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Area and Perimeter of Triangles

Area is half the base times the height while the perimeter is the sum of the sides.

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Find the Dimensions and Area of Triangles

Jessie saw this median as she rode to school. The height of each triangle in the median is 7 feet and the base is 5 feet. Jessie wondered about the area of each triangle. If there are seven triangles in a row, what is the total area of all seven triangles?

In this concept, you will learn to find the dimensions and area of triangles.

Area of a Triangle

Area is the amount of two-dimensional space a figure covers. To find the area of a triangle, you multiply the dimensions, or sides, of the figure. In a triangle, those dimensions are its height, h\begin{align*}h\end{align*}, and its base, b\begin{align*}b\end{align*}. The area formula for triangles is:

A=12bh\begin{align*}A = \frac{1}{2}bh\end{align*}

The base is the area at the bottom of the triangle opposite the vertex or top point. When finding the area of triangles, remember that the height of a triangle is always perpendicular to the base. The height is not necessarily a side of the triangle; this happens only in right triangles, because the two sides joined by a right angle are perpendicular.

You can see in the right triangle above that the left side is also the height of the triangle. It is perpendicular to the base. The equilateral triangle has a dotted line to show you the measurement for the height of the triangle.

Let’s look at an example.

Find the area of the triangle below.

First, substitute 11 for b\begin{align*}b\end{align*} (the base) and 16 for h\begin{align*}h\end{align*} (the height) into the formula for area.

AA==12bh12(11)(16)\begin{align*}\begin{array}{rcl} A &=& \frac{1}{2} bh \\ A &=& \frac{1}{2}(11)(16) \end{array}\end{align*}
Next, solve for A\begin{align*}A\end{align*}.

AA==12(11)(16)88\begin{align*}\begin{array}{rcl} A &=& \frac{1}{2}(11)(16) \\ A &=& 88 \end{array}\end{align*}

The area of the triangle is 88 cm2\begin{align*}88 \ cm^2\end{align*}.

You can also use this same formula to figure out a missing dimension of the triangle. This is possible if you are given the area and one dimension to start with. Then you can use the formula, substitute the values, and solve for the missing dimension.

Let’s take a look at an example.

A triangle has an area of 44 m2\begin{align*}44 \ m^2\end{align*}. The base of the triangle is 8 m\begin{align*}8 \ m\end{align*}. What is its height?

First, substitute 8 for b\begin{align*}b \end{align*} (the base) and 44 for A\begin{align*}A\end{align*} (the area) into the formula for area.

A4444===12bh12(8)(h)4h\begin{align*}\begin{array}{rcl} A &=& \frac{1}{2} bh \\ 44 &=& \frac{1}{2} (8)(h) \\ 44 &=& 4h \end{array}\end{align*}

Next, divide both sides by 4 to solve for h\begin{align*}h\end{align*}.

44444h===4h4h411\begin{align*}\begin{array}{rcl} 44 &=& 4h \\ \frac{44}{4} &=& \frac{4h}{4} \\ h &=& 11 \end{array}\end{align*}

The height of the triangle is 11 cm\begin{align*}11 \ cm\end{align*}.

Examples

Example 1

Earlier, you were given a problem about the triangles.

Each of the seven triangles has a base length of 5 feet and a height of 7 feet.

First, substitute 5 for b\begin{align*}b\end{align*} (the base) and 7 for h\begin{align*}h\end{align*} (the height) into the formula for area.

AA==12bh12(5)(7)\begin{align*}\begin{array}{rcl} A &=& \frac{1}{2} bh \\ A &=& \frac{1}{2} (5)(7) \end{array}\end{align*}

Next, solve for A\begin{align*}A\end{align*}.

AA==12(5)(7)17.5\begin{align*}\begin{array}{rcl} A &=& \frac{1}{2}(5)(7) \\ A &=& 17.5 \end{array}\end{align*}

Then, since there are 7 triangles, multiply this area by 7.

Total AreaTotal Area==(17.5)(7)122.5\begin{align*}\begin{array}{rcl} \text{Total Area} &=& (17.5)(7) \\ \text{Total Area} &=& 122.5 \end{array} \end{align*}

The total area of the media is 122.5 ft2\begin{align*}122.5 \ ft^2\end{align*}.

Example 2

First, substitute 4.5 for b\begin{align*}b\end{align*} (the base) and 12 for h\begin{align*}h\end{align*} (the height) into the formula for area.

AA==12bh12(4.5)(12)\begin{align*}\begin{array}{rcl} A &=& \frac{1}{2} bh \\ A &=& \frac{1}{2} (4.5)(12) \end{array} \end{align*}

Next, solve for A\begin{align*}A\end{align*}.

AA==12(4.5)(12)27\begin{align*}\begin{array}{rcl} A &=& \frac{1}{2} (4.5)(12) \\ A &=& 27 \end{array}\end{align*}

The area of the triangle is 27 cm2\begin{align*}27 \ cm^2\end{align*}.

Example 3

Find the area of each triangle given the base=6 inches\begin{align*}\text{base} = 6 \ \text{inches}\end{align*} and the height=4 inches\begin{align*}\text{height} = 4 \ \text{inches}\end{align*}.

