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Common Core: High School - Geometry : Similarity, Right Triangles, & Trigonometry

Study concepts, example questions & explanations for Common Core: High School - Geometry

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All Common Core: High School - Geometry Resources

6 Diagnostic Tests 114 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #2 : Derive The Formula A = 1/2 Ab Sin(C) For The Area Of A Triangle By Drawing An Auxiliary Line From A Vertex Perpendicular To The Opposite Side

Is the following statement True or False?

In order to use this formula, you only need to know two of the triangles sides' lengths.

Possible Answers:

True

False

Correct answer:

False

Explanation:

In order to use this formula you need to either the height (which can be used to find the angle) or the angle (which can be used to find the height).

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Example Question #3 : Derive The Formula A = 1/2 Ab Sin(C) For The Area Of A Triangle By Drawing An Auxiliary Line From A Vertex Perpendicular To The Opposite Side

Given the triangle below, what is the formula for finding the area?

Screen shot 2020 08 07 at 2.48.01 pm

Possible Answers:

The area of triangle ABC is $A = \frac{1}{2}ah$ . Considering the entire triangle, $sin(C) = \frac{h}{a}$ . Solving for , h=a sin(C) . Therefore, $A = \frac{1}{2}(a^2)sin(C)$ .

The area of triangle ABC is $A = \frac{1}{2}ah$ . Only considering the right triangle ACD , $sin(C) = \frac{h}{b}$ . Solving for , h=b sin(C) . Therefore, $A = \frac{1}{2}(ab)sin(C)$ .

The area of triangle ABC is $A = \frac{1}{2}bh$ . Considering the right triangle ACD , $sin(C) = \frac{h}{a}$ . Solving for , h=a sin(C) . Therefore, $A = \frac{1}{2}(ab)sin(C)$ .

The area of triangle ABC is $A = \frac{1}{2}ah$ . Considering the entire triangle, $sin(C) = \frac{h}{c}$ . Solving for , h=c sin(C) . Therefore, $A = \frac{1}{2}(ac)sin(C)$ .

Correct answer:

The area of triangle ABC is $A = \frac{1}{2}ah$ . Only considering the right triangle ACD , $sin(C) = \frac{h}{b}$ . Solving for , h=b sin(C) . Therefore, $A = \frac{1}{2}(ab)sin(C)$ .

Explanation:

The formula for the area of a triangle is $\frac{1}{2}$ (base)(height). Since this is an obtuse triangle we need to break it into two right triangles by drawing the line down from the vertex perpendicular to the opposite side.

Now that we have two right triangles we can solve for the area of this triangle. Notice our base is and our height is . Plugging into the formula for the area of a triangle gives us $A = \frac{1}{2}ah$ . Most of the time, we will not have an exact length for , but we may have, or will be able to solve for, the lengths of a, b, c and the angles A, B, C . Using our sin relationship for right triangles, we know $sin(\theta) = \frac{opposite}{hypotenuse}$ . In this case we will use as our angle. So

$sin(C) = \frac{h}{b}$

b sin(C) = h

We can plug b sin(C) in for in our formula for area and we are left with $A = \frac{1}{2}ab(sin(C))$ .

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Example Question #4 : Derive The Formula A = 1/2 Ab Sin(C) For The Area Of A Triangle By Drawing An Auxiliary Line From A Vertex Perpendicular To The Opposite Side

Stephanie is building a triangular garden in her backyard. She has cut wood for the sides with lengths of 5, 4, and 3 feet. Knowing that the height from the vertex $A$ to the base formed by the line $CB$ is 1.778 feet, first find the area that will be in the garden and then calculate each angle that Stephanie will have to put the boards together in order for the lengths to fit just right. Round your answer to the third decimal place if needed.

