How to find the length of the diagonal of a square

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Math › How to find the length of the diagonal of a square

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The perimeter of a square is 48. What is the length of its diagonal?

$12\sqrt{2}$

$\sqrt{12}$

$24\sqrt{2}$

$48\sqrt{2}$

$2\sqrt{12}$

Explanation

Perimeter = side * 4

48 = side * 4

Side = 12

We can break up the square into two equal right triangles. The diagonal of the sqaure is then the hypotenuse of these two triangles.

Therefore, we can use the Pythagorean Theorem to solve for the diagonal:

$\sqrt{144+144}=\sqrt{hypotenuse^2}$

$hypotenuse=\sqrt{2\times 144}=\sqrt{2\times 12\times 12}=12\sqrt{2}$

Find the length of the square's diagonal.

Square_8

$8\sqrt2$

$4\sqrt5$

None of the other answers are correct.

Explanation

The diagonal line cuts the square into two equal triangles. Their hypotenuse is the diagonal of the square, so we can solve for the hypotenuse.

We need to use the Pythagorean Theorem: , where a and b are the legs and c is the hypotenuse.

The two legs have lengths of 8. Plug this in and solve for c:

$c=\sqrt{128}=\sqrt{2\cdot 64}=8\sqrt2$

The perimeter of a square is 48. What is the length of its diagonal?

$12\sqrt{2}$

$\sqrt{12}$

$24\sqrt{2}$

$48\sqrt{2}$

$2\sqrt{12}$

Explanation

Perimeter = side * 4

48 = side * 4

Side = 12

We can break up the square into two equal right triangles. The diagonal of the sqaure is then the hypotenuse of these two triangles.

Therefore, we can use the Pythagorean Theorem to solve for the diagonal:

$\sqrt{144+144}=\sqrt{hypotenuse^2}$

$hypotenuse=\sqrt{2\times 144}=\sqrt{2\times 12\times 12}=12\sqrt{2}$

Which is the greater quantity?

(a) The length of a diagonal of a square with sidelength 10 inches

(b) The hypotenuse of an isosceles right triangle with legs 10 inches each

(a) and (b) are equal.

(a) is greater.

(b) is greater.

It is impossible to tell which is greater from the information given.

Explanation

A diagonal of a square cuts the square into two isosceles right triangles, of which the diagonal is the common hypotenuse. Therefore, each figure is the hypotenuse of an isosceles right triangle with legs 10 inches, making them equal in length.

Track

The track at Franklin Pierce High School is a perfect square, as seen above, with sides of length 700 feet and a diagonal path connecting Points A and C.

Ellen wants to run three miles. Her plan is to begin at Point A, run along the diagonal path, run clockwise around the square track once, run along the diagonal path, run clockwise around the square track once, then repeat this pattern until she has run three miles. Where will she be when she is done?

A hint: $\sqrt{2} \approx 1.414$

On the diagonal path between Point A and Point C

On the square path between Point A and Point B

On the square path between Point B and Point C

On the square path between Point C and Point D

On the square path between Point D and Point A

Explanation

The diagonal of a square has length $\sqrt{2}$ , or about 1.414, times the length of a side, which here is 700 feet; this makes the diagonal path about

$700 \times 1.414 \approx 990$ feet long.

We will call one complete circuit one running of the diagonal, which is 990 feet long, and one running around the square; the completion of one complete circuit amounts to running a distance of

$990 + 4 \times 700 = 990 + 2,800 = 3,790$ feet.

Ellen seeks to run three miles, or

$5,280 \times 3 = 15, 840$ feet, which, divided by 3,790 feet, is about:

$15, 840 \div 3,790 \approx 4.17$ ,

or four complete circuits and 0.17 of a fifth.

After four complete circuits, Ellen is backat Point A. She has yet to run

$3,790 \times 0.17 = 629$ feet.

She will now run along the diagonal from Point A to Point C, but since the diagonal has length 990 feet, which is greater than 629 feet, she will finish running three miles when she is on this diagonal path.

Track

The track at Peter Stuyvesant High School is a perfect square, as seen above, with sides of length 600 feet and a diagonal connecting two of the corners.

Les begins at Point A, takes the diagonal path directly to Point B, then runs counterclockwise around the square track twice. He then takes the diagonal from Point B back to Point A. Which of the following is closest to the distance he runs?

A hint: $\sqrt{2} \approx 1.414$

$1\frac{1}{4} \textup{ mi}$

$1 \textup{ mi}$

$1\frac{1}{2} \textup{ mi}$

$1\frac{3}{4} \textup{ mi}$

$2 \textup{ mi}$

Explanation

The diagonal of a square has length $\sqrt{2}$ , or about 1.414, times the length of a side, which here is 600 feet; this makes the diagonal path about

$600 \times 1.414 \approx 848$ feet long.

Les runs around the square track twice, meaning that he runs the length of one side eight times; he also runs the length of the diagonal twice, This is a total of about

$600 \times 8 + 848 \times 2 = 4,800 + 1,696 = 6,496$ feet.

Divide by 5,280 to convert to miles:

$6,496 \div 5,280 \approx 1.23$

Of the given responses, $1\frac{1}{4}$ miles comes closest to the correct distance.

Find the length of the diagonal of a square that has side lengths of cm.

$4\sqrt{2}$ $\text{cm}$

$16\text{ cm}$

$4\text{ cm}$

$25\text{ cm}$

$10\text{ cm}$

Explanation

You can do this problem in two different ways that lead to the final answer:

1. Pythagorean Theorem

2. Special Triangles (45-45-90)

1. For the first idea, use the Pythagorean Theorem: , where a and b are the side lengths of the square and c is the length of the diagonal.

$c=\sqrt{32}$

$c=\sqrt{16*2}=\sqrt{16}, \times \sqrt{2} = 4\sqrt{2}$

2. If you know that ALL squares can be made into two special right triangles such that their angles are 45-45-90, then there's a formula you could use:

Let's say that your side length of the square is "a". Then the diagonal of the square (or the hypotenuse of the right triangle) will be $a\sqrt{2}$ .

So using this with a=4:

$4\sqrt{2}$

True or false: The length of a diagonal of a square with sides of length 1 is $\sqrt{3}$ .

False

True

Explanation

A square is shown below with its diagonal.

Square 1

Each of the triangles formed is an isosceles right triangle with congruent legs - by the 45-45-90 Triangle Theorem, they are 45-45-90 triangles. Also by the 45-45-90 Triangle Theorem, the diagonal, the hypotenuse of each triangle, measures $\sqrt{2}$ times the length of a leg. Since each side of the square measures 1, the diagonal has length $\sqrt{2}$ , not $\sqrt{3}$ .

Find the length of the diagonal of a square whose side length is 3.

$3\sqrt2$

Explanation

To find a diagonal of a square recall that the diagonal will create a triangle in the square for which it is the hypotenuse and the side lengths will be the other two lengths of the triangle.

To solve, simply use the Pythagorean Theorem to solve.

Thus,

$c=\sqrt{a^2+b^2}=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt2$