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,

\(\displaystyle c=\sqrt{a^2+b^2}=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt2\)

To solve, simply use the Pythagorean Theorem. Thus,

\(\displaystyle c=\sqrt{a^2+b^2}=\sqrt{2^2+2^2}=\sqrt{2(2^2)}=2\sqrt{2}\)

Remember, that when simplifying square roots, you can only pull a number out if you have two factors of it. That is why I grouped the end of this answer the way I did so you could see that since I had two squared, I could pull one out, one disappears, and the two on the outside of the parenthesis remains under the radical.

Since the area is 25 square meters and we know it's a square, the length of each side is 5 meters, since 5 x 5 = 25.

To find the length of the diagonal, use the Pythagorean theorem:

\(\displaystyle 5^ 2 + 5^ 2 = d^2\)

\(\displaystyle 25 + 25 = d^2\)

\(\displaystyle 50 = d^2\)

\(\displaystyle \sqrt{50} = d\)

To solve, use the Pythagorean Theorem:

\(\displaystyle 3^2 + 3^ 2 = d^2\)

\(\displaystyle 9 + 9 = d^2\)

\(\displaystyle 18 = d^2\)

\(\displaystyle \sqrt{18 } = d\)

A square is shown below with its diagonal.

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 \(\displaystyle \sqrt{2}\) times the length of a leg. Since each side of the square measures 1, the diagonal has length \(\displaystyle \sqrt{2}\), not \(\displaystyle \sqrt{3}\) .

Recall that the side lengths of a square are each congruent; recall further that the angles in a square are each right angles. Hence, the diagonal \(\displaystyle d\) of the square \(\displaystyle ABCD\) may be considered the hypotenuse of a right isosceles triangle whose other two sides are \(\displaystyle AB\) and \(\displaystyle CB\). The side lengths of \(\displaystyle AB\) and \(\displaystyle CB\) are given as \(\displaystyle 2\) units; hence, we can deduce the diagonal \(\displaystyle d\) of the square \(\displaystyle ABCD\) by the Pythagorean Theorem, as shown:

\(\displaystyle a^2+b^2=c^2\)

\(\displaystyle (AB)^2+(CB)^2=d^2\)

\(\displaystyle (2)^2+(2)^2=d^2\)

\(\displaystyle 8=d^2\)

\(\displaystyle \Rightarrow d = \sqrt{8} = 2\sqrt{2}\).

Hence, the length of the diagonal \(\displaystyle d\) of the square \(\displaystyle ABCD\) is \(\displaystyle 2\sqrt{2}\) units.

Squares have four congruent sides. If one side is 12, the other sides must be as well. Since the diagonal makes a right triangle with the other sides we can use Pythagorean theorem to find the diagonal. 

\(\displaystyle a^2+b^2=c^2\)

\(\displaystyle 12^2+12^2=c^2\)

\(\displaystyle c^2=288\)

Take the square root of both sides:

\(\displaystyle \mathbf{c\approx17}\)

The length of the diagonal is 17 inches.

The formula for the area of a square is

\(\displaystyle Area = s^{2}\)

where \(\displaystyle s\) is the length of the sides. So the solution can be found by

\(\displaystyle Area=(10)^{2}=100 ft^{2}\)

Since \(\displaystyle ABCD\) is a square, the side \(\displaystyle AB\) is the same length as all of the other three sides.

We get area by multiplying length by width (or base by height, if you prefer), since all sides are equal, it looks like this:

\(\displaystyle 15\times15=225\)

Don't forget, the units are SQUARE INCHES.

Square ABCD contains both the red and green shapes. The red shape is equal to the area of one-fourth of the circle. Finding the area of square ABCD and subtracting only the area of the red shape will give the area of only the green shape.

Since ABCD is a square, angle BAC is a right angle that sits at the center of the circle (point A). Since a right angle is 90^o and a circle is 360^o, the red shape's area must be one quarter (or \(\displaystyle \frac{1}{4}\)) of the entire circle's area. Use the equation \(\displaystyle A = \pi r^2\) to find the area of the entire circle, then multiply this by \(\displaystyle \frac{1}{4}\) to find the area of only the red shape.

 \(\displaystyle A_{red} = (\pi (4)^2) * (\frac{1}{4})\)

\(\displaystyle A_{red} = 4\pi\)

Subtracting this from the area of the square gives the area of the green area outside of the circle.

\(\displaystyle A_{square}=s^2=4^2=16\)

\(\displaystyle A_{green}=A_{square}-A_{red}\)

\(\displaystyle A_{green}=16-4\pi\)