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# 5.2: Area of a Triangle

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

A student will be able to:

• Understand how the area formula is derived.
• Apply the area formula to triangles where you know two sides and the included angle (SAS).
• Apply the are formula to triangles where you know all three sides (SSS).
• Understand Heron’s Formula.
• Use the area formulas in real-world and applied problems.

## Introduction

Real-World Application: The Pyramid Hotel recently installed a triangular pool. One side of the pool is $24\;\mathrm{feet}$, another side is $26\;\mathrm{feet}$, and the angle in between the two sides is $87^\circ$. If the hotel manager needs to order a cover for the pool, and the cost is $\35$ per square foot, how much can he expect to spend?

In this situation, we need to find the area of the surface of the pool in order to calculate the cost of the cover. We have already learned that the formula for the area of a triangle is $A = \frac{1} {2}\ bh$ where $b$ is the base of the triangle and $h$ is the height. The problem with this formula is that it can only be used when the height of the triangle is known. In this situation, we don’t know the height of the triangle formed by the sides of the pool. How do we find the area if we don’t know the height?

We will refer back to this application later on.

In this section, we will look at how we can derive a new formula using the area formula that we already know and the sine function. This new formula will allow us to find the area of a triangle when we don’t know the height. We will also look at when we can use this formula and how to apply it to real-world situations.

## Derive Area = 1/2 bcsinA

$\mathrm{Derive\ Area} = \frac{1} {2}\;\mathrm{bcsinA}$

We can use the area formula from above $(A = \frac{1} {2}\ \text{bh})$, as well as the sine function, to derive a new formula that can be used when the height is unknown.

In $\triangle ABC$ at the right, $BD$ is altitude from $B$ to $AC$. We will refer to the length of $BD$ as $h$ since It represents the height of the triangle. Also, we will refer to the area of the triangle as $K$ to avoid confusing the area with $\angle {A}.$

$K & = \frac{1} {2}\ bh & & \text{Area of a triangle}\\K & = \frac{1} {2}\ b(c\ \sin A) & & \sin A = \frac{h} {c}\ \text{therefore}\ c\ \sin A = h \\K & = \frac{1} {2}\ bc\ \sin A & & \text{Simplify}$

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

$K & = \frac{1} {2} bc\ \sin A \\K & = \frac{1} {2} ac\ \sin B \\K & = \frac{1} {2} ab\ \sin C$

## Find the Area Using Two Sides and an Included Angle--SAS (side-angle-side)

The formula $K = \frac{1} {2}\ bc\ \sin A$ 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 1:

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

$K & = \frac{1} {2} (31.6)(23.9) \sin 62 & & K = \frac{1} {2}\ ab\ \sin C \\K & \approx 333.4 & & \text{Evaluate}$

Answer: The area of the triangle is approximately $333.4.$

We will now refer back to the application at the beginning of the chapter.

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 $K = \frac{1} {2}\ bc\ \sin A$.

$K & = \frac{1} {2}\ (24)(26)\ \sin 87 & & K = \frac{1} {2}\ ab\ \sin C \\K & \approx 311.6 & & \text{Evaluate}$

$311.6\ \text{square\ feet} \times \35\ \text{per square foot} = \10,905.03$

Answer: The area of the pool cover is $311.6\;\mathrm{square\ feet.}$ The cost of the cover will be $\10,905.03.$

## Find the Area Using Three Sides--SSS (side-side-side) Heron’s Formula

In the last section, we learned how to find the area of an oblique triangle when we know two sides and the included angle using the formula $K = \frac{1} {2} bc\ \sin A$. We could also find the area of a triangle in which we know all three sides by first using the Law of Cosines to find one of the angles and then using the formula $K = \frac{1} {2} bc\ \sin A$. While this process works, it is time-consuming and requires a lot of calculation. Fortunately, we have another formula, called Heron’s Formula, which allows us to calculate the area of a triangle when we know all three sides.

