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Basic Geometry : How to find circumference

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 #81 : How To Find Circumference

Find the circumference of the circle if the lengths of the legs of the inscribed isosceles triangle are $5\sqrt2$ $\displaystyle 5\sqrt2$.

Possible Answers:

$20\pi$ $\displaystyle 20\pi$

$30\pi$ $\displaystyle 30\pi$

$25\pi$ $\displaystyle 25\pi$

$10\pi$ $\displaystyle 10\pi$

Correct answer:

$10\pi$ $\displaystyle 10\pi$

Explanation:

Notice that the hypotenuse of the triangle in the figure is also the diameter of the circle.

Use the Pythagorean theorem to find the length of the hypotenuse.

$\text{Hypotenuse}^2=\text{leg}^2+\text{leg}^2$ $\displaystyle \text{Hypotenuse}^2=\text{leg}^2+\text{leg}^2$

$\text{Hypotenuse}^2=2(\text{leg}^2)$ $\displaystyle \text{Hypotenuse}^2=2(\text{leg}^2)$

$\text{Hypotenuse}=\sqrt{2(\text{leg}^2)}=\text{leg}\sqrt2$ $\displaystyle \text{Hypotenuse}=\sqrt{2(\text{leg}^2)}=\text{leg}\sqrt2$

Substitute in the length of triangle’s legs to find the missing length of the hypotenuse.

$\text{Hypotenuse}=5\sqrt2(\sqrt2)$ $\displaystyle \text{Hypotenuse}=5\sqrt2(\sqrt2)$

Simplify.

$\text{Hypotenuse}=10$ $\displaystyle \text{Hypotenuse}=10$

Now, recall that the hypotenuse of the triangle and the diameter of the circle are the same:

$\text{Hypotenuse}=\text{Diameter}=10$ $\displaystyle \text{Hypotenuse}=\text{Diameter}=10$

Now, recall how to find the circumference of a circle:

$\text{Circumference}=\text{Diameter}\times\pi$ $\displaystyle \text{Circumference}=\text{Diameter}\times\pi$

Substitute in the value for the diameter to find the circumference of the circle.

$\text{Circumference}=10\pi$ $\displaystyle \text{Circumference}=10\pi$

Report an Error

Example Question #271 : Radius

Find the circumference of the circle if the lengths of the legs of the inscribed isosceles triangle are $6\sqrt2$ $\displaystyle 6\sqrt2$.

Possible Answers:

$24\pi$ $\displaystyle 24\pi$

144 $\displaystyle 144$

$18\pi$ $\displaystyle 18\pi$

$12\pi$ $\displaystyle 12\pi$

Correct answer:

$12\pi$ $\displaystyle 12\pi$

Explanation:

Notice that the hypotenuse of the triangle in the figure is also the diameter of the circle.

Use the Pythagorean theorem to find the length of the hypotenuse.

$\text{Hypotenuse}^2=\text{leg}^2+\text{leg}^2$ $\displaystyle \text{Hypotenuse}^2=\text{leg}^2+\text{leg}^2$

$\text{Hypotenuse}^2=2(\text{leg}^2)$ $\displaystyle \text{Hypotenuse}^2=2(\text{leg}^2)$

$\text{Hypotenuse}=\sqrt{2(\text{leg}^2)}=\text{leg}\sqrt2$ $\displaystyle \text{Hypotenuse}=\sqrt{2(\text{leg}^2)}=\text{leg}\sqrt2$

Substitute in the length of triangle’s legs to find the missing length of the hypotenuse.

$\text{Hypotenuse}=6\sqrt2(\sqrt2)$ $\displaystyle \text{Hypotenuse}=6\sqrt2(\sqrt2)$

Simplify.

$\text{Hypotenuse}=12$ $\displaystyle \text{Hypotenuse}=12$

Now, recall that the hypotenuse of the triangle and the diameter of the circle are the same:

$\text{Hypotenuse}=\text{Diameter}=12$ $\displaystyle \text{Hypotenuse}=\text{Diameter}=12$

Now, recall how to find the circumference of a circle:

$\text{Circumference}=\text{Diameter}\times\pi$ $\displaystyle \text{Circumference}=\text{Diameter}\times\pi$

Substitute in the value for the diameter to find the circumference of the circle.

$\text{Circumference}=12\pi$ $\displaystyle \text{Circumference}=12\pi$

Report an Error

Example Question #81 : How To Find Circumference

Find the circumference of the circle if the lengths of the legs of the inscribed isosceles triangle are $7\sqrt2$ $\displaystyle 7\sqrt2$.

