscribed in a rhombus

This lesson is focused on one problem.

### Problem

For a given rhombus, construct the circle inscribed in the rhombus.

To solve this problem, we subdivide it into smaller sub-problems.

Problem 1
Where is located the center of the circle inscribed in the rhombus?
 Solution - Mmmm... It is located in the center of the rhombus.                                                 - OK. Then, where is located the center of the rhombus? - Mmmm... At the intersection of its diagonals. - OK. This is correct. Now, answer the question: why the center of the inscribed circle is located at the intersection of the rhombus diagonals? - Mmmm...    Oooo!!! I have an idea! Figure 1. To the Problem 1

If the circle is inscribed in the rhombus, it is inscribed in each of four of the rhombus interior angles.
Therefore, the center of the inscribed circle is located at the angle bisectors of the rhombus interior angles. But the angle bisector of the rhombus is its diagonal.
Hence, the center of the inscribed circle lies at the intersection point of the rhombus diagonals.

- Very good. You got the major idea to solve this construction problem.

To complete this part of the lesson, I'd like to give the references to the lessons where the relevant geometric facts were proved.

The fact that the center of the circle inscribed in the angle is located at the angle bisector was proved in the lesson An angle bisector properties under the topic
Triangles of the section Geometry in this site.
The fact that in the rhombus the diagonals are the angle bisectors was proved in the lesson Diagonals of a rhombus bisect its angles under the topic Parallelograms
of the section Geometry in this site.

Problem 2
Construct the radius of the circle inscribed in the given rhombus.
 Solution If the circle is inscribed in the rhombus, then the rhombus side is                           tangent to the circle. It is visually clear from the Figure 2. Let us accept it without the formal proof now. (It will be proved later in one of the lessons that follow). Therefore, in order to construct the radius of the inscribed circle, we should to construct the perpendicular from the center of the circle to the rhombus side. It doesn't matter to which one of four sides of the rhombus. To any. Figure 2. To the Problem 2

In other words, we should to construct the perpendicular from the rhombus center, which is the diagonal intersection point, to the rhombus side.

The algorithm of constructing the perpendicular from the point outside the straight line to the line using the ruler and the compass was described in the lesson
How to bisect a segment using a compass and a ruler. This lesson is under the topic Triangles of the section Geometry in this site.

Thus you have the algorithm to construct the radius of the inscribed circle to the rhombus.
The construction Problem 2 is solved.

As the last step in this lesson, I'd like you to solve the following

Problem 3
Calculate the radius of the circle inscribed in the given rhombus.
The rhombus is given by its side measure and its diagonals measures and .
 Solution As we agreed above in the solution of the Problem 2, the radius                               of the circle inscribed in the given rhombus is the perpendicular drawn from the center of the circle (which is the diagonals intersection point) to the side of the rhombus at the tangent point. The diagonals of the rhombus are perpendicular and bisect each other. This was proved in the lessons Diagonals of a rhombus are perpendicular and Properties of diagonals of parallelograms under the topic Parallelograms of the section Geometry in this site. Therefore, the diagonals of the rhombus divide it in four congruent Figure 3. To the Problem 3
right triangles, and we need to find the length of the altitude of any of four right triangles drawn from the right angle vertex to the hypotenuse.

So, our original problem is reduced to the following one:

find the altitude of the right triangle drawn from the right angle vertex to the hypotenuse.

The parameters of this right triangle are its legs measures a and b and the hypotenuse measure c (Figure 4) linked to the parameters of the original rhombus by the equalities
 , , . Let x and y be the measures of the segments the altitude cuts                                   the hypotenuse. Let z be the length of the altitude drawn to the hypotenuse (Figure 4). We want to derive the formula expressing z via a, b and c. There is more than one way to derive such a formula. Figure 4. The altitude of the right triangle
For example, one can write the equality for the area of the triangle as

and to derive from it that

.

But I do not suppose in this lesson that you are familiar with the notion of the area of triangles.

By acting differently, one can note that the triangle ACD is similar to the triangle ABC (Figure 4), then to obtain the proportion from this fact

and finally to get the same formula from the proportion

.

But I do not suppose in this lesson that you are familiar with the notion of similarity of triangles.

