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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 #81 : Squares

If the diagonal of a square is , what is the area of the square?

Possible Answers:

100

Correct answer:

Explanation:

The diagonal of a square is also the hypotenuse of a triangle whose legs are the sides of the square.

Thus, from knowing the length of the diagonal, we can use Pythagorean's Theorem to figure out the side lengths of the square.

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

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

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

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

We can now find the side length of the square in question.

$\text{side length}=\frac{10}{\sqrt2}$

Multiply the fraction by one in the form: $\frac{\sqrt{2}}\sqrt{2}{}$ .

$\text{side length}=\frac{10}{\sqrt2}\times\frac{\sqrt{2}}\sqrt{2}{}$

Simplify.

$\text{side length}=\frac{10\sqrt2}{2}$

Reduce.

$\text{side length}=5\sqrt2$

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

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

For the square in question,

$\text{Area }=(5\sqrt2) ^2$

Solve.

$\text{Area }=50$

Report an Error

Example Question #82 : Squares

If the diagonal of a square is , what is the area of the square?

Possible Answers:

Correct answer:

Explanation:

The diagonal of a square is also the hypotenuse of a triangle whose legs are the sides of the square.

Thus, from knowing the length of the diagonal, we can use Pythagorean's Theorem to figure out the side lengths of the square.

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

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

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

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

We can now find the side length of the square in question.

$\text{side length}=\frac{12}{\sqrt2}$

Multiply the fraction by one in the form: $\frac{\sqrt{2}}\sqrt{2}{}$ .

$\text{side length}=\frac{12}{\sqrt2}\times\frac{\sqrt{2}}\sqrt{2}{}$

Simplify.

$\text{side length}=\frac{12\sqrt2}{2}$

Reduce.

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

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

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

For the square in question,

$\text{Area }=(6\sqrt2) ^2$

Solve.

$\text{Area }=72$

Report an Error

Example Question #83 : Squares

If the diagonal of a square is , what is the area of the square?

Possible Answers:

105

Correct answer:

Explanation:

The diagonal of a square is also the hypotenuse of a triangle whose legs are the sides of the square.

Thus, from knowing the length of the diagonal, we can use Pythagorean's Theorem to figure out the side lengths of the square.

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

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

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

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

We can now find the side length of the square in question.

$\text{side length}=\frac{14}{\sqrt2}$

Multiply the fraction by one in the form: $\frac{\sqrt{2}}\sqrt{2}$ .

$\text{side length}=\frac{14}{\sqrt2}\times\frac{\sqrt{2}}\sqrt{2}$

Simplify.

$\text{side length}=\frac{14\sqrt2}{2}$

Reduce.

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

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

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

For the square in question,

$\text{Area }=(7\sqrt2) ^2$

Solve.

$\text{Area }=98$

Report an Error

Example Question #81 : Squares

If the diagonal of a square is $30\sqrt2$ , what is the area of the square?

Possible Answers:

300

900

600

120

Correct answer:

900

Explanation:

The diagonal of a square is also the hypotenuse of a triangle whose legs are the sides of the square.

Thus, from knowing the length of the diagonal, we can use Pythagorean's Theorem to figure out the side lengths of the square.

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

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

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

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

We can now find the side length of the square in question.

$\text{side length}=\frac{30\sqrt2}{\sqrt2}$

Simplify.

$\text{side length}=30$

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

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

For the square in question,

$\text{Area }=30 ^2$

Solve.

$\text{Area }=900$

Report an Error

Example Question #81 : Squares

If the diagonal of a square is $31\sqrt2$ , what is the area of the square?

Possible Answers:

661

961

696

1204

Correct answer:

961

Explanation:

The diagonal of a square is also the hypotenuse of a triangle whose legs are the sides of the square.

Thus, from knowing the length of the diagonal, we can use Pythagorean's Theorem to figure out the side lengths of the square.

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

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

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

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

We can now find the side length of the square in question.

$\text{side length}=\frac{31\sqrt2}{\sqrt2}$

Simplify.

$\text{side length}=31$

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

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

For the square in question,

$\text{Area }=31 ^2$

Solve.