Possible Answers:

$35\pi$ $\displaystyle 35\pi$

$\displaystyle 49$

$28\pi$ $\displaystyle 28\pi$

$14\pi$ $\displaystyle 14\pi$

Correct answer:

$14\pi$ $\displaystyle 14\pi$

Explanation:

Notice that the hypotenuse of the triangle in the figure is also the diameter of the circle.

Use the Pythagorean theorem to find the length of the hypotenuse.

$\text{Hypotenuse}^2=\text{leg}^2+\text{leg}^2$ $\displaystyle \text{Hypotenuse}^2=\text{leg}^2+\text{leg}^2$

$\text{Hypotenuse}^2=2(\text{leg}^2)$ $\displaystyle \text{Hypotenuse}^2=2(\text{leg}^2)$

$\text{Hypotenuse}=\sqrt{2(\text{leg}^2)}=\text{leg}\sqrt2$ $\displaystyle \text{Hypotenuse}=\sqrt{2(\text{leg}^2)}=\text{leg}\sqrt2$

Substitute in the length of triangle’s legs to find the missing length of the hypotenuse.

$\text{Hypotenuse}=7\sqrt2(\sqrt2)$ $\displaystyle \text{Hypotenuse}=7\sqrt2(\sqrt2)$

Simplify.

$\text{Hypotenuse}=14$ $\displaystyle \text{Hypotenuse}=14$

Now, recall that the hypotenuse of the triangle and the diameter of the circle are the same:

$\text{Hypotenuse}=\text{Diameter}=14$ $\displaystyle \text{Hypotenuse}=\text{Diameter}=14$

Now, recall how to find the circumference of a circle:

$\text{Circumference}=\text{Diameter}\times\pi$ $\displaystyle \text{Circumference}=\text{Diameter}\times\pi$

Substitute in the value for the diameter to find the circumference of the circle.

$\text{Circumference}=14\pi$ $\displaystyle \text{Circumference}=14\pi$

Report an Error

Example Question #273 : Radius

Find the circumference of the circle if the lengths of the legs of the inscribed isosceles triangle are $9\sqrt2$ $\displaystyle 9\sqrt2$.

Possible Answers:

$\displaystyle 81$

$54\pi$ $\displaystyle 54\pi$

$36\pi$ $\displaystyle 36\pi$

$18\pi$ $\displaystyle 18\pi$

Correct answer:

$18\pi$ $\displaystyle 18\pi$

Explanation:

Notice that the hypotenuse of the triangle in the figure is also the diameter of the circle.

Use the Pythagorean theorem to find the length of the hypotenuse.

$\text{Hypotenuse}^2=\text{leg}^2+\text{leg}^2$ $\displaystyle \text{Hypotenuse}^2=\text{leg}^2+\text{leg}^2$

$\text{Hypotenuse}^2=2(\text{leg}^2)$ $\displaystyle \text{Hypotenuse}^2=2(\text{leg}^2)$

$\text{Hypotenuse}=\sqrt{2(\text{leg}^2)}=\text{leg}\sqrt2$ $\displaystyle \text{Hypotenuse}=\sqrt{2(\text{leg}^2)}=\text{leg}\sqrt2$

Substitute in the length of triangle’s legs to find the missing length of the hypotenuse.

$\text{Hypotenuse}=9\sqrt2(\sqrt2)$ $\displaystyle \text{Hypotenuse}=9\sqrt2(\sqrt2)$

Simplify.

$\text{Hypotenuse}=18$ $\displaystyle \text{Hypotenuse}=18$

Now, recall that the hypotenuse of the triangle and the diameter of the circle are the same:

$\text{Hypotenuse}=\text{Diameter}=18$ $\displaystyle \text{Hypotenuse}=\text{Diameter}=18$

Now, recall how to find the circumference of a circle:

$\text{Circumference}=\text{Diameter}\times\pi$ $\displaystyle \text{Circumference}=\text{Diameter}\times\pi$

Substitute in the value for the diameter to find the circumference of the circle.

$\text{Circumference}=18\pi$ $\displaystyle \text{Circumference}=18\pi$

Report an Error

Example Question #274 : Radius

Find the circumference of the circle if the lengths of the legs of the inscribed isosceles triangle are $8\sqrt2$ $\displaystyle 8\sqrt2$.

