Area of a right triangle

For example:

If we have a right triangle whose legs measure $5~cm$ and $6~cm$ and we are asked to find its area, we should multiply $5$ by $6$, giving us a result of 30 and then divide the product by $2$.

That is, the area of the given triangle is $15~cm^2$. 

Homework:

In front of you is a right triangle, calculate its area.

Solution:

Calculate the area of the triangle using the formula for calculating the area of a right triangle.

$\frac{leg\times leg}{2}$

$\frac{AB\cdot BC}{2}=\frac{8\cdot6}{2}=\frac{48}{2}=24$

Answer:

The answer is $24~cm²$.

Homework:

Given the right triangle $\triangle ADB$

The perimeter of the triangle is equal to $30\operatorname{cm}$.

Given:

$AB=15$

$AC=13$

$DC=5$

$CB=4$

Homework:

Calculate the area of the triangle$\triangle~ABC$

Solution:

Given the perimeter of the triangle $Δ~ADC$ equal to $30\operatorname{cm}$.

From here we can calculate $AD$.

$AD+DC+AD=PerimeterΔ~ADC$

$AD+5+13=30$

$AD+18=30$ /$-18$

$AD=12$

Now we can calculate the area of the triangle $Δ~ABC$

Pay attention: we are talking about an obtuse triangle therefore its height is $AD$.

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

$\frac{height\times side}{2}=$

$\frac{AD\cdot BC}{2}=\frac{12\cdot4}{2}=\frac{48}{2}=24$

Answer:

The area of the triangle $ΔABC$ is equal to $24~cm²$.

Homework:

Given the right triangle $Δ~ABC$

The area of the triangle is equal to $38~cm²$, $AC=8$

Find the measure of the leg $BC$

Solution:

We will calculate the length of $BC$ using the formula for calculating the area of the right triangle:

$\frac{leg\times leg}{2}$

$\frac{AC\cdot BC}{2}=\frac{8\cdot BC}{2}=38$

We multiply the equation by the common denominator

/ $\times2$

Then we divide the equation by the coefficient of $BC$

8\timesBC=76 /$:8$

$BC=9.5$

Answer:

The length of the leg $BC$ is equal to $9.5$ centimeters.

In front of you, there is a right triangle $Δ~ABC$.

Given that $BC=6$. The length of the leg $AB$ is greater by $33\frac{1}{3}\%$ than the length of $BD$.

The area of the triangle $\triangle~ADC$ is greater by $25\%$ than the area of the triangle $\triangle~ABD$.

Task:

What is the area of the triangle $\triangle~ABC$?

Solution:

To find the measure of the leg $AB$ we will use the data that its length is greater by $33.33$ than the length of $BD$.

$AB=1.33333\cdot BD$

$(\frac{100}{100}+\frac{33.33}{100}=\frac{133.33}{100}=1.333)$

$AB=1.333\cdot6=8$

Now we will calculate the area of the triangle $ΔABD$.

$A~Δ\text{ABD}=\frac{AB\cdot BD}{2}=\frac{8\cdot6}{2}=\frac{48}{2}=24$

Answer:

$24~cm²$.

Homework:

Which data in the graph is incorrect?

For the area of the triangle to be $24~cm²$, what is the data that should be in place of the error?

Solution:

Explanation: area of the right triangle.

$AΔEDF=\frac{ED\cdot EF}{2}=\frac{8\cdot6}{2}=\frac{48}{2}=24$

According to the formula:

$\frac{leg\times leg}{2}$

If the area of the triangle can also be calculated from the formula of:

$\frac{side\times side~height}{2}$

$\frac{EG\times10}{2}=24$ /$\times2$

$10EG=48$ /$:10$

$EG=4.8$

Answer:

The incorrect data is $EG$.

The length of $EG$ should be $4.8\operatorname{cm}$.

If you are interested in learning more about other triangle topics, you can enter one of the following articles:

• Acute triangle
• Obtuse triangle
• Scalene triangle
• Equilateral triangle
• Isosceles triangle
• The edges of a triangle
• Height of the triangle
• How to calculate the area of a triangle
• How is the perimeter of a triangle calculated?
• From a quadrilateral to a rectangle

In the blog of Tutorela you will find a variety of articles about mathematics.