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Basic Geometry : Quadrilaterals

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Example Questions

Example Question #261 : Quadrilaterals

Find the length of the diagonal of a square with side lengths of $\displaystyle 22$.

Possible Answers:

$11\sqrt2$ $\displaystyle 11\sqrt2$

$16\sqrt2$ $\displaystyle 16\sqrt2$

$22\sqrt2$ $\displaystyle 22\sqrt2$

$15\sqrt2$ $\displaystyle 15\sqrt2$

Correct answer:

$22\sqrt2$ $\displaystyle 22\sqrt2$

Explanation:

The diagonal of a square is also a hypotenuse of a right triangle with the side lengths as legs of the triangle.

Use the Pythagorean Theorem to find the length of the diagonal.

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

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

$\text{Diagonal}=(\text{side length})\sqrt2$ $\displaystyle \text{Diagonal}=(\text{side length})\sqrt2$

For the square given in the question,

$\text{Diagonal}=22\sqrt2$ $\displaystyle \text{Diagonal}=22\sqrt2$

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Example Question #262 : Quadrilaterals

Find the length of the diagonal of a square with a side length of $\displaystyle 12$.

Possible Answers:

$\displaystyle 12$

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

$16\sqrt2$ $\displaystyle 16\sqrt2$

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

Correct answer:

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

Explanation:

The diagonal of a square is also a hypotenuse of a right triangle with the side lengths as legs of the triangle.

Use the Pythagorean Theorem to find the length of the diagonal.

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

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

$\text{Diagonal}=(\text{side length})\sqrt2$ $\displaystyle \text{Diagonal}=(\text{side length})\sqrt2$

For the square given in the question,

$\text{Diagonal}=12\sqrt2$ $\displaystyle \text{Diagonal}=12\sqrt2$

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Example Question #261 : Quadrilaterals

Find the length of the diagonal of a square that has a side length of $\displaystyle 5$.

Possible Answers:

2.5 $\displaystyle 2.5$

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

$\displaystyle 5$

$10\sqrt2$ $\displaystyle 10\sqrt2$

Correct answer:

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

Explanation:

The diagonal of a square is also a hypotenuse of a right triangle with the side lengths as legs of the triangle.

Use the Pythagorean Theorem to find the length of the diagonal.

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

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

$\text{Diagonal}=(\text{side length})\sqrt2$ $\displaystyle \text{Diagonal}=(\text{side length})\sqrt2$

For the square given in the question,

$\text{Diagonal}=5\sqrt2$ $\displaystyle \text{Diagonal}=5\sqrt2$

Report an Error

Example Question #11 : How To Find The Length Of The Diagonal Of A Square

Find the length of the diagonal of a square that has a side length of 159 $\displaystyle 159$.

Possible Answers:

159 $\displaystyle 159$

The length of the diagonal cannot be determined.

$159\sqrt2$ $\displaystyle 159\sqrt2$

$169\sqrt2$ $\displaystyle 169\sqrt2$

Correct answer:

$159\sqrt2$ $\displaystyle 159\sqrt2$

Explanation:

The diagonal of a square is also a hypotenuse of a right triangle with the side lengths as legs of the triangle.

Use the Pythagorean Theorem to find the length of the diagonal.

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

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

$\text{Diagonal}=(\text{side length})\sqrt2$ $\displaystyle \text{Diagonal}=(\text{side length})\sqrt2$

For the square given in the question,

$\text{Diagonal}=159\sqrt2$ $\displaystyle \text{Diagonal}=159\sqrt2$

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Example Question #262 : Quadrilaterals

Find the length of the diagonal of a square with side lengths of $\displaystyle 27$.

Possible Answers:

$25\sqrt2$ $\displaystyle 25\sqrt2$

$9\sqrt2$ $\displaystyle 9\sqrt2$

$28\sqrt2$ $\displaystyle 28\sqrt2$

$27\sqrt2$ $\displaystyle 27\sqrt2$

Correct answer:

$27\sqrt2$ $\displaystyle 27\sqrt2$

Explanation:

The diagonal of a square is also a hypotenuse of a right triangle with the side lengths as legs of the triangle.

Use the Pythagorean Theorem to find the length of the diagonal.

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

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

$\text{Diagonal}=(\text{side length})\sqrt2$ $\displaystyle \text{Diagonal}=(\text{side length})\sqrt2$

For the square given in the question,

$\text{Diagonal}=27\sqrt2$ $\displaystyle \text{Diagonal}=27\sqrt2$

Report an Error

Example Question #15 : Squares

Find the length of the diagonal of a square that has a side length of $\displaystyle 23$.

Possible Answers:

$46\sqrt2$ $\displaystyle 46\sqrt2$

$54\sqrt2$ $\displaystyle 54\sqrt2$

$\displaystyle 23$

$23\sqrt2$ $\displaystyle 23\sqrt2$

Correct answer:

$23\sqrt2$ $\displaystyle 23\sqrt2$

Explanation:

The diagonal of a square is also a hypotenuse of a right triangle with the side lengths as legs of the triangle.

