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° 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° 90° degrees.

Start practice

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

einstein

True or false

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

Practice more now

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?

Exercise 3 on parallel lines

Explanation

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

86°+4°=90° 86°+4°=90°

Exercise 4 on parallel lines solution

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 ∡ABC and what type of angle is it in relation to CBF ∡CBF ?
  2. How many degrees is the angle BDE ∡BDE and what type of angle is it in relation to ADC ∡ADC ?
Exercise 5 on parallel lines

Answer 1:

A. The angle of ABC ∡ABC is equal to 180º130º=50º 180º-130º=50º

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

Answer 2:

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

Do you know what the answer is?

Exercise 3

Given the triangle ABC \triangle ABC :

Task:

Find the length of BC BC

Solution:

Pythagorean Theorem - Apply the formula

Given the triangle ABC \triangle ABC in the drawing.

Assignment:

Find the length of BC BC

Solution:

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

AB2+BC2=AC2 AB²+BC²=AC²

We place the known lengths:

52+BC2=132 5²+BC²=13²

25+BC2=169 25+BC²=169

BC2=16925=144 BC²=169-25=144 , \sqrt{}

BC=12 BC=12

Answer:

12 12 cm.


Exercise 4

Homework:

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

a right triangle, calculate its area

Solution:

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

leg×leg2 \frac{leg\times leg}{2}

ABBC2=862=482=24 \frac{AB\cdot BC}{2}=\frac{8\cdot6}{2}=\frac{48}{2}=24

Answer:

The answer is 24 24 cm².


Check your understanding

Exercise 5

Homework:

Given the right triangle ADB \triangle ADB

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

Given:

AB=15 AB=15

AC=13 AC=13

DC=5 DC=5

CB=4 CB=4

Calculate the area of the triangle ABC \triangle ABC

Given the right triangle ADB

Solution:

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

From here we can calculate AD AD.

AD+DC+AD=PerimeterΔADC AD+DC+AD=PerimeterΔADC

AD+5+13=30 AD+5+13=30

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

AD=12 AD=12

Now we can calculate the area of the triangle ΔABC ΔABC

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

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

sideheight×side2= \frac{sideheight\times side}{2}=

ADBC2=1242=482=24 \frac{AD\cdot BC}{2}=\frac{12\cdot4}{2}=\frac{48}{2}=24

Answer:

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


Exercise 6

Homework:

Given the right triangle ΔABC ΔABC

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

Find the measure of the leg BC BC

5.a The area of the triangle is equal to 38 cm²

Solution:

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

leg×leg2 \frac{leg\times leg}{2}

ACBC2=8BC2=38 \frac{AC\cdot BC}{2}=\frac{8\cdot BC}{2}=38

Multiply the equation by the common denominator

/ ×2 \times2

Then divide the equation by the coefficient of BC BC

8×BC=76 8\times BC=76 /:8 :8

BC=9.5 BC=9.5

Answer:

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


Do you think you will be able to solve it?

Exercise 7

Exercise 4 In front of you, there is a right triangle ABC

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

Given that BC=6 BC=6 The length of the leg AB AB is greater by 3313% 33\frac{1}{3}\% than the length of BD BD .

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

Task:

What is the area of the triangle ΔABC ΔABC ?

Solution:

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

AB=1.33333BD AB=1.33333\cdot BD

(100100+33.33100=133.33100=1.333)(\frac{100}{100}+\frac{33.33}{100}=\frac{133.33}{100}=1.333)

AB=1.3336=8 AB=1.333\cdot6=8

Now we will calculate the area of the triangle ΔABD.

SΔABD=ABBD2=862=482=24 SΔ\text{ABD}=\frac{AB\cdot BD}{2}=\frac{8\cdot6}{2}=\frac{48}{2}=24

Answer:

24 24 cm².


Exercise 8

the area of the triangle is 24 cm²

Homework:

What data in the graph is incorrect?

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

Solution:

Explanation: area of the right triangle.

SΔEDF=EDEF2=862=482=24 SΔEDF=\frac{ED\cdot EF}{2}=\frac{8\cdot6}{2}=\frac{48}{2}=24

According to the formula:

leg×leg2 \frac{leg\times leg}{2}

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

side×heightofside2 \frac{side\times heightofside}{2}

EG×102=24 \frac{EG\times10}{2}=24 /×2 \times2

10EG=48 10EG=48 /:10 :10

EG=4.8 EG=4.8

Answer:

The incorrect data is EG EG .

The length of EG EG should be 4.8 4.8 cm.


Test your knowledge

Exercise 9

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

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

Solution to exercise 2:

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

Answer:

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


Do you know what the answer is?
Start practice