First, substitute 6 for b\begin{align*}b\end{align*} (the base) and 4 for h\begin{align*}h\end{align*} (the height) into the formula for area.
AA==12bh12(6)(4)\begin{align*}\begin{array}{rcl} A &=& \frac{1}{2} bh \\ A &=& \frac{1}{2} (6)(4) \end{array} \end{align*}Next, solve for A\begin{align*}A\end{align*}.

AA==12(6)(4)12\begin{align*}\begin{array}{rcl} A &=& \frac{1}{2} (6)(4) \\ A &=& 12 \end{array}\end{align*}The answer is 12.

The area of the triangle is 12 in2\begin{align*}12 \ in^2\end{align*}.

Example 4

Find the area of each triangle given the base=3.5 feet\begin{align*}\text{base} = 3.5 \ \text{feet}\end{align*} and the height=4 feet\begin{align*}\text{height} = 4 \ \text{feet}\end{align*}.

First, substitute 3.5 for b\begin{align*}b\end{align*} (the base) and 4 for h\begin{align*}h\end{align*} (the height) into the formula for area.

AA==12bh12(3.5)(4)\begin{align*}\begin{array}{rcl} A &=& \frac{1}{2} bh \\ A &=& \frac{1}{2} (3.5)(4) \end{array} \end{align*}Next, solve for A\begin{align*}A\end{align*}.

AA==12(3.5)(4)7\begin{align*}\begin{array}{rcl} A &=& \frac{1}{2} (3.5)(4) \\ A &=& 7 \end{array}\end{align*} The answer is 7.

The area of the triangle is 7 ft2\begin{align*}7 \ ft^2\end{align*}.

Example 5

Find the area of each triangle given the base=8 mm\begin{align*}\text{base} = 8 \ mm\end{align*} and the height=9 mm\begin{align*}\text{height} = 9 \ mm\end{align*}.

First, substitute 8 for b\begin{align*}b\end{align*} (the base) and 9 for h\begin{align*}h\end{align*} (the height) into the formula for area.

\begin{align*}\begin{array}{rcl} A &=& \frac{1}{2} bh \\ A &=& \frac{1}{2} (8)(9) \end{array} \end{align*}

Next, solve for \begin{align*}A\end{align*}.

\begin{align*}\begin{array}{rcl} A &=& \frac{1}{2} (8)(9) \\ A &=& 36 \end{array}\end{align*} The answer is 36.

The area of the triangle is \begin{align*}36 \ mm^2\end{align*}.

Review

Find the area of each triangle described below.

1. \begin{align*}b = 10 \ \text{inches}, h = 5 \ \text{inches}\end{align*}

2. \begin{align*}b = 7 \ \text{inches}, h = 5.5 \ \text{inches}\end{align*}

3. \begin{align*}b = 8 \ \text{feet}, \text{height} = 6 \ \text{feet}\end{align*}

4. \begin{align*}b = 9 \ \text{feet}, \text{height} = 7.5 \ \text{feet}\end{align*}

5. \begin{align*}b = 12 \ \text{meters}, h = 9 \ \text{meters}\end{align*}

6. \begin{align*}b = 15 \ \text{feet}, h = 12 \ \text{feet}\end{align*}

7. \begin{align*}b = 12.5 \ \text{feet}, h = 3.5 \ \text{feet}\end{align*}

8. \begin{align*} b = 15.25 \ \text{feet}, h = 8.5 \ \text{feet}\end{align*}

9. \begin{align*}b = 25.75 \ \text{feet}, h = 13.5 \ \text{feet}\end{align*}

Find the missing dimension for each triangle given the area and one other dimension.

10. \begin{align*}A = 4.5 \ sq.\ in , b = 4.5 \ in , h =?\end{align*}

11. \begin{align*}A = 21 \ sq.\ ft, b = 7 \ ft, h =?\end{align*}

12. \begin{align*}A = 60 \ sq. \ in, h = 10 \ in, b =?\end{align*}

13. \begin{align*}A = 97.5 \ sq. \ ft, h = 13 \ ft, b =?\end{align*}

14. \begin{align*}A = 187 \ sq. \ ft, b = 22 \ ft, h =?\end{align*}

15. \begin{align*}A = 405 \ sq.\ ft, b = 30 \ ft, h =?\end{align*}

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Color Highlighted Text Notes

Vocabulary Language: English

Area

Area is the space within the perimeter of a two-dimensional figure.

Base

The side of a triangle parallel with the bottom edge of the paper or screen is commonly called the base. The base of an isosceles triangle is the non-congruent side in the triangle.

Height

The height of a triangle is the perpendicular distance from the base of the triangle to the opposite vertex of the triangle.

Perimeter

Perimeter is the distance around a two-dimensional figure.

Perpendicular

Perpendicular lines are lines that intersect at a $90^{\circ}$ angle. The product of the slopes of two perpendicular lines is -1.

Polygon

A polygon is a simple closed figure with at least three straight sides.

Right Angle

A right angle is an angle equal to 90 degrees.

Right Triangle

A right triangle is a triangle with one 90 degree angle.

Triangle

A triangle is a polygon with three sides and three angles.

Area of a Parallelogram

The area of a parallelogram is equal to the base multiplied by the height: A = bh. The height of a parallelogram is always perpendicular to the base (the sides are not the height).

Area of a Triangle

The area of a triangle is half the area of a parallelogram. Hence the formula: $A = \frac{1}{2}bh \text{ or } A = \frac{bh}{2}$.