Triangles

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Questions 1 - 10

Find the perimeter.

Explanation

How do you find the perimeter of a right triangle?

There are three primary methods used to find the perimeter of a right triangle.

When side lengths are given, add them together.
Solve for a missing side using the Pythagorean theorem.
If we know side-angle-side information, solve for the missing side using the Law of Cosines.

Method 1:

This method will show you how to calculate the perimeter of a triangle when all sides lengths are known. Consider the following figure:

Screen shot 2016 07 07 at 11.31.44 am

If we know the lengths of sides , , and , then we can simply add them together to find the perimeter of the triangle. It is important to note several things. First, we need to make sure that all the units given match one another. Second, when all the side lengths are known, then the perimeter formula may be used on all types of triangles (e.g. right, acute, obtuse, equilateral, isosceles, and scalene). The perimeter formula is written formally in the following format:

$\textup{Perimeter}=a+b+c$

Method 2:

In right triangles, we can calculate the perimeter of a triangle when we are provided only two sides. We can do this by using the Pythagorean theorem. Let's first discuss right triangles in a general sense. A right triangle is a triangle that has one $90^{\circ}$ angle. It is a special triangle and needs to be labeled accordingly. The legs of the triangle form the $90^{\circ}$ angle and they are labeled and . The side of the triangle that is opposite of the $90^{\circ}$ angle and connects the two legs is known as the hypotenuse. The hypotenuse is the longest side of the triangle and is labeled as .

Screen shot 2016 07 07 at 10.13.54 am

If a triangle appears in this format, then we can use the Pythagorean theorem to solve for any missing side. This formula is written in the following manner:

We can rearrange it in a number of ways to solve for each of the sides of the triangle. Let's rearrange it to solve for the hypotenuse, .

Rearrange and take the square root of both sides.

$\sqrt{c^2}=\sqrt{a^2+b^2}$

Simplify.

$c=\sqrt{a^2+b^2}$

Now, let's use the Pythagorean theorem to solve for one of the legs, .

Subtract from both sides of the equation.

Take the square root of both sides.

$\sqrt{a^2}=\sqrt{c^2-b^2}$

Simplify.

$a=\sqrt{c^2-b^2}$

Last, let's use the Pythagorean theorem to solve for the adjacent leg, .

Subtract from both sides of the equation.

Take the square root of both sides.

$\sqrt{b^2}=\sqrt{c^2-a^2}$

Simplify.

$b=\sqrt{c^2-a^2}$

It is important to note that we can only use the following formulas to solve for the missing side of a right triangle when two other sides are known:

$c=\sqrt{a^2+b^2}$

$a=\sqrt{c^2-b^2}$

$b=\sqrt{c^2-a^2}$

After we find the missing side, we can use the perimeter formula to calculate the triangle's perimeter.

Method 3:

This method is the most complicated method and can only be used when we know two side lengths of a triangle as well as the measure of the angle that is between them. When we know side-angle-side (SAS) information, we can use the Law of Cosines to find the missing side. In order for this formula to accurately calculate the missing side we need to label the triangle in the following manner:

Screen shot 2016 07 07 at 12.58.14 pm

When the triangle is labeled in this way each side directly corresponds to the angle directly opposite of it. If we label our triangle carefully, then we can use the following formulas to find missing sides in any triangle given SAS information:

$a^2=b^2+c^2-2bc\cdot \cos A$

$b^2=a^2+c^2-2ac\cdot \cos B$

$c^2=a^2+b^2-2ab\cdot \cos C$

After, we calculate the right side of the equation, we need to take the square root of both sides in order to obtain the final side length of the missing side. Last, we need to use the perimeter formula to obtain the distance of the side lengths of the polygon.

Solution:

Now, that we have discussed the three methods used to calculate the perimeter of a triangle, we can use this information to solve the problem.

Recall how to find the perimeter of a triangle:

$\text{Perimeter}=a+b+c$

The given triangle has of the three sides needed. Use the Pythagorean theorem to find the length of the third side.

Recall the Pythagorean theorem:

Since we are finding the length of the hypotenuse, , rewrite the equation.

$c=\sqrt{a^2+b^2}$

Plug in the values of and .

$c=\sqrt{2^2+6^2}=\sqrt{40}$

Now, plug in all three values into the equation to find the perimeter. Use a calculator and round to decimal places.

$\text{Perimeter}=2+6+\sqrt{40}=14.32$

In ΔABC: a = 10, c = 15, and B = 50°.

Find b (to the nearest tenth).

11.5

9.7

12.1

11.9

14.3

Explanation

This problem requires us to use either the Law of Sines or the Law of Cosines. To figure out which one we should use, let's write down all the information we have in this format:

A = ? a = 10

B = 50° b = ?

