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Basic Geometry : Squares

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Example Questions

Example Question #241 : Squares

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

Possible Answers:

$36\sqrt2$

$48\sqrt2$

Correct answer:

$36\sqrt2$

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.

$\text{Perimeter}=36\sqrt2$

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Example Question #899 : Basic Geometry

Find the perimeter of a square inscribed in a circle that has a diameter of $14\sqrt3$ .

Possible Answers:

$28\sqrt5$

$28\sqrt2$

$28\sqrt6$

Correct answer:

$28\sqrt6$

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.

$\text{side}=\frac{14\sqrt3(\sqrt2)}{2}$

Simplify.

$\text{side}=7\sqrt6$

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.

$\text{Perimeter}=4(7\sqrt6)$

Solve.

$\text{Perimeter}=28\sqrt6$

Report an Error

Example Question #900 : Basic Geometry

Find the perimeter of a square inscribed in a circle that has a diameter of 100 .

Possible Answers:

$400\sqrt2$

$200\sqrt2$

10000

$100\sqrt2$

Correct answer:

$200\sqrt2$

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.

$\text{side}=\frac{100(\sqrt2)}{2}$

Simplify.

$\text{side}=50\sqrt2$

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.

$\text{Perimeter}=4(50\sqrt2)$

Solve.

$\text{Perimeter}=200\sqrt2$

Report an Error

Example Question #491 : Quadrilaterals

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

Possible Answers:

200

160

$80\sqrt2$

$40\sqrt2$

Correct answer:

$80\sqrt2$

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.

$\text{side}=\frac{40(\sqrt2)}{2}$

Simplify.

$\text{side}=20\sqrt2$

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.

$\text{Perimeter}=4(20\sqrt2)$

Solve.

$\text{Perimeter}=80\sqrt2$

Report an Error

Example Question #492 : Quadrilaterals

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

Possible Answers:

176

$44\sqrt2$

$88\sqrt2$

$132\sqrt2$

Correct answer:

$88\sqrt2$

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.

$\text{side}=\frac{44(\sqrt2)}{2}$

Simplify.

$\text{side}=22\sqrt2$

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.

$\text{Perimeter}=4(22\sqrt2)$

Solve.

$\text{Perimeter}=88\sqrt2$

Report an Error

Example Question #493 : Quadrilaterals

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

Possible Answers:

$120\sqrt2$

$180\sqrt2$

$60\sqrt2$

240

Correct answer:

$120\sqrt2$

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.

$\text{side}=\frac{60(\sqrt2)}{2}$

Simplify.

$\text{side}=30\sqrt2$

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.

$\text{Perimeter}=4(30\sqrt2)$

Solve.

$\text{Perimeter}=120\sqrt2$

Report an Error

Example Question #494 : Quadrilaterals

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

Possible Answers:

$152\sqrt2$

$144\sqrt2$

304

$76\sqrt2$

Correct answer:

$152\sqrt2$

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.

$\text{side}=\frac{76(\sqrt2)}{2}$

Simplify.

$\text{side}=38\sqrt2$

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.

$\text{Perimeter}=4(38\sqrt2)$

Solve.

$\text{Perimeter}=152\sqrt2$

Report an Error

Example Question #901 : Basic Geometry

Find the perimeter of a square whose side length is 7.

Possible Answers:

Correct answer:

Explanation:

The perimeter of a square is found by adding all the side lengths together. In a square there are two widths, two lengths, and they are equal.

In other words, simply use the formula for the perimeter of a square and let,

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

s=7

thus,

P=4s=4*7=28 .

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Example Question #61 : How To Find The Perimeter Of A Square

Find the perimeter of a square with side length 2.

Possible Answers:

Correct answer:

Explanation:

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

A=4s=4*2=8

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Example Question #243 : Squares

Find the perimeter of a square with side length of 2.

Possible Answers:

$2\sqrt2$

Correct answer:

Explanation:

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

P=4s=4*2=8

If the formula escapes you, simply sum the sides. However, you must know that all the sides in a square are equal in order to add them up. Since they are equal, the formula just shows you that you can multiply one side by 4 instead of adding them all together.

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Basic Geometry : Squares

Study concepts, example questions & explanations for Basic Geometry

All Basic Geometry Resources

Example Questions

Example Question #241 : Squares

Example Question #899 : Basic Geometry

Example Question #900 : Basic Geometry

Example Question #491 : Quadrilaterals

Example Question #492 : Quadrilaterals

Example Question #493 : Quadrilaterals

Example Question #494 : Quadrilaterals

Example Question #901 : Basic Geometry

Example Question #61 : How To Find The Perimeter Of A Square

Example Question #243 : Squares

All Basic Geometry Resources

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