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Right triangle Area of a square Area of a rectangleHere you will learn how to find the area of a right triangle, including what formula to use and why it works.
Students will first learn about the area of right triangles as part of geometry in 6th grade.
The area of a right triangle is the amount of the space inside a right triangle.
A right triangle is a type of triangle that has a right angle. The right angle is always opposite of the longest side of the triangle.
In order to find the area of a right triangle, start with the area of a rectangle.
The area of the rectangle is calculated by multiplying the \text { base } \times \text { height}.
Within a rectangle, there are two congruent right triangles. They can be shown by connecting two opposite vertices.
The area of the rectangle is the same as two congruent right triangles. This means the area of each right triangle is exactly half the area of the rectangle.
The general formula to find the area of any triangle is:
\text { Area of a triangle } = \cfrac{\text { base } \times \text { height }}{2}
This can also be written as the following formula:
A=\cfrac{1}{2} \, b h
where b is the base length and h is the height of the triangle.
The space inside of shapes is measured in square units, so the final answer must be given in \text {units}^2 . For example, \mathrm{cm}^2, \mathrm{~m}^2, \mathrm{~mm}^2 .
For example,
What is the area of the right triangle?
A=\cfrac{1}{2} \, b h
The base is 4 and the height is 3 .
A few things to note:
\begin{aligned} & A=\cfrac{1}{2} \times 4 \times 3 \\\\ & A=2 \times 3 \\\\ & A=6 \end{aligned}
The area of the right triangle is 6 ft^2 .
Use this quiz to check your grade 4 to 6 studentsβ understanding of area. 15+ questions with answers covering a range of 4th, 5th and 6th grade area topics to identify areas of strength and support!
DOWNLOAD FREEUse this quiz to check your grade 4 to 6 studentsβ understanding of area. 15+ questions with answers covering a range of 4th, 5th and 6th grade area topics to identify areas of strength and support!
DOWNLOAD FREEHow does this relate to 6th grade math?
In order to find the area of a right triangle:
Find the area of the triangle below:
b = 8 \, cm
h = 7 \, cm
2Write the area formula.
A=\cfrac{1}{2} \, b h
3Substitute known values into the area formula.
\begin{aligned} A &=\cfrac{1}{2} \, b h \\\\ &=\cfrac{1}{2} \, (8)(7) \end{aligned}
4Solve the equation.
\begin{aligned} A &=\cfrac{1}{2} \, b h \\\\ &=\cfrac{1}{2} \, (8)(7)\\\\ &=28 \end{aligned}
5Write the answer, including the units.
A=28 \mathrm{~cm}^2
Remember: Your final answer must be in units squared.
Find the area of the triangle below:
Identify the base and perpendicular height of the triangle.
b = 6 \, m
h = 200 \, cm
Notice, there are 2 different units here. You must convert them to a common unit:
200 \, cm = 2 \, m
b = 6 \, m
h = 2 \, m
Write the area formula.
A=\cfrac{1}{2} \, b h
Substitute known values into the area formula.
\begin{aligned} A &=\cfrac{1}{2} \, b h \\\\ &=\cfrac{1}{2} \, (6)(2) \end{aligned}
Solve the equation.
\begin{aligned} A &=\cfrac{1}{2} \, b h \\\\ &=\cfrac{1}{2} \, (6)(2)\\\\ &=6\end{aligned}
Write the answer, including the units.
A=6 \, m^2
Find the area of triangle BAC below:
Identify the base and perpendicular height of the triangle.
b = 12 \, cm
h = 5 \, cm
Note: The 13 \, cm is not needed to solve. It is extra information.
Write the area formula.
A=\cfrac{1}{2} \, b h
Substitute known values into the area formula.
\begin{aligned} A &=\cfrac{1}{2} \, b h \\\\ &=\cfrac{1}{2} \, (12)(5) \end{aligned}
Solve the equation.
\begin{aligned} A &=\cfrac{1}{2} \, b h \\\\ &=\cfrac{1}{2} \, (5)(12) \\\\ &=30 \end{aligned}
Write the answer, including the units.
