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Intermediate Geometry : Intermediate Geometry

Study concepts, example questions & explanations for Intermediate Geometry

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8 Diagnostic Tests 250 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #8 : Triangles

The perimeter of an isosceles triangle is . If the length of the base is five more than one-fourth the length of a leg, what is the height of the triangle?

Possible Answers:

18.45

15.20

19.36

19.01

Correct answer:

19.36

Explanation:

First, find the lengths of the triangle.

Let be the length of each leg. Then, the length of the base must be $\frac{1}{4}x+5$ .

Use the information given about the perimeter to solve for .

$x+x+\frac{1}{4}x+5=50$

$\frac{9}{4}x+5=50$

$\frac{9}{4}x=45$

x=20

Plug this value in to find the length of the base.

$\text{Length of Base}=\frac{1}{4}(20)+5=10$

Now, recall that the height of an isosceles triangle can split the entire triangle into two congruent right triangle as shown by the figure below.

Thus, we can use the Pythagorean Theorem to find the length of the height.

$\text{Height}=\sqrt{\text{leg}^2-\text{half of base}^2}$

Plug in the given values to find the height of the triangle.

$\text{Height}=\sqrt{20^2-5^2}=\sqrt{375}=19.36$

Make sure to round to places after the decimal.

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Example Question #9 : Triangles

The perimeter of an isosceles triangle is . If the length of the base is ten more than one-eighth the length of a leg, what is the height of the triangle?

Possible Answers:

10.21

15.28

13.41

14.83

Correct answer:

14.83

Explanation:

First, find the lengths of the triangle.

Let be the length of each leg. Then, the length of the base must be $\frac{1}{8}x+10$ .

Use the information given about the perimeter to solve for .

$x+x+\frac{1}{8}x+10=44$

$\frac{17}{8}x+10=44$

$\frac{17}{8}x=34$

x=16

Plug this value in to find the length of the base.

$\text{Length of Base}=\frac{1}{8}(16)+10=12$

Now, recall that the height of an isosceles triangle can split the entire triangle into two congruent right triangle as shown by the figure below.

Thus, we can use the Pythagorean Theorem to find the length of the height.

$\text{Height}=\sqrt{\text{leg}^2-\text{half of base}^2}$

Plug in the given values to find the height of the triangle.

$\text{Height}=\sqrt{16^2-6^2}=\sqrt{220}=14.83$

Make sure to round to places after the decimal.

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Example Question #11 : Triangles

The perimeter of an isosceles triangle is . If the length of the base is eleven less than twice the length of a leg, what is the height of the triangle?

Possible Answers:

6.84

6.65

6.73

6.99

Correct answer:

6.84

Explanation:

First, find the lengths of the triangle.

Let be the length of each leg. Then, the length of the base must be 2x-11 .

Use the information given about the perimeter to solve for .

x+x+2x-11=17

4x-11=17

4x=28

x=7

Plug this value in to find the length of the base.

$\text{Length of Base}=2(7)-11=3$

Now, recall that the height of an isosceles triangle can split the entire triangle into two congruent right triangle as shown by the figure below.

Thus, we can use the Pythagorean Theorem to find the length of the height.

$\text{Height}=\sqrt{\text{leg}^2-\text{half of base}^2}$

Plug in the given values to find the height of the triangle.

$\text{Height}=\sqrt{7^2-1.5^2}=\sqrt{46.75}=6.84$

Make sure to round to places after the decimal.

Report an Error

Example Question #12 : Triangles

The perimeter of an isosceles triangle is . If the length of the base is fifteen more than one-third the length of a leg, what is the height of the triangle?

Possible Answers:

11.18

10.09

13.09

12.76

Correct answer:

11.18

Explanation:

First, find the lengths of the triangle.

Let be the length of each leg. Then, the length of the base must be $\frac{1}{3}x+15$ .

Use the information given about the perimeter to solve for .

$x+x+\frac{1}{3}x+15=50$

$\frac{7}{3}x+15=50$

$\frac{7}{3}x=35$

x=15

Plug this value in to find the length of the base.

$\text{Length of Base}=\frac{1}{3}(15)+15=20$

Now, recall that the height of an isosceles triangle can split the entire triangle into two congruent right triangle as shown by the figure below.

Thus, we can use the Pythagorean Theorem to find the length of the height.

$\text{Height}=\sqrt{\text{leg}^2-\text{half of base}^2}$

Plug in the given values to find the height of the triangle.

$\text{Height}=\sqrt{15^2-10^2}=\sqrt{125}=11.18$

Make sure to round to places after the decimal.

Report an Error

Example Question #13 : Triangles

An isosceles triangle has a perimeter of . If the length of the base is seven more than one-seventh of the length of a leg, what is the height of the triangle?

Possible Answers:

5.60

6.09

6.93

5.82

Correct answer:

6.93

Explanation:

First, find the lengths of the triangle.

Let be the length of each leg. Then, the length of the base must be $\frac{1}{7}x+7$ .

Use the information given about the perimeter to solve for .

$x+x+\frac{1}{7}x+7=22$

$\frac{15}{7}x+7=22$

$\frac{15}{7}x=15$

x=7

Plug this value in to find the length of the base.

$\text{Length of Base}=\frac{1}{7}(7)+1=2$

Now, recall that the height of an isosceles triangle can split the entire triangle into two congruent right triangle as shown by the figure below.

