Three feet make a yard, so the length and width of the pool are \(\displaystyle \frac{50}{3} = 16 \frac{2}{3}\) yards and \(\displaystyle \frac{35}{3} = 11 \frac{2}{3}\) yards, respectively. Since the dimensions of the tarp must exceed those of the pool by at least one yard, the tarp must be at least \(\displaystyle 17 \frac{2}{3}\) yards by \(\displaystyle 12 \frac{2}{3}\) yards; but since both dimensions must be multiples of three yards, we take the next multiple of three for each.

The tarp must be 18 yards by 15 yards, so the area of the tarp is the product of these dimensions, or

\(\displaystyle 18 \times 15 = 270\) square yards.

The shaded portion of the entire triangle is similar to the entire large triangle by the Angle-Angle postulate, so sides are in proportion. The short leg of the blue triangle has length 20; that of the large triangle, 30. Therefore, the similarity ratio is \(\displaystyle 3:2\). The ratio of the areas is the square of this, or \(\displaystyle 3^{2}:2^{2}\), or \(\displaystyle 9:4\). 

The blue triangle is therefore \(\displaystyle \frac{4}{9}\) of the entire triangle, or \(\displaystyle \frac{4}{9} \times 100 = 44 \frac{4}{9} \%\) of it.

If we see hypotenuse \(\displaystyle \overline{AC}\) as the base of the large triangle, then we can look at the segment perpendicular to it, \(\displaystyle \overline{BD}\), as its altitude. Therefore, the area of \(\displaystyle \Delta ABC\) is

\(\displaystyle A = \frac{1}{2} (AC) (BD)\)

\(\displaystyle AC = AD + DC = 6 + 24 = 30\).

\(\displaystyle BD\), 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, 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 area of \(\displaystyle \Delta ABC\) is therefore

\(\displaystyle A = \frac{1}{2} (AC) (BD) = \frac{1}{2} \cdot 30 \cdot 12 = 180\)

Three feet make a yard, so the length and width of the pool are \(\displaystyle \frac{50}{3}\) yards and \(\displaystyle \frac{35}{3}\) yards; the area of the pool, and that of the tarp needed to cover it, must be the product of these dimensions, or

\(\displaystyle \frac{50}{3} \times \frac{35}{3}= \frac{1,750}{9} = 194 \frac{4}{9}\) square yards. 

The manager will need to buy a number of square yards of tarp equal to the next highest multiple of ten, which is 200 square yards.

The area of \(\displaystyle \Delta ADB\), being right, is half the products of its legs, which is:

\(\displaystyle \frac{1}{2} (AD) (BD) = \frac{1}{2} \times 6 (BD) = 3 (BD)\)

The area of \(\displaystyle \Delta ABC\) is one half the product of its base and height; we can use its hypotenuse \(\displaystyle AC = AD + DC = 6 + 24 = 30\) as the base and \(\displaystyle BD\) as the height, so this area is

\(\displaystyle \frac{1}{2} (AC) (BD) = \frac{1}{2} \times 30 (BD) = 15 (BD)\)

Therefore, in terms of area, \(\displaystyle \Delta ADB\) is \(\displaystyle \frac{3 (BD)}{15 (BD)} = \frac{3}{15} = \frac{1}{5}\) of \(\displaystyle \Delta ABC\).

\(\displaystyle BD\), 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 areas of \(\displaystyle \Delta BDC\) and \(\displaystyle \Delta ADB\), each being right, are half the products of their legs, so:

The area of \(\displaystyle \Delta ADB\) is \(\displaystyle \frac{1}{2} \times (AD) (BD) = \frac{1}{2} \times 6 \times (BD)= 3(BD)\)

The area of \(\displaystyle \Delta BDC\) is \(\displaystyle \frac{1}{2} \times (CD) (BD) = \frac{1}{2} \times 24 \times (BD)= 12(BD)\)

The ratio of the areas is \(\displaystyle \frac{12 (BD)}{3 (BD)} = \frac{12 }{3 } =\frac{4}{1}\) - that is, 4 to 1.

Polygon \(\displaystyle ABCDE\) is a composite of \(\displaystyle \Delta ABE\) and Square \(\displaystyle BCDE\); its area is the sum of the areas of the two figures.

\(\displaystyle \Delta ABE\) is a right triangle; its area is half the product of its legs, which is 

\(\displaystyle A = \frac{1}{2} \cdot 5 \cdot 8 = 20\)

\(\displaystyle \overline{BE }\) 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:

\(\displaystyle BE = \sqrt{5^{2}+8^{2}} = \sqrt{25 + 64 } = \sqrt{89}\).

Square \(\displaystyle BCDE\) has area the square of this, which is 89.

Polygon \(\displaystyle ABCDE\) has as its area the sum of these two areas:

\(\displaystyle 20 + 89 = 109\).

Write the formula for the area of a circle.

\(\displaystyle A_{circle}= \pi r^2 = \frac{\pi d^2}{4}\)

Substitute the diameter and solve.

\(\displaystyle \frac{\pi (16)^2}{4} = \frac{\pi (16)(16)}{4}=\pi (16)(4)= 64\pi\)

Because we know the perimeter is 36 inches, we can determine the length of side w based on the equation of the perimeter of a rectangle:

\(\displaystyle \small Perimeter= 2(length)+2(width)\)

\(\displaystyle \small 36=2(12)+2w\)

\(\displaystyle \small 36=24+2w\)

\(\displaystyle \small 12=2w\)

\(\displaystyle \small w=6\)

Side w is 6 in long.

Now that we know that side w is 6 inches long, we have everythinng we need to calculate the area of the rectangle. 

\(\displaystyle \small Area=(length)(width)\)

\(\displaystyle \small (12)(6)=72\)

The area of the rectangle is 72 in²

A kite is a four-sided shape with straight sides that has two pairs of sides. Each pair of adjacent sides are equal in length. A square is also considered a kite.