C = ? c = 15

Now we can easily see that we do not have any complete pairs (such as A and a, or B and b) but we do have one term from each pair (a, B, and c). This tells us that we should use the Law of Cosines. (We use the Law of Sines when we have one complete pair, such as B and b).

Law of Cosines:
$a^2 = b^2 + c^2 - 2bc\cos(A)$
$b^2 = a^2 + c^2 - 2ac\cos(B)$
$c^2 = a^2 + b^2 - 2ab\cos(C)$

Since we are solving for b, let's use the second version of the Law of Cosines.
This gives us:

$b^2 = 10^2 + 15^2 - 2(10)(15)\cos(50)$

An equilateral triangle is placed together with a semicircle as shown by the figure below.

Find the perimeter of the figure.

Explanation

In order to find the perimeter of the entire figure, we will need to find the lengths of the segments highlighted in red.

Notice that the side length of the equilateral triangle is equal to the diameter of the semicircle.

Next, you should recall that the height of an equilateral triangle splits the triangle into two congruent triangles.

Recall that the side lengths in a triangle are in a $1:\sqrt3:2$ ratio. Thus, the radius of the circle, which is also the base of the triangle, the height of the triangle, and the side length of the triangle are in the same ratio.

We can then set up the following to determine the length of the side of the equilateral triangle:

$\frac{\text{height}}{\sqrt3}=\frac{\text{side}}{2}$

Rearrange the equation to solve for the length of the side.

$\text{side}=\frac{2(\text{height})}{\sqrt3}=\frac{2\sqrt3(\text{height})}{3}$

Plug in the length of the height to find the length of the side.

$\text{side}=\frac{2\sqrt3(3)}{3}=2\sqrt3$

Since the diameter of the semi-circle and the length of a side of the equilateral triangle are the same, we can write the following equation:

$\text{side}=\text{diameter}$

We have two sides of the equilateral triangle and the circumference of a semi-circle.

$\text{Perimeter}=2(\text{side})+\frac{\pi(\text{side})}{2}$

Plug in the length of the side to find the perimeter.

$\text{Perimeter}=2(2\sqrt3)+\frac{\pi(2\sqrt3)}{2}=4\sqrt3+\pi\sqrt3=12.37$

Equilateral

Refer to the above diagram. $\bigtriangleup ABC$ has perimeter 56.

True or false: $m \angle A = 60^{\circ }$

False

True

Explanation

Assume $m \angle A = 60^{\circ }$ . Then, since , it follows by the Isosceles Triangle Theorem that their opposite angles are also congruent. Since the measures of the angles of a triangle total $180^{\circ }$ , letting $t = m \angle B = m \angle C$ :

$m \angle A + m \angle B + m \angle C = 180^{\circ }$

$60^{\circ }+ t+t = 180^{\circ }$

$60^{\circ }+2t = 180^{\circ }$

$60^{\circ }+2t - 60^{\circ } = 180^{\circ } - 60^{\circ }$

$2t = 120^{\circ }$

$\frac{2t }{2}= \frac{120^{\circ }}{2}$

$t = 60^{\circ }$

All three angles have measure $60^{\circ }$ , making $\bigtriangleup ABC$ equiangular and, as a consequence, equilateral. Therefore, , and the perimeter, or the sum of the lengths of the sides, is

However, the perimeter is given to be 56. Therefore, $m \angle A \ne 60^{\circ }$ .

A circle with a radius of is inscribed in an equilateral triangle with side lengths of as shown in the figure below.

Find the area of the shaded region.

Explanation

In order to find the area of the shaded region, we must first find the area of the circle and the area of the equilateral triangle.

Recall how to find the area of a circle:

$\text{Area of Circle}=\pi\times\text{radius}^2$

Plug in the given radius to find the area of the circle.

$\text{Area of Circle}=\pi\times 2^2=4\pi$

Next, recall how to find the area of an equilateral triangle:

$\text{Area of Equilateral Triangle}=\frac{\sqrt3}{4}(\text{side})^2$

Plug in the length of the side of the triangle to find the area.

$\text{Area of Equilateral Triangle}=\frac{\sqrt3}{4}(6)^2=9\sqrt3$

In order to find the area of the shaded region, we will need to subtract the area of the circle from the area of the triangle.

$\text{Area of Shaded Region}=\text{Area of Triangle}-\text{Area of Circle}$

$\text{Area of Shaded Region}=9\sqrt3-4\pi=3.02$

Make sure to round to places after the decimal.

A circle is placed in an equilateral triangle as shown by the figure.

If the radius of the circle is , what is the area of the shaded region?

