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Basic Geometry : Squares

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Example Questions

Example Question #461 : Quadrilaterals

Find the perimeter of a square given area is $\displaystyle 49$.

Possible Answers:

$\displaystyle 28$

$\displaystyle 24$

$\displaystyle 7$

$\displaystyle 49$

Correct answer:

$\displaystyle 28$

Explanation:

To find perimeter, we must first find the side length and then multiply is by $\displaystyle 4$.

$A=s^2\Rightarrow s=\sqrt{A}=\sqrt{49}=7$ $\displaystyle A=s^2\Rightarrow s=\sqrt{A}=\sqrt{49}=7$

$P=4s=4\cdot7=28$ $\displaystyle P=4s=4\cdot7=28$

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Example Question #872 : Basic Geometry

Find the area of the square.

Possible Answers:

900 $\displaystyle 900$

1200 $\displaystyle 1200$

1600 $\displaystyle 1600$

800 $\displaystyle 800$

Correct answer:

800 $\displaystyle 800$

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.

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

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

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

Recall how to find the area of a square.

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

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

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

Solve.

$\text{Area}=800$ $\displaystyle \text{Area}=800$

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Example Question #35 : How To Find The Perimeter Of A Square

A square has an area of 49in^2 $\displaystyle 49in^2$. Find the perimeter.

Possible Answers:

25in $\displaystyle 25in$

49in $\displaystyle 49in$

14in $\displaystyle 14in$

21in $\displaystyle 21in$

28in $\displaystyle 28in$

Correct answer:

28in $\displaystyle 28in$

Explanation:

Since a square has four equal sides, finding the side lengths from the area is quite simple.

$\displaystyle Area = Length(Width)$

Since all the sides are equal in a square, the length and width are equal, meaning that $\displaystyle Area = side ^2$. Plugging in the numbers gives us:

$\displaystyle 49 = side^2$

$\sqrt{}49 = side$ $\displaystyle \sqrt{}49 = side$

$\displaystyle 7 = side$

Now that we have found the length of each of the sides of the square, we simply add all the sides together, or multiply by 4 (because there are 4 sides in a square).

$\displaystyle 7(4) = 28in$

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Example Question #881 : Plane Geometry

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

Possible Answers:

$8\sqrt2$ $\displaystyle 8\sqrt2$

$\displaystyle 8$

$\displaystyle 12$

$\displaystyle 16$

Correct answer:

$\displaystyle 8$

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.

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

Simplify.

$\text{side}=2$ $\displaystyle \text{side}=2$

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.

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

Solve.

$\text{Perimeter}=8$ $\displaystyle \text{Perimeter}=8$

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Example Question #37 : How To Find The Perimeter Of A Square

Find the perimeter of a square inscribed in a circle that has a diameter of $3\sqrt2$ $\displaystyle 3\sqrt2$.

Possible Answers:

$\displaystyle 12$

$12\sqrt2$ $\displaystyle 12\sqrt2$

$\displaystyle 24$

$\displaystyle 16$

Correct answer:

$\displaystyle 12$

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.

$\text{side}=\frac{3\sqrt2(\sqrt2)}{2}$ $\displaystyle \text{side}=\frac{3\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=3$ $\displaystyle \text{side}=3$

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.

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

Solve.

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

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Example Question #38 : How To Find The Perimeter Of A Square

Find the perimeter of a square inscribed in a circle that has a diameter of $4\sqrt2$ $\displaystyle 4\sqrt2$.

Possible Answers:

$\displaystyle 16$

$\displaystyle 12$

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

$\displaystyle 24$

Correct answer:

$\displaystyle 16$

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.

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

Simplify.

$\text{side}=4$ $\displaystyle \text{side}=4$

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.

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

Solve.

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

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Example Question #39 : How To Find The Perimeter Of A Square

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

Possible Answers:

$\displaystyle 20$

$20\sqrt2$ $\displaystyle 20\sqrt2$

$\displaystyle 50$

$25\sqrt2$ $\displaystyle 25\sqrt2$

Correct answer:

$\displaystyle 20$

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.

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

Simplify.

$\text{side}=5$ $\displaystyle \text{side}=5$

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.

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

Solve.

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

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Example Question #40 : How To Find The Perimeter Of A Square

Find the perimeter of a square inscribed in a circle that has a diameter of $7\sqrt2$ $\displaystyle 7\sqrt2$.

Possible Answers:

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

$\displaystyle 98$

$\displaystyle 56$

$\displaystyle 28$

Correct answer:

$\displaystyle 28$

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.

$\text{side}=\frac{7\sqrt2(\sqrt2)}{2}$ $\displaystyle \text{side}=\frac{7\sqrt2(\sqrt2)}{2}$

Simplify.

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

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.

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

Solve.

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

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Example Question #881 : Basic Geometry

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

Possible Answers:

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

$\displaystyle 72$

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

$\displaystyle 36$

Correct answer:

$\displaystyle 36$

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.

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

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.

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

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Example Question #882 : Plane Geometry

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

Possible Answers:

$\displaystyle 22$

$\displaystyle 44$

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

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

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.

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

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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 #461 : Quadrilaterals

Example Question #872 : Basic Geometry

Example Question #35 : How To Find The Perimeter Of A Square

Example Question #881 : Plane Geometry

Example Question #37 : How To Find The Perimeter Of A Square

Example Question #38 : How To Find The Perimeter Of A Square

Example Question #39 : How To Find The Perimeter Of A Square

Example Question #40 : How To Find The Perimeter Of A Square

Example Question #881 : Basic Geometry

Example Question #882 : Plane Geometry

All Basic Geometry Resources

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