A=30 \, cm^2
Find the area of triangle below:
Identify the base and perpendicular height of the triangle.
Rotating the triangle makes it easier to identify the base and height.
\begin{aligned} & \mathrm{b}=2 \, \cfrac{1}{2} \\\\ & \mathrm{h}=2 \, \cfrac{1}{2} \end{aligned}
Write the area formula.
A=\cfrac{1}{2} \, b h
Substitute known values into the area formula.
\begin{aligned} A & = \cfrac{1}{2} \, b h \\\\ & = \cfrac{1}{2} \, \left(2 \, \cfrac{1}{2} \, \right) \left(2 \, \cfrac{1}{2} \, \right) \end{aligned}
Solve the equation.
\begin{aligned} A & =\cfrac{1}{2} \, b h \\\\ & =\cfrac{1}{2} \, \left(2 \, \cfrac{1}{2} \, \right) \left(2 \, \cfrac{1}{2} \, \right) \\\\ & =3 \, \cfrac{1}{8} \end{aligned}
Write the answer, including the units.
A=3 \, \cfrac{1}{8} \text { units}^2
Note, when no units are given the answer can be labeled with the general term \text {units}^2.
Below is the floor plan for a new deck that needs to be painted. One can of paint costs \$12 and covers an area of 2.5 \, m^2. How much would it cost to paint the entire deck?
Identify the base and perpendicular height of the triangle.
Split the plan into 2 shapes; a rectangle and a right triangle.
For the triangle:
b = 3.5 \, m
h = 2.5 \, m
Write the area formula.
A=\cfrac{1}{2} \, b h
Substitute known values into the area formula.
\begin{aligned} A &=\cfrac{1}{2} \, b h \\\\ &=\cfrac{1}{2} \, (3.5)(2.5) \end{aligned}
Solve the equation.
\begin{aligned} A &=\cfrac{1}{2} \, b h \\\\ &=\cfrac{1}{2} \, (2.5)(3.5)\\\\ &=4.375 \end{aligned}
Now you must find the area of the rectangle.
\begin{aligned}
\text { Area of a rectangle }&=l \times w \\\\
&=2.5 \times 3 \\\\
&=7.5 \end{aligned}
Write the answer, including the units.
Total Area = 4.375+7.5=11.875 \, m^2
Now divide 11.875 by 2.5 since one can of paint only covers an area of 2.5 \, m^2 .
11.875 \div 2.5=4.75
Since you canβt buy 0.75 paint cans and 4.75 is not quite enough, you have to buy 5 cans of paint.
Multiply to find the cost: 5 \text { cans of paint } \times \$ 12=\$ 60 .
The total cost to paint the deck is \$60 .
Triangle XYZ is a right triangle with an area of 15 \, cm^2 . The height of the triangle is 5 \, cm . Find the base length of the triangle.
Identify the base and perpendicular height of the triangle.
\begin{aligned} A &=15 \mathrm{~cm}^2 \\\\ b &= \, ? \\\\ h &=5 \mathrm{~cm} \end{aligned}
Write the area formula.
A=\cfrac{1}{2} \, b h
Substitute known values into the area formula.
\begin{aligned} & A=\cfrac{1}{2} \, b h \\\\ & 15=\cfrac{1}{2} \, \times b \times 5 \\\\ & 15=\cfrac{1}{2} \, \times 5 \times b \\\\ & 15=2.5 \times b \end{aligned}
Solve the equation.
15=2.5 \times b
To solve, find what times 2.5 is equal to 15 . Since 2.5 \times 6=15 , the base is 6 .
Write the answer, including the units.
b=6 \mathrm{~cm}
The units are not squared, because the base is the length of the bottom of the triangle (one-dimensional), not the square units within the triangle (two-dimensional).
1. What is the area of this triangle?
The area of a triangle is calculated with the formula: A=\cfrac{1}{2} \, b h .
Rotating the triangle makes it easier to identify the base and height.
