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Basic Geometry : Plane Geometry

Study concepts, example questions & explanations for Basic Geometry

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All Basic Geometry Resources

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

Example Questions

Example Question #1281 : Basic Geometry

Find the length of the hypotenuse.

Possible Answers:

$3\sqrt3$

$\sqrt{15}$

$\sqrt{14}$

$4\sqrt2$

Correct answer:

$\sqrt{14}$

Explanation:

Recall how to find the length of the hypotenuse, , of a right triangle by using the Pythagorean Theorem.

a^2+b^2=c^2

Substitute in the given values.

$c^2=(\sqrt6)^2+(2\sqrt2)^2$

Simplify.

c^2=6+8

Solve.

c^2=14

Now, because we want to solve for just , take the square root of the value you found above.

$\sqrt{c^2}=c=\sqrt{14}$

Report an Error

Example Question #1281 : Basic Geometry

Find the length of the hypotenuse.

Possible Answers:

$6\sqrt2$

$\sqrt{26}$

$4\sqrt5$

Correct answer:

Explanation:

Recall how to find the length of the hypotenuse, , of a right triangle by using the Pythagorean Theorem.

a^2+b^2=c^2

Substitute in the given values.

$c^2=(\sqrt6)^2+(\sqrt{19})^2$

Simplify.

c^2=6+19

Solve.

c^2=25

Now, because we want to solve for just , take the square root of the value you found above.

$\sqrt{c^2}=\sqrt{25}$

Simplify.

c=5

Report an Error

Example Question #293 : Triangles

Find the length of the hypotenuse.

Possible Answers:

$2\sqrt3$

$2\sqrt6$

$5\sqrt2$

Correct answer:

Explanation:

Recall how to find the length of the hypotenuse, , of a right triangle by using the Pythagorean Theorem.

a^2+b^2=c^2

Substitute in the given values.

$c^2=(\sqrt{11})^2+(\sqrt{14})^2$

Simplify.

c^2=11+14

Solve.

c^2=25

Now, because we want to solve for just , take the square root of the value you found above.

$\sqrt{c^2}=\sqrt{25}$

Simplify.

c=5

Report an Error

Example Question #301 : Triangles

Find the length of the hypotenuse.

Possible Answers:

$4\sqrt2$

$\sqrt{26}$

$3\sqrt3$

Correct answer:

$\sqrt{26}$

Explanation:

Recall how to find the length of the hypotenuse, , of a right triangle by using the Pythagorean Theorem.

a^2+b^2=c^2

Substitute in the given values.

$c^2=(\sqrt{10})^2+(4)^2$

Simplify.

c^2=10+16

Solve.

c^2=26

Now, because we want to solve for just , take the square root of the value you found above.

$\sqrt{c^2}=c=\sqrt{26}$

Report an Error

Example Question #302 : Triangles

Find the length of the hypotenuse.

Possible Answers:

$6\sqrt5$

$8\sqrt3$

Correct answer:

Explanation:

Recall how to find the length of the hypotenuse, , of a right triangle by using the Pythagorean Theorem.

a^2+b^2=c^2

Substitute in the given values.

$c^2=(\sqrt{10})^2+(\sqrt{39})^2$

Simplify.

c^2=10+39

Solve.

c^2=49

Now, because we want to solve for just , take the square root of the value you found above.

$\sqrt{c^2}=\sqrt{49}$

Simplify.

c=7

Report an Error

Example Question #303 : Triangles

Find the length of the hypotenuse.

Possible Answers:

$4\sqrt6$

The length of the hypotenuse cannot be determined.

$16\sqrt2$

Correct answer:

Explanation:

Recall how to find the length of the hypotenuse, , of a right triangle by using the Pythagorean Theorem.

a^2+b^2=c^2

Substitute in the given values.

$c^2=(2\sqrt{5})^2+(\sqrt{29})^2$

Simplify.

c^2=20+29

Solve.

c^2=49

Now, because we want to solve for just , take the square root of the value you found above.

$\sqrt{c^2}=\sqrt{49}$

Simplify.

c=7

Report an Error

Example Question #304 : Triangles

Find the length of the hypotenuse.

Possible Answers:

$4\sqrt{14}$

$14\sqrt2$

$2\sqrt{14}$

Correct answer:

Explanation:

Recall how to find the length of the hypotenuse, , of a right triangle by using the Pythagorean Theorem.

a^2+b^2=c^2

Substitute in the given values.

$c^2=(\sqrt{2})^2+(\sqrt{14})^2$

Simplify.

c^2=2+14

Solve.

c^2=16

Now, because we want to solve for just , take the square root of the value you found above.

$\sqrt{c^2}=\sqrt{16}$

Simplify.

c=4

Report an Error

Example Question #305 : Triangles

Find the length of the hypotenuse.

Possible Answers:

$\sqrt{39}$

$3\sqrt{13}$

Correct answer:

Explanation:

Recall how to find the length of the hypotenuse, , of a right triangle by using the Pythagorean Theorem.

a^2+b^2=c^2

Substitute in the given values.

$c^2=(\sqrt{3})^2+(\sqrt{13})^2$

Simplify.

c^2=3+13

Solve.

c^2=16

Now, because we want to solve for just , take the square root of the value you found above.

$\sqrt{c^2}=\sqrt{16}$

Simplify.

c=4

Report an Error

Example Question #111 : Right Triangles

Find the length of the hypotenuse.

Possible Answers:

$89\sqrt{11}$

$11\sqrt{89}$

Correct answer:

Explanation:

Recall how to find the length of the hypotenuse, , of a right triangle by using the Pythagorean Theorem.

a^2+b^2=c^2

Substitute in the given values.

$c^2=(\sqrt{89})^2+(\sqrt{11})^2$

Simplify.

c^2=89+11

Solve.

c^2=100

Now, because we want to solve for just , take the square root of the value you found above.

$\sqrt{c^2}=\sqrt{100}$

Simplify.

c=10

Report an Error

Example Question #1281 : Basic Geometry

Find the length of the hypotenuse.

Possible Answers:

$12\sqrt2$

$9\sqrt3$

Correct answer:

Explanation:

Recall how to find the length of the hypotenuse, , of a right triangle by using the Pythagorean Theorem.

a^2+b^2=c^2

Substitute in the given values.

$c^2=(2\sqrt{3})^2+(2\sqrt{6})^2$

Simplify.

c^2=12+24

Solve.

c^2=36

Now, because we want to solve for just , take the square root of the value you found above.

$\sqrt{c^2}=\sqrt{36}$

Simplify.

c=6

Report an Error

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All Basic Geometry Resources

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Basic Geometry : Plane Geometry

Study concepts, example questions & explanations for Basic Geometry

All Basic Geometry Resources

Example Questions

Example Question #1281 : Basic Geometry

Example Question #1281 : Basic Geometry

Example Question #293 : Triangles

Example Question #301 : Triangles

Example Question #302 : Triangles

Example Question #303 : Triangles

Example Question #304 : Triangles

Example Question #305 : Triangles

Example Question #111 : Right Triangles

Example Question #1281 : Basic Geometry

All Basic Geometry Resources

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