What I really want to do in this lesson is to obtain the formula for z from the "first principles".
 These "first principles" are the following three equations for x, y and z:                             (1)    (Pythagorean formula for the right triangle ADC),           (2)    (Pythagorean formula for the right triangle BDC), and                (3)    (obvious). Let us square the equation (3) (both sides). We get . (4) Figure 5. The altitude of the right triangle

Let us add the equations (1) and (2) (both sides). We get
,
or
(5),

because

due to Pythagorean theorem for the right triangle ABC.

Comparing the equations (4) and (5), we get
.               (6)

Square both sides of the equation (6). You get
.          (7)

Substitute the expressions
and

to the equation (7). (Note that these expressions are rewritten equations (1) and (2)). You get
.

Open the brackets and cancel the like terms. You get
,

or
.

Now take the square root of both sides. You get
,

which is our target.

Now, returning to our original rhombus, the formula for the radius of the inscribed circle is

.

Summary
In this lesson you learned that the center of the circle inscribed in the rhombus lies at the intersection of its diagonals.
You learned how to construct the radius of the inscribed circle using a ruler and a compass.
You learned how to calculate the radius of the circle inscribed in the rhombus.
You learned that there are different ways to get the target formula, and learned how to get this formula from the "first principles".

On the way you learned how to calculate the altitude of a right triangle drawn to the hypotenuse via its legs and the hypotenuse measures.

Below are couple of examples that show how the formula works.

Example 1
The rhombus has the diagonal measures of 40 cm and 30 cm.
Find the radius of the circle inscribed in the rhombus.

Solution
Let us find the rhombus side length first.

Note that the diagonals of the rhombus are perpendicular and bisect each other. So, the diagonals of the rhombus divide it in four congruent right triangles
with the leg measures and , where and are measures of the diagonals of the rhombus.

Find the rhombus side measure as the length of the hypotenuse of the right triangle with the legs of 40/2 = 20 cm and 30/2 = 15 cm long:
.

You get the side length of 25 cm.

Or you can directly apply the formula from the lesson The length of diagonals of a rhombus which is under the current topic Geometry of the section Word problems in this site:
,

where and are measures of the diagonals of the rhombus and is its side length.

Substitute the given data for the rhombus diagonals into this formula
,

and you get the same 25 cm for the rhombus side length.

So, you know now that the rhombus diagonal measures are 40 cm and 30 cm, and the side length is 25 cm.

OK, very good.
Now, use the formula of the Theorem in this lesson to calculate the radius of the circle inscribed to the rhombus:
.

You get 12 cm for the radius of the circle inscribed to the rhombus.

Answer. The radius of the circle inscribed to the rhombus is 12 cm.

Example 2
The rhombus has one diagonal of 18 cm long and the perimeter of 60 cm.
Find the radius of the circle inscribed in the rhombus.

Solution
Let us find the length of the side of the rhombus first.
It is equal to one fourth of the perimeter, that is 60/4 = 15 cm.

Note that the diagonals of the rhombus are perpendicular and bisect each other. So, the diagonals of the rhombus divide it in four congruent right triangles. For these triangles you know the hypotenuse length, which is equal to the rhombus side measure of 15 cm, and one leg measure, which is half of the known diagonal length 18/2 = 9 cm.

Find the unknown length of the leg of one of these right triangle by applying the Pythagorean formula to the triangle with the hypotenuse of 15 cm and the leg measure of 9 cm:
.

You get the length of the second leg of the right triangle of 12 cm.
Since it is half of the rhombus diagonal measure, the whole diagonal measure is equal to 2*12 cm = 24 cm.

Or you can apply the formula derived in the lesson The length of diagonals of a rhombus which is under the current topic Geometry of the section Word problems in this site:
,

where is the given and is an unknown measure of the diagonals of the rhombus and is its side length.

Substitute the given data for the rhombus into this formula
,

and you get the same 24 cm for the rhombus second diagonal length.

So, you know now that the rhombus diagonal measures are 18 cm and 24 cm, and the side length is 15 cm.

OK, very good.
Now, use the formula of the Theorem in this lesson to calculate the radius of the circle inscribed to the rhombus:
.

You get 7.2 cm for the radius of the circle inscribed to the rhombus.

Answer. The radius of the circle inscribed to the rhombus is 7.2 cm.

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