# Area of a right triangle - Examples, Exercises and Solutions

## Formula to find the area of a right triangle

The area of a right triangle is an important subtopic that is repeated over and over again in exercises that include any right triangle.

It is calculated by multiplying the two sides that form the right angle (called legs) and dividing the result by 2.

## Practice Area of a right triangle

### Exercise #1

Calculate the area of the right triangle below:

### Step-by-Step Solution

As we see that AB is perpendicular to BC and forms a 90-degree angle

It can be argued that AB is the height of the triangle.

Then we can calculate the area as follows:

$\frac{AB\times BC}{2}=\frac{8\times6}{2}=\frac{48}{2}=24$

24 cm²

### Exercise #2

Calculate the area of the triangle ABC using the data in the figure.

### Step-by-Step Solution

First, let's remember the formula for the area of a triangle:

(the side * the height that descends to the side) /2

In the question, we have three pieces of data, but one of them is redundant!

We only have one height, the line that forms a 90-degree angle - AD,

The side to which the height descends is CB,

Therefore, we can use them in our calculation:

$\frac{CB\times AD}{2}$

$\frac{8\times9}{2}=\frac{72}{2}=36$

36 cm²

### Exercise #3

Calculate the area of the following triangle:

### Step-by-Step Solution

The formula for calculating the area of a triangle is:

(the side * the height from the side down to the base) /2

That is:

$\frac{BC\times AE}{2}$

Now we replace the existing data:

$\frac{4\times5}{2}=\frac{20}{2}=10$

10

### Exercise #4

Calculate the area of the triangle below, if possible.

### Step-by-Step Solution

The formula to calculate the area of a triangle is:

(side * height corresponding to the side) / 2

Note that in the triangle provided to us, we have the length of the side but not the height.

That is, we do not have enough data to perform the calculation.

Cannot be calculated

### Exercise #5

What is the area of the given triangle?

### Step-by-Step Solution

This question is a bit confusing. We need start by identifying which parts of the data are relevant to us.

Remember the formula for the area of a triangle:

The height is a straight line that comes out of an angle and forms a right angle with the opposite side.

In the drawing we have a height of 6.

It goes down to the opposite side whose length is 5.

And therefore, these are the data points that we will use.

We replace in the formula:

$\frac{6\times5}{2}=\frac{30}{2}=15$

15

### Exercise #1

What is the area of the triangle in the drawing?

### Step-by-Step Solution

First, we will identify the data points we need to be able to find the area of the triangle.

the formula for the area of the triangle: height*opposite side / 2

Since it is a right triangle, we know that the straight sides are actually also the heights between each other, that is, the side that measures 5 and the side that measures 7.

We multiply the legs and divide by 2

$\frac{5\times7}{2}=\frac{35}{2}=17.5$

17.5

### Exercise #2

Calculate X using the data in the figure below.

### Step-by-Step Solution

The formula to calculate the area of a triangle is:

(side * height descending from the side) /2

We place the data we have into the formula to find X:

$20=\frac{AB\times AC}{2}$

$20=\frac{x\times5}{2}$

Multiply by 2 to get rid of the fraction:

$5x=40$

Divide both sections by 5:

$\frac{5x}{5}=\frac{40}{5}$

$x=8$

8

### Exercise #3

The triangle ABC is given below.
AC = 10 cm

BC = 11.6 cm
What is the area of the triangle?

### Step-by-Step Solution

The triangle we are looking at is the large triangle - ABC

The triangle is formed by three sides AB, BC, and CA.

Now let's remember what we need for the calculation of a triangular area:

(side x the height that descends from the side)/2

Therefore, the first thing we must find is a suitable height and side.

We are given the side AC, but there is no descending height, so it is not useful to us.

The side AB is not given,

And so we are left with the side BC, which is given.

From the side BC descends the height AD (the two form a 90-degree angle).

It can be argued that BC is also a height, but if we delve deeper it seems that CD can be a height in the triangle ADC,

and BD is a height in the triangle ADB (both are the sides of a right triangle, therefore they are the height and the side).

As we do not know if the triangle is isosceles or not, it is also not possible to know if CD=DB, or what their ratio is, and this theory fails.

