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Right triangle Area of a triangle Area of a square Area of a rectangleHere you will learn about how to find the area of obtuse triangles, including what formula to use and how to apply it to different situations.

Students will first learn about the area of obtuse triangles as part of geometry in 6 th grade.

The **area of an obtuse triangle** is the amount of space inside an obtuse triangle.

An obtuse-angled triangle is a type of triangle with one obtuse angle.

The general formula to find the area of any triangle is half of the product of the base times height:

\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 obtuse triangle?

A=\cfrac{1}{2} \, b h

The height is always the perpendicular distance from the base to the top of the triangle.

\begin{aligned} & A=\cfrac{1}{2} \times 5 \times 3 \\\\ & A=7.5 \end{aligned}

The area of the obtuse triangle is 7.5 \mathrm{ft}^2.

For example,

What is the area of the obtuse triangle?

A=\cfrac{1}{2} \, b h

The height is always the perpendicular distance from the base to the top of the triangle.

\begin{aligned} & A=\frac{1}{2} \times 12 \times 4.1 \\\\ & A=24.6 \end{aligned}

The area of the obtuse triangle is 24.6~{ft}^2.

How does this relate to 6 th grade math?

**Grade 6 – Geometry (6.G.A.1)**

Find the area of right triangles, other triangles, special quadrilaterals, and polygons by composing into rectangles or decomposing into triangles and other shapes; apply these techniques in the context of solving real-world and mathematical problems.

Use this quiz to check your grade 4 to 6 students’ understanding of area. 15+ questions with answers covering a range of 4th, 6th and 7th grade area topics to identify areas of strength and support!

DOWNLOAD FREEUse this quiz to check your grade 4 to 6 students’ understanding of area. 15+ questions with answers covering a range of 4th, 6th and 7th grade area topics to identify areas of strength and support!

DOWNLOAD FREEIn order to find the area of obtuse triangles:

**Identify the base and perpendicular height of the triangle.****Write the area formula.****Substitute known values into the area formula.****Solve the equation.****Write the answer, including the units.**

Find the area of this triangle:

**Identify the base and perpendicular height of the triangle.**

b = 11 \, m

h = 7 \, m

2**Write the area formula.**

A=\cfrac{1}{2} \, b h

3**Substitute known values into the area formula.**

\begin{aligned} A & =\cfrac{1}{2} \, b h \\\\ & =\cfrac{1}{2} \, (11)(7) \end{aligned}

4**Solve the equation.**

\begin{aligned} A & =\cfrac{1}{2} \, b h \\\\ & =\cfrac{1}{2} \, (11)(7) \\\\ & =38.5 \end{aligned}

5**Write the answer, including the units.**

A=38.5 \mathrm{~m}^2

Remember: Your final answer must be in square units.

Calculate the area of the triangle below:

**Identify the base and perpendicular height of the triangle.**

Rotating the triangle makes it easier to identify the base and height.

b = 4~{m}

h = 90~{cm}

Notice there are 2 different units here. You must convert them to a common unit:

90~{cm} = 0.9~{m}

b = 4~{m}

h = 0.9~{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} \, (4)(0.9) \end{aligned}

**Solve the equation.**

\begin{aligned} A & =\cfrac{1}{2} \, b h \\\\ & =\cfrac{1}{2} \, (4)(0.9) \\\\ & =1.8 \end{aligned}

**Write the answer, including the units.**

A=1.8~{m}^2

Shown below is a triangular-shaped field. Each horse needs 2,000 \mathrm{~m}^2 to graze. How many horses can fit into this field?

**Identify the base and perpendicular height of the triangle.**

b = 412~{m}

h = 237.6~{m}

Note: The 308~{m} is not needed to solve this question. 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} \, (412)(237.6) \end{aligned}

**Solve the equation.**

\begin{aligned} A & =\cfrac{1}{2} \, b h \\\\ & =\cfrac{1}{2} \, (412)(237.6) \\\\ & =48,945.6 \end{aligned}

**Write the answer, including the units.**

A=48,945.6~{m}^2

Now to calculate how many horses will fit into the field, take 48,945.6~{m}^2 and divide it by 2,000~{m}^2, because each horse needs that much space to graze on the field.

48,945.6 \div 2,000=24.4728

The quotient has 24 wholes and 4,728 ten-thousandths. There cannot be part of a horse and 24.4728 is not enough for 25 horses. So, 24 horses will fit in the field.

Find the area of the triangle below:

**Identify the base and perpendicular height of the triangle.**

b = 23 \text{ inches}

h = 19.9 \text{ inches}

**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} \, (23)(19.9) \end{aligned}

**Solve the equation.**

\begin{aligned} A & =\cfrac{1}{2} \, b h \\\\ & =\cfrac{1}{2} \, (23)(19.9) \\\\ & =228.85 \end{aligned}

**Write the answer, including the units.**

A=228.85 \text { inches}^2

Below is the outline of an arrow head formed by two congruent triangles. Calculate the area of the arrow head.

**Identify the base and perpendicular height of the triangle.**

b = 8.1~{cm}

h = 6~{cm}

**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} \, (8.1)(6) \end{aligned}

**Solve the equation.**

\begin{aligned} A & =\frac{1}{2} \, b h \\\\ & =\frac{1}{2} \, (8.1)(6) \\\\ & =24.3 \end{aligned}

**Write the answer, including the units.**

The area of one triangle is 24.3~{cm}^2, so multiply by 2 to calculate the area of both triangles.

24.3 \mathrm{~cm}^2 \times 2=48.6 \mathrm{~cm}^2

Triangle ABC is an obtuse triangle with an area of 21~{cm}^2. The height of the triangle is 7~{cm}. Find the length of the base of the triangle.

