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Intermediate Geometry : Triangles

Study concepts, example questions & explanations for Intermediate Geometry

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Example Questions

Example Question #591 : Plane Geometry

A triangle is placed in a parallelogram so that they share a base.

If the height of the triangle is half the height of the parallelogram, find the area of the shaded region.

Possible Answers:

Correct answer:

Explanation:

In order to find the area of the shaded region, we will need to find the areas of the triangle and of the parallelogram.

First, recall how to find the area of a parallelogram.

$\text{Area of Parallelogram}=\text{base}\times\text{height}$

$\text{Area of Parallelogram}=10 \times 6 = 60$

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

$\text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})$

Now, find the height of the triangle.

$\text{Height of Triangle}=\frac{\text{Height of Parallelogram}}{2}$

$\text{Height of Triangle}=\frac{6}{2}=3$

Plug this value in to find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(10 \times 3)=15$

Subtract the two areas to find the area of the shaded region.

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

$\text{Area of Shaded Region}=60-15=45$

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Example Question #592 : Plane Geometry

A triangle is placed in a parallelogram so that they share a base.

If the height of the triangle is half the height of the parallelogram, find the area of the shaded region.

Possible Answers:

508

496

522

532

Correct answer:

522

Explanation:

In order to find the area of the shaded region, we will need to find the areas of the triangle and of the parallelogram.

First, recall how to find the area of a parallelogram.

$\text{Area of Parallelogram}=\text{base}\times\text{height}$

$\text{Area of Parallelogram}=29\times 24 = 696$

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

$\text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})$

Now, find the height of the triangle.

$\text{Height of Triangle}=\frac{\text{Height of Parallelogram}}{2}$

$\text{Height of Triangle}=\frac{24}{2}=12$

Plug this value in to find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(29\times 12)=174$

Subtract the two areas to find the area of the shaded region.

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

$\text{Area of Shaded Region}=696-174=522$

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Example Question #591 : Intermediate Geometry

A triangle is placed in a parallelogram so that they share a base.

If the height of the triangle is one-fifth that of the parallelogram, find the area of the shaded region.

Possible Answers:

650

675

700

625

Correct answer:

675

Explanation:

In order to find the area of the shaded region, we will need to find the areas of the triangle and of the parallelogram.

First, recall how to find the area of a parallelogram.

$\text{Area of Parallelogram}=\text{base}\times\text{height}$

$\text{Area of Parallelogram}=30 \times 25 = 750$

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

$\text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})$

Now, find the height of the triangle.

$\text{Height of Triangle}=\frac{\text{Height of Parallelogram}}{5}$

$\text{Height of Triangle}=\frac{25}{5}=5$

Plug this value in to find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(30 \times 5)=75$

Subtract the two areas to find the area of the shaded region.

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

$\text{Area of Shaded Region}=750-75=675$

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Example Question #593 : Plane Geometry

A triangle is placed in a parallelogram so that they share a base.

If the height of the triangle is half that of the parallelogram, find the area of the shaded region.

Possible Answers:

795

775

755

765

Correct answer:

765

Explanation:

In order to find the area of the shaded region, we will need to find the areas of the triangle and of the parallelogram.

First, recall how to find the area of a parallelogram.

$\text{Area of Parallelogram}=\text{base}\times\text{height}$

$\text{Area of Parallelogram}=34\times 30 =1020$

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

$\text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})$

Now, find the height of the triangle.

$\text{Height of Triangle}=\frac{\text{Height of Parallelogram}}{2}$

$\text{Height of Triangle}=\frac{30}{2}=15$

Plug this value in to find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(34 \times 15)=255$

Subtract the two areas to find the area of the shaded region.

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

$\text{Area of Shaded Region}=1020-255=765$

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Example Question #155 : Triangles

A triangle is placed in a parallelogram so that they share a base.

If the height of the triangle is one-ninth that of the parallelogram, find the area of the shaded region.

Possible Answers:

2100

2150

2175

2125

Correct answer:

2125

Explanation:

In order to find the area of the shaded region, we will need to find the areas of the triangle and of the parallelogram.

First, recall how to find the area of a parallelogram.

$\text{Area of Parallelogram}=\text{base}\times\text{height}$

$\text{Area of Parallelogram}=50 \times45 = 2250$

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

$\text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})$

Now, find the height of the triangle.