Heron’s Formula:

$K = \sqrt{s (s - a)(s - b)(s - c)}$ where $s = \frac{1} {2}\ (a + b + c)$ or half of the perimeter of the triangle.

Example 2:

In $\triangle ABC, a = 23, b = 46$, and $C = 41$. Find the area of the triangle.

$s & = \frac{1} {2} (23 + 41 + 46) = 55 & & s = \frac{1} {2} (a + b + c) \\K & = \sqrt{55(55 - 23)(55 - 46)(55 - 41)} & & \text{Heron's Formula}\\K & = \sqrt{55(32)(9)(14)} & & \text{Subtract}\\K & = \sqrt{221760} & & \text{Multiply}\\K & \approx 470.9 & & \text{Square root}$

Answer: The area is approximately $470.9$.

Real-World Application:

Tile: A handyman is installing a tile floor in a kitchen. Since the corners of the kitchen are not exactly square, he needs to have special triangular shaped triangles made for the corners. One side of the tile needs to be $11.3 \;\mathrm(inch)$, the second side needs to be $11.9\;\mathrm(inch)$,and the third side is $13.6\;\mathrm(inch)$ If the tile costs $\4.89$ per square foot, and he needs four of them, how much will it cost to have the tiles made?

In order to find the cost of the tiles, we first need to find the area of one tile. Since we know the measurements of all three sides, we can use Heron’s Formula to calculate the area.

$s & = \frac{1} {2} (11.3 + 11.6 + 13.6) = 18.4 & & s = \frac{1} {2} (a + b + c) \\K & = \sqrt{18.4 (18.4 - 11.3)(18.4 - 11.9)(18.4 - 13.6)} & & \text{Heron's Formula}\\K & = \sqrt{18.4(7.1)(6.5)(4.8)} & & \text{Subtract}\\K & = \sqrt{40759.68} & & \text{Multiply}\\K & \approx 201.9\ \text{in}^2 & & \text{square root}$

The area of one tile would be $201.9\;\mathrm{square\ inches.}$ The cost of the tile is given to us in square feet, while the area of the tile is in square inches. In order to find the cost of one tile, we must first convert the area of the tile into square feet.

$\text {1 square foot} & = 12\ \text{in} \times 12\ \text{in} = 144\ \text{in}^2 & &\\\frac{201.9} {144} & = 1.4\ \text{ft}^2 & & \text{Covert square inches into square feet}\\1.4\ \text{ft}^2 \times 4.89 & = 6.89 & & \text{Cost for one tile is}\ \6.86 \\6.84 \times 4 & = 27.42 & & \text{The cost for four tiles is}\ \27.42$

Answer: The cost for four tiles would be $\27.42$.

## Applications, Technological Tools

We have already looked at two examples of situations where we can apply the two new area formulas we learned in this section. In this section, we will look at another real-world application where we know the area but need to find another part of the triangle, as well as an application involving a quadrilateral.

Real-World Application:

The jib sail on a sailboat came untied and the rope securing it was lost. If the area of the jib sail is $56.1\;\mathrm{square\ feet}$, use the figure and information at the right to find the length of the rope.

Since we know the area, one of the sides, and one angle of the jib sale, we can use the formula $K = \frac{1} {2}\ bc\ \sin A$ to find the side of the jib sale that is attached to the mast. We will call this side $y$.

$56.1 & = \frac{1} {2}\ 28(y)\ \sin 11 & & K = \frac{1} {2}\ bc\ \sin A \\56.1 & = 2.671325935\ y & & \text{Multiply}\\21.0 & = y & & \text{Divide}$

Now that we know side $y$, we know two sides and the included angle in the triangle formed by the mast, the rope, and the jib sail. We can now use the Law of Cosines to calculate the length of the rope.