Possible Answers:

$96\pi$ $\displaystyle 96\pi$

$16\pi$ $\displaystyle 16\pi$

$32\pi$ $\displaystyle 32\pi$

$64\pi$ $\displaystyle 64\pi$

Correct answer:

$16\pi$ $\displaystyle 16\pi$

Explanation:

Notice that the hypotenuse of the triangle in the figure is also the diameter of the circle.

Use the Pythagorean theorem to find the length of the hypotenuse.

$\text{Hypotenuse}^2=\text{leg}^2+\text{leg}^2$ $\displaystyle \text{Hypotenuse}^2=\text{leg}^2+\text{leg}^2$

$\text{Hypotenuse}^2=2(\text{leg}^2)$ $\displaystyle \text{Hypotenuse}^2=2(\text{leg}^2)$

$\text{Hypotenuse}=\sqrt{2(\text{leg}^2)}=\text{leg}\sqrt2$ $\displaystyle \text{Hypotenuse}=\sqrt{2(\text{leg}^2)}=\text{leg}\sqrt2$

Substitute in the length of triangle’s legs to find the missing length of the hypotenuse.

$\text{Hypotenuse}=8\sqrt2(\sqrt2)$ $\displaystyle \text{Hypotenuse}=8\sqrt2(\sqrt2)$

Simplify.

$\text{Hypotenuse}=16$ $\displaystyle \text{Hypotenuse}=16$

Now, recall that the hypotenuse of the triangle and the diameter of the circle are the same:

$\text{Hypotenuse}=\text{Diameter}=16$ $\displaystyle \text{Hypotenuse}=\text{Diameter}=16$

Now, recall how to find the circumference of a circle:

$\text{Circumference}=\text{Diameter}\times\pi$ $\displaystyle \text{Circumference}=\text{Diameter}\times\pi$

Substitute in the value for the diameter to find the circumference of the circle.

$\text{Circumference}=16\pi$ $\displaystyle \text{Circumference}=16\pi$

Report an Error

Example Question #82 : How To Find Circumference

Find the circumference of the circle if the lengths of the legs of the inscribed isosceles triangle are $10\sqrt2$ $\displaystyle 10\sqrt2$.

Possible Answers:

$40\pi$ $\displaystyle 40\pi$

$80\pi$ $\displaystyle 80\pi$

$160\pi$ $\displaystyle 160\pi$

$20\pi$ $\displaystyle 20\pi$

Correct answer:

$20\pi$ $\displaystyle 20\pi$

Explanation:

Notice that the hypotenuse of the triangle in the figure is also the diameter of the circle.

Use the Pythagorean theorem to find the length of the hypotenuse.

$\text{Hypotenuse}^2=\text{leg}^2+\text{leg}^2$ $\displaystyle \text{Hypotenuse}^2=\text{leg}^2+\text{leg}^2$

$\text{Hypotenuse}^2=2(\text{leg}^2)$ $\displaystyle \text{Hypotenuse}^2=2(\text{leg}^2)$

$\text{Hypotenuse}=\sqrt{2(\text{leg}^2)}=\text{leg}\sqrt2$ $\displaystyle \text{Hypotenuse}=\sqrt{2(\text{leg}^2)}=\text{leg}\sqrt2$

Substitute in the length of triangle’s legs to find the missing length of the hypotenuse.

$\text{Hypotenuse}=10\sqrt2(\sqrt2)$ $\displaystyle \text{Hypotenuse}=10\sqrt2(\sqrt2)$

Simplify.

$\text{Hypotenuse}=20$ $\displaystyle \text{Hypotenuse}=20$

Now, recall that the hypotenuse of the triangle and the diameter of the circle are the same:

$\text{Hypotenuse}=\text{Diameter}=20$ $\displaystyle \text{Hypotenuse}=\text{Diameter}=20$

Now, recall how to find the circumference of a circle:

$\text{Circumference}=\text{Diameter}\times\pi$ $\displaystyle \text{Circumference}=\text{Diameter}\times\pi$

Substitute in the value for the diameter to find the circumference of the circle.

$\text{Circumference}=20\pi$ $\displaystyle \text{Circumference}=20\pi$

Report an Error

Example Question #275 : Radius

Find the circumference of the circle if the lengths of the legs of the inscribed isosceles triangle are $11\sqrt2$ $\displaystyle 11\sqrt2$.

