Triangle Height

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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:

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AB is a side in triangle ADB

AAABBBCCCDDDEEE

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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 ABCD

CE is the altitude from side AB AB

CB=5 CB=5

AE=7 AE=7

EB=2 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 EBC \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: a2+b2=c2 a^2+b^2=c^2

In this case: EB2+EC2=BC2 EB^2+EC^2=BC^2

Substituting the given information: 22+EC2=52 2^2+EC^2=5^2

Isolating the variable: EC2=5222 EC^2=5^2-2^2

And solving: EC2=254=21 EC^2=25-4=21

EC=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 AB ,

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

21×9=41.24 \sqrt{21}\times9=41.24

Answer:

41.24 41.24


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Exercise 2

Given the right triangle:

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 c²=a²+b²

In our right triangle

a=3 a=3

b=4 b=4

c=x 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=3²+4²

x2=9+16 x²=9+16

x2=25 x²=25

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

x=25 x=\sqrt{25}

x=5 x=5

Answer:

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:

We are given the following trapezoid with these features

What is its height?

Solution

Trapezoid area formula:

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

The formula is not displaying correctly on the page.

9+62×h=30 \frac{9+6}{2}\times h=30

And we solve:

152×h=30 \frac{15}{2}\times h=30

712×h=30 7\frac{1}{2}\times h=30

h=30152 h=\frac{30}{\frac{15}{2}}

h=6015 h=\frac{60}{15}

h=4 h=4

Answer:

Height BE BE is equal to 4 4 cm.


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Do you know what the answer is?

Exercise 4

Given the isosceles triangle ABC \triangle ABC .

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

Given the isosceles triangle ABC

AF=5 AF=5

AB=17 AB=17

AG=3 AG=3

AD=8 AD=8

A A is the height of the triangle.

What is the area of EFBC EFBC ?

Solution:

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

We focus on finding the bases.

To find GF GF , we will use the theorem of Pythagoras: A2+B2=C2 A^2+B^2=C^2 in triangle AFG \triangle AFG

Replace:

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

Isolate GF GF and solve:

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

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

GF=4 GF=4

We proceed with the same process with side DB DB in triangle ABD \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 GFBD and verify that it is equal to trapezoid EGDC 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 GDGD :

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

Now, let's reveal EF EF and CB 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.

Therefore:

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 95


Exercise 5

Given the isosceles triangle ABD \triangle ABD ,

Within it, EF is drawn:

AF=5 AF=5

AB=17 AB=17

AG=3 AG=3

AD=8 AD=8

Given the isosceles triangle ABC

Task:

What is the perimeter of the trapezoid EFBCEFBC ?

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 GF , we will use the theorem of Pythagoras: A2+B2=C2 A^2+B^2=C^2 in triangle AFG.

We substitute:

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

We isolate GF GF and solve:

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

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

GF=4 GF=4

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

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

Now, we reveal EF EF and CB 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.

Therefore:

EF=GF×2=4×2=8 EF=GF\times2=4\times2=8

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

What remains is to calculate:

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

Answer:

62 62


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