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# 5.4: Derivation of the Triangle Area Formula

Difficulty Level: At Grade Created by: CK-12
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Practice Derivation of the Triangle Area Formula
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While in the lunch room with your friends one day, you're discussing different ways you can use the things you've learned in math class. You tell your friends that you've been learning a lot about triangles, such as how to find their area. One of your friends looks down at your plate and starts to smile.

"Alright," he says. "If you're so good at things involving triangles, I dare you to find something simple. Tell me the area of your slice of pizza." He points down at the pizza on your plate.

The pizza is shaped like a triangle. But unfortunately its not a right triangle. The outer edge is 5 inches long, and the long sides are 7 inches long. The angle between the edge and the long side of the slice is 69\begin{align*}69^\circ\end{align*}. Is there any way to tell the area of your pizza slice?

### Watch This

James Sousa Example: Determine the Area of a Triangle Using the Sine Function

### Guidance

We can use the area formula from Geometry, A=12bh\begin{align*}A = \frac{1}{2} bh\end{align*}, as well as the sine function, to derive a new formula that can be used when the height, or altitude, of a triangle is unknown.

In ABC\begin{align*}\triangle{ABC}\end{align*} below, BD\begin{align*}BD\end{align*} is altitude from B\begin{align*}B\end{align*} to AC\begin{align*}AC\end{align*}. We will refer to the length of BD\begin{align*}BD\end{align*} as h\begin{align*}h\end{align*} since it also represents the height of the triangle. Also, we will refer to the area of the triangle as K\begin{align*}K\end{align*} to avoid confusing the area with A\begin{align*}\angle{A}\end{align*}.

kkk=12bh=12b(csinA)=12bcsinAArea of a trianglesinA=hc therefore csinA=hSimplify

We can use a similar method to derive all three forms of the area formula, regardless of the angle:

KKK=12bcsinA=12acsinB=12absinC

The formula K=12 bcsinA\begin{align*}K = \frac{1}{2} \ bc \sin A \end{align*} requires us to know two sides and the included angle (SAS) in a triangle. Once we know these three things, we can easily calculate the area of an oblique triangle.

#### Example A

In ABC,C=62,b=23.9\begin{align*}\triangle{ABC}, \angle{C} = 62^\circ, b = 23.9\end{align*}, and a=31.6\begin{align*}a = 31.6\end{align*}. Find the area of the triangle.

Solution: Using our new formula, K=12 absinC\begin{align*}K = \frac{1}{2} \ ab \sin C \end{align*}, plug in what is known and solve for the area.

KK=12(31.6)(23.9)sin62333.4

#### Example B

The Pyramid Hotel recently installed a triangular pool. One side of the pool is 24 feet, another side is 26 feet, and the angle in between the two sides is 87\begin{align*}87^\circ\end{align*}. If the hotel manager needs to order a cover for the pool, and the cost is 35\begin{align*}\ 35\end{align*} per square foot, how much can he expect to spend? Solution: In order to find the cost of the cover, we first need to know the area of the cover. Once we know how many square feet the cover is, we can calculate the cost. In the illustration above, you can see that we know two of the sides and the included angle. This means we can use the formula K=12 bcsinA\begin{align*}K = \frac{1}{2} \ bc \sin A\end{align*}. K=12 (24)(26)sin87K311.6311.6 sq.ft.×35/sq.ft.=$10,905.03 The cost of the cover will be$10,905.03\begin{align*}\10,905.03\end{align*}.

#### Example C

In GHI,I=15,g=14.2\begin{align*}\triangle{GHI}, \angle{I} = 15^\circ, g = 14.2\end{align*}, and h=7.9\begin{align*}h = 7.9\end{align*}. Find the area of the triangle.

Solution: Using our new formula, K=12 absinC\begin{align*}K = \frac{1}{2} \ ab \sin C \end{align*}, which is the same as K=12 ghsinI\begin{align*}K = \frac{1}{2} \ gh \sin I \end{align*}, plug in what is known and solve for the area.

KK=12(14.2)(7.9)sin1514.52

### Guided Practice

1. A farmer needs to replant a triangular section of crops that died unexpectedly. One side of the triangle measures 186 yards, another measures 205 yards, and the angle formed by these two sides is 148\begin{align*}148^\circ\end{align*}.

What is the area of the section of crops that needs to be replanted?

2. The farmer goes out a few days later to discover that more crops have died. The side that used to measure 205 yards now measures 288 yards. How much has the area that needs to be replanted increased by?

