# 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:

## Practice Triangle Height

### Exercise #1

ABC is an isosceles triangle.

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

### 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.

90

### Exercise #2

Given the following triangle:

Write down the height of the triangle ABC.

### 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.

AE

### Exercise #3

Which of the following is the height in triangle ABC?

### 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.

AB

### Exercise #4

True or false?

$\alpha+\beta=180$

### 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.

True

### Exercise #5

Find the measure of the angle $\alpha$

### Step-by-Step Solution

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

Therefore, we will use the following formula:

$A+B+C=180$

Now let's insert the known data:

$\alpha+50+50=180$

$\alpha+100=180$

We will simplify the expression and keep the appropriate sign:

$\alpha=180-100$

$\alpha=80$

80

### Exercise #1

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

Is it possible that these are angles in a triangle?

### 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:

$60+50+70=180$

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

Possible.

### Exercise #2

ABC Right triangle

Since BD is the median

Given AC=10.

Find the length of the side BD.

### 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: $AC=10$

$BD=10:2=5$

5

### Exercise #3

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

AF=5 AB=17
AD the height in the triangle

What is the area of the trapezoid EFBC?

### Step-by-Step Solution

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

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

We replace:

$3^2+GF^2=5^2$

We isolate GF and solve:

$9+GF^2=25$

$GF^2=25-9=16$

$GF=4$

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

$8^2+DB^2=17^2$

$64+DB^2=289$

$DB^2=289-64=225$

$DB=15$

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:

$GD=AD-AG=8-3=5$

Now we reveal that EF and CB:

$GF=GE=4$

$DB=DC=15$

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

$EF=GF\times2=4\times2=8$

$CB=DB\times2=15\times2=30$

We replace the data in the trapezoid formula:

$\frac{8+30}{2}\times5=\frac{38}{2}\times5=19\times5=95$

95

### Exercise #4

ABC is an isosceles triangle.

AD is the height of triangle ABC.

AF = 5

AB = 17
AG = 3

What is the perimeter of the trapezoid EFBC?

### 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: $A^2+B^2=C^2$in the triangle AFG

We replace

$3^2+GF^2=5^2$

We isolate GF and solve:

$9+GF^2=25$

$GF^2=25-9=16$

$GF=4$

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

$8^2+DB^2=17^2$

$64+DB^2=289$

$DB^2=289-64=225$

$DB=15$

We start by finding FB:

$FB=AB-AF=17-5=12$

Now we reveal EF and CB:

$GF=GE=4$

$DB=DC=15$

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

$EF=GF\times2=4\times2=8$

$CB=DB\times2=15\times2=30$

All that's left is to calculate:

$30+8+12\times2=30+8+24=62$

62

### Exercise #5

True or false:

DE not a side in any of the triangles.

True

### Exercise #1

Is DE side in one of the triangles?

Not true

### Exercise #2

True or false:

AB is a side of the triangle ABC.

True

### Exercise #3

True or false:

AD is a side of triangle ABC.

Not true

### Exercise #4

True or false:

BC is a side of triangle ABC.

True

### Exercise #5

AD is the median in triangle ABC.

BD = 4

Find the length of DC.