Possible Answers:

$44\pi$ $\displaystyle 44\pi$

121 $\displaystyle 121$

$22\pi$ $\displaystyle 22\pi$

$66\pi$ $\displaystyle 66\pi$

Correct answer:

$22\pi$ $\displaystyle 22\pi$

Explanation:

Notice that the hypotenuse of the triangle in the figure is also the diameter of the circle.

Use the Pythagorean theorem to find the length of the hypotenuse.

$\text{Hypotenuse}^2=\text{leg}^2+\text{leg}^2$ $\displaystyle \text{Hypotenuse}^2=\text{leg}^2+\text{leg}^2$

$\text{Hypotenuse}^2=2(\text{leg}^2)$ $\displaystyle \text{Hypotenuse}^2=2(\text{leg}^2)$

$\text{Hypotenuse}=\sqrt{2(\text{leg}^2)}=\text{leg}\sqrt2$ $\displaystyle \text{Hypotenuse}=\sqrt{2(\text{leg}^2)}=\text{leg}\sqrt2$

Substitute in the length of triangle’s legs to find the missing length of the hypotenuse.

$\text{Hypotenuse}=11\sqrt2(\sqrt2)$ $\displaystyle \text{Hypotenuse}=11\sqrt2(\sqrt2)$

Simplify.

$\text{Hypotenuse}=22$ $\displaystyle \text{Hypotenuse}=22$

Now, recall that the hypotenuse of the triangle and the diameter of the circle are the same:

$\text{Hypotenuse}=\text{Diameter}=22$ $\displaystyle \text{Hypotenuse}=\text{Diameter}=22$

Now, recall how to find the circumference of a circle:

$\text{Circumference}=\text{Diameter}\times\pi$ $\displaystyle \text{Circumference}=\text{Diameter}\times\pi$

Substitute in the value for the diameter to find the circumference of the circle.

$\text{Circumference}=22\pi$ $\displaystyle \text{Circumference}=22\pi$

Report an Error

Example Question #276 : Radius

Find the circumference of the circle if the lengths of the legs of the inscribed isosceles triangle are $12\sqrt2$ $\displaystyle 12\sqrt2$.

Possible Answers:

$\displaystyle 36$

$24\pi$ $\displaystyle 24\pi$

$48\pi$ $\displaystyle 48\pi$

$12\pi$ $\displaystyle 12\pi$

Correct answer:

$24\pi$ $\displaystyle 24\pi$

Explanation:

Notice that the hypotenuse of the triangle in the figure is also the diameter of the circle.

Use the Pythagorean theorem to find the length of the hypotenuse.

$\text{Hypotenuse}^2=\text{leg}^2+\text{leg}^2$ $\displaystyle \text{Hypotenuse}^2=\text{leg}^2+\text{leg}^2$

$\text{Hypotenuse}^2=2(\text{leg}^2)$ $\displaystyle \text{Hypotenuse}^2=2(\text{leg}^2)$

$\text{Hypotenuse}=\sqrt{2(\text{leg}^2)}=\text{leg}\sqrt2$ $\displaystyle \text{Hypotenuse}=\sqrt{2(\text{leg}^2)}=\text{leg}\sqrt2$

Substitute in the length of triangle’s legs to find the missing length of the hypotenuse.

$\text{Hypotenuse}=12\sqrt2(\sqrt2)$ $\displaystyle \text{Hypotenuse}=12\sqrt2(\sqrt2)$

Simplify.

$\text{Hypotenuse}=24$ $\displaystyle \text{Hypotenuse}=24$

Now, recall that the hypotenuse of the triangle and the diameter of the circle are the same:

$\text{Hypotenuse}=\text{Diameter}=24$ $\displaystyle \text{Hypotenuse}=\text{Diameter}=24$

Now, recall how to find the circumference of a circle:

$\text{Circumference}=\text{Diameter}\times\pi$ $\displaystyle \text{Circumference}=\text{Diameter}\times\pi$

Substitute in the value for the diameter to find the circumference of the circle.

$\text{Circumference}=24\pi$ $\displaystyle \text{Circumference}=24\pi$

Report an Error

Example Question #272 : Radius

Find the circumference of the circle if the lengths of the legs of the inscribed isosceles triangle are $13\sqrt2$ $\displaystyle 13\sqrt2$.

