# Right angle

🏆Practice types of angles (right, acute, obtuse, flat)

## Definition of Right Angle

A right angle is one of the types of angles that we will encounter during engineering studies.

A right angle is one that measures $90°$. We generally mark it with a small square at the area where it is formed.

Right angles can appear in triangles, squares, rectangles, and other geometric shapes with angles of $90°$ degrees.

## Test yourself on types of angles (right, acute, obtuse, flat)!

True or false

One of the angles of the rectangle can be an acute angle.

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
Test your knowledge

## 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:

1. How many degrees is the angle of $∡ABC$ and what type of angle is it in relation to $∡CBF$?
2. 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:

1. 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.

Check your understanding

### 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.

Test your knowledge

### 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?
Related Subjects