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Basic Geometry : Basic Geometry

Study concepts, example questions & explanations for Basic Geometry

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All Basic Geometry Resources

9 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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 #761 : Basic Geometry

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

Possible Answers:

$18\sqrt2$

$4\sqrt6$

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 #101 : Squares

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

Example Question #358 : Quadrilaterals

Find the area of a square if it has a diagonal of $2\sqrt7$ .

Possible Answers:

$7\sqrt2$

$7\sqrt3$

$14\sqrt2$

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.

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

Simplify.

$\text{Area}=\frac{28}{2}$

$\text{Area}=14$

Report an Error

Example Question #359 : Quadrilaterals

Find the area of a square if it has a diagonal of $3\sqrt7$ .

Possible Answers:

$63\sqrt2$

$7\sqrt{10}$

$9\sqrt3$

$\frac{63}{2}$

Correct answer:

$\frac{63}{2}$

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.

$\text{Area}=\frac{(3\sqrt7)^2}{2}$

Simplify.

$\text{Area}=\frac{63}{2}$

Report an Error

Example Question #761 : Plane Geometry

Find the area of a square if it has a diagonal of $\sqrt{26}$ .

Possible Answers:

$13\sqrt3$

$26\sqrt2$

$\frac{13}{2}$

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.

$\text{Area}=\frac{(\sqrt{26})^2}{2}$

Simplify.

$\text{Area}=\frac{26}{2}$

$\text{Area}=13$

Report an Error

Example Question #762 : Plane Geometry

Find the area of a square if it has a diagonal of $\sqrt{34}$ .

Possible Answers:

$17\sqrt3$

$17\sqrt2$

$\frac{17\sqrt2}{2}$

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.

$\text{Area}=\frac{(\sqrt{34})^2}{2}$

Simplify.

$\text{Area}=\frac{34}{2}$

$\text{Area}=17$

Report an Error

Example Question #763 : Plane Geometry

Find the area of a square if it has a diagonal of $12\sqrt3$ .

Possible Answers:

$108\sqrt2$

$216\sqrt3$

216

$432\sqrt2$

Correct answer:

216

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.

$\text{Area}=\frac{(12\sqrt3)^2}{2}$

Simplify.

$\text{Area}=\frac{432}{2}$

$\text{Area}=216$

Report an Error

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All Basic Geometry Resources

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Basic Geometry : Basic Geometry

Study concepts, example questions & explanations for Basic Geometry

All Basic Geometry Resources

Example Questions

Example Question #353 : Quadrilaterals

Example Question #354 : Quadrilaterals

Example Question #761 : Basic Geometry

Example Question #101 : Squares

Example Question #357 : Quadrilaterals

Example Question #358 : Quadrilaterals

Example Question #359 : Quadrilaterals

Example Question #761 : Plane Geometry

Example Question #762 : Plane Geometry

Example Question #763 : Plane Geometry

All Basic Geometry Resources

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