Screen shot 2020 08 12 at 9.51.18 am

Possible Answers:

$A = 5.334, \angle A = 117.263, \angle B = 36.346, \angle C = 26.391$

$A = 5.334, \angle A = 110, \angle B = 45, \angle C = 25$

$A = 6.12, \angle A = 110, \angle B = 45, \angle C = 25$

$A = 6.12, \angle A = 117.263, \angle B = 36.346, \angle C = 26.391$

Correct answer:

$A = 5.334, \angle A = 117.263, \angle B = 36.346, \angle C = 26.391$

Explanation:

We must begin by finding the area first. To find the area we begin by using our equation $A = \frac{1}{2}ab(sin(C))$ and recall that $sin(C) = \frac{h}{b}$ . We will let:

a = 6

b = 4

$A = \frac{1}{2}ab(sin(C))$

$A = \frac{1}{2}ab\frac{h}{b} \text{because $sin(C) = \frac{h}{b}$}$

$A = \frac{1}{2}ah$

$A = \frac{1}{2}(6)(1.778)$

A = 5.334

So our area is 5.34ft^2

How to find $\angle C$ : a=6 , b=4 , A=5.334

$A = \frac{1}{2}ab(sin(C))$

2A = ab(sin(C))

$\frac{2A}{ab} = sin(C)$

$sin^{-1}(\frac{2A}{ab}) = C$

$sin^{-1}(\frac{2(5.334)}{(6)(4)} = C$

$sin^{-1}(\frac{10.668}{24} = C$

26.391 = C

How to find $\angle B$ : a=6 , b=3 , A=5.334

$A = \frac{1}{2}ab(sin(C))$

2A = ab(sin(C))

$\frac{2A}{ab} = sin(C)$

$sin^{-1}(\frac{2A}{ab}) = C$

$sin^{-1}(\frac{2(5.334)}{(6)(3)}) = C$

36.346 = C

To find $\angle A$ , we know the angles of a triangle should add up to 180 so:

$180 = \angle A + \angle B + \angle C$

$180 = 26.391 + 36.346 + \angle C$

$117.263 = \angle C$

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Example Question #5 : Derive The Formula A = 1/2 Ab Sin(C) For The Area Of A Triangle By Drawing An Auxiliary Line From A Vertex Perpendicular To The Opposite Side

Using the triangle below, find the area then solve for .

Screen shot 2020 08 13 at 8.23.50 am

Possible Answers:

A = 22, h = 6

A = 22, h = 4

A = 24, h = 6

A = 24, h = 4

Correct answer:

A = 24, h = 4

Explanation:

To solve for , we first need to find the area using our formula $A = \frac{1}{2}ab(sin(C))$ , considering the triangle ACD . We will begin by labeling our variables:

a = 12

b = 8

C = 30

To find the area, we plug the values for the variables into the formula.

$A = \frac{1}{2}(12)(8)sin(30)$

A = 24

Knowing that our area is 24, we can now solve for . Recall that:

$sin(C) = \frac{h}{b}$

$\Rightarrow bsin(c) = h$

It follows that:

$A = \frac{1}{2}ab(sin(C))$

$A = \frac{1}{2}ah$ (because bsin(C) = h )

$24 = \frac{1}{2}(12)h$ (plugging in our values for area and )

24 = (6)h

4 = h

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Example Question #6 : Derive The Formula A = 1/2 Ab Sin(C) For The Area Of A Triangle By Drawing An Auxiliary Line From A Vertex Perpendicular To The Opposite Side

Complete the statement by choosing from one of the following

Regarding the triangle below, to find the area using the formula $A = \frac{1}{2}ab(sin(C))$ , I assume that sin(C) = ______

Screen shot 2020 08 19 at 4.39.16 pm

Possible Answers:

$\frac{h}{B}$

$\frac{b}{h}$

$\frac{h}{c}$

$\frac{h}{b}$

Correct answer:

$\frac{h}{b}$

Explanation:

Recall that $sin(\theta) = \frac{opposite}{hypotenuse}$ for right triangles. When we draw a line down from the vertex to the opposite side this creates two right triangles within the original obtuse triangle. If we are solving for sin(C) , then becomes our hypotenuse and becomes the opposite side we are working with. Therefore $sin(C) = \frac{h}{b}$ .

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Example Question #1 : Derive The Formula A = 1/2 Ab Sin(C) For The Area Of A Triangle By Drawing An Auxiliary Line From A Vertex Perpendicular To The Opposite Side

Using the information from the following triangle, find the area. Round to the second decimal place.