Use the Pythagorean Theorem to find the length of the diagonal.

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

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

$\text{Diagonal}=(\text{side length})\sqrt2$ $\displaystyle \text{Diagonal}=(\text{side length})\sqrt2$

For the square given in the question,

$\text{Diagonal}=23\sqrt2$ $\displaystyle \text{Diagonal}=23\sqrt2$

Report an Error

Example Question #16 : Squares

Find the length of the diagonal of a square with side lengths of 122 $\displaystyle 122$.

Possible Answers:

$122\sqrt2$ $\displaystyle 122\sqrt2$

$244\sqrt2$ $\displaystyle 244\sqrt2$

$61\sqrt2$ $\displaystyle 61\sqrt2$

122 $\displaystyle 122$

Correct answer:

$122\sqrt2$ $\displaystyle 122\sqrt2$

Explanation:

The diagonal of a square is also a hypotenuse of a right triangle with the side lengths as legs of the triangle.

Use the Pythagorean Theorem to find the length of the diagonal.

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

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

$\text{Diagonal}=(\text{side length})\sqrt2$ $\displaystyle \text{Diagonal}=(\text{side length})\sqrt2$

For the square given in the question,

$\text{Diagonal}=122\sqrt2$ $\displaystyle \text{Diagonal}=122\sqrt2$

Report an Error

Example Question #17 : Squares

Find the length of the diagonal of a square that has side lengths of $\displaystyle 35$.

Possible Answers:

$17\sqrt2$ $\displaystyle 17\sqrt2$

$\displaystyle 35$

$70\sqrt2$ $\displaystyle 70\sqrt2$

$35\sqrt2$ $\displaystyle 35\sqrt2$

Correct answer:

$35\sqrt2$ $\displaystyle 35\sqrt2$

Explanation:

The diagonal of a square is also a hypotenuse of a right triangle with the side lengths as legs of the triangle.

Use the Pythagorean Theorem to find the length of the diagonal.

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

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

$\text{Diagonal}=(\text{side length})\sqrt2$ $\displaystyle \text{Diagonal}=(\text{side length})\sqrt2$

For the square given in the question,

$\text{Diagonal}=35\sqrt2$ $\displaystyle \text{Diagonal}=35\sqrt2$

Report an Error

Example Question #262 : Quadrilaterals

Find the length of the diagonal of a square whose side length is $\displaystyle 8$.

Possible Answers:

$8\sqrt{2}$ $\displaystyle 8\sqrt{2}$

$16\sqrt{2}$ $\displaystyle 16\sqrt{2}$

$\displaystyle 16$

$\displaystyle 64$

Correct answer:

$8\sqrt{2}$ $\displaystyle 8\sqrt{2}$

Explanation:

To find the diagonal, you case use the pythagorean theorem or realize that this in isosceles triangle, and therefore the hypotenuse is

$s\sqrt{2}=8\sqrt{2}$ $\displaystyle s\sqrt{2}=8\sqrt{2}$

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Example Question #21 : How To Find The Length Of The Diagonal Of A Square

A square has side lengths of 15ft $\displaystyle 15ft$. Find the length of the diagonal.

Possible Answers:

27.4ft $\displaystyle 27.4ft$

21.2ft $\displaystyle 21.2ft$

20ft $\displaystyle 20ft$

23.2ft $\displaystyle 23.2ft$

24.5ft $\displaystyle 24.5ft$

Correct answer:

21.2ft $\displaystyle 21.2ft$

Explanation:

Finding the diagonal of a square is the same as finding the hypotenuse of a triangle, and uses the Pythagorean Theorem. (Imagine the square being sliced diagonally into two triangles, for you visual learners.) Within this theorem, both a and b are the same number, since the sides of a square are equal.

$\displaystyle a^2 + b^2 = c^2$

$\displaystyle 15^2 + 15^2 = c^2$

$\displaystyle 225 + 225 = c^2$

$\displaystyle 450 = c^2$

$\sqrt{}450 = c$ $\displaystyle \sqrt{}450 = c$

$\displaystyle 21.2 ft = c$

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Basic Geometry : Quadrilaterals

Study concepts, example questions & explanations for Basic Geometry

All Basic Geometry Resources

Example Questions

Example Question #261 : Quadrilaterals

Example Question #262 : Quadrilaterals

Example Question #261 : Quadrilaterals

Example Question #11 : How To Find The Length Of The Diagonal Of A Square

Example Question #262 : Quadrilaterals

Example Question #15 : Squares

Example Question #16 : Squares

Example Question #17 : Squares

Example Question #262 : Quadrilaterals

Example Question #21 : How To Find The Length Of The Diagonal Of A Square

All Basic Geometry Resources

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