Explanation

In order to find the area of the shaded region, you must first find the areas of the triangle and circle.

First, recall how to find the area of an equilateral triangle:

$\text{Area of Equilateral Triangle}=\frac{\sqrt3}{4}(\text{side})^2$

Substitute in the value of the side to find the area of the triangle.

$\text{Area of Equilateral Triangle}=\frac{\sqrt3}{4}(12)^2$

$\text{Area of Equilateral Triangle}=48\sqrt3$

Next, recall how to find the area of a circle.

$\text{Area of Circle}=\pi\times\text{radius}^2$

Substitue in the value of the radius to find the area of the circle.

$\text{Area of Circle}=\pi\times 5^2$

$\text{Area of Circle}=25\pi$

Finally, find the area of the shaded region.

$\text{Area of Shaded Region}=\text{Area of Equilateral Triangle}-\text{Area of Circle}$

$\text{Area of Shaded Region}=48\sqrt3-25\pi$

Solve and round to two decimal places.

$\text{Area of Shaded Region}=4.60$

An equilateral triangle is placed on top of a regular hexagon as shown by the figure below.

Find the area of the entire figure.

Explanation

Recall that a regular hexagon can be divided into congruent equilateral triangles.

Since the extra equilateral triangle on top has the same side lengths as of the equilateral triangles made by dividing the hexagon, the entire figure has a total of congruent equilateral triangles.

Thus, to find the area of the entire figure, we must first find the area of the equilateral triangle.

Recall how to find the area of an equilateral triangle:

$\text{Area of Equilateral Triangle}=\frac{\sqrt3(\text{side})^2}{4}$

Plug in the length of a side of the equilateral triangle.

$\text{Area of Equilateral Triangle}=\frac{\sqrt3(\text{2})^2}{4}=\frac{4\sqrt3}{4}=\sqrt3$

Now, multiply this area by to find the area of the entire figure.

$\text{Area of Figure}=7(\text{Area of Equilateral Triangle})$

$\text{Area of Figure}=7(\sqrt3)=12.12$

Make sure to round to places after the decimal.

Right triangle

True or false: The provided triangle has the following perimeter:

True

False

Explanation

How do you find the perimeter of a right triangle?

There are three primary methods used to find the perimeter of a right triangle.

When side lengths are given, add them together.
Solve for a missing side using the Pythagorean theorem.
If we know side-angle-side information, solve for the missing side using the Law of Cosines.

Method 1:

This method will show you how to calculate the perimeter of a triangle when all sides lengths are known. Consider the following figure:

Screen shot 2016 07 07 at 11.31.44 am

$\textup{Perimeter}=a+b+c$

Method 2:

Screen shot 2016 07 07 at 10.13.54 am

If a triangle appears in this format, then we can use the Pythagorean theorem to solve for any missing side. This formula is written in the following manner:

We can rearrange it in a number of ways to solve for each of the sides of the triangle. Let's rearrange it to solve for the hypotenuse, .

Rearrange and take the square root of both sides.

$\sqrt{c^2}=\sqrt{a^2+b^2}$

Simplify.

$c=\sqrt{a^2+b^2}$

Now, let's use the Pythagorean theorem to solve for one of the legs, .

Subtract from both sides of the equation.

Take the square root of both sides.

$\sqrt{a^2}=\sqrt{c^2-b^2}$

Simplify.

$a=\sqrt{c^2-b^2}$

Last, let's use the Pythagorean theorem to solve for the adjacent leg, .

Subtract from both sides of the equation.

Take the square root of both sides.

$\sqrt{b^2}=\sqrt{c^2-a^2}$

Simplify.

$b=\sqrt{c^2-a^2}$

It is important to note that we can only use the following formulas to solve for the missing side of a right triangle when two other sides are known:

$c=\sqrt{a^2+b^2}$

$a=\sqrt{c^2-b^2}$

$b=\sqrt{c^2-a^2}$

After we find the missing side, we can use the perimeter formula to calculate the triangle's perimeter.

Method 3:

Screen shot 2016 07 07 at 12.58.14 pm

$a^2=b^2+c^2-2bc\cdot \cos A$

$b^2=a^2+c^2-2ac\cdot \cos B$

$c^2=a^2+b^2-2ab\cdot \cos C$

Solution:

Now, that we have discussed the three methods used to calculate the perimeter of a triangle, we can use this information to solve the problem.

The perimeter of a polygon is the sum of the lengths of its sides. We know the lengths of two sides of the triangle - the lengths of its legs, 9 and 12. We can find the length of the hypotenuse by way of the Pythagorean theorem:

$a^{2}+b^{2}=c^2$

Set , and solve for :

$c ^{2} = 9^{2}+12^{2}$

$c ^{2} = 81 + 144$

$c ^{2} = 225$

$c = \sqrt{225} = 15$

The perimeter is the sum of the lengths of the sides:

Which of these could NOT be the perimeter of a right triangle with sides of length and ?