Screen shot 2020 08 19 at 4.48.00 pm

Possible Answers:

A = 31. 64

A = 29.49

A = 25.67

A = 30.64

Correct answer:

A = 30.64

Explanation:

Using the formula $A = \frac{1}{2}ab(sin(C))$ , we will let a = 8, b = 10 and C = 130

$A = \frac{1}{2}ab(sin(C))$

$A = \frac{1}{2}(8)(10)sin(130)$

A = 30.64

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Example Question #2 : Derive The Formula A = 1/2 Ab Sin(C) For The Area Of A Triangle By Drawing An Auxiliary Line From A Vertex Perpendicular To The Opposite Side

Given the triangle below, find the area and round to the second decimal place (in degrees).

Screen shot 2020 08 20 at 8.46.15 am

Possible Answers:

A = 12.49

A = 20.19

A = 15.32

A = 18.39

Correct answer:

A = 18.39

Explanation:

Screen shot 2020 08 20 at 8.46.52 am

Following from the figure above, we draw an auxiliary line from the vertex perpendicular to the opposite side. We will call this line for height and will label the new vertex . The formula we will use is $A = \frac{1}{2}ab(sin(C))$ . First we begin by labeling our variables. We will let be the base of our triangle and be the hypotenuse of the triangle formed by BCD . The values for the variables are as follows:

a=8

b=6

C=50

Plugging these into our formula, it follows that:

$A = \frac{1}{2}(8)(6)sin(50)$

A = 18.39

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Example Question #1 : Derive The Formula A = 1/2 Ab Sin(C) For The Area Of A Triangle By Drawing An Auxiliary Line From A Vertex Perpendicular To The Opposite Side

You are making your friend a triangular fabric decoration for their home. It has ripped so you need to go buy new fabric. The fabric you are buying is $3 per square foot. Using the dimensions below, how much money do you need to buy enough fabric?

Screen shot 2020 08 20 at 11.11.05 am

Possible Answers:

$A = \$156$

$A = \$540$

$A = \$162$

$A = \$324$

Correct answer:

$A = \$162$

Explanation:

Using the formula $A = \frac{1}{2}ab(sin(C))$ , we will let a=18 and b=10 .

$A = \frac{1}{2}ab(sin(C))$

$A = \frac{1}{2}ab\frac{h}{b}$ (because $sin(C) = \frac{h}{b}$ )

$A = \frac{1}{2}(18)(10)\frac{6}{10}$

A = 54

Then we must account for each square foot costing $3.

54*3 = 162

The answer is $162.

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Common Core: High School - Geometry : Similarity, Right Triangles, & Trigonometry

Study concepts, example questions & explanations for Common Core: High School - Geometry

All Common Core: High School - Geometry Resources

Example Questions

Example Question #2 : Derive The Formula A = 1/2 Ab Sin(C) For The Area Of A Triangle By Drawing An Auxiliary Line From A Vertex Perpendicular To The Opposite Side

Example Question #3 : Derive The Formula A = 1/2 Ab Sin(C) For The Area Of A Triangle By Drawing An Auxiliary Line From A Vertex Perpendicular To The Opposite Side

Example Question #4 : Derive The Formula A = 1/2 Ab Sin(C) For The Area Of A Triangle By Drawing An Auxiliary Line From A Vertex Perpendicular To The Opposite Side

Example Question #5 : Derive The Formula A = 1/2 Ab Sin(C) For The Area Of A Triangle By Drawing An Auxiliary Line From A Vertex Perpendicular To The Opposite Side

Example Question #6 : Derive The Formula A = 1/2 Ab Sin(C) For The Area Of A Triangle By Drawing An Auxiliary Line From A Vertex Perpendicular To The Opposite Side

Example Question #1 : Derive The Formula A = 1/2 Ab Sin(C) For The Area Of A Triangle By Drawing An Auxiliary Line From A Vertex Perpendicular To The Opposite Side

Example Question #2 : Derive The Formula A = 1/2 Ab Sin(C) For The Area Of A Triangle By Drawing An Auxiliary Line From A Vertex Perpendicular To The Opposite Side

Example Question #1 : Derive The Formula A = 1/2 Ab Sin(C) For The Area Of A Triangle By Drawing An Auxiliary Line From A Vertex Perpendicular To The Opposite Side

All Common Core: High School - Geometry Resources

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