Let's remember again the formula for triangular area and replace the data we have in the formula:

(side* the height that descends from the side)/2

Now we replace the existing data in this formula:

$\frac{CB\times AD}{2}$

$\frac{11.6\times3}{2}$

$\frac{34.8}{2}=17.4$

17.4

### Exercise #4

Which of the following triangles have the same areas?

### Step-by-Step Solution

We calculate the area of triangle ABC:

$\frac{12\times5}{2}=\frac{60}{2}=30$

We calculate the area of triangle EFG:

$\frac{6\times10}{2}=\frac{60}{2}=30$

We calculate the area of triangle JIK:

$\frac{6\times5}{2}=\frac{30}{2}=15$

Therefore, the triangles that have the same areas are ABC and EFG.

EFG and ABC

### Exercise #5

The area of triangle ABC is 20 cm².

Calculate the length of the side BC.

### Step-by-Step Solution

We can present the data in the formula to calculate the area of the triangle:

$S=\frac{AD\times BC}{2}$

$20=\frac{8\times BC}{2}$

Cross multiplication:

$40=8BC$

Divide both sides by 8:

$\frac{40}{8}=\frac{8BC}{8}$

$BC=5$

5 cm

### Exercise #1

PRS is a triangle.

The length of side SR is 4 cm.
The area of triangle PSR is 30 cm².

Calculate the height PQ.

### Step-by-Step Solution

We use the formula to calculate the area of the triangle.

Pay attention: in an obtuse triangle, the height is located outside of the triangle!

$\frac{Side\cdot Height}{2}=\text{Triangular Area}$

Double the equation by a common denominator:

$\frac{4\cdot PQ}{2}=30$

$\cdot2$

Divide the equation by the coefficient of $PQ$.

$4PQ=60$ / $:4$

$PQ=15$

15 cm

### Exercise #2

Below is the right triangle ABD, which has a perimeter of 36 cm.

AB = 15

AC = 13

DC = 5

CB = 4

Work out the area of the triangle.

### Step-by-Step Solution

According to the data:

$BD=4+5=9$

Now that we are given the perimeter of triangle ABD we can find the missing side AD:

$AD+15+9=36$

$AD+24=36$

$AD=36-24=12$

Now we can calculate the area of triangle ABD:

$\frac{AD\times BD}{2}=\frac{12\times9}{2}=\frac{108}{2}=54$

54 cm²

### Exercise #3

The triangle ABC has a perimeter measuring 42 cm.

AC = 15

AB = 13

Calculate the area of the triangle.

### Step-by-Step Solution

Given that the perimeter of triangle ABC is 42.

We will use this data to find side CB:

$13+15+CB=42$

$CB+28=42$

$CB=42-28=14$

Now we can calculate the area of triangle ABC:

$\frac{AD\times BC}{2}=\frac{12\times14}{2}=\frac{168}{2}=84$

84 cm²

### Exercise #4

Calculate the area of the

triangle ABC given that its

perimeter equals 26.

### Step-by-Step Solution

Remember that the perimeter of a triangle is equal to the sum of all sides together,

Now find side BC:

$26=9+7+BC$

$26=16+BC$

We move the 16 to the left section and keep the corresponding sign:

$26-16=BC$

$10=BC$

We use the formula to calculate the area of a triangle:

(the side * the height) /2

That is:

$\frac{BC\times AE}{2}$

We replace the existing data:

$\frac{10\times6}{2}=\frac{60}{2}=30$

30

### Exercise #5

Triangle ABC is a right triangle.

The area of the triangle is 6 cm².

Calculate X and the length of the side BC.

### Step-by-Step Solution

We use the formula to calculate the area of the right triangle:

$\frac{AC\cdot BC}{2}=\frac{cateto\times cateto}{2}$

And compare the expression with the area of the triangle $6$

$\frac{4\cdot(X-1)}{2}=6$

Multiplying the equation by the common denominator means that we multiply by $2$

$4(X-1)=12$

We distribute the parentheses before the distributive property

$4X-4=12$ / $+4$

$4X=16$ / $:4$

$X=4$

We replace $X=4$ in the expression $BC$ and

find:

$BC=X-1=4-1=3$