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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 #348 : Quadrilaterals

Find the area of a square if its diagonal is $12\sqrt{14}$ .

Possible Answers:

$504\sqrt{3}$

$1008\sqrt5$

1008

$1008\sqrt2$

Correct answer:

1008

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.

$\text{Area}=\frac{(12\sqrt{14})^2}{2}=\frac{2016}{2}=1008$

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Example Question #342 : Quadrilaterals

Find the area of a square if its diagonal is $\sqrt{30}$ .

Possible Answers:

$30\sqrt2$

$15\sqrt2$

Correct answer:

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.

$\text{Area}=\frac{(\sqrt{30})^2}{2}=\frac{30}{2}=15$

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Example Question #350 : Quadrilaterals

Find the area of a square if its diagonal is $2\sqrt{15}$ .

Possible Answers:

$30\sqrt2$

$30\sqrt3$

Correct answer:

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.

$\text{Area}=\frac{(2\sqrt{15})^2}{2}=\frac{60}{2}=30$

Report an Error

Example Question #351 : Quadrilaterals

Find the area of a square if it has a diagonal of $\sqrt5$ .

Possible Answers:

$\frac{5}{2}$

$5\sqrt2$

$\frac{15}{2}$

Correct answer:

$\frac{5}{2}$

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.

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

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

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

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

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

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

Simplify.

$\text{Area}=\frac{5}{2}$

Report an Error

Example Question #101 : Squares

Find the area of a square if it has a diagonal of $\sqrt3$ .

Possible Answers:

$\frac{3}{2}$

$\frac{3\sqrt2}{2}$

$3\sqrt2$

$\sqrt3$

Correct answer:

$\frac{3}{2}$

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.

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

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

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

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

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

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

Simplify.

$\text{Area}=\frac{3}{2}$

Report an Error

Example Question #353 : Quadrilaterals

Find the area of a square if it has a diagonal of $\sqrt6$ .

Possible Answers:

$\sqrt6$

$3\sqrt2$

$6\sqrt2$

Correct answer:

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.

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

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

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

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

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

$\text{Area}=\frac{(\sqrt6)^2}{2}$

Simplify.

$\text{Area}=\frac{6}{2}$

$\text{Area}=3$

Report an Error

Example Question #354 : Quadrilaterals

Find the area of a square if it has a diagonal of $\sqrt{10}$ .

Possible Answers:

$5\sqrt2$

Correct answer:

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.

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

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

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

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

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

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

Simplify.

$\text{Area}=\frac{10}{2}$

$\text{Area}=5$

Report an Error

Example Question #101 : Squares

Find the area of a square if it has a diagonal of $4\sqrt3$ .

Possible Answers:

$4\sqrt6$

$18\sqrt2$

Correct answer:

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.

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

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

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

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

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

$\text{Area}=\frac{(4\sqrt3)^2}{2}$

Simplify.

$\text{Area}=\frac{48}{2}$

$\text{Area}=24$

Report an Error

Example Question #761 : Plane Geometry

Find the area of a square if it has a diagonal of $2\sqrt3$ .

Possible Answers:

$8\sqrt2$

$12\sqrt2$

Correct answer:

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.

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

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

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

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

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

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

Simplify.

$\text{Area}=\frac{12}{2}$

$\text{Area}=6$

Report an Error

Example Question #357 : Quadrilaterals

Find the area of a square if it has a diagonal of $3\sqrt2$ .

Possible Answers:

Correct answer:

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.

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

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

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

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

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

$\text{Area}=\frac{(3\sqrt2)^2}{2}$

Simplify.

$\text{Area}=\frac{18}{2}$

$\text{Area}=9$

Report an Error

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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 #348 : Quadrilaterals

Example Question #342 : Quadrilaterals

Example Question #350 : Quadrilaterals

Example Question #351 : Quadrilaterals

Example Question #101 : Squares

Example Question #353 : Quadrilaterals

Example Question #354 : Quadrilaterals

Example Question #101 : Squares

Example Question #761 : Plane Geometry

Example Question #357 : Quadrilaterals

All Basic Geometry Resources

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