$\text{Area }=961$

Report an Error

Example Question #82 : Squares

If the diagonal of a square is $32\sqrt2$ , what is the area of the square?

Possible Answers:

900

964

1224

1024

Correct answer:

1024

Explanation:

The diagonal of a square is also the hypotenuse of a triangle whose legs are the sides of the square.

Thus, from knowing the length of the diagonal, we can use Pythagorean's Theorem to figure out the side lengths of the square.

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

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

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

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

We can now find the side length of the square in question.

$\text{side length}=\frac{32\sqrt2}{\sqrt2}$

Simplify.

$\text{side length}=32$

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

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

For the square in question,

$\text{Area }=32 ^2$

Solve.

$\text{Area }=1024$

Report an Error

Example Question #87 : Squares

If the diagonal of a square is $34\sqrt2$ , what is the area of the square?

Possible Answers:

1113

1156

1354

1264

Correct answer:

1156

Explanation:

The diagonal of a square is also the hypotenuse of a triangle whose legs are the sides of the square.

Thus, from knowing the length of the diagonal, we can use Pythagorean's Theorem to figure out the side lengths of the square.

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

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

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

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

We can now find the side length of the square in question.

$\text{side length}=\frac{34\sqrt2}{\sqrt2}$

Simplify.

$\text{side length}=34$

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

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

For the square in question,

$\text{Area }=34 ^2$

Solve.

$\text{Area }=1156$

Report an Error

Example Question #88 : Squares

If the diagonal of a square is $\frac{\sqrt2}{4}$ , what is the area of the square?

Possible Answers:

$\frac{1}{2}$

$\frac{1}{8}$

$\frac{1}{32}$

$\frac{1}{16}$

Correct answer:

$\frac{1}{16}$

Explanation:

The diagonal of a square is also the hypotenuse of a triangle whose legs are the sides of the square.

Thus, from knowing the length of the diagonal, we can use Pythagorean's Theorem to figure out the side lengths of the square.

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

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

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

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

We can now find the side length of the square in question.

$\text{side length}=\frac{\frac{\sqrt2}{4}}{\sqrt2}=\frac{1}{4}$

Dividing by a number is the same as multiplying by its reciprocal. Rewrite.

$\text{side length}=\frac{\sqrt2}{4}} \times \frac{1}{\sqrt{2}}$

Simplify.

$\text{side length}=\frac{\sqrt{2}}{4\sqrt{2}}$

Multiply the fraction by one in the form: $\frac{\sqrt{2}}\sqrt{2}{}$ .

$\text{side length}=\frac{\sqrt{2}}{4\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}$

Simplify.

$\text{side length}=\frac{2}{8}$

Reduce.

$\text{side length}=\frac{1}{4}$

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

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

For the square in question,

$\text{Area }=\left ( \frac{1}{4} \right )^2$

Solve.

$\text{Area }=\frac{1}{16}$

Report an Error

Example Question #81 : Squares

Find the area of a square whose side length is .

Possible Answers:

Correct answer:

Explanation:

To find the area of a square, use the following formula where is side length.

A=s^2=6^2=36

Report an Error

Example Question #90 : Squares

Find the area of a square if its diagonal is $5\sqrt3$

Possible Answers:

$\frac{75\sqrt3}{2}$

$75\sqrt6$

$75\sqrt2$

$\frac{75}{2}$

Correct answer:

$\frac{75}{2}$

Explanation:

The diagonal of a square is also the hypotenuse of a 45-45-90 triangle.

Recall how to find the area of a square:

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

Now, use the Pythagorean theorem to find the area of the square.

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

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

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

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

Plug in the length of the diagonal to find the area of the square.

$\text{Area}=\frac{(5\sqrt3)^2}{2}=\frac{75}{2}$

Report an Error

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Basic Geometry : Squares

Study concepts, example questions & explanations for Basic Geometry

All Basic Geometry Resources

Example Questions

Example Question #81 : Squares

Example Question #82 : Squares

Example Question #83 : Squares

Example Question #81 : Squares

Example Question #81 : Squares

Example Question #82 : Squares

Example Question #87 : Squares

Example Question #88 : Squares

Example Question #81 : Squares

Example Question #90 : Squares

All Basic Geometry Resources

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