Triangle Height - Examples, Exercises and Solutions

Set the Height of a Triangle

The height of a triangle is the segment that connects a vertex to the opposite side such that it creates a 90-degree angle.

In every triangle, there are three heights, as there are three vertices from which the height can be calculated relative to the side that is opposite to each of them.

The height can be found either inside or outside of the triangle. If it does not run through the interior of the triangle, it is called an external height.

Below, we provide you with some examples of triangle heights:

A1 - triangle height

Practice Triangle Height

Exercise #1

ABC is an isosceles triangle.

AD is the median.

What is the size of angle ADC ∢\text{ADC} ?

AAABBBCCCDDD

Video Solution

Step-by-Step Solution

In an isosceles triangle, the median to the base is also the height to the base.

That is, side AD forms a 90° angle with side BC.

That is, two right triangles are created.

Therefore, angle ADC is equal to 90 degrees.

Answer

90

Exercise #2

Given the following triangle:

Write down the height of the triangle ABC.

AAABBBCCCEEEDDD

Video Solution

Step-by-Step Solution

An altitude in a triangle is the segment that connects the vertex and the opposite side, in such a way that the segment forms a 90-degree angle with the side.

If we look at the drawing, we can notice that the previous theorem is true for the line AE that crosses BC and forms a 90-degree angle, comes out of vertex A and therefore is the altitude of the triangle.

Answer

AE

Exercise #3

Which of the following is the height in triangle ABC?

AAABBBCCCDDD

Video Solution

Step-by-Step Solution

Let's remember the definition of height of a triangle:

A height is a straight line that descends from the vertex of a triangle and forms a 90-degree angle with the opposite side.

The sides that form a 90-degree angle are sides AB and BC. Therefore, the height is AB.

Answer

AB

Exercise #4

True or false?

α+β=180 \alpha+\beta=180

αβ

Video Solution

Step-by-Step Solution

Given that the angles alpha and beta are on the same straight line and given that they are adjacent angles. Together they are equal to 180 degrees and the statement is true.

Answer

True

Exercise #5

Find the measure of the angle α \alpha

505050AAABBBCCC50

Video Solution

Step-by-Step Solution

Recall that the sum of angles in a triangle equals 180 degrees.

Therefore, we will use the following formula:

A+B+C=180 A+B+C=180

Now let's insert the known data:

α+50+50=180 \alpha+50+50=180

α+100=180 \alpha+100=180

We will simplify the expression and keep the appropriate sign:

α=180100 \alpha=180-100

α=80 \alpha=80

Answer

80

Exercise #1

Three angles measure as follows: 60°, 50°, and 70°.

Is it possible that these are angles in a triangle?

Video Solution

Step-by-Step Solution

Recall that the sum of angles in a triangle equals 180 degrees.

Let's add the three angles to see if their sum equals 180:

60+50+70=180 60+50+70=180

Therefore, it is possible that these are the values of angles in some triangle.

Answer

Possible.

Exercise #2

ABC Right triangle

Since BD is the median

Given AC=10.

Find the length of the side BD.

AAABBBCCCDDD10

Video Solution

Step-by-Step Solution

We calculate BD according to the rule:

In a right triangle, the midpoint of the hypotenuse is equal to half of the hypotenuse.

That is:

BD is equal to half of AC:

Given that: AC=10 AC=10

BD=10:2=5 BD=10:2=5

Answer

5

Exercise #3

Given that the triangle ABC is isosceles,
and inside it we draw EF parallel to CB:

171717888AAABBBCCCDDDEEEFFFGGG53 AF=5 AB=17
AG=3 AD=8
AD the height in the triangle

What is the area of the trapezoid EFBC?

Video Solution

Step-by-Step Solution

To find the area of the trapezoid, you must remember its formula:(base+base)2+altura \frac{(base+base)}{2}+\text{altura} We will focus on finding the bases.