Possible Answers:

$\displaystyle 90$

$39\pi$ $\displaystyle 39\pi$

$52\pi$ $\displaystyle 52\pi$

$26\pi$ $\displaystyle 26\pi$

Correct answer:

$26\pi$ $\displaystyle 26\pi$

Explanation:

Notice that the hypotenuse of the triangle in the figure is also the diameter of the circle.

Use the Pythagorean theorem to find the length of the hypotenuse.

$\text{Hypotenuse}^2=\text{leg}^2+\text{leg}^2$ $\displaystyle \text{Hypotenuse}^2=\text{leg}^2+\text{leg}^2$

$\text{Hypotenuse}^2=2(\text{leg}^2)$ $\displaystyle \text{Hypotenuse}^2=2(\text{leg}^2)$

$\text{Hypotenuse}=\sqrt{2(\text{leg}^2)}=\text{leg}\sqrt2$ $\displaystyle \text{Hypotenuse}=\sqrt{2(\text{leg}^2)}=\text{leg}\sqrt2$

Substitute in the length of triangle’s legs to find the missing length of the hypotenuse.

$\text{Hypotenuse}=13\sqrt2(\sqrt2)$ $\displaystyle \text{Hypotenuse}=13\sqrt2(\sqrt2)$

Simplify.

$\text{Hypotenuse}=26$ $\displaystyle \text{Hypotenuse}=26$

Now, recall that the hypotenuse of the triangle and the diameter of the circle are the same:

$\text{Hypotenuse}=\text{Diameter}=26$ $\displaystyle \text{Hypotenuse}=\text{Diameter}=26$

Now, recall how to find the circumference of a circle:

$\text{Circumference}=\text{Diameter}\times\pi$ $\displaystyle \text{Circumference}=\text{Diameter}\times\pi$

Substitute in the value for the diameter to find the circumference of the circle.

$\text{Circumference}=26\pi$ $\displaystyle \text{Circumference}=26\pi$

Report an Error

Example Question #281 : Circles

Find the circumference of the circle if the lengths of the legs of the inscribed isosceles triangle are $\displaystyle 13$.

Possible Answers:

$39\sqrt2\pi$ $\displaystyle 39\sqrt2\pi$

$13\sqrt2\pi$ $\displaystyle 13\sqrt2\pi$

$\displaystyle 13$

$26\sqrt2\pi$ $\displaystyle 26\sqrt2\pi$

Correct answer:

$13\sqrt2\pi$ $\displaystyle 13\sqrt2\pi$

Explanation:

Notice that the hypotenuse of the triangle in the figure is also the diameter of the circle.

Use the Pythagorean theorem to find the length of the hypotenuse.

$\text{Hypotenuse}^2=\text{leg}^2+\text{leg}^2$ $\displaystyle \text{Hypotenuse}^2=\text{leg}^2+\text{leg}^2$

$\text{Hypotenuse}^2=2(\text{leg}^2)$ $\displaystyle \text{Hypotenuse}^2=2(\text{leg}^2)$

$\text{Hypotenuse}=\sqrt{2(\text{leg}^2)}=\text{leg}\sqrt2$ $\displaystyle \text{Hypotenuse}=\sqrt{2(\text{leg}^2)}=\text{leg}\sqrt2$

Substitute in the length of triangle’s legs to find the missing length of the hypotenuse.

$\text{Hypotenuse}=13\sqrt2$ $\displaystyle \text{Hypotenuse}=13\sqrt2$

Now, recall that the hypotenuse of the triangle and the diameter of the circle are the same:

$\text{Hypotenuse}=\text{Diameter}=13\sqrt2$ $\displaystyle \text{Hypotenuse}=\text{Diameter}=13\sqrt2$

Now, recall how to find the circumference of a circle:

$\text{Circumference}=\text{Diameter}\times\pi$ $\displaystyle \text{Circumference}=\text{Diameter}\times\pi$

Substitute in the value for the diameter to find the circumference of the circle.

$\text{Circumference}=13\sqrt2\pi$ $\displaystyle \text{Circumference}=13\sqrt2\pi$

Report an Error

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Basic Geometry : How to find circumference

Study concepts, example questions & explanations for Basic Geometry

All Basic Geometry Resources

Example Questions

Example Question #81 : How To Find Circumference

Example Question #271 : Radius

Example Question #81 : How To Find Circumference

Example Question #273 : Radius

Example Question #274 : Radius

Example Question #82 : How To Find Circumference

Example Question #275 : Radius

Example Question #276 : Radius

Example Question #272 : Radius

Example Question #281 : Circles

All Basic Geometry Resources

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