$11 + \sqrt{51}$

$11 + \sqrt{11}$

$11 + \sqrt{61}$

All of these work

None of these work.

Explanation

How do you find the perimeter of a right triangle?

There are three primary methods used to find the perimeter of a right triangle.

When side lengths are given, add them together.
Solve for a missing side using the Pythagorean theorem.
If we know side-angle-side information, solve for the missing side using the Law of Cosines.

Method 1:

This method will show you how to calculate the perimeter of a triangle when all sides lengths are known. Consider the following figure:

Screen shot 2016 07 07 at 11.31.44 am

$\textup{Perimeter}=a+b+c$

Method 2:

Screen shot 2016 07 07 at 10.13.54 am

If a triangle appears in this format, then we can use the Pythagorean theorem to solve for any missing side. This formula is written in the following manner:

We can rearrange it in a number of ways to solve for each of the sides of the triangle. Let's rearrange it to solve for the hypotenuse, .

Rearrange and take the square root of both sides.

$\sqrt{c^2}=\sqrt{a^2+b^2}$

Simplify.

$c=\sqrt{a^2+b^2}$

Now, let's use the Pythagorean theorem to solve for one of the legs, .

Subtract from both sides of the equation.

Take the square root of both sides.

$\sqrt{a^2}=\sqrt{c^2-b^2}$

Simplify.

$a=\sqrt{c^2-b^2}$

Last, let's use the Pythagorean theorem to solve for the adjacent leg, .

Subtract from both sides of the equation.

Take the square root of both sides.

$\sqrt{b^2}=\sqrt{c^2-a^2}$

Simplify.

$b=\sqrt{c^2-a^2}$

It is important to note that we can only use the following formulas to solve for the missing side of a right triangle when two other sides are known:

$c=\sqrt{a^2+b^2}$

$a=\sqrt{c^2-b^2}$

$b=\sqrt{c^2-a^2}$

After we find the missing side, we can use the perimeter formula to calculate the triangle's perimeter.

Method 3:

Screen shot 2016 07 07 at 12.58.14 pm

$a^2=b^2+c^2-2bc\cdot \cos A$

$b^2=a^2+c^2-2ac\cdot \cos B$

$c^2=a^2+b^2-2ab\cdot \cos C$

Solution:

Now, that we have discussed the three methods used to calculate the perimeter of a triangle, we can use this information to solve the problem.

Since it wasn't specified if the two lengths given were the legs, or a leg and the hypotenuse, there are 2 options for the third side length.

Option 1:

$\sqrt{61 } = c$

To find the perimeter, add the following:

$5+6+ \sqrt{61} = 11 + \sqrt{61}$

Option 2:

$\sqrt{11 } = b$

To find the perimeter, add the following:

$5 +6 + \sqrt{11} = 11 + \sqrt{11}$

The only option listed that isn't either of those is $11 + \sqrt{51}$ , so that is the answer.

Which of the following could NOT be the perimeter of a right triangle with sides of length and ?

$11 + \sqrt{21}$

$11 + \sqrt{65}$

$11 + \sqrt{33}$

None of these work

All of these work

Explanation

How do you find the perimeter of a right triangle?

There are three primary methods used to find the perimeter of a right triangle.

When side lengths are given, add them together.
Solve for a missing side using the Pythagorean theorem.
If we know side-angle-side information, solve for the missing side using the Law of Cosines.

Method 1:

This method will show you how to calculate the perimeter of a triangle when all sides lengths are known. Consider the following figure:

Screen shot 2016 07 07 at 11.31.44 am

$\textup{Perimeter}=a+b+c$

Method 2:

Screen shot 2016 07 07 at 10.13.54 am

If a triangle appears in this format, then we can use the Pythagorean theorem to solve for any missing side. This formula is written in the following manner:

We can rearrange it in a number of ways to solve for each of the sides of the triangle. Let's rearrange it to solve for the hypotenuse, .

Rearrange and take the square root of both sides.

$\sqrt{c^2}=\sqrt{a^2+b^2}$

Simplify.

$c=\sqrt{a^2+b^2}$

Now, let's use the Pythagorean theorem to solve for one of the legs, .

Subtract from both sides of the equation.

Take the square root of both sides.

$\sqrt{a^2}=\sqrt{c^2-b^2}$

Simplify.

$a=\sqrt{c^2-b^2}$

Last, let's use the Pythagorean theorem to solve for the adjacent leg, .