How to find the perimeter of a square

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Geometry › How to find the perimeter of a square

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1

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

$24\sqrt2$

CORRECT

$48\sqrt2$

0

$36\sqrt2$

0

0

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.

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

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.

$\text{Perimeter}=24\sqrt2$

2

Find the perimeter of a square inscribed in a circle that has a diameter of $2\sqrt2$ .

CORRECT

0

0

$8\sqrt2$

0

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.

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

Simplify.

$\text{side}=2$

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.

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

Solve.

$\text{Perimeter}=8$

3

Find the area of the square.

CORRECT

0

0

0

Explanation

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

Thus, we can use the Pythagorean Theorem to find the length of the sides of the square.

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

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

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

Recall how to find the area of a square.

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

Now, substitute in the value of the diagonal to find the area of the square.

$\text{Area}=\frac{40^2}{2}$

Solve.

$\text{Area}=800$

4

Find the perimeter of a square inscribed in a circle that has a diameter of $5\sqrt2$ .

CORRECT

$20\sqrt2$

0

$25\sqrt2$

0

0

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.

$\text{side}=\frac{5\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=5$

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.

$\text{Perimeter}=4(5)$

Solve.

$\text{Perimeter}=20$

5

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

$12\sqrt2$

CORRECT

$24\sqrt2$

0

0

0

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.

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

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.

$\text{Perimeter}=12\sqrt2$

6

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

$40\sqrt2$

CORRECT

0

$160\sqrt2$

0

0

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.

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

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.

$\text{Perimeter}=40\sqrt2$

7

Find the perimeter of a square with side length 2.

CORRECT

0

0

0

Explanation

To solve, simply use the formula for the perimeter of a square. Thus,

8

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

$36\sqrt2$

CORRECT

0

$48\sqrt2$

0

0

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.

$\text{side}=\frac{18(\sqrt2)}{2}$

Simplify.

$\text{side}=9\sqrt2$

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.

$\text{Perimeter}=4(9\sqrt2)$

Solve.

$\text{Perimeter}=36\sqrt2$

9

Find the perimeter of a square that has side lengths of .

CORRECT

0

0

0

Explanation

Use the following formula to find the perimeter of a square:

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

For the given square,

$\text{Perimeter}=4(102)=408$

10

If the diagonal of a square is $236\sqrt2$ , what is the perimeter of the square?

CORRECT

0

0

0

Explanation

The diagonal of a square is also a hypotenuse of a right triangle with the side lengths as legs of the triangle.

Use the Pythagorean Theorem to find the length of one side of the square.

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

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

$\text{Diagonal}=(\text{side length})\sqrt2$

$\text{Side length}=\frac{\text{Diagonal}}{\sqrt2}$

For the square given in the question,

$\text{Side length}=\frac{236\sqrt2}{\sqrt{2}}$

Simplify.

$\text{Side length}=236$

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

$\text{Perimeter}=4\times\text{side length}$

For the square in question,

$\text{Perimeter}=4\times 236$

Solve.

$\text{Perimeter}=944$

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