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Basic Geometry : How to find the area of a square

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 #61 : How To Find The Area Of A Square

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

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Example Question #82 : Squares

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

Possible Answers:

$\frac{75}{2}$

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

$75\sqrt6$

$75\sqrt2$

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.

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

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Example Question #332 : Quadrilaterals

Find the area of a square if its diagonal is $6\sqrt7$ .

Possible Answers:

126

$252\sqrt2$

$63\sqrt3$

$126\sqrt3$

Correct answer:

126

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.

$\text{Area}=\frac{(6\sqrt7)^2}{2}=\frac{252}{2}=126$

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Example Question #341 : Quadrilaterals

Find the area of a square if its diagonal is $7\sqrt5$ .

Possible Answers:

$46\sqrt5$

$71\sqrt3$

$\frac{245\sqrt2}{4}$

$\frac{245}{2}$

Correct answer:

$\frac{245}{2}$

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.

$\text{Area}=\frac{(7\sqrt5)^2}{2}=\frac{245}{2}$

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Example Question #342 : Quadrilaterals

Find the area of a square if its diagonal is $9\sqrt{10}$ .

Possible Answers:

$405\sqrt2$

$\frac{405}{2}$

$405\sqrt3$

405

Correct answer:

405

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.

$\text{Area}=\frac{(9\sqrt{10})^2}{2}=\frac{810}{2}=405$

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Example Question #91 : Squares

Find the area of a square if its diagonal is $7\sqrt{10}$ .

Possible Answers:

245

$490\sqrt2$

$\frac{245}{2}$

$\frac{245\sqrt3}{2}$

Correct answer:

245

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.

$\text{Area}=\frac{(7\sqrt{10})^2}{2}=\frac{490}{2}=245$

Report an Error

Example Question #344 : Quadrilaterals

Find the area of a square if its diagonal is $6\sqrt{10}$ .

Possible Answers:

180

$360\sqrt3$

$90\sqrt3$

$180\sqrt2$

Correct answer:

180

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.

$\text{Area}=\frac{(6\sqrt{10})^2}{2}=\frac{360}{2}=180$

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Example Question #345 : Quadrilaterals

Find the area of a square if its diagonal is $8\sqrt{10}$ .

Possible Answers:

$320\sqrt2$

$320\sqrt3$

320

$160\sqrt3$

Correct answer:

320

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.

$\text{Area}=\frac{(8\sqrt{10})^2}{2}=\frac{640}{2}=320$

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Example Question #346 : Quadrilaterals

Find the area of a square if its diagonal is $11\sqrt2$ .

Possible Answers:

121

$\frac{121}{2}$

$121\sqrt3$

$\frac{121\sqrt2}{2}$

Correct answer:

121

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.

$\text{Area}=\frac{(11\sqrt{2})^2}{2}=\frac{242}{2}=121$

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Example Question #341 : Quadrilaterals

Find the area of a square if its diagonal is $15\sqrt{10}$ .

Possible Answers:

2250

1125

875

775

Correct answer:

1125

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.

$\text{Area}=\frac{(15\sqrt{10})^2}{2}=\frac{2250}{2}=1125$

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Basic Geometry : How to find the area of a square

Study concepts, example questions & explanations for Basic Geometry

All Basic Geometry Resources

Example Questions

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

Example Question #82 : Squares

Example Question #332 : Quadrilaterals

Example Question #341 : Quadrilaterals

Example Question #342 : Quadrilaterals

Example Question #91 : Squares

Example Question #344 : Quadrilaterals

Example Question #345 : Quadrilaterals

Example Question #346 : Quadrilaterals

Example Question #341 : Quadrilaterals

All Basic Geometry Resources

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