For this triangle:
b = 12 \, cm
h = 9 \, m
Substitute the values into the formula.
\begin{aligned} A & =\cfrac{1}{2} \, b h \\\\ & =\cfrac{1}{2} \, (12)(9) \\\\ & =54 \mathrm{~cm}^2 \end{aligned}
2. Find the area of the triangle below:
The area of a triangle is calculated with the formula: A=\cfrac{1}{2} \, b h.
For this triangle:
b = 0.9 \, m
h = 50 \, cm
Since the measurements are in different units (m and cm), they need to be converted to the same unit:
0.9 \, m = 90 \, cm.
b = 90 \, cm
h = 50 \, cm
Note: 103 \, cm is not needed to solve. It is extra information.
Substitute the values into the formula.
\begin{aligned} A & =\cfrac{1}{2} \, b h \\\\ & =\cfrac{1}{2} \, (90)(50) \\\\ & =2,250 \mathrm{~cm}^2 \end{aligned}
3. Shown below is a compound shaped meerkat enclosure. Each meerkat needs a minimum of 9 \, m^2 to roam around. What is the maximum number of meerkats that can fit into this cage?
2 meerkats
3 meerkats
24 meerkats
25 meerkats
The shape can be split into a rectangle and a triangle. The opposite sides of the rectangle are congruent.
The area of the rectangle is 4.5 \times 3.2 = 14.4 \, m^{2}
The area of the triangle is \cfrac{1}{2} \, \times 4.5 \times 4.5 = 10.125 \, m^{2}
This means the total area is 14.4+ 10.125 = 24.525 \, m^{2}
Now to calculate how many meerkats will fit into the enclosure, take 24.525 \, m^2 and divide it by 9 \, m^2, because each meerkat needs that much space to roam.
24.525 \div 9 = 2.725
The quotient has Β 2 wholes and 725 thousandths.
There cannot be part of a meerkat and 2.725 is not enough for 3 meerkats.
So, Β 2 meerkats will fit in the enclosure.
4. Find the area of the triangle below:
The area of a triangle is calculated with the formula:
A=\cfrac{1}{2} \, b h .
Rotating the triangle makes it easier to identify the base and height.
For this triangle:
b = 6 \, cm
h = 17 \, cm
Substitute the values into the formula.
\begin{aligned} A & =\cfrac{1}{2} \, b h \\\\ & =\cfrac{1}{2} \, (6)(17) \\\\ & =51 \mathrm{~cm}^2 \end{aligned}
5. Find the area of the triangle below:
The area of a triangle is calculated with the formula:
A=\cfrac{1}{2} \, b h .
For this triangle:
b = 11 \, ft
h = 4 \, ft
Substitute the values into the formula.
\begin{aligned} A & =\cfrac{1}{2} \, b h \\\\ & =\cfrac{1}{2} \, (11)(4) \\\\ & =22 \, f t^2 \end{aligned}
6. Triangle PQR is a right triangle with an area of 40{cm}^2 . The base length of the triangle is 0.1 \, m . Find the height of the triangle.
The area of a triangle is calculated with the formula:
A=\cfrac{1}{2} \, b h .
\begin{aligned} & A=40 \mathrm{~cm}^2 \\\\ & b=0.1 \mathrm{~m} \\\\ & h= \, ? \end{aligned}
Since the measurements are in different units ( m and cm ), they need to be converted to the same unit:
0.1 \, m = 10 \, cm .
\begin{aligned} & A=40 \mathrm{~cm}^2 \\\\ & b=10 \mathrm{~cm} \end{aligned}
\begin{aligned} & A=\cfrac{1}{2} \, b h \\\\ & 40=\cfrac{1}{2} \, \times 10 \times h \\\\ & 40=5 \times h \end{aligned}
To solve, find what times 5 is equal to 40 .
Since 5 \times 8=40 , the base is 8 \, cm .
The perimeter is the distance around the right triangle. To calculate perimeter, add the lengths of the three sides together.
The pythagorean theorem is shown by the formula a^2+b^2=c^2 , where c is the hypotenuse of a right triangle (longest side of a right triangle) and a and b are adjacent sides of the triangle.
The theorem explains a relationship between the sides that make up right triangles and can be used to find a missing third side of a right triangle, if the two other side lengths are given.
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