# Triangle Height

🏆Practice parts of a triangle

## 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:

## Test yourself on parts of a triangle!

Is the straight line in the figure the height of the triangle?

If you're interested in learning more about other triangle topics, you can check out one of the following articles:

• Acute Triangle
• Obtuse Triangle
• Scalene Triangle
• Equilateral Triangle
• Isosceles Triangle
• Edges of a Triangle
• Area of a Right Triangle
• How to Calculate the Area of a Triangle
• How is the Perimeter of a Triangle Calculated?

On the Tutorela blog, you'll find a variety of mathematics articles.

## Triangle Height Calculation Exercises:

### Exercise 1

Given the parallelogram $ABCD$

CE is the altitude from side $AB$

$CB=5$

$AE=7$

$EB=2$

What is the area of the parallelogram?

Solution:

To find the area, you must first determine the height of the parallelogram.

For this, let's take a look at the triangle $\triangle EBC$,

Why do we know it's a right triangle? Because it's the height of the parallelogram.

We can use the Pythagorean theorem: $a^2+b^2=c^2$

In this case: $EB^2+EC^2=BC^2$

Substituting the given information: $2^2+EC^2=5^2$

Isolating the variable: $EC^2=5^2-2^2$

And solving: $EC^2=25-4=21$

$EC=\sqrt{21}$

Now, all we have to do is calculate the area.

It's important to remember that this requires using the length of side $AB$,

That is, $AE+EB=7+2=9$

$\sqrt{21}\times9=41.24$

$41.24$

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### Exercise 2

Given the right triangle:

What is the length of the third side?

Solution:

The image shows a triangle of which we know the length of two of its sides and we want to find the value of the third side.

We also know that the triangle shown is a right triangle because a small square indicates which angle is the right angle.

The Pythagorean theorem states that in a right triangle the following applies:

$c²=a²+b²$

In our right triangle

$a=3$

$b=4$

$c=x$

When we replace the values of our triangle into the algebraic expression of the Pythagorean theorem, we get the following equation:

$x²=3²+4²$

$x²=9+16$

$x²=25$

If we now take the square root of both sides of the equation we can solve for $x$ and obtain the desired value

$x=\sqrt{25}$

$x=5$

$x=5$

### Exercise 3

Homework:

How do we calculate the area of a trapezoid?

We are given the following trapezoid with these features:

What is its height?

Solution

Trapezoid area formula:

$\frac{(Base+Base)}{2}\times height$

The formula is not displaying correctly on the page.

$\frac{9+6}{2}\times h=30$

And we solve:

$\frac{15}{2}\times h=30$

$7\frac{1}{2}\times h=30$

$h=\frac{30}{\frac{15}{2}}$

$h=\frac{60}{15}$

$h=4$

Height $BE$ is equal to $4$ cm.

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Do you know what the answer is?

### Exercise 4

Given the isosceles triangle $\triangle ABC$.

And within it, we draw $EF$, parallel to $CB$:

$AF=5$

$AB=17$

$AG=3$

$AD=8$

$A$ is the height of the triangle.

What is the area of $EFBC$?

Solution:

To find the area of the trapezoid, it is worth remembering the formula for its area: $\frac{(base + base)}{2}\times height$

We focus on finding the bases.

To find $GF$, we will use the theorem of Pythagoras: $A^2+B^2=C^2$ in triangle $\triangle AFG$

Replace:

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

Isolate $GF$ and solve:

$9+GF^2=25$

$GF^2=25-9=16$

$GF=4$

We proceed with the same process with side $DB$ in triangle $\triangle ABD$:

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

$64+DB^2=289$

$DB^2=289-64=225$

$DB=15$

From here there are two ways to finish the exercise:

1. Calculate the area of the trapezoid $GFBD$ and verify that it is equal to trapezoid $EGDC$ and add them together.
2. Use the data we have discovered so far to find the parts of the trapezoid and solve.

We start by finding the height $GD$:

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

Now, let's 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.

Therefore:

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

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

We replace the data in the trapezoid formula:

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

$95$

### Exercise 5

Given the isosceles triangle $\triangle ABD$,

Within it, EF is drawn:

$AF=5$

$AB=17$

$AG=3$

$AD=8$

What is the perimeter of the trapezoid $EFBC$ ?

Solution:

To find the perimeter of the trapezoid, we need to add up all its sides.

We will focus on finding the bases.

To find $GF$, we will use the theorem of Pythagoras: $A^2+B^2=C^2$ in triangle AFG.

We substitute:

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

We isolate $GF$ and solve:

$9+GF^2=25$

$GF^2=25-9=16$

$GF=4$

We operate the same process with side DB in triangle $\triangle ABD$:

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

$64+DB^2=289$

$DB^2=289-64=225$

$DB=15$

We start by finding side $FB$:

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

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.

Therefore:

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

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

What remains is to calculate:

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

$62$

## Examples with solutions for Triangle Height

### Exercise #1

Can a triangle have two right angles?

### Step-by-Step Solution

The sum of angles in a triangle is 180 degrees. Since two angles of 90 degrees equal 180, a triangle can never have two right angles.

No

### Exercise #2

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 #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

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 image it is clear that the above theorem is true for the line AE. AE not only connects the A vertex with the opposite side. It also crosses BC forming a 90-degree angle. Undoubtedly making AE the altitude.

AE

### Exercise #5

Tree angles have the sizes:

50°, 41°, and 81.

Is it possible that these angles are in a triangle?

### Step-by-Step Solution

Let's remember that the sum of angles in a triangle is equal to 180 degrees.

We'll add the three angles to see if their sum equals 180:

$50+41+81=172$

Therefore, these cannot be the values of angles in any triangle.