3. Find the perimeter of the quadrilateral at the left If the area of DEG=56.5\begin{align*}\triangle{DEG} = 56.5\end{align*} and the area of EGF=84.7\begin{align*}\triangle{EGF} = 84.7\end{align*}.

Solutions:

1. Use K=12 bcsinA,K=12(186)(205)sin148\begin{align*}K = \frac{1}{2} \ bc \sin A, K= \frac{1}{2}(186)(205)\sin 148^\circ\end{align*}. So, the area that needs to be replaced is 10102.9 square yards.

2. K=12(186)(288)sin148=14193.4\begin{align*}K = \frac{1}{2}(186)(288)\sin 148^\circ = 14193.4\end{align*}, the area has increased by 4090.5 yards.

3. You need to use the K=12 bcsinA\begin{align*}K = \frac{1}{2} \ bc \sin A\end{align*} formula to find DE\begin{align*}DE\end{align*} and GF\begin{align*}GF\end{align*}.

56.5=12(13.6)DEsin39DE=13.284.7=12(13.6)EFsin60EF=14.4

Second, you need to find sides DG\begin{align*}DG\end{align*} and GF\begin{align*}GF\end{align*} using the Law of Cosines.

DG2GF2=13.22+13.62213.213.6cos39DG=8.95=14.42+13.62214.413.6cos60GF=14.0

The perimeter of the quadrilateral is 50.55.

### Concept Problem Solution

K=12 bcsinA\begin{align*}K = \frac{1}{2} \ bc \sin A\end{align*}

where in this case, one of the sides is equal to 5, the other is equal to 7, and the angle is 69\begin{align*}69^\circ\end{align*}.

K=12(5)(7)sin69=16.34in2\begin{align*}K = \frac{1}{2} (5)(7) \sin 69^\circ = 16.34 in^2\end{align*}

### Explore More

Find the area of each triangle.

1. ABC\begin{align*}\triangle ABC\end{align*} if a=13, b=15, and mC=71\begin{align*}m\angle C=71^\circ\end{align*}.
2. ABC\begin{align*}\triangle ABC\end{align*} if b=8, c=4, and mA=67\begin{align*}m\angle A=67^\circ\end{align*}.
3. ABC\begin{align*}\triangle ABC\end{align*} if b=34, c=29, and mA=138\begin{align*}m\angle A=138^\circ\end{align*}.
4. ABC\begin{align*}\triangle ABC\end{align*} if a=3, b=7, and mC=80\begin{align*}m\angle C=80^\circ\end{align*}.
5. ABC\begin{align*}\triangle ABC\end{align*} if a=4.8, c=3.7, and mB=43\begin{align*}m\angle B=43^\circ\end{align*}.
6. ABC\begin{align*}\triangle ABC\end{align*} if a=12, b=5, and mC=20\begin{align*}m\angle C=20^\circ\end{align*}.
7. ABC\begin{align*}\triangle ABC\end{align*} if a=3, b=10, and mC=50\begin{align*}m\angle C=50^\circ\end{align*}.
8. ABC\begin{align*}\triangle ABC\end{align*} if a=5, b=9, and mC=14\begin{align*}m\angle C=14^\circ\end{align*}.
9. ABC\begin{align*}\triangle ABC\end{align*} if a=5, b=7, and c=11.
10. ABC\begin{align*}\triangle ABC\end{align*} if a=7, b=8, and c=9.
11. ABC\begin{align*}\triangle ABC\end{align*} if a=12, b=14, and c=4.
12. A farmer measures the three sides of a triangular field and gets 114, 165, and 257 feet. What is the measure of the largest angle of the triangle?
13. Using the information from the previous problem, what is the area of the field?

Another field is a quadrilateral where three sides measure 30, 50, and 60 yards, and two angles measure 130\begin{align*}130^\circ\end{align*} and 140\begin{align*}140^\circ\end{align*}, as shown below.

1. Find the area of the quadrilateral. Hint: divide the quadrilateral into two triangles and find the area of each.
2. Find the length of the fourth side.
3. Find the measures of the other two angles.

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 5.4.

### Vocabulary Language: English

Oblique Triangle

Oblique Triangle

An oblique triangle is a triangle without a right angle as one of its internal angles.
SAS Triangle

SAS Triangle

An SAS triangle is a triangle where two sides and the angle in between them are known quantities.

## Date Created:

Sep 26, 2012

Feb 26, 2015
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