ISEE Upper Level Quantitative Reasoning - How to find the length of the diagonal of a square

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 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.

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.

Your new friend has a very small, square-shaped dorm room. She tells you that it is only 225 square feet. Assuming this is true, what is the diagonal distance from one corner of her room to the other?

$15\sqrt{2}ft$

$15\sqrt{3}ft$

Explanation

So, we need to find the diagonal of a square. First, we need to find the side length.

Let's begin with our formula for the area of a square:

where s is our side length and A is our area.

With this formula, we can solve for our side length by plugging in our area and square rooting both sides.

$s=\sqrt{225ft^2}=15ft$

Now, to find the diagonal, we can think of an isosceles right triangle, where the two equal sides are 15 ft. This is also a 45/45/90 triangle, which means the side lengths follow the ratio of $x, x , x \sqrt{2}$ .

This means our answer is $15 \sqrt{2}ft$ .

We could also find this by using Pythagorean Theorem.

$c^2=\sqrt{450ft^2}=\sqrt{15^2*2}ft=15\sqrt{2}ft$

While out walking, you find a strange, square-shaped piece of metal. If the side length of the piece is 26 inches, what is the length of the diagonal?

$26\sqrt{2} in$

$26\sqrt{3} in$

Explanation

While out walking, you find a strange, square-shaped piece of metal. If the side length of the piece is 26 inches, what is the length of the diagonal?

To find the diagonal of a square, we can recognize one of two things.

The diagonal of a square creates a right triangle, and we can use Pythagorean theorem to find our diagonal.
The diagonal of a square creates two 45/45/90 triangles, with side length ratios of $x/x/\sqrt{2}x$

Using 2), we can find that the diagonal of the square must be $26\sqrt{2} in$

What is the diagonal of a square with a side of 4?

$4\sqrt{2}$

$\sqrt{2}$

Explanation

Squares have all congruent sides. To find the diagonal, first recognize that you're dealing with an isoceles triangle when you draw the diagonal in the square. That means that two of the sides are congruent in the triangle. Thus, it's a special 45-45-90 triangle. In such triangles, the sides are x and the hypotenuse is $x\sqrt{2}$ . Since we know x is 4, we can plug in 4 to the expression $x\sqrt{2}$ . Thus, the answer is $4\sqrt{2}$ .

Track

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

Kenny begins at Point A, runs the path to Point C, and proceeds to run counterclockwise around the square track one complete time. He then runs again along the diagonal path from Point C to Point A.

Which is the greater quantity?

(a) The length of Kenny's run

(b) One mile

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

(b) is greater

(a) and (b) are equal

(a) is greater

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

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.

Kenny runs along this path twice, and he runs along the entire perimeter of the square path, so his run is about

$848 \times 2 + 600 \times 4 = 1,696 +2,400 = 4,096$ feet. Since one mile is equal to 5,280 feet, the greater quantity is (b).

You recently bought some special filter paper for a laboratory apparatus. The paper comes in square sheets, but you want to cut it into two equal triangle-shaped pieces. If the square sheets have a side length of , what will the length of the hypotenuse of the triangles be?

$15\sqrt{2} in$

Explanation

This problem is trying to distract you by thinking of triangles. What we are really asked to find here is the length of the diagonal of a square with sides of 15 inches.

Splitting a square along its diagonal yields two 45/45/90 triangles. If you know the ratios for 45/45/90 triangles, you can find the answer very quickly.

Think:

$45/45/90\rightarrow x/x/x\sqrt{2}$