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Basic Geometry : Squares

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 #883 : Plane Geometry

Find the perimeter of a square inscribed in a circle that has a diameter of $13\sqrt2$ .

Possible Answers:

$52\sqrt3$

104

$52\sqrt2$

Correct answer:

Explanation:

Notice that the diameter of the circle is also the diagonal of the square. The diagonal of the square is also the hypotenuse of a right isosceles triangle that has the sides of the square as its legs.

$\text{diameter}=\text{diagonal}$

Now, use the Pythagorean theorem to find the length of the sides of the square.

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

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

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

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

Now, substitute in the value of the diagonal to find the length of a side of the square.

$\text{side}=\frac{13\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=13$

Now, recall how to find the perimeter of a square:

$\text{Perimeter}=4(\text{side})$

Substitute in the value of the side to find the perimeter of the square.

$\text{Perimeter}=4(13)$

Solve.

$\text{Perimeter}=52$

Report an Error

Example Question #231 : Squares

Find the perimeter of a square inscribed in a circle that has a diameter of $15\sqrt2$ .

Possible Answers:

180

120

Correct answer:

Explanation:

$\text{diameter}=\text{diagonal}$

Now, use the Pythagorean theorem to find the length of the sides of the square.

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

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

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

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

Now, substitute in the value of the diagonal to find the length of a side of the square.

$\text{side}=\frac{15\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=15$

Now, recall how to find the perimeter of a square:

$\text{Perimeter}=4(\text{side})$

Substitute in the value of the side to find the perimeter of the square.

$\text{Perimeter}=4(15)$

Solve.

$\text{Perimeter}=60$

Report an Error

Example Question #881 : Basic Geometry

Find the perimeter of a square inscribed in a circle that has a diameter of $12\sqrt2$ .

Possible Answers:

Correct answer:

Explanation:

$\text{diameter}=\text{diagonal}$

Now, use the Pythagorean theorem to find the length of the sides of the square.

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

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

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

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

Now, substitute in the value of the diagonal to find the length of a side of the square.

$\text{side}=\frac{12\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=12$

Now, recall how to find the perimeter of a square:

$\text{Perimeter}=4(\text{side})$

Substitute in the value of the side to find the perimeter of the square.

$\text{Perimeter}=4(12)$

Solve.

$\text{Perimeter}=48$

Report an Error

Example Question #41 : How To Find The Perimeter Of A Square

Find the perimeter of a square inscribed in a circle that has a diameter of $19\sqrt2$ .

Possible Answers:

104

$76\sqrt2$

Correct answer:

Explanation:

$\text{diameter}=\text{diagonal}$

Now, use the Pythagorean theorem to find the length of the sides of the square.

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

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

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

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

Now, substitute in the value of the diagonal to find the length of a side of the square.

$\text{side}=\frac{19\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=19$

Now, recall how to find the perimeter of a square:

$\text{Perimeter}=4(\text{side})$

Substitute in the value of the side to find the perimeter of the square.

$\text{Perimeter}=4(19)$

Solve.

$\text{Perimeter}=76$

Report an Error

Example Question #484 : Quadrilaterals

Find the perimeter of a square inscribed in a circle that has a diameter of .

Possible Answers:

$40\sqrt2$

$160\sqrt2$

120

Correct answer:

$40\sqrt2$

Explanation:

$\text{diameter}=\text{diagonal}$

Now, use the Pythagorean theorem to find the length of the sides of the square.

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

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

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

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

Now, substitute in the value of the diagonal to find the length of a side of the square.

$\text{side}=\frac{20(\sqrt2)}{2}$

Simplify.

$\text{side}=10\sqrt2$

Now, recall how to find the perimeter of a square:

$\text{Perimeter}=4(\text{side})$

Substitute in the value of the side to find the perimeter of the square.

$\text{Perimeter}=4(10\sqrt2)$

Solve.

$\text{Perimeter}=40\sqrt2$

Report an Error

Example Question #485 : Quadrilaterals

Find the perimeter of a square inscribed in a circle that has a diameter of .

Possible Answers:

144

$24\sqrt2$

$36\sqrt2$

$48\sqrt2$

Correct answer:

$24\sqrt2$

Explanation:

$\text{diameter}=\text{diagonal}$

Now, use the Pythagorean theorem to find the length of the sides of the square.

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

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

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

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

Now, substitute in the value of the diagonal to find the length of a side of the square.

$\text{side}=\frac{12(\sqrt2)}{2}$

Simplify.