Thus, we can use the Pythagorean Theorem to find the length of the height.

$\text{Height}=\sqrt{\text{leg}^2-\text{half of base}^2}$

Plug in the given values to find the height of the triangle.

$\text{Height}=\sqrt{7^2-1^2}=\sqrt{33}=6.93$

Make sure to round to places after the decimal.

Report an Error

Example Question #11 : Acute / Obtuse Isosceles Triangles

An isosceles triangle has two legs of length 10" with a base of unknown length, and has a height of 6". Find the area.

Possible Answers:

$48\ in^2$

$64\ in^2$

$36\ in^2$

$60\ in^2$

$32\ in^2$

Correct answer:

$48\ in^2$

Explanation:

To find the area:

$\frac{1}{2}\cdot base\cdot height$

The information giving us the two sides helps us find the base.

Using the fact that an isosceles triangle can be split vertically down the middle (note: the length of this extra line will be equal to the height) to form two identical right triangles, we then use the Pythagorean Theorem to find the base:

a^2 + b^2 = c^2

For our problem, is the height, is the base (of ONE of the right triangles; the base of the isosceles triangle will be twice as big) and is the hypotenuse, or the leg of length 10".

We find that the base is 8", so the base of the isosceles triangle is 16".

Plugging in our numbers we get:

$16\cdot 6\cdot \frac{1}{2}=48\ in^{2}$

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Example Question #2 : How To Find The Area Of An Acute / Obtuse Isosceles Triangle

An isosceles triangle is placed in a circle as shown by the figure below.

If diameter of the circle is , find the area of the shaded region.

Possible Answers:

4.22

2.14

3.19

1.93

Correct answer:

2.14

Explanation:

From the given image, you should notice that the base of the triangle is also the diameter of the circle. In addition, the height of the triangle is also the radius of the circle.

Thus, we can find the area of the triangle.

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

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

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

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

$\text{Area of Circle}=\pi\times 1^2= \pi$

To find the area of the shaded region, subtract the two areas.

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

$\text{Area of Shaded Region}=\pi-1=2.14$

Make sure to round to places after the decimal.

Report an Error

Example Question #3 : How To Find The Area Of An Acute / Obtuse Isosceles Triangle

An isosceles triangle is placed in a circle as shown by the figure below.

If the diameter of the circle is , find the area of the shaded region.

Possible Answers:

60.92

77.10

75.59

91.21

Correct answer:

77.10

Explanation:

From the given image, you should notice that the base of the triangle is also the diameter of the circle. In addition, the height of the triangle is also the radius of the circle.

Thus, we can find the area of the triangle.

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

$\text{Area of Triangle}=\frac{1}{2}(12 \times 6)=36$

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

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

$\text{Area of Circle}=\pi\times 6^2= 36\pi$

To find the area of the shaded region, subtract the two areas.

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

$\text{Area of Shaded Region}=36\pi-36=77.10$

Make sure to round to places after the decimal.

Report an Error

Example Question #4 : How To Find The Area Of An Acute / Obtuse Isosceles Triangle

An isosceles triangle is placed in a circle as shown by the figure below.

If the diameter of the circle is , find the area of the shaded region.

Possible Answers:

98.87

112.52

104.94

102.05

Correct answer:

104.94

Explanation:

From the given image, you should notice that the base of the triangle is also the diameter of the circle. In addition, the height of the triangle is also the radius of the circle.

Thus, we can find the area of the triangle.

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

$\text{Area of Triangle}=\frac{1}{2}(14 \times 7)=49$

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

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

$\text{Area of Circle}=\pi\times 7^2=49 \pi$

To find the area of the shaded region, subtract the two areas.

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

$\text{Area of Shaded Region}=49\pi-49=104.94$

Make sure to round to places after the decimal.

Report an Error

Example Question #5 : How To Find The Area Of An Acute / Obtuse Isosceles Triangle

An isosceles triangle is placed in a circle as shown by the figure below.

If the diameter of the circle is , find the area of the shaded region.

Possible Answers:

29.95

34.27

40.29

31.02

Correct answer:

34.27

Explanation:

From the given image, you should notice that the base of the triangle is also the diameter of the circle. In addition, the height of the triangle is also the radius of the circle.

Thus, we can find the area of the triangle.

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

$\text{Area of Triangle}=\frac{1}{2}(8 \times 4)=16$

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

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

$\text{Area of Circle}=\pi\times 4^2= 16\pi$

To find the area of the shaded region, subtract the two areas.

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

$\text{Area of Shaded Region}=16\pi-16=34.27$

Make sure to round to places after the decimal.

Report an Error

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Intermediate Geometry : Intermediate Geometry

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

Example Question #8 : Triangles

Example Question #9 : Triangles

Example Question #11 : Triangles

Example Question #12 : Triangles

Example Question #13 : Triangles

Example Question #11 : Acute / Obtuse Isosceles Triangles

Example Question #2 : How To Find The Area Of An Acute / Obtuse Isosceles Triangle

Example Question #3 : How To Find The Area Of An Acute / Obtuse Isosceles Triangle

Example Question #4 : How To Find The Area Of An Acute / Obtuse Isosceles Triangle

Example Question #5 : How To Find The Area Of An Acute / Obtuse Isosceles Triangle

All Intermediate Geometry Resources

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