**Identify the base and perpendicular height of the triangle.**

\begin{aligned} & A=21 \mathrm{~cm}^2 \\\\ & b= \, ? \\\\ & h=7 \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 \\\\ & 21=\cfrac{1}{2} \times b \times 7 \\\\ & 21=\cfrac{1}{2} \times 7 \times b \\\\ & 21=3.5 \times b \end{aligned}

**Solve the equation.**

21=3.5 \times b

To solve, find what times 3.5 is equal to 21. Since 3.5 \times 6=21, 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).

- Be sure to expose students to all types of obtuse triangles (including isosceles triangles and scalene triangles) and in all different orientations. It is crucial that students learn how to identify the base and the height no matter what type of triangle or its position. Always look for worksheets, quizzes or study materials that provide this type of variety.

**Using the wrong measurement as the base or the height**Since none of the sides of the triangle in an obtuse triangle are perpendicular, the height will not correlate with a side – but will either be inside or outside the triangle. When a side of the triangle is chosen for a base, there must be a measurement that goes from the base to the top of the triangle at a 90 degree angle.

For example,

**Forgetting to write the units**

It is common to forget the units for area in the final answer. When calculating the area, the answer must always have units squared.

**Thinking that obtuse triangles have all obtuse angles**

All interior angles of a triangle add up to 180 degrees. Since an obtuse angle measures more than 90 degrees, the other two angles of the triangle together measure less than 90 degrees, making both of them acute.

- Area of a triangle
- Area of isosceles triangle
- Area of a right triangle
- Area of equilateral triangle
- Area of composite shapes
- Area of a hexagon
- Area of a pentagon
- Area of rectilinear figures

1) Find the area of the triangle below:

62~{cm}^{2}

114 \mathrm{~cm}^2

228 \mathrm{~cm}^2

118 \mathrm{~cm}^2

Identify the base and perpendicular height of the triangle.

\begin{aligned} A & =\cfrac{1}{2} \, b h \\\\ & =\cfrac{1}{2} \, (19)(12) \\\\ & =114 \mathrm{~cm}^2 \end{aligned}

2) Find the area of the triangle below giving your answer in cm^2\text{:}

1.48 \mathrm{~cm}^2

2.96 \mathrm{~cm}^2

14,800 \mathrm{~cm}^2

29,600 \mathrm{~cm}^2

Notice there are 2 different units here. You must convert them to a common unit:

3.7~{m} = 370~{cm}

\begin{aligned} A & =\frac{1}{2} \, b h \\\\ & =\frac{1}{2} \, (370)(80) \\\\ & =14,800 \mathrm{~cm}^2 \end{aligned}

3) What is the area of the triangle?

57.5 \text { inches}^2

66.1 \text { inches}^2

78 \text { inches}^2

145 \text { inches}^2

Identify the base and perpendicular height of the triangle.

\begin{aligned} A & =\cfrac{1}{2} \, b h \\\\ & =\cfrac{1}{2} \, (11.5)(10) \\\\ & =57.5 \text { inches}^2 \end{aligned}

4) The shape below is made of two congruent triangles. Find the total area of the shape below.

51.46 \mathrm{ft}^2

25.73 \mathrm{ft}^2

12.4 \mathrm{ft}^2

24.8 \mathrm{ft}^2

Remember, the height and base are always perpendicular to each other.

\begin{aligned} A & =\frac{1}{2} \, b h \\\\ & =\frac{1}{2} \, (6.2)(4) \\\\ & =12.4 \mathrm{ft}^2 \end{aligned}

The area of one triangle is 12.4~{ft}^2, so multiply by 2 to calculate the area of the entire shape.

12.4~{ft}^2 \times 2=24.8~{ft}^2

5) Kiki has a small pool in her backyard – shown below. What is the area of the pool? Round to the nearest tenth.

8 \mathrm{~m}^2

7.3 \mathrm{~m}^2

14.6 \mathrm{~m}^2

11.4 \mathrm{~m}^2

\begin{aligned} A & =\frac{1}{2} \, b h \\\\ & =\frac{1}{2} \, (3.65)(4) \\\\ & =7.3 \mathrm{~m}^2 \end{aligned}

6) Triangle MNP is an obtuse triangle with an area of 30~{cm}^2. The base length of the triangle is 4~{cm}. Find the height of the triangle.

26 \mathrm{~cm}

7.5 \mathrm{~cm}

2 \mathrm{~cm}

15 \mathrm{~cm}

The area of a triangle is calculated with the formula: A=\cfrac{1}{2} \, b h

\begin{aligned} & A=30 \mathrm{~cm}^2 \\\\ & b= \, ? \\\\ & h=4 \mathrm{~cm} \end{aligned}

\begin{aligned} & A=\cfrac{1}{2} \, b h \\\\ & 30=\cfrac{1}{2} \times b \times 4 \\\\ & 30=\cfrac{1}{2} \times 4 \times b \\\\ & 30=2 \times b \end{aligned}

To solve, find what times 2 is equal to 30. Since 2 \times 15=30, the base is 15.

To find the perimeter, add up all three sides of an obtuse triangle.

The lengths of the sides of an equilateral triangle are always the same. This causes all the angles to be equal and therefore acute angles, making it an acute triangle. Since the sides of a triangle that has an obtuse angle will never be equilateral, an obtuse triangle can only be an isosceles triangle or a scalene triangle.

The pythagorean theorem explains a relationship between the sides of all right triangles. It is used in trigonometry to find a missing third side measurement.

For a right-angled triangle, the longest side of the triangle is called the hypotenuse. It is opposite of the vertex angle measuring 90 degrees.

It is a formula that can be used to calculate the area of a triangle, given the measurements of each side.

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