$x^2 & = 21^2 + 27^2 - 2(21)(27) \cos 18 & & \text{Law of Cosines}\\x^2 & = 91.50191052 & & \text{Evaluate}\\x & \approx 9.6\ \text{ft} & & \text{Square root}$

Answer: The length of the rope is approximately $9.6\;\mathrm{feet}$.

Quadrilaterals: In quadrilateral $QUAD$ at the right, The area of $\triangle QUA = 5.64$, the area of $\triangle UAD = 6.39, \angle {QUD} = 31^\circ, \angle {UAD} = 40^\circ$, and $UD = 7.8$. Find the perimeter of $QUAD.$

In order to find the perimeter of $QUAD$, we need to know sides $QU, QD, UA,$ and $AD$. Since we know the area, one side, and one angle in each of the triangles, we can use $K = \frac{1} {2}\ bc\ \sin A$ to figure out $QU$ and $UA$.

$5.64 & = \frac{1} {2} (7.8)(QU) \text{sin 31} & & 6.39 = \frac{1} {2} (7.8) UA\ \text{sin 40} \\2.8 & \approx\ QU & & 2.5 \approx\ UA$

Now that we know $QU$ and $UA$, we know two sides and the included angle in each triangle (SAS). This means that we can use the Law of Cosines to find the other two sides, $QD$ and $DA$. First we will find $QD$.

$QD^2 & = 2.8^2 + 7.8^2 - 2(2.8)(7.8) \cos 31 & & \text{Law of Cosines}\\QD^2 & = 31.23893231 & & \text{Evaluate}\\QD & \approx 5.6 & & \text{Square root}$

Now, we will find $DA$.

$DA^2 & = 2.5^2 + 7.8^2 - 2(2.5)(7.8) \cos 40 & & \text{Law of Cosines}\\DA^2 & = 37.21426672 & & \text{\text{Evaluate}}\\DA & \approx 6.1 & & \text{Square root}$

Finally, we can calculate the perimeter since we have found all four sides of the quadrilateral.

$pQUAD = 2.8 + 5.6 + 6.1 + 2.5 = 17$

Answer: The perimeter of $QUAD$ is $17$.

## Points to Consider

1. Why can’t s (half of the perimeter) in Heron’s Formula be smaller than any of the three sides in the triangle?
2. How could we find the area of a triangle is AAS, SSA, and ASA cases?
3. Is it possible to figure out the length of the third side of a triangle if we know the other two sides and the area?

## Lesson Summary

• In an oblique triangle where we know two sides and the included angle, we can use the formula $K = \frac{1} {2}\ bc\ \sin A$ to calculate the area of the triangle.
• In an oblique triangle where we know all three sides of the triangle, we can calculate the area using Heron’s Formula.
• Given the area, we can use these two area formulas to find an unknown side or angle.
• We have explored three scenarios where we can use these area formulas in real-world situations. We will look at more applications in the review questions.

## Review Questions

1. Using the figures and given information below, determine which formula you would need to use in order to find the area of each triangle ($A = 1/2\;\mathrm{bh}$, $K = 1/2\;\mathrm{bc\ sinA}$, or Heron’s Formula).
Given Figure Formula
a. $CF = 3, FM = 8,$ and $CO = 5$
b. $HC = 4.1, CE = 7.4,$ and $HE = 9.6$
c. $AP = 59.8, PH = 86.3, \angle {\mathrm{APH}} = 103^\circ$
d. $RX = 11.1, XE = 18.9, \angle {\mathrm{R}} = 41^\circ$
1. Find the area of all of the triangles in the chart above to the nearest tenth.
2. Using the given information and the figures below, decide which area formula you would need to use to find each side, angle, or area.
Given Figure Find Formula
a. $\mathrm{Area} = 1618.98, b = 36.3$ $h$
b. $\mathrm{Area} = 387.6, b = 25.6, c = 32.9$ $\angle {\mathrm{A}}$
c. $\mathrm{Area} \triangle ABD = 16.96, AD = 3.2, \angle {\mathrm{DBC}} = 49.6^\circ$ $\mathrm{Area \ of} \triangle ABC$
1. Using the figures and information from the table above, find the angle, side, or area requested.
2. The Pyramid Hotel is planning on repainting the exterior of the building. The building has four sides that are isosceles triangles with bases measuring $590\;\mathrm{ft}$ and legs measuring $375\;\mathrm{ft}$.
1. What is the total area that needs to be painted?
2. If one gallon of paint covers $25\;\mathrm{square\ feet}$, how many gallons of paint are needed?

