**If you are interested in learning more about other angle topics, you can enter one of the following articles:**

- Angle Notation
- Sides, Vertices, and Angles
- Bisector
- Acute Angle
- Obtuse Angle
- Plane Angle
- Adjacent Angles
- Vertically Opposite Angles
- Alternate Angles
- Corresponding Angles
- Sum of the Angles of a Triangle
- Sum of the Angles of a Polygon
- Sum and Difference of Angles

**In the Mathematics Blog of** **Tutorela**** you will find a wide variety of articles about mathematics**

## Several Examples of Right Angles

**Right angles within a** **circle**

**Right angles within a triangle**

**Right angles within a square**

**Right angles within a rectangle**

Join Over 30,000 Students Excelling in Math!

Endless Practice, Expert Guidance - Elevate Your Math Skills Today

## Exercise

### Exercise 1

**How many degrees do we need to add to angle β so that there is another** **parallel line**** in the following graph?**

**Explanation**

By adding 4° degrees to the angle $∡β$ we get an angle of $90º$ degrees and basically another parallel line will be created below the two of them.

$86°+4°=90°$

**Solution:**

The correct answer is: $4º$

### Exercise 5 (on parallel lines)

**This question is divided into several parts:**

- How many degrees is the angle of $∡ABC$ and what type of angle is it in relation to $∡CBF$?
- How many degrees is the angle $∡BDE$ and what type of angle is it in relation to $∡ADC$?

**Answer 1:**

A. The angle of $∡ABC$ is equal to $180º-130º=50º$

B. The angle of $∡ABC$ in relation to the angle of $∡CBF$ is called Adjacent angles

**Answer 2:**

- The angle $∡BDE$ is equal to $90º$ since it is a vertex opposite angle in relation to the angle $∡ADC=90º$

Do you know what the answer is?

### Exercise 3

Given the triangle $\triangle ABC$:

**Task:**

Find the length of $BC$

**Solution:**

Pythagorean Theorem - Apply the formula

Given the triangle $\triangle ABC$ in the drawing.

Assignment:

Find the length of $BC$

**Solution:**

Write the Pythagorean Theorem for the right triangle $\triangle ABC$

$AB²+BC²=AC²$

We place the known lengths:

$5²+BC²=13²$

$25+BC²=169$

$BC²=169-25=144$,$\sqrt{}$

$BC=12$

**Answer:**

$12$ cm.

### Exercise 4

**Homework:**

In front of you is a right triangle, calculate its area.

**Solution:**

Calculate the area of the triangle using the formula for calculating the area of a right triangle.

$\frac{leg\times leg}{2}$

$\frac{AB\cdot BC}{2}=\frac{8\cdot6}{2}=\frac{48}{2}=24$

**Answer:**

The answer is $24$ cm².

### Exercise 5

**Homework:**

Given the right triangle $\triangle ADB$

The perimeter of the triangle is equal to $30$ cm.

Given:

$AB=15$

$AC=13$

$DC=5$

$CB=4$

Calculate the area of the triangle $\triangle ABC$

**Solution:**

Given the perimeter of the triangle $\triangle ADC$ equal to $30$ cm.

From here we can calculate $AD$.

$AD+DC+AD=PerimeterΔADC$

$AD+5+13=30$

$AD+18=30$ /$-18$

$AD=12$

Now we can calculate the area of the triangle $ΔABC$

Pay attention: we are talking about an obtuse triangle therefore its height is $AD$.

We use the formula to calculate the area of the triangle:

$\frac{sideheight\times side}{2}=$

$\frac{AD\cdot BC}{2}=\frac{12\cdot4}{2}=\frac{48}{2}=24$

**Answer:**

The area of the triangle $ΔABC$ is equal to $24$ cm².

### Exercise 6

**Homework:**

Given the right triangle $ΔABC$

The area of the triangle is equal to $38$ cm², $AC=8$

Find the measure of the leg $BC$

**Solution:**

We will calculate the length of $BC$ using the formula for calculating the area of a right triangle:

$\frac{leg\times leg}{2}$

$\frac{AC\cdot BC}{2}=\frac{8\cdot BC}{2}=38$

Multiply the equation by the common denominator

/ $\times2$

Then divide the equation by the coefficient of $BC$

$8\times BC=76$ /$:8$

$BC=9.5$

**Answer:**

The length of the leg $BC$ is equal to $9.5$ centimeters.

Do you think you will be able to solve it?

### Exercise 7

In front of you, there is a right triangle $ΔABC$.

Given that $BC=6$ The length of the leg $AB$ is greater by $33\frac{1}{3}\%$ than the length of $BD$.

The area of the triangle $ΔADC$ is greater by $25%$ than the area of the triangle $ΔABD$.

**Task:**

What is the area of the triangle $ΔABC$?

**Solution:**

To find the measure of the leg $AB$ we will use the data that its length is greater by $33.33$ than the length of $BD$.

$AB=1.33333\cdot BD$

$(\frac{100}{100}+\frac{33.33}{100}=\frac{133.33}{100}=1.333)$

$AB=1.333\cdot6=8$

Now we will calculate the area of the triangle ΔABD.

$SΔ\text{ABD}=\frac{AB\cdot BD}{2}=\frac{8\cdot6}{2}=\frac{48}{2}=24$

**Answer:**

$24$ cm².

### Exercise 8

**Homework:**

What data in the graph is incorrect?

For the area of the triangle to be $24$ cm², what is the data that should replace the error?

**Solution:**

Explanation: area of the right triangle.

$SΔEDF=\frac{ED\cdot EF}{2}=\frac{8\cdot6}{2}=\frac{48}{2}=24$

According to the formula:

$\frac{leg\times leg}{2}$

If the area of the triangle can also be calculated from the formula of:

$\frac{side\times heightofside}{2}$

$\frac{EG\times10}{2}=24$ /$\times2$

$10EG=48$ /$:10$

$EG=4.8$

**Answer:**

The incorrect data is $EG$.

The length of $EG$ should be $4.8$ cm.

### Exercise 9

In the following example, a square $ABCD$ is presented.

A. Is the angle $∡ABC$ equal to the angle of $∡ADC$? Can it be said that $BD$ serves as the bisector of the angle $∡ABC$?

**Solution to exercise 2:**

The line $BD$ created $2$ points where the angle was divided into $2$ equal angles.

**Answer:**

Therefore, $DB$ is a bisector of the two angles $∡ADC$ and $∡ABC$

Do you know what the answer is?