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Basic Geometry : Quadrilaterals

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

Example Questions

Example Question #102 : Squares

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

Possible Answers:

$\displaystyle 5$

$5\sqrt2$ $\displaystyle 5\sqrt2$

$\frac{5}{2}$ $\displaystyle \frac{5}{2}$

$\frac{15}{2}$ $\displaystyle \frac{15}{2}$

Correct answer:

$\frac{5}{2}$ $\displaystyle \frac{5}{2}$

Explanation:

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

Recall how to find the area of a square:

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

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

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

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

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

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

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

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

Simplify.

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

Report an Error

Example Question #103 : Squares

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

Possible Answers:

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

$3\sqrt2$ $\displaystyle 3\sqrt2$

$\frac{3}{2}$ $\displaystyle \frac{3}{2}$

$\sqrt3$ $\displaystyle \sqrt3$

Correct answer:

$\frac{3}{2}$ $\displaystyle \frac{3}{2}$

Explanation:

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

Recall how to find the area of a square:

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

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

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

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

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

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

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

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

Simplify.

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

Report an Error

Example Question #104 : Squares

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

Possible Answers:

$\sqrt6$ $\displaystyle \sqrt6$

$6\sqrt2$ $\displaystyle 6\sqrt2$

$3\sqrt2$ $\displaystyle 3\sqrt2$

$\displaystyle 3$

Correct answer:

$\displaystyle 3$

Explanation:

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

Recall how to find the area of a square:

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

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

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

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

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

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

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

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

Simplify.

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

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

Report an Error

Example Question #105 : Squares

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

Possible Answers:

$\displaystyle 15$

$\displaystyle 10$

$5\sqrt2$ $\displaystyle 5\sqrt2$

$\displaystyle 5$

Correct answer:

$\displaystyle 5$

Explanation:

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

Recall how to find the area of a square:

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

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

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

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

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

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

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

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

Simplify.

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

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

Report an Error

Example Question #106 : Squares

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

Possible Answers:

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

$\displaystyle 18$

$4\sqrt6$ $\displaystyle 4\sqrt6$

$\displaystyle 24$

Correct answer:

$\displaystyle 24$

Explanation:

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

Recall how to find the area of a square:

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

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

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

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

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

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

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

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

Simplify.

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

$\text{Area}=24$ $\displaystyle \text{Area}=24$

Report an Error

Example Question #107 : Squares

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

Possible Answers:

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

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

$\displaystyle 10$

$\displaystyle 6$

Correct answer:

$\displaystyle 6$

Explanation:

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

Recall how to find the area of a square:

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

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

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

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

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

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

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

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

Simplify.

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

$\text{Area}=6$ $\displaystyle \text{Area}=6$

Report an Error

Example Question #108 : Squares

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

Possible Answers:

$\displaystyle 9$

$\displaystyle 18$

$\displaystyle 6$

$\displaystyle 12$

Correct answer:

$\displaystyle 9$

Explanation:

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

Recall how to find the area of a square:

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

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

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

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

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

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

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

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

Simplify.

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

$\text{Area}=9$ $\displaystyle \text{Area}=9$

Report an Error

Example Question #81 : How To Find The Area Of A Square

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

Possible Answers:

$14\sqrt2$ $\displaystyle 14\sqrt2$

$7\sqrt2$ $\displaystyle 7\sqrt2$

$\displaystyle 14$

$7\sqrt3$ $\displaystyle 7\sqrt3$

Correct answer:

$\displaystyle 14$

Explanation:

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

Recall how to find the area of a square:

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

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

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

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

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

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

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

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

Simplify.

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

$\text{Area}=14$ $\displaystyle \text{Area}=14$

Report an Error

Example Question #82 : How To Find The Area Of A Square

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

Possible Answers:

$63\sqrt2$ $\displaystyle 63\sqrt2$

$9\sqrt3$ $\displaystyle 9\sqrt3$

$7\sqrt{10}$ $\displaystyle 7\sqrt{10}$

$\frac{63}{2}$ $\displaystyle \frac{63}{2}$

Correct answer:

$\frac{63}{2}$ $\displaystyle \frac{63}{2}$

Explanation:

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

Recall how to find the area of a square:

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

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

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

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

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

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

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

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

Simplify.

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

Report an Error

Example Question #111 : Squares

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

Possible Answers:

$\frac{13}{2}$ $\displaystyle \frac{13}{2}$

$\displaystyle 13$

$26\sqrt2$ $\displaystyle 26\sqrt2$

$13\sqrt3$ $\displaystyle 13\sqrt3$

Correct answer:

$\displaystyle 13$

Explanation:

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

Recall how to find the area of a square:

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

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

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

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

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

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

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

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

Simplify.

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

$\text{Area}=13$ $\displaystyle \text{Area}=13$

Report an Error

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Basic Geometry : Quadrilaterals

Study concepts, example questions & explanations for Basic Geometry

All Basic Geometry Resources

Example Questions

Example Question #102 : Squares

Example Question #103 : Squares

Example Question #104 : Squares

Example Question #105 : Squares

Example Question #106 : Squares

Example Question #107 : Squares

Example Question #108 : Squares

Example Question #81 : How To Find The Area Of A Square

Example Question #82 : How To Find The Area Of A Square

Example Question #111 : Squares

All Basic Geometry Resources

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