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 $AB$

$CB=5$

$AE=7$

$EB=2$

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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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?

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:

$62$