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Basic Geometry : Triangles

Study concepts, example questions & explanations for Basic Geometry

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9 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1132 : Plane Geometry

Find the perimeter.

Possible Answers:

111.67

115.29

112.67

109.76

Correct answer:

112.67

Explanation:

Notice that the given triangle is a right isosceles triangle. The two legs with the tick marks are the same length.

The lengths of the legs in the given triangle are then .

Next, find the length of the hypotenuse by using the Pythagorean Theorem.

a^2+a^2=b^2

b^2=2a^2

$b=a\sqrt2$

Plug in the value of the length of a leg to find the length of the hypotenuse.

$b=33\sqrt2$

Finally, recall how to find the perimeter of a triangle:

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

Plug in the values for this triangle to find its perimeter.

$\text{Perimeter}=33+33+33\sqrt2=112.67$

Make sure to round to two places after the decimal.

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Example Question #152 : Triangles

A right isosceles triangle is stacked on top of a square as shown in the figure. Find the perimeter of the compound shape.

Possible Answers:

55.29

49.08

52.97

54.20

Correct answer:

52.97

Explanation:

Notice that the hypotenuse of the right isosceles triangle is also the length of a side of the square.

First, we will need to find the length of the legs of the triangle by using the Pythagorean theorem.

$\text{leg}^2+\text{leg}^2=\text{side}^2$

$2(\text{leg})^2=\text{side}^2$

$\text{leg}^2=\frac{\text{side}^2}{2}$

$\text{leg}=\sqrt{\frac{\text{side}^2}{2}}=\frac{\text{side}\sqrt2}{2}$

Substitute in the given length of the hypotenuse to find the length of the leg of the triangle.

$\text{leg}=\frac{12\sqrt2}{2}$

$\text{leg}=6\sqrt2$

In order to find the perimeter, add up the lengths outlined in red. The perimeter includes the two legs of the triangle and three sides of the square.

Therefore:

$\text{Perimeter}=12+12+12+6\sqrt2+6\sqrt2$

$\text{Perimeter}=52.97$

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Example Question #31 : How To Find The Perimeter Of A 45/45/90 Right Isosceles Triangle

A right isosceles triangle is stacked on top of a square as shown in the figure. Find the perimeter of the compound shape.

Possible Answers:

8.10

8.83

9.41

10.06

Correct answer:

8.83

Explanation:

Notice that the hypotenuse of the right isosceles triangle is also the length of a side of the square.

First, we will need to find the length of the legs of the triangle by using the Pythagorean theorem.

$\text{leg}^2+\text{leg}^2=\text{side}^2$

$2(\text{leg})^2=\text{side}^2$

$\text{leg}^2=\frac{\text{side}^2}{2}$

$\text{leg}=\sqrt{\frac{\text{side}^2}{2}}=\frac{\text{side}\sqrt2}{2}$

Substitute in the given length of the hypotenuse to find the length of the leg of the triangle.

$\text{leg}=\frac{2\sqrt2}{2}$

$\text{leg}=\sqrt2$

In order to find the perimeter, add up the lengths outlined in red. The perimeter includes the two legs of the triangle and three sides of the square.

Therefore:

$\text{Perimeter}=2+2+2+\sqrt2+\sqrt2$

$\text{Perimeter}=8.83$

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Example Question #31 : How To Find The Perimeter Of A 45/45/90 Right Isosceles Triangle

A right isosceles triangle is stacked on top of a square as shown in the figure. Find the perimeter of the compound shape.

Possible Answers:

19.07

17.66

16.92

18.92

Correct answer:

17.66

Explanation:

Notice that the hypotenuse of the right isosceles triangle is also the length of a side of the square.

First, we will need to find the length of the legs of the triangle by using the Pythagorean theorem.

$\text{leg}^2+\text{leg}^2=\text{side}^2$

$2(\text{leg})^2=\text{side}^2$

$\text{leg}^2=\frac{\text{side}^2}{2}$

$\text{leg}=\sqrt{\frac{\text{side}^2}{2}}=\frac{\text{side}\sqrt2}{2}$

Substitute in the given length of the hypotenuse to find the length of the leg of the triangle.