$\text{side}=6\sqrt2$

Now, recall how to find the perimeter of a square:

$\text{Perimeter}=4(\text{side})$

Substitute in the value of the side to find the perimeter of the square.

$\text{Perimeter}=4(6\sqrt2)$

Solve.

$\text{Perimeter}=24\sqrt2$

Report an Error

Example Question #486 : Quadrilaterals

Find the perimeter of a square inscribed in a circle that has a diameter of .

Possible Answers:

$35\sqrt2$

$42\sqrt2$

$28\sqrt2$

$14\sqrt2$

Correct answer:

$28\sqrt2$

Explanation:

$\text{diameter}=\text{diagonal}$

Now, use the Pythagorean theorem to find the length of the sides of the square.

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

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

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

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

Now, substitute in the value of the diagonal to find the length of a side of the square.

$\text{side}=\frac{14(\sqrt2)}{2}$

Simplify.

$\text{side}=7\sqrt2$

Now, recall how to find the perimeter of a square:

$\text{Perimeter}=4(\text{side})$

Substitute in the value of the side to find the perimeter of the square.

$\text{Perimeter}=4(7\sqrt2)$

Solve.

$\text{Perimeter}=28\sqrt2$

Report an Error

Example Question #487 : Quadrilaterals

Find the perimeter of a square inscribed in a circle that has a diameter of .

Possible Answers:

$24\sqrt2$

$16\sqrt2$

$32\sqrt2$

Correct answer:

$32\sqrt2$

Explanation:

$\text{diameter}=\text{diagonal}$

Now, use the Pythagorean theorem to find the length of the sides of the square.

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

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

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

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

Now, substitute in the value of the diagonal to find the length of a side of the square.

$\text{side}=\frac{16(\sqrt2)}{2}$

Simplify.

$\text{side}=8\sqrt2$

Now, recall how to find the perimeter of a square:

$\text{Perimeter}=4(\text{side})$

Substitute in the value of the side to find the perimeter of the square.

$\text{Perimeter}=4(8\sqrt2)$

Solve.

$\text{Perimeter}=32\sqrt2$

Report an Error

Example Question #488 : Quadrilaterals

Find the perimeter of a square inscribed in a circle that has a diameter of .

Possible Answers:

$8\sqrt2$

$4\sqrt2$

$10\sqrt2$

Correct answer:

$8\sqrt2$

Explanation:

$\text{diameter}=\text{diagonal}$

Now, use the Pythagorean theorem to find the length of the sides of the square.

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

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

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

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

Now, substitute in the value of the diagonal to find the length of a side of the square.

$\text{side}=\frac{4(\sqrt2)}{2}$

Simplify.

$\text{side}=2\sqrt2$

Now, recall how to find the perimeter of a square:

$\text{Perimeter}=4(\text{side})$

Substitute in the value of the side to find the perimeter of the square.

$\text{Perimeter}=4(2\sqrt2)$

Solve.

$\text{Perimeter}=8\sqrt2$

Report an Error

Example Question #489 : Quadrilaterals

Find the perimeter of a square inscribed in a circle that has a diameter of .

Possible Answers:

$24\sqrt2$

$12\sqrt2$

Correct answer:

$12\sqrt2$

Explanation:

$\text{diameter}=\text{diagonal}$

Now, use the Pythagorean theorem to find the length of the sides of the square.

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

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

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

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

Now, substitute in the value of the diagonal to find the length of a side of the square.

$\text{side}=\frac{6(\sqrt2)}{2}$

Simplify.

$\text{side}=3\sqrt2$

Now, recall how to find the perimeter of a square:

$\text{Perimeter}=4(\text{side})$

Substitute in the value of the side to find the perimeter of the square.

$\text{Perimeter}=4(3\sqrt2)$

Solve.

$\text{Perimeter}=12\sqrt2$

Report an Error

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All Basic Geometry Resources

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Basic Geometry : Squares

Study concepts, example questions & explanations for Basic Geometry

All Basic Geometry Resources

Example Questions

Example Question #883 : Plane Geometry

Example Question #231 : Squares

Example Question #881 : Basic Geometry

Example Question #41 : How To Find The Perimeter Of A Square

Example Question #484 : Quadrilaterals

Example Question #485 : Quadrilaterals

Example Question #486 : Quadrilaterals

Example Question #487 : Quadrilaterals

Example Question #488 : Quadrilaterals

Example Question #489 : Quadrilaterals

All Basic Geometry Resources

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