To find GF we use the Pythagorean theorem: A2+B2=C2 A^2+B^2=C^2  In triangle AFG

We replace:

32+GF2=52 3^2+GF^2=5^2

We isolate GF and solve:

9+GF2=25 9+GF^2=25

GF2=259=16 GF^2=25-9=16

GF=4 GF=4

We will do the same process with side DB in triangle ABD:

82+DB2=172 8^2+DB^2=17^2

64+DB2=289 64+DB^2=289

DB2=28964=225 DB^2=289-64=225

DB=15 DB=15

From here there are two ways to finish the exercise:

  1. Calculate the area of the trapezoid GFBD, prove that it is equal to the trapezoid EGDC and add them up.

  2. Use the data we have revealed so far to find the parts of the trapezoid EFBC and solve.

Let's start by finding the height of GD:

GD=ADAG=83=5 GD=AD-AG=8-3=5

Now we reveal that EF and CB:

GF=GE=4 GF=GE=4

DB=DC=15 DB=DC=15

This is because in an isosceles triangle, the height divides the base into two equal parts then:

EF=GF×2=4×2=8 EF=GF\times2=4\times2=8

CB=DB×2=15×2=30 CB=DB\times2=15\times2=30

We replace the data in the trapezoid formula:

8+302×5=382×5=19×5=95 \frac{8+30}{2}\times5=\frac{38}{2}\times5=19\times5=95

Answer

95

Exercise #4

ABC is an isosceles triangle.

AD is the height of triangle ABC.

555333171717888AAABBBCCCDDDEEEFFFGGG

AF = 5

AB = 17
AG = 3

AD = 8

What is the perimeter of the trapezoid EFBC?

Video Solution

Step-by-Step Solution

To find the perimeter of the trapezoid, all its sides must be added:

We will focus on finding the bases.

To find GF we use the Pythagorean theorem: A2+B2=C2 A^2+B^2=C^2 in the triangle AFG

We replace

32+GF2=52 3^2+GF^2=5^2

We isolate GF and solve:

9+GF2=25 9+GF^2=25

GF2=259=16 GF^2=25-9=16

GF=4 GF=4

We perform the same process with the side DB of the triangle ABD:

82+DB2=172 8^2+DB^2=17^2

64+DB2=289 64+DB^2=289

DB2=28964=225 DB^2=289-64=225

DB=15 DB=15

We start by finding FB:

FB=ABAF=175=12 FB=AB-AF=17-5=12

Now we reveal EF and CB:

GF=GE=4 GF=GE=4

DB=DC=15 DB=DC=15

This is because in an isosceles triangle, the height divides the base into two equal parts so:

EF=GF×2=4×2=8 EF=GF\times2=4\times2=8

CB=DB×2=15×2=30 CB=DB\times2=15\times2=30

All that's left is to calculate:

30+8+12×2=30+8+24=62 30+8+12\times2=30+8+24=62

Answer

62

Exercise #5

True or false:

DE not a side in any of the triangles.
AAABBBCCCDDDEEE

Video Solution

Answer

True

Exercise #1

Is DE side in one of the triangles?
AAABBBCCCDDDEEE

Video Solution

Answer

Not true

Exercise #2

True or false:

AB is a side of the triangle ABC.

AAABBBCCC

Video Solution

Answer

True

Exercise #3

True or false:

AD is a side of triangle ABC.

AAABBBCCC

Video Solution

Answer

Not true

Exercise #4

True or false:

BC is a side of triangle ABC.

AAABBBCCC

Video Solution

Answer

True

Exercise #5

AD is the median in triangle ABC.

BD = 4

Find the length of DC.

AAABBBCCCDDD4

Video Solution

Answer

4

Topics learned in later sections

  1. Area
  2. The Sum of the Interior Angles of a Triangle
  3. The sides or edges of a triangle
  4. Exterior angles of a triangle
  5. Types of Triangles
  6. Obtuse Triangle
  7. Equilateral triangle
  8. Identification of an Isosceles Triangle
  9. Scalene triangle
  10. Acute triangle
  11. Isosceles triangle
  12. The Area of a Triangle
  13. Area of a right triangle
  14. Area of Isosceles Triangles
  15. Area of a Scalene Triangle
  16. Area of Equilateral Triangles
  17. Perimeter
  18. Triangle
  19. Perimeter of a triangle