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Intermediate Geometry : Plane Geometry

Study concepts, example questions & explanations for Intermediate Geometry

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All Intermediate Geometry Resources

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

Example Questions

Example Question #11 : How To Find The Perimeter Of An Acute / Obtuse Triangle

Find the perimeter of the triangle below. Round to the nearest tenths place.

Possible Answers:

9.6

8.7

11.2

12.3

Correct answer:

11.2

Explanation:

Draw in the height to create a right triangle.

Now, using the relationship between the lengths of sides in a 30-60-90 triangle, where the long leg is the short leg times $\sqrt 3$ and the hypotenuse is two times the short leg, we can find out that the height of the triangle is $\frac{3}{2}$ and the hypotenuse is .

The dashes on two sides of the triangle indicate that these two sides are congruent. The three side lengths of the triangle are $3, 3, \text{and }3\sqrt3$ .

Now, add up these side lengths to find the perimeter.

$\text{Perimeter}=3+3+3\sqrt3=11.2$

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Example Question #511 : Intermediate Geometry

Find the perimeter of the triangle below. Round to the nearest tenths place.

Possible Answers:

32.4

23.8

26.4

29.9

Correct answer:

29.9

Explanation:

Draw in the height to create a right triangle.

Now, using the relationship between the lengths of sides in a 30-60-90 triangle, where the long leg is the short leg times $\sqrt3$ and the hypotenuse is twice the length of the short leg, we can find out that the height of the triangle is and the hypotenuse is .

The dashes on two sides of the triangle indicate that these two sides are congruent. The three side lengths of the triangle are $8, 8, \text{ and }8\sqrt3$ .

Now, add up these side lengths to find the perimeter.

$\text{Perimeter}=8+8+8\sqrt3=29.9$

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Example Question #71 : Triangles

In terms of , find the perimeter.

Possible Answers:

$6x+2x\sqrt3$

$4x+2x\sqrt3$

$8x+2x\sqrt3$

$2x+2x\sqrt3$

Correct answer:

$4x+2x\sqrt3$

Explanation:

Draw in the height to create a right triangle.

Now, using the relationship between the lengths of sides in a 30-60-90 triangle, we can find out that the height of the triangle is and the hypotenuse is .

The ratio of the sides in a 30-60-90 triangle are: $x:x\sqrt3:2x$ .

The dashes on two sides of the triangle indicate that these two sides are congruent. The three side lengths of the triangle are $2x, 2x, \text{ and }2x\sqrt3$ .

Now, add up these side lengths to find the perimeter.

$\text{Perimeter}=2x+2x+2x\sqrt3=4x+2x\sqrt3$

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Example Question #14 : How To Find The Perimeter Of An Acute / Obtuse Triangle

In terms of , find the perimeter.

Possible Answers:

$62x+42x\sqrt3$

$84x+42x\sqrt3$

$42x+42x\sqrt3$

$84+42x\sqrt3$

Correct answer:

$84x+42x\sqrt3$

Explanation:

Draw in the height to create a right triangle.

Now, using the relationship between the lengths of sides in a 30-60-90 triangle, we can find out that the height of the triangle is 21x and the hypotenuse is 42x .

The ratio of the sides in a 30-60-90 triangle are: $x:x\sqrt3:2x$ .

The dashes on two sides of the triangle indicate that these two sides are congruent. The three side lengths of the triangle are $42x, 42x, \text{ and }42x\sqrt3$ .

Now, add up these side lengths to find the perimeter.

$\text{Perimeter}=42x+42x+42x\sqrt3=84x+42x\sqrt3$

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Example Question #72 : Triangles

Find the perimeter of the triangle below. Round to the nearest tenths place.

Possible Answers:

59.7

59.1

58.3

47.8

Correct answer:

59.7

Explanation:

Draw in the height to create a right triangle.

Now, using the relationship between the lengths of sides in a 30-60-90 triangle, we can find out that the height of the triangle is and the hypotenuse is .

The ratio of the sides in a 30-60-90 triangle are: $x:x\sqrt3:2x$ .

The dashes on two sides of the triangle indicate that these two sides are congruent. The three side lengths of the triangle are $16, 16, \text{ and }16\sqrt3$ .

Now, add up these side lengths to find the perimeter.

$\text{Perimeter}=16+16+16\sqrt3=59.7$

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Example Question #73 : Triangles

In terms of , find the perimeter of the triangle.

Possible Answers:

$108x+36x\sqrt3$

$72x+36x\sqrt3$

$36x+36x\sqrt3$

$72+36x\sqrt3$

Correct answer:

$72x+36x\sqrt3$

Explanation:

Draw in the height to create a right triangle.

Now, using the relationship between the lengths of sides in a 30-60-90 triangle, we can find out that the height of the triangle is 18x and the hypotenuse is 36x .

The ratio of the sides in a 30-60-90 triangle are: $x:x\sqrt3:2x$ .

The dashes on two sides of the triangle indicate that these two sides are congruent. The three side lengths of the triangle are $36x, 36x, \text{ and }36x\sqrt3$ .

Now, add up these side lengths to find the perimeter.

$\text{Perimeter}=36x+36x+36x\sqrt3=72x+36x\sqrt3$

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Example Question #512 : Intermediate Geometry

A triangle is defined by the following points on a coordinate plane: (1, 5), (2, 12), (-3, 4)

What is the perimeter of the triangle?

Possible Answers:

21.67

25.06

22.49

20.63

Correct answer:

20.63

Explanation:

In order to find the perimeter of the triangle, we will first need to find the length of each side of the triangle by using the distance formula.