3. A contractor needs to replace a triangular section of roof on the front of a house. The sides of the triangle are $8.2\;\mathrm{feet},14.6\;\mathrm{feet}$, and $16.3\;\mathrm{feet}$. If one bundle of shingles covers $33\ \frac{1} {3}\;\mathrm{square\ feet}$ and costs $\15.45$, how many bundles does he need to purchase? How much will the shingles cost him? How much of the bundle will go to waste?
4. A farmer needs to replant a triangular section of crops that died unexpectedly. One side of the triangle measures $186\;\mathrm{yards}$ , another measures $205\;\mathrm{yards}$, and the angle formed by these two sides is $148^\circ$.
1. 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 $205\;\mathrm{yards}$ now measures $288\;\mathrm{yards}$. How much has the area that needs to be replanted increased by?

5. Find the perimeter of the quadrilateral at the left If the area of $\triangle DEG = 56.5$ and the area of $\triangle EGF = 84.7$.
6. In $\triangle ABC, BD$ is an altitude from $B$ to $AC$. The area of $\triangle ABC = 232.96, AB = 16.2$, and $AD = 14.4$. Find $DC$.
7. Show that in any triangle $DEF$, $d^2 + e^2 + f^2 = 2(ef \cos D + df \cos E + de \cos F)$.

1. $A = 1/2\;\mathrm{bh}$
2. Heron’s formula
3. $K = 1/2\;\mathrm{bc\ sinA}$
4. $A = 1/2\;\mathrm{bh}$
1. $A = 22$
2. $A = 14.3$
3. $= 2514.2$
4. $= 144.7$
1. $A = 1/2\;\mathrm{bh}$
2. $K = 1/2\;\mathrm{bc\ sinA}$
3. $A = 1/2\;\mathrm{bh}$
1. $h = 89.2$
2. $\angle {\mathrm{A}} = 67^\circ$
3. $\mathrm{Area\ of}\ \triangle ABC = 82.5$
1. The total area is $419,550\;\mathrm{square\ feet.}$
2. $16,782\;\mathrm{gallons}$ of paint are needed.
1. He will need $2\;\mathrm{bundles}$. The shingles will cost him $\30.90$. $6.9\;\mathrm{square\ feet}$ will go to waste.
1. The area that needs to be replaced is $10.876.4\;\mathrm{square\ yards.}$
2. The area has increased by $4079.2\;\mathrm{yards.}$
2. The perimeter of the quadrilateral is $50.5$.
3. $DC$ is approximately $24.94$.
4. $d^2 & = e^2 + f - 2ef \cos D \\e^2 & = d^2 + f - 2df \cos E \\f^2 & = d^2 + e^2 - 2de \cos F \\d^2 + e^2 + f^2 & = e^2 + f^2 - 2ef \cos D + d^2 + f^2 - 2df \cos E + d^2 + e^2 - 2de \cos F \\d^2 + e^2 + f^2 & = 2(d^2 + e^2 + f^2) - 2(ef \cos D + df \cos E + de \cos F) \\- (d^2 + e^2 + f^2) & = -2(ef \cos D + df \cos E + de \cos F) \\d^2 + e^2 + f^2 & = 2(ef \cos D + df \cos E + de \cos F)$

## Vocabulary

Heron’s Formula
A formula used to calculate the area of a triangle when all three sides are known.

## Date Created:

Feb 23, 2012

Aug 22, 2014
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