ABC is a triangle. What is the median of the triangle? A A A B B B C C C E E E F F F D D D

There is no median in the figure.

Triangle Height | Tutorela

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:

Detailed explanation

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Test yourself on parts of a triangle!

Is DE side in one of the triangles?

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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 $AB$

$CB=5$

$AE=7$

$EB=2$

Image of Exercise 1 Given the parallelogram ABCD

Task:

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

$\sqrt{21}\times9=41.24$

Answer:

$41.24$

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Test your knowledge

Question 1

The triangle ABC is shown below.

To which side(s) are the median and the altitude drawn?

Question 2

Look at triangle ABC below.

What is the median of the triangle and to which side is it drawn?

Question 3

Look at triangle ABC below.

Which is the median?

Exercise 2

Given the right triangle:

Task:

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

$x=\sqrt{25}$

$x=5$

Answer:

$x=5$

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:

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

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

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

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

$h=4$

Answer:

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

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Do you know what the answer is?

Question 1

Look at the triangle ABC below.

$AD=\frac{1}{2}AB$

$BE=\frac{1}{2}EC$

What is the median in the triangle?

Question 2

ABC is a triangle.

What is the median of the triangle?

Question 3

Look at the triangles in the figure.

Which line is the median of triangle ABC?

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:

Calculate the area of the trapezoid $GFBD$ and verify that it is equal to trapezoid $EGDC$ and add them together.
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:

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

Answer:

$95$

Exercise 5

Given the isosceles triangle $\triangle ABD$ ,

Within it, EF is drawn:

$AF=5$

$AB=17$

$AG=3$

$AD=8$

Task:

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:

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

Answer:

$62$

Check your understanding

Question 1

What is the median of triangle ABC?

Question 2

What is the median of triangle ABC.

Question 3

Look at the triangle ABC below.

Which of the following lines is the median of the triangle?

Examples with solutions for Triangle Height

Exercise #1

Is DE side in one of the triangles?

Video Solution

Step-by-Step Solution

Since line segment DE does not correspond to a full side of any of the triangles present within the given geometry, we conclude that the statement “DE is a side in one of the triangles” is Not true.

Answer

Not true

Exercise #2

The triangle ABC is shown below.

To which side(s) are the median and the altitude drawn?

Step-by-Step Solution

To solve the problem of identifying to which side of triangle $ABC$ the median and the altitude are drawn, let's analyze the diagram given for triangle $ABC$ .

We acknowledge that a median is a line segment drawn from a vertex to the midpoint of the opposite side. An altitude is a line segment drawn from a vertex perpendicular to the opposite side.
Upon reviewing the diagram of triangle $ABC$ , line segment $AD$ is a reference term. It appears to meet point $C$ in the middle, suggesting it's a median, but it also forms right angles suggesting it is an altitude.
Given the placement and orientation of $AD$ , it is perpendicular to line $BC$ (the opposite base for the median from $A$ ). Therefore, this line is both the median and the altitude to side $BC$ .

Thus, the side to which both the median and the altitude are drawn is BC.

Therefore, the correct answer to the problem is the side $BC$ , corresponding with choice $\text{Option 2: BC}$ .

Answer

Exercise #3

Look at triangle ABC below.

What is the median of the triangle and to which side is it drawn?

Step-by-Step Solution

A median of a triangle is a line segment that connects a vertex to the midpoint of the opposite side. In triangle $\triangle ABC$ , we need to identify such a median from the diagram provided.

Step 1: Observe the diagram to identify the midpoint of each side.

Step 2: It is given that point $E$ is located on side $AC$ . If $E$ is the midpoint of $AC$ , then any line from a vertex to point $E$ would be a median.

Step 3: Check line segment $BE$ . This line runs from vertex $B$ to point $E$ .

Step 4: Since $E$ is labeled as the midpoint of $AC$ , line $BE$ is the median of $\triangle ABC$ drawn to side $AC$ .

Therefore, the median of the triangle is $BE$ for $AC$ .

Answer

BE for AC

Exercise #4

Look at triangle ABC below.

Which is the median?

Step-by-Step Solution

To solve this problem, we must identify which line segment in triangle ABC is the median.

First, review the definition: a median in a triangle connects a vertex to the midpoint of the opposite side. Now, in triangle ABC:

Point A represents the vertex.
Point E lies on line segment AB.
Line segment EC needs to be checked to see if it connects vertex E to point C.

From the diagram, it appears that E is indeed the midpoint of side AB. Thus, line segment EC connects vertex C to this midpoint.

This fits the definition of a median, verifying that EC is the median line segment in triangle ABC.

Therefore, the solution to the problem is: $\text{EC}$ .

Answer

Exercise #5

Look at the triangle ABC below.

$AD=\frac{1}{2}AB$

$BE=\frac{1}{2}EC$

What is the median in the triangle?

Step-by-Step Solution

A median in a triangle is a line segment connecting a vertex to the midpoint of the opposite side. Here, we need to find such a segment in triangle $\triangle ABC$ .

Let's analyze the given conditions:

$AD = \frac{1}{2}AB$ : Point $D$ is the midpoint of $AB$ .
$BE = \frac{1}{2}EC$ : Point $E$ is the midpoint of $EC$ .

Given that $D$ is the midpoint of $AB$ , if we consider the line segment $DC$ , it starts from vertex $D$ and ends at $C$ , passing through the midpoint of $AB$ (which is $D$ ), fulfilling the condition for a median.

Therefore, the line segment $DC$ is the median from vertex $A$ to side $BC$ .

In summary, the correct answer is the segment $DC$ .

Answer

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Triangle Height

Set the Height of a Triangle

Test yourself on parts of a triangle!

Triangle Height Calculation Exercises:

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Examples with solutions for Triangle Height

Exercise #1

Video Solution

Step-by-Step Solution

Answer

Exercise #2

Step-by-Step Solution

Answer

Exercise #3

Step-by-Step Solution

Answer

Exercise #4

Step-by-Step Solution

Answer

Exercise #5

Step-by-Step Solution

Answer

More Questions

Parts of a Triangle