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.

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

einstein

True or false?

One of the angles in a rectangle may be an acute angle.

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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


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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?
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