# 5.6: Determination of Unknown Triangle Measures Given Area

**At Grade**Created by: CK-12

**Practice**Determination of Unknown Triangle Measures Given Area

You are working on creating a mobile for your art class. A mobile is a piece of art that has a rod with different shapes hanging from it, so they can spin.

To create your project, you need to cut a set of triangles that have a variety of sizes. You are about to start cutting triangles, when your friend, who is helping you with the project, comes over. She tells you that each piece needs to have a rod through the side of it to balance the shape in a certain way. She wants you to make a piece that looks like this:

You have already cut a triangle by cutting a piece out of construction paper. You know that one side of your triangle is 6 inches long, but you don't know the length of the other two sides. Can you use the information you have to find the length of \begin{align*}b\end{align*}

By the end of this Concept, you'll be able to solve this problem.

### Watch This

How do you find the height of a triangle if you know the area and the base

### Guidance

In this section, we will look at situations where we know the area but need to find another part of the triangle, as well as an application involving a quadrilateral. All of this will involve the use of the Law of Cosines, Law of Sines, and the Alternate Formula for the Area of a Triangle.

#### Example A

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 square feet, use the figure and information below to find the length of the rope.

**Solution:** Since we know the area, one of the sides, and one angle of the jib sail, we can use the formula \begin{align*}K = \frac{1}{2} \ bc \sin A\end{align*}

\begin{align*}56.1 & = \frac{1}{2} \ 28(y)\sin 11 \\
56.1 & = 2.671325935\ y \\
21.0 & = y\end{align*}

Now that we know side \begin{align*}y\end{align*}

\begin{align*}x^2 & = 21^2 + 27^2 - 2(21)(27) \cos 18 \\
x^2 & = 91.50191052 \\
x & \approx 9.6\ ft\end{align*}

The length of the rope is approximately 9.6 feet.

#### Example B

In quadrilateral \begin{align*}QUAD\end{align*}

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

\begin{align*}5.64 & = \frac{1}{2}(7.8)(QU)\sin 31 && 6.39 = \frac{1}{2}(7.8)UA\sin 40 \\ 2.8 & \approx QU && 2.5 \approx UA\end{align*}

Now that we know \begin{align*}QU\end{align*} and \begin{align*}UA\end{align*}, 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, \begin{align*}QD\end{align*} and \begin{align*}DA\end{align*}. First we will find \begin{align*}QD\end{align*} and \begin{align*}DA\end{align*}.

\begin{align*}QD^2 & = 2.8^2 + 7.8^2 - 2(2.8)(7.8) \cos 31 && DA^2 = 2.5^2 + 7.8^2 - 2(2.5)(7.8) \cos 40 \\ QD^2 & = 31.23893231 && DA^2 = 37.21426672 \\ QD & \approx 5.6 && DA \approx 6.1\end{align*}

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

\begin{align*}pQUAD = 2.8 + 5.6 + 6.1 + 2.5 = 17 \end{align*}

#### Example C

In \begin{align*}\triangle{ABC}, BD\end{align*} is an altitude from \begin{align*}B\end{align*} to \begin{align*}AC\end{align*}. The area of \begin{align*}\triangle{ABC} = 232.96, AB = 16.2\end{align*}, and \begin{align*}AD = 14.4\end{align*}. Find \begin{align*}DC\end{align*}.

**Solution:**

First, find \begin{align*}BD\end{align*} by using the Pythagorean Theorem. \begin{align*}BD = \sqrt{16.2^2 - 14.4^2} = 7.42\end{align*}. Then, using the area and formula \begin{align*}(A = \frac{1}{2}bh)\end{align*}, you can find \begin{align*}AC\end{align*}. \begin{align*}232.96 = \frac{1}{2}(7.42)AC \rightarrow AC = 62.78\end{align*}. \begin{align*}DC = 62.78 - 14.4 = 48.38\end{align*}.

### Vocabulary

**Law of Cosines:** The ** law of cosines** is an equation relating the length of one side of a triangle to the lengths of the other two sides and the sine of the angle included between the other two sides.

**Law of Sines:** The ** law of sines** is an equation relating the sine of an interior angle of a triangle divided by the side opposite that angle to a different interior angle of the same triangle divided by the side opposite that second angle.

### Guided Practice

1. Find "h" in the triangle below: Area \begin{align*}= 1618.98, b = 36.3\end{align*}

2. Find \begin{align*}\angle A \end{align*} in the triangle below: Area \begin{align*}= 387.6, b = 25.6, c = 32.9\end{align*}

3. Find the area of \begin{align*}\triangle ABC \end{align*} below: Area \begin{align*}\triangle{ABD} = 16.96, AD = 3.2, \angle{DBC} = 49.6^\circ\end{align*}

**Solutions:**

1. Since we know the area, one of the sides (18.15), and one angle of the triangle (\begin{align*}45^\circ\end{align*}), we can use the formula \begin{align*}K = \frac{1}{2} \ bc \sin A\end{align*} to find the other side of the triangle. We can then use the Pythagorean Theorem to find the height of the triangle.

This gives a result of:

\begin{align*}h = 89.2\end{align*}

2. Since we know the area and the lengths of two of the sides of the triangle, we can use the formula \begin{align*}K = \frac{1}{2} \ bc \sin A\end{align*} to solve for the included angle, which gives:

\begin{align*}\angle{A} = 67^\circ\end{align*}

3. Area of \begin{align*}\triangle{ABC} = 83.0\end{align*}

### Concept Problem Solution

Since you know that the mobile piece is six inches on one side, and that the area of the triangle is \begin{align*}25 in^2\end{align*}, you can use the formula \begin{align*}K = \frac{1}{2}ab\sin C\end{align*} to find the length of the other side:

\begin{align*} K = \frac{1}{2}ab\sin C\\ 25 = \frac{1}{2}(6)(b) \sin 35^\circ\\ 25 = 1.72b\\ b = \frac{25}{1.72}\\ b = 14.53 in \end{align*}

### Practice

- The area of the triangle below is \begin{align*}138 in^2\end{align*}. Solve for x, the height.
- The area of the triangle below is \begin{align*}250 cm^2\end{align*}. A height is given on the diagram. Solve for x.

Use the triangle below for questions 3-5. The area of the large triangle is \begin{align*}65 cm^2\end{align*}.

- Solve for x.
- Find the perimeter of the large triangle.
- Find the measure of all three angles of the large triangle.

Use the triangle below for questions 6-8. The area of the triangle is \begin{align*}244 cm^2\end{align*}.

- Solve for \begin{align*}\theta\end{align*}.
- Solve for x.
- Find the perimeter of the triangle.

Use the triangle below for questions 9-11. The area of the triangle is \begin{align*}299.8 in^2\end{align*}.

- Solve for x.
- Solve for y.
- Find the measure of the other two angles of the triangle.

Use the triangle below for questions 12-15. The area of the large triangle is \begin{align*}84 in^2\end{align*}.

- Solve for x.
- Solve for y.
- Solve for z.
- Solve for \begin{align*}\theta\end{align*}.

### Notes/Highlights Having trouble? Report an issue.

Color | Highlighted Text | Notes | |
---|---|---|---|

Please Sign In to create your own Highlights / Notes | |||

Show More |

law of cosines

The law of cosines is a rule relating the sides of a triangle to the cosine of one of its angles. The law of cosines states that , where is the angle across from side .law of sines

The law of sines is a rule applied to triangles stating that the ratio of the sine of an angle to the side opposite that angle is equal to the ratio of the sine of another angle in the triangle to the side opposite that angle.### Image Attributions

Here you'll learn how to apply rules for finding unknown triangle quantities when the area is known.