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Basic Geometry : How to find the perimeter of a square

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Example Questions

Example Question #471 : Quadrilaterals

Find the perimeter of a square inscribed in a circle that has a diameter of $9\sqrt2$ $\displaystyle 9\sqrt2$.

Possible Answers:

$36\sqrt2$ $\displaystyle 36\sqrt2$

$\displaystyle 72$

$18\sqrt2$ $\displaystyle 18\sqrt2$

$\displaystyle 36$

Correct answer:

$\displaystyle 36$

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.

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

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

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

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

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

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

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

$\text{side}=\frac{9\sqrt2(\sqrt2)}{2}$ $\displaystyle \text{side}=\frac{9\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=9$ $\displaystyle \text{side}=9$

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

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

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

$\text{Perimeter}=4(9)$ $\displaystyle \text{Perimeter}=4(9)$

Solve.

$\text{Perimeter}=36$ $\displaystyle \text{Perimeter}=36$

Report an Error

Example Question #472 : Quadrilaterals

Find the perimeter of a square inscribed in a circle that has a diameter of $11\sqrt2$ $\displaystyle 11\sqrt2$.

Possible Answers:

$121\sqrt2$ $\displaystyle 121\sqrt2$

$22\sqrt2$ $\displaystyle 22\sqrt2$

$\displaystyle 44$

$\displaystyle 22$

Correct answer:

$\displaystyle 44$

Explanation:

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

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

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

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

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

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

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

$\text{side}=\frac{11\sqrt2(\sqrt2)}{2}$ $\displaystyle \text{side}=\frac{11\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=11$ $\displaystyle \text{side}=11$

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

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

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

$\text{Perimeter}=4(11)$ $\displaystyle \text{Perimeter}=4(11)$

Solve.

$\text{Perimeter}=44$ $\displaystyle \text{Perimeter}=44$

Report an Error

Example Question #480 : Quadrilaterals

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

Possible Answers:

$52\sqrt3$ $\displaystyle 52\sqrt3$

$\displaystyle 52$

$52\sqrt2$ $\displaystyle 52\sqrt2$

104 $\displaystyle 104$

Correct answer:

$\displaystyle 52$

Explanation:

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

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

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

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

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

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

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

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

Simplify.

$\text{side}=13$ $\displaystyle \text{side}=13$

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

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

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

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

Solve.

$\text{Perimeter}=52$ $\displaystyle \text{Perimeter}=52$

Report an Error

Example Question #481 : Quadrilaterals

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

Possible Answers:

120 $\displaystyle 120$

$\displaystyle 90$

$\displaystyle 60$

180 $\displaystyle 180$

Correct answer:

$\displaystyle 60$

Explanation:

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

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

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

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

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

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

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

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

Simplify.

$\text{side}=15$ $\displaystyle \text{side}=15$

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

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

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

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

Solve.

$\text{Perimeter}=60$ $\displaystyle \text{Perimeter}=60$

Report an Error

Example Question #482 : Quadrilaterals

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

Possible Answers:

$\displaystyle 60$

$\displaystyle 36$

$\displaystyle 48$

$\displaystyle 24$

Correct answer:

$\displaystyle 48$

Explanation:

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

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

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

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

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

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

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

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

Simplify.

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

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

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

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

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

Solve.

$\text{Perimeter}=48$ $\displaystyle \text{Perimeter}=48$

Report an Error

Example Question #483 : Quadrilaterals

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

Possible Answers:

$\displaystyle 90$

$76\sqrt2$ $\displaystyle 76\sqrt2$

$\displaystyle 76$

104 $\displaystyle 104$

Correct answer:

$\displaystyle 76$

Explanation:

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

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

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

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

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

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

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

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

Simplify.

$\text{side}=19$ $\displaystyle \text{side}=19$

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

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

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

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

Solve.

$\text{Perimeter}=76$ $\displaystyle \text{Perimeter}=76$

Report an Error

Example Question #484 : Quadrilaterals

Find the perimeter of a square inscribed in a circle that has a diameter of $\displaystyle 20$.

Possible Answers:

$40\sqrt2$ $\displaystyle 40\sqrt2$

120 $\displaystyle 120$

$\displaystyle 80$

$160\sqrt2$ $\displaystyle 160\sqrt2$

Correct answer:

$40\sqrt2$ $\displaystyle 40\sqrt2$

Explanation:

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

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

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

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

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

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

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

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

Simplify.

$\text{side}=10\sqrt2$ $\displaystyle \text{side}=10\sqrt2$

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

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

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

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

Solve.

$\text{Perimeter}=40\sqrt2$ $\displaystyle \text{Perimeter}=40\sqrt2$

Report an Error

Example Question #485 : Quadrilaterals

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

Possible Answers:

144 $\displaystyle 144$

$36\sqrt2$ $\displaystyle 36\sqrt2$

$24\sqrt2$ $\displaystyle 24\sqrt2$

$48\sqrt2$ $\displaystyle 48\sqrt2$

Correct answer:

$24\sqrt2$ $\displaystyle 24\sqrt2$

Explanation:

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

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

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

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

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

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

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

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

Simplify.

$\text{side}=6\sqrt2$ $\displaystyle \text{side}=6\sqrt2$

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

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

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

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

Solve.

$\text{Perimeter}=24\sqrt2$ $\displaystyle \text{Perimeter}=24\sqrt2$

Report an Error

Example Question #486 : Quadrilaterals

Find the perimeter of a square inscribed in a circle that has a diameter of $\displaystyle 14$.

Possible Answers:

$28\sqrt2$ $\displaystyle 28\sqrt2$

$35\sqrt2$ $\displaystyle 35\sqrt2$

$14\sqrt2$ $\displaystyle 14\sqrt2$

$42\sqrt2$ $\displaystyle 42\sqrt2$

Correct answer:

$28\sqrt2$ $\displaystyle 28\sqrt2$

Explanation:

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

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

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

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

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

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

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

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

Simplify.

$\text{side}=7\sqrt2$ $\displaystyle \text{side}=7\sqrt2$

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

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

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

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

Solve.

$\text{Perimeter}=28\sqrt2$ $\displaystyle \text{Perimeter}=28\sqrt2$

Report an Error

Example Question #487 : Quadrilaterals

Find the perimeter of a square inscribed in a circle that has a diameter of $\displaystyle 16$.

Possible Answers:

$24\sqrt2$ $\displaystyle 24\sqrt2$

$32\sqrt2$ $\displaystyle 32\sqrt2$

$\displaystyle 64$

$16\sqrt2$ $\displaystyle 16\sqrt2$

Correct answer:

$32\sqrt2$ $\displaystyle 32\sqrt2$

Explanation:

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

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

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

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

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

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

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

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

Simplify.

$\text{side}=8\sqrt2$ $\displaystyle \text{side}=8\sqrt2$

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

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

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

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

Solve.

$\text{Perimeter}=32\sqrt2$ $\displaystyle \text{Perimeter}=32\sqrt2$

Report an Error

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Basic Geometry : How to find the perimeter of a square

Study concepts, example questions & explanations for Basic Geometry

All Basic Geometry Resources

Example Questions

Example Question #471 : Quadrilaterals

Example Question #472 : Quadrilaterals

Example Question #480 : Quadrilaterals

Example Question #481 : Quadrilaterals

Example Question #482 : Quadrilaterals

Example Question #483 : Quadrilaterals

Example Question #484 : Quadrilaterals

Example Question #485 : Quadrilaterals

Example Question #486 : Quadrilaterals

Example Question #487 : Quadrilaterals

All Basic Geometry Resources

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