$\text{leg}=\frac{4\sqrt2}{2}$

$\text{leg}=2\sqrt2$

In order to find the perimeter, add up the lengths outlined in red. The perimeter includes the two legs of the triangle and three sides of the square.

Therefore:

$\text{Perimeter}=4+4+4+2\sqrt2+2\sqrt2$

$\text{Perimeter}=17.66$

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Example Question #152 : Triangles

A right isosceles triangle is stacked on top of a square as shown in the figure. Find the perimeter of the compound shape.

Possible Answers:

31.22

26.49

30.09

27.81

Correct answer:

26.49

Explanation:

Notice that the hypotenuse of the right isosceles triangle is also the length of a side of the square.

First, we will need to find the length of the legs of the triangle by using the Pythagorean theorem.

$\text{leg}^2+\text{leg}^2=\text{side}^2$

$2(\text{leg})^2=\text{side}^2$

$\text{leg}^2=\frac{\text{side}^2}{2}$

$\text{leg}=\sqrt{\frac{\text{side}^2}{2}}=\frac{\text{side}\sqrt2}{2}$

Substitute in the given length of the hypotenuse to find the length of the leg of the triangle.

$\text{leg}=\frac{6\sqrt2}{2}$

$\text{leg}=3\sqrt2$

In order to find the perimeter, add up the lengths outlined in red. The perimeter includes the two legs of the triangle and three sides of the square.

Therefore:

$\text{Perimeter}=6+6+6+3\sqrt2+3\sqrt2$

$\text{Perimeter}=26.49$

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Example Question #153 : Triangles

A right isosceles triangle is stacked on top of a square as shown in the figure. Find the perimeter of the compound shape.

Possible Answers:

36.19

35.31

35.59

34.10

Correct answer:

35.31

Explanation:

Notice that the hypotenuse of the right isosceles triangle is also the length of a side of the square.

First, we will need to find the length of the legs of the triangle by using the Pythagorean theorem.

$\text{leg}^2+\text{leg}^2=\text{side}^2$

$2(\text{leg})^2=\text{side}^2$

$\text{leg}^2=\frac{\text{side}^2}{2}$

$\text{leg}=\sqrt{\frac{\text{side}^2}{2}}=\frac{\text{side}\sqrt2}{2}$

Substitute in the given length of the hypotenuse to find the length of the leg of the triangle.

$\text{leg}=\frac{8\sqrt2}{2}$

$\text{leg}=4\sqrt2$

In order to find the perimeter, add up the lengths outlined in red. The perimeter includes the two legs of the triangle and three sides of the square.

Therefore:

$\text{Perimeter}=8+8+8+4\sqrt2+4\sqrt2$

$\text{Perimeter}=35.31$

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Example Question #154 : Triangles

A right isosceles triangle is stacked on top of a square as shown in the figure. Find the perimeter of the compound shape.

Possible Answers:

42.99

44.14

45.61

48.20

Correct answer:

44.14

Explanation:

Notice that the hypotenuse of the right isosceles triangle is also the length of a side of the square.

First, we will need to find the length of the legs of the triangle by using the Pythagorean theorem.

$\text{leg}^2+\text{leg}^2=\text{side}^2$

$2(\text{leg})^2=\text{side}^2$

$\text{leg}^2=\frac{\text{side}^2}{2}$

$\text{leg}=\sqrt{\frac{\text{side}^2}{2}}=\frac{\text{side}\sqrt2}{2}$

Substitute in the given length of the hypotenuse to find the length of the leg of the triangle.

$\text{leg}=\frac{10\sqrt2}{2}$

$\text{leg}=5\sqrt2$

In order to find the perimeter, add up the lengths outlined in red. The perimeter includes the two legs of the triangle and three sides of the square.

Therefore:

$\text{Perimeter}=10+10+10+5\sqrt2+5\sqrt2$

$\text{Perimeter}=44.14$

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Example Question #1141 : Plane Geometry

A right isosceles triangle is stacked on top of a square as shown in the figure. Find the perimeter of the compound shape.

