Since the question is asking for area of the square with side length AB, recall the formula for the area of a square.

\(\displaystyle A=\text{side}^2\)

Now, use the distance formula to calculate the length of AB.

\(\displaystyle d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

let 

\(\displaystyle \\(x_1,y_1)=(0,a) \\(x_2,y_2)=(b,0)\)

Now substitute the values into the distance formula.

\(\displaystyle \\d=\sqrt{(b-0)^2+(0-2)^2} \\d=\sqrt{b^2+a^2}\)

From here substitute the side length value into the area formula.

\(\displaystyle \\A=\text{side}^2 \\A=\left(\sqrt{b^2+a^2}\right)^2 \\A=b^2+a^2\)

The area of a triangle, given its three sidelengths, can be calculated using Heron's formula:

\(\displaystyle A = \sqrt{s(s-a)(s-b)(s-c)}\),

where \(\displaystyle a\), \(\displaystyle b\), and \(\displaystyle c\) are the lengths of the sides, and \(\displaystyle s = \frac{a+b+ c}{2}\).

Setting, \(\displaystyle a = AB = 13\), \(\displaystyle b = AC = 17\), and \(\displaystyle c= BC = 25\),

\(\displaystyle s = \frac{13+17+25}{2} = \frac{55}{2} = 27.5\)

and, substituting in Heron's formula:

\(\displaystyle A = \sqrt{27.5(27.5-13)(27.5-17)(27.5-25 )}\)

\(\displaystyle = \sqrt{27.5(14.5)(10.5)(2.5 )}\)

\(\displaystyle = \sqrt{10,467.1875}\)

\(\displaystyle \approx 102\)

One side of the white trapezoid is the hypotenuse of the small top triangle, which has legs 10 and 20. Therefore, the length of this side can be determined using the Pythagorean Theorem:

\(\displaystyle c = \sqrt{10^{2}+ 20^{2}} = \sqrt{100 + 400} = \sqrt{500} = \sqrt{100} \cdot \sqrt{5} = 10 \sqrt{5}\)

The trapezoid has perimeter 

\(\displaystyle 30 + 20 + 20 + 10 \sqrt{5} = 70 + 10 \sqrt{5}\).

The small top triangle has legs 10 and 20 and hypotenuse \(\displaystyle 10 \sqrt{5}\), and, therefore, perimeter

\(\displaystyle 10 + 20 + 10 \sqrt{5} = 30 + 10 \sqrt{5}\). The blue triangle, which is similar as a result of the parallelism of the opposite sides of the square, has short leg 20. Since the perimeter of two similar triangles is directly proportional to a side, we can set up and solve the proportion statement to find the perimeter of the blue triangle:

\(\displaystyle \frac{P}{30 + 10 \sqrt{5}} = \frac{20}{10}\)

\(\displaystyle \frac{P}{30 + 10 \sqrt{5}}\left ( 30 + 10 \sqrt{5} \right ) = \frac{20}{10}\left ( 30 + 10 \sqrt{5} \right )\)

\(\displaystyle P =60 + 20 \sqrt{5}\)

The difference between the perimeters is

\(\displaystyle \left (60 + 20 \sqrt{5} \right )- \left (70 + 10 \sqrt{5} \right )=-10 +10 \sqrt{5}\)

The altitude of a right triangle from the vertex of its right triangle to its hypotenuse divides it into two similar triangles.

\(\displaystyle BD\), as the length of the altitude corresponding to the hypotenuse, is the geometric mean of the lengths of the parts of the hypotenuse it forms; that is, it is the square root of the product of the two:

\(\displaystyle BD =\sqrt{ \left ( AD\right ) \left ( DC\right )} = \sqrt{6 \cdot 24} = \sqrt{144} = 12\).

The ratio of the smaller side of \(\displaystyle \Delta BDC\) to that of \(\displaystyle \Delta ADB\) is 

\(\displaystyle \frac{BD}{AD}=\frac{12}{6} = \frac{2}{1}\) or 2:1, making this the similarity ratio. Theratio of the perimeters is always equal to the similarity ratio, so 2:1 is also the ratio of the perimeter of \(\displaystyle \Delta BDC\) to that of \(\displaystyle \Delta ADB\).