$\text{Height of Triangle}=\frac{\text{Height of Parallelogram}}{9}$

$\text{Height of Triangle}=\frac{45}{9}=5$

Plug this value in to find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(50 \times 5)=125$

Subtract the two areas to find the area of the shaded region.

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

$\text{Area of Shaded Region}=2250-125=2125$

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Example Question #156 : Triangles

A triangle is placed in a parallelogram so that they share a base.

If the height of the triangle is half that of the parallelogram, find the area of the shaded region.

Possible Answers:

1700

1900

1812

1716

Correct answer:

1716

Explanation:

In order to find the area of the shaded region, we will need to find the areas of the triangle and of the parallelogram.

First, recall how to find the area of a parallelogram.

$\text{Area of Parallelogram}=\text{base}\times\text{height}$

$\text{Area of Parallelogram}=52\times 44 = 2288$

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

$\text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})$

Now, find the height of the triangle.

$\text{Height of Triangle}=\frac{\text{Height of Parallelogram}}{2}$

$\text{Height of Triangle}=\frac{44}{2}=22$

Plug this value in to find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(52\times 22)=572$

Subtract the two areas to find the area of the shaded region.

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

$\text{Area of Shaded Region}=2288-572=1716$

Report an Error

Example Question #157 : Triangles

A triangle is placed in a parallelogram so that they share a base.

If the height of the triangle is half that of the parallelogram, find the area of the shaded region.

Possible Answers:

510

490

480

500

Correct answer:

510

Explanation:

In order to find the area of the shaded region, we will need to find the areas of the triangle and of the parallelogram.

First, recall how to find the area of a parallelogram.

$\text{Area of Parallelogram}=\text{base}\times\text{height}$

$\text{Area of Parallelogram}=34\times 20 = 680$

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

$\text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})$

Now, find the height of the triangle.

$\text{Height of Triangle}=\frac{\text{Height of Parallelogram}}{2}$

$\text{Height of Triangle}=\frac{20}{2}=10$

Plug this value in to find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(340 \times 10)=170$

Subtract the two areas to find the area of the shaded region.

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

$\text{Area of Shaded Region}=680-170=510$

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Example Question #1 : How To Find The Length Of The Side Of An Acute / Obtuse Triangle

In ΔABC: a = 10, c = 15, and B = 50°.

Find b (to the nearest tenth).

Possible Answers:

9.7

14.3

12.1

11.9

11.5

Correct answer:

11.5

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:

$b^2 = 10^2 + 15^2 - 2(10)(15)\cos(50)$
b^2 = 100 + 225 - 192.836
b = 11.5

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Example Question #1 : How To Find The Length Of The Side Of An Acute / Obtuse Triangle

In ΔABC, A = 35°, b = 8, c = 12.

Find a (to the nearest tenth).

Possible Answers:

6.8

8.3

5.3

7.1

9.0

Correct answer:

7.1

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 = 35° a = ?

B = ? b = 8

C = ? c = 12

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 a, let's use the first version of the Law of Cosines.
This gives us:

$a^2 = 8^2 + 12^2 - 2(8)(12)\cos(35)$
b^2 = 64 + 144 - 157.277
b = 7.1

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Example Question #159 : Triangles

In ΔABC, A = 25°, B = 50°, a = 17.

Find b (to the nearest tenth).

Possible Answers:

33.2

28.5

25.1

29.8

30.8

Correct answer:

30.8

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 = 25° a = 17

B = 50° b = ?

C = ? c = ?

Now we can easily see that we have a complete pair, A and a. This tells us that we can use the Law of Sines. (We use the Law of Cosines when we do not have a complete pair).

Law of Sines:
$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

To solve for b, we can use the first two terms which gives us:
$\frac{\sin 25°}{17}=\frac{\sin 50°}{b}$
$b=\sin 50*\frac{17}{\sin 25}$
b=30.8

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Intermediate Geometry : Triangles

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

Example Question #591 : Plane Geometry

Example Question #592 : Plane Geometry

Example Question #591 : Intermediate Geometry

Example Question #593 : Plane Geometry

Example Question #155 : Triangles

Example Question #156 : Triangles

Example Question #157 : Triangles

Example Question #1 : How To Find The Length Of The Side Of An Acute / Obtuse Triangle

Example Question #1 : How To Find The Length Of The Side Of An Acute / Obtuse Triangle

Example Question #159 : Triangles

All Intermediate Geometry Resources

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