Possible Answers:

72.59

74.09

69.39

61.80

Correct answer:

61.80

Explanation:

Notice that the hypotenuse of the right isosceles triangle is also the length of a side of the square.

First, we will need to find the length of the legs of the triangle by using the Pythagorean theorem.

$\text{leg}^2+\text{leg}^2=\text{side}^2$

$2(\text{leg})^2=\text{side}^2$

$\text{leg}^2=\frac{\text{side}^2}{2}$

$\text{leg}=\sqrt{\frac{\text{side}^2}{2}}=\frac{\text{side}\sqrt2}{2}$

Substitute in the given length of the hypotenuse to find the length of the leg of the triangle.

$\text{leg}=\frac{14\sqrt2}{2}$

$\text{leg}=7\sqrt2$

In order to find the perimeter, add up the lengths outlined in red. The perimeter includes the two legs of the triangle and three sides of the square.

Therefore:

$\text{Perimeter}=14+14+14+7\sqrt2+7\sqrt2$

$\text{Perimeter}=61.80$

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Example Question #1142 : Basic Geometry

A right isosceles triangle is stacked on top of a square as shown in the figure. Find the perimeter of the compound shape.

Possible Answers:

48.20

69.59

70.63

80.05

Correct answer:

70.63

Explanation:

Notice that the hypotenuse of the right isosceles triangle is also the length of a side of the square.

First, we will need to find the length of the legs of the triangle by using the Pythagorean theorem.

$\text{leg}^2+\text{leg}^2=\text{side}^2$

$2(\text{leg})^2=\text{side}^2$

$\text{leg}^2=\frac{\text{side}^2}{2}$

$\text{leg}=\sqrt{\frac{\text{side}^2}{2}}=\frac{\text{side}\sqrt2}{2}$

Substitute in the given length of the hypotenuse to find the length of the leg of the triangle.

$\text{leg}=\frac{16\sqrt2}{2}$

$\text{leg}=8\sqrt2$

In order to find the perimeter, add up the lengths outlined in red. The perimeter includes the two legs of the triangle and three sides of the square.

Therefore:

$\text{Perimeter}=16+16+16+8\sqrt2+8\sqrt2$

$\text{Perimeter}=70.63$

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Example Question #1142 : Plane Geometry

A right isosceles triangle is stacked on top of a square as shown in the figure. Find the perimeter of the compound shape.

Possible Answers:

77.65

79.46

84.19

81.20

Correct answer:

79.46

Explanation:

Notice that the hypotenuse of the right isosceles triangle is also the length of a side of the square.

First, we will need to find the length of the legs of the triangle by using the Pythagorean theorem.

$\text{leg}^2+\text{leg}^2=\text{side}^2$

$2(\text{leg})^2=\text{side}^2$

$\text{leg}^2=\frac{\text{side}^2}{2}$

$\text{leg}=\sqrt{\frac{\text{side}^2}{2}}=\frac{\text{side}\sqrt2}{2}$

Substitute in the given length of the hypotenuse to find the length of the leg of the triangle.

$\text{leg}=\frac{18\sqrt2}{2}$

$\text{leg}=9\sqrt2$

In order to find the perimeter, add up the lengths outlined in red. The perimeter includes the two legs of the triangle and three sides of the square.

Therefore:

$\text{Perimeter}=18+18+18+9\sqrt2+9\sqrt2$

$\text{Perimeter}=79.46$

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Basic Geometry : Triangles

Study concepts, example questions & explanations for Basic Geometry

All Basic Geometry Resources

Example Questions

Example Question #1132 : Plane Geometry

Example Question #152 : Triangles

Example Question #31 : How To Find The Perimeter Of A 45/45/90 Right Isosceles Triangle

Example Question #31 : How To Find The Perimeter Of A 45/45/90 Right Isosceles Triangle

Example Question #152 : Triangles

Example Question #153 : Triangles

Example Question #154 : Triangles

Example Question #1141 : Plane Geometry

Example Question #1142 : Basic Geometry

Example Question #1142 : Plane Geometry

All Basic Geometry Resources

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