\(\displaystyle \overline{BE }\) is both one side of Square \(\displaystyle BCDE\) and the hypotenuse of \(\displaystyle \Delta ABE\);  its hypotenuse can be calculated from the lengths of the legs using the Pythagorean Theorem:

\(\displaystyle BE = \sqrt{(AB)^{2}+ \left ( AE\right )^{2}}\)

\(\displaystyle BE = \sqrt{x^{2}+ \left ( 2x\right )^{2}} = \sqrt{x^{2}+ 4x ^{2}} = \sqrt{5x ^{2}} = x\sqrt{5}\)

Therefore, \(\displaystyle BC = CD = DE = x\sqrt{5}\).

The perimeter of Polygon \(\displaystyle ABCDE\) is

\(\displaystyle P = AB +BC + CD + DE + EA\)

\(\displaystyle P=x + x\sqrt{5} + x\sqrt{5} + x\sqrt{5}+ 2x\)

\(\displaystyle P=3x + 3x\sqrt{5} = \left ( 3 + 3\sqrt{5}\right )x\)

The area of a rectangle is length times width.

\(\displaystyle A=LW\)

Determine all the integer combinations that will multiply to an area of 16.  The numbers can represent length or width interchangably.

\(\displaystyle (1,16),(2,8),(4,4)\)

Write the perimeter formula for a rectangle.

\(\displaystyle P=2(L+W)\)

Substitute all the following combinations to determine the perimeters.

\(\displaystyle P=2(1+16)= 2(17)=34\)

\(\displaystyle P=2(2+8)= 2(10)=20\)

\(\displaystyle P=2(4+4)= 2(8)=16\)

The maximum allowable perimeter is \(\displaystyle 34.\)

The perimeter that is not possible is \(\displaystyle 42\).

Step 1: Define Perimeter...

The perimeter of any shape is the sum of all the sides of that figure..

Step 2: Find how many sides a square has...

A square has \(\displaystyle 4\) sides.

Step 3: Calculate the perimeter...

\(\displaystyle side (cm)*4\)

\(\displaystyle 5cm*4=20cm\)

Write the formula for the area of the rectangle.  We will need the length of the other side of the rectangle.

\(\displaystyle A=bh\)

Let the known side length be the base.

\(\displaystyle 60=15h\)

Divide by 15 on both sides to get the height.

\(\displaystyle \frac{60}{15}=\frac{15h}{15}\)

\(\displaystyle h=4\)

There are two bases and two heights in a rectangle.

The perimeter is:

\(\displaystyle P=2b+2h = 2(15)+2(4) = 30+8 = 38\)

The answer is:  \(\displaystyle 38\)

An equilateral triangle has three congruent side lengths.

This means that the perimeter has to be triple the side length.

\(\displaystyle P=3s = 3(-2x+1)\)

Distribute the three with both terms of the binomial.

The answer is:  \(\displaystyle -6x+3\)

First, you will need to work backward from the circumference to find the radius of the circular enclosure.

\(\displaystyle Circumference=2r\pi\)

\(\displaystyle \small \frac{75.36}{3.14}=\frac{2r(3.14)}{3.14}\)

\(\displaystyle \small \frac{24}{2}=\frac{2r}{2}\)

\(\displaystyle \small \small r=12ft\)

Now we know what the radius is, we can calculate the surface area of the floor of the enclosure. 

\(\displaystyle Surface Area=\pi r^{^{2}}\)

\(\displaystyle \small \small (3.14)(12^{^{2}})=452.16 ft^{^{2}}\)

Finally, we need to find the number of units of sand needed to cover the floor of the enclosure. 

\(\displaystyle \small 452.15ft^{^{2}} \left (\frac{1unit}{27ft^{2}} \right )=16.75 units\)

Because we need to round up to the nearest unit, the final is 17 units of sand.