Recall the distance formula for a line:

$\text{distance}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

The first side of the triangle is the line segment made with $(1, 5)\text{ and }(2, 12)$ as its endpoints.

$\text{side 1}=\sqrt{(2-1)^2+(12-5)^2}=\sqrt{50}=5\sqrt2$

The second side of the triangle is the line segment that has $(2, 12)\text{ and }(-3, 4)$ as its endpoints.

$\text{side 2}=\sqrt{(2-(-3))^2+(12-4)^2}=\sqrt{89}$

The third side of the triangle is the line segment that has $(-3, 4)\text{ and }(1, 5)$ as its endpoints.

$\text{side 3}=\sqrt{(1-(-3))^2+(5-4)^2}=\sqrt{17}$

Now, add up these three sides with a calculator to find the perimeter of the triangle.

$\text{Perimeter}=5\sqrt{2} +\sqrt{89}+\sqrt{17}=20.63$

Make sure to round to places after the decimal.

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Example Question #513 : Intermediate Geometry

A triangle is defined by the following points in a coordinate plane: (5, 6), (7, 2), (9, 11) .

What is the perimeter of the triangle?

Possible Answers:

26.78

20.09

22.13

24.59

Correct answer:

20.09

Explanation:

In order to find the perimeter of the triangle, we will first need to find the length of each side of the triangle by using the distance formula.

Recall the distance formula for a line:

$\text{distance}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

The first side of the triangle is the line segment made with $(5, 6)\text{ and }(7, 2)$ as its endpoints.

$\text{side 1}=\sqrt{(7-5)^2+(2-6)^2}=\sqrt{20}=2\sqrt5$

The second side of the triangle is the line segment that has $(7, 2)\text{ and }(9, 11)$ as its endpoints.

$\text{side 2}=\sqrt{(9-7)^2+(11-2)^2}=\sqrt{85}$

The third side of the triangle is the line segment that has $(5, 6)\text{ and }(9, 11)$ as its endpoints.

$\text{side 3}=\sqrt{(9-5)^2+(11-6)^2}=\sqrt{41}$

Now, add up these three sides with a calculator to find the perimeter of the triangle.

$\text{Perimeter}=2\sqrt{5} +\sqrt{85}+\sqrt{41}=20.09$

Make sure to round to places after the decimal.

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Example Question #514 : Intermediate Geometry

A triangle is defined by the following points on a coordinate plane: (-5, 1), (6, -2), (7, 0) .

What is the perimeter of the triangle?

Possible Answers:

27.09

28.81

25.68

30.01

Correct answer:

25.68

Explanation:

In order to find the perimeter of the triangle, we will first need to find the length of each side of the triangle by using the distance formula.

Recall the distance formula for a line:

$\text{distance}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

The first side of the triangle is the line segment made with $(-5, 1)\text{ and }(6, -2)$ as its endpoints.

$\text{side 1}=\sqrt{(6-(-5))^2+(-2-1)^2}=\sqrt{130}$

The second side of the triangle is the line segment that has $(6, -2)\text{ and }(7, 0)$ as its endpoints.

$\text{side 2}=\sqrt{(7-6)^2+(0-(-2))^2}=\sqrt{5}$

The third side of the triangle is the line segment that has $(-5, 1)\text{ and }(7, 0)$ as its endpoints.

$\text{side 3}=\sqrt{(7-(-5))^2+(0-1)^2}=\sqrt{145}$

Now, add up these three sides with a calculator to find the perimeter of the triangle.

$\text{Perimeter}=\sqrt{130} +\sqrt{5}+\sqrt{145}=25.68$

Make sure to round to places after the decimal.

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Example Question #515 : Intermediate Geometry

A triangle is defined by the following points on a coordinate plane: (9, 8), (12, 0), (-1, -2) .

What is the perimeter of the triangle?

Possible Answers:

36.09

35.84

33.87

31.45

Correct answer:

35.84

Explanation:

In order to find the perimeter of the triangle, we will first need to find the length of each side of the triangle by using the distance formula.

Recall the distance formula for a line:

$\text{distance}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

The first side of the triangle is the line segment made with $(9, 8)\text{ and }(12, 0)$ as its endpoints.

$\text{side 1}=\sqrt{(12-9)^2+(0-8)^2}=\sqrt{73}$

The second side of the triangle is the line segment that has $(12, 0)\text{ and }(-1, -2)$ as its endpoints.

$\text{side 2}=\sqrt{(-1-12)^2+(-2-0)^2}=\sqrt{173}$

The third side of the triangle is the line segment that has $(9, 8)\text{ and }(-1, -2)$ as its endpoints.

$\text{side 3}=\sqrt{(-1-9))^2+(-2-8)^2}=\sqrt{200}=10\sqrt2$

Now, add up these three sides with a calculator to find the perimeter of the triangle.

$\text{Perimeter}=\sqrt{73} +\sqrt{173}+10\sqrt{2}=35.84$

Make sure to round to places after the decimal.

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All Intermediate Geometry Resources

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Intermediate Geometry : Plane Geometry

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

Example Question #11 : How To Find The Perimeter Of An Acute / Obtuse Triangle

Example Question #511 : Intermediate Geometry

Example Question #71 : Triangles

Example Question #14 : How To Find The Perimeter Of An Acute / Obtuse Triangle

Example Question #72 : Triangles

Example Question #73 : Triangles

Example Question #512 : Intermediate Geometry

Example Question #513 : Intermediate Geometry

Example Question #514 : Intermediate Geometry

Example Question #515 : Intermediate Geometry

All Intermediate Geometry Resources

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