\(\displaystyle BD\), 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 area of \(\displaystyle \Delta ADB\), the shaded region, is half the products of its legs:

\(\displaystyle A= \frac{1}{2} (AD) \times (DB ) =\frac{1}{2} \times 6 \times 12 = 36\)

The area of \(\displaystyle \Delta ABC\) is half the product of its hypoteuse, which we can see as the base, and the length of corresponding altitude  \(\displaystyle \overline{BD}\):

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

\(\displaystyle \Delta ADB\) comprises

\(\displaystyle \frac{36}{180} \times \100 = 20 \%\)

of \(\displaystyle \Delta ADC\).

The area of the dirt path is the area of the outer square minus that of the inner square.

The outer square has sidelength 75 feet and therefore has area

\(\displaystyle 75 \times 75 = 5,625\) square feet.

The inner square has sidelength \(\displaystyle 75 - 2 \times 7 = 61\) feet and therefore has area 

\(\displaystyle 61 \times 61 = 3,721\) square feet.

Subtract to get the area of the dirt path:

\(\displaystyle 5,625 - 3,721 = 1,904\) square feet.

The length of the garden is \(\displaystyle 2 \times 6= 12\) feet less than that of the entire lot, or 

\(\displaystyle L = 2x-12\);

The width of the garden is \(\displaystyle 2 \times 6= 12\) less than that of the entire lot, or 

\(\displaystyle W = x-12\);

The area of the garden is their product:

\(\displaystyle A = LW = (2x-12)(x - 12)\)

\(\displaystyle =2x \cdot x - 2x \cdot 12 - 12 \cdot x + 12 \cdot 12\)

\(\displaystyle = 2x ^{2} - 24x - 12 x + 144\)

\(\displaystyle = 2x ^{2} - 36 x + 144\)

As an interior angle of a regular decagon, \(\displaystyle \angle B\) measures

\(\displaystyle m \angle B = \left [\frac{180(10-2)}{10} \right ] ^{\circ }= 144^{\circ }\).

\(\displaystyle AB = BC = 100\).

\(\displaystyle AC\) can be found using the Law of Cosines:

\(\displaystyle \left ( AC\right )^{2} = \left ( AB\right )^{2}+ \left ( BC\right )^{2} - 2 \left ( AB\right ) \left ( BC\right ) \cos m\angle B\)

\(\displaystyle \left ( AC\right )^{2} = 100^{2}+ 100^{2} - 2 \cdot 100 \cdot 100 \cos 144^{\circ }\)

\(\displaystyle \left ( AC\right )^{2} \approx 10,000 + 10,000 - 20,000 (-0.8090)\)

\(\displaystyle \left ( AC\right )^{2} \approx 20,000 - (-16,180) \approx 36,180\)

\(\displaystyle AC \approx \sqrt{36,180} \approx 190\)

First, it is necessary to determine the radius of the circle. This is the distance between \(\displaystyle (0,0)\) and \(\displaystyle (-8,6)\), so we apply the distance formula:

\(\displaystyle r = \sqrt{(x_{2} - x _{1})^{2}+(y_{2} - y_{1})^{2}}\)

\(\displaystyle r = \sqrt{\left ( -8 -0 \right )^{2} +\left ( 6-0\right )^{2} }=\sqrt{ 8^{2}+6^{2}} = \sqrt{64+36 }= \sqrt{100} = 10\)

Subsequently, the area of the circle is 

\(\displaystyle A = \pi r^{2} = \pi \cdot 10^{2} = 100 \pi\)

Now, we need to find the central angle of the shaded sector. This is found using the relationship

\(\displaystyle \tan \theta = \frac{y}{x}\)

\(\displaystyle \tan \theta = \frac{6}{-8} = -0.75\)

Using a calculator, we find that \(\displaystyle \tan^{-1}(-0.75) \approx -36.87^{\circ }\); since we want a degree measure between \(\displaystyle 90^{\circ }\) and \(\displaystyle 180 ^{\circ }\), we adjust by adding \(\displaystyle 180 ^{\circ }\), so

\(\displaystyle \theta \approx \left (-36.87+ 180 \right ) ^{\circ } \approx 143.13^{\circ }\)

The area of the sector is calculated as follows:

\(\displaystyle A \approx \frac{143.13 }{360} \cdot 100 \pi \approx 124.9\)

You own a mug with a circular bottom. If the distance around the outside of the mug's base is  \(\displaystyle 20 \pi cm\) what is the area of the base?

Begin by solving for the radius:

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

\(\displaystyle 20 \pi cm=2\pi r\)

\(\displaystyle r=10cm\)

Next, plug the radius back into the area formula and solve:

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

\(\displaystyle A_{circle}=\pi (10cm)^2=100 \pi cm^2\)

So our answer is:

\(\displaystyle 100 \pi cm^2\)

You have a right triangle with a hypotenuse of 13 inches and a leg of 5 inches, what is the area of the triangle?

So find the area of a triangle, we need the following formula:

\(\displaystyle A_{tri}=\frac{1}{2}bh\)

However, we only know one leg, so we only know b or h.

To find the other leg, we can either use Pythagorean Theorem, or recognize that this is a 5-12-13 triangle. Meaning, our final leg is 12 inches long.

To prove this:

\(\displaystyle (13in)^2=(5in)^2+(xin)^2\)

\(\displaystyle 169in^2-25in^2=144in^2=(xin)^2\)

\(\displaystyle 144in^2=x^2in\)

\(\displaystyle x=\sqrt{144in^2}=12in\)

Now, we know both legs, let's just plug in and solve for area:

\(\displaystyle A_{tri}=\frac{1}{2}(5in)(12in)=6in*5in=30in^2\)

You have a rectangular-shaped rug which you want to put in your living room. If the rug is 12.5 feet long and 18 inches wide, what is the area of the rug?

To begin, we need to realize two things. 

1) Our given measurements are not in equivalent units, so we need to convert one of them before doing any solving.

2) The area of a rectangle is given by: \(\displaystyle A=l*w\)

Now, let's convert 18 inches to feet, because it seems easier than 12.5 feet to inches:

\(\displaystyle 18inches*\frac{1ft}{12in}=1.5ft\)

Now, using what we know from 2) we can find our answer

\(\displaystyle A=12.5ft*1.5ft=18.75ft^2\)

The area of a triangle with two sides of lengths \(\displaystyle a\) and \(\displaystyle b\) and included angle of measure \(\displaystyle \gamma\) can be calculated using the formula

\(\displaystyle A = \frac{1}{2} ab \sin \gamma\).

Setting \(\displaystyle a = AB = 34, b= AC = 27 , \gamma = m \angle A = 124^{\circ }\) and evaluating \(\displaystyle A\):

\(\displaystyle A = \frac{1}{2} (34 )(27)\sin 124 ^{\circ }\)

\(\displaystyle \approx 459 ( 0.8290 )\)

\(\displaystyle \approx 381\)

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 \(\displaystyle a = AB = 20\), \(\displaystyle b = BC = 26\), and \(\displaystyle c = AC = 32\), evaluate \(\displaystyle s\):

\(\displaystyle s = \frac{20+26+32}{2} = \frac{78}{2} = 39\)

and, substituting in Heron's formula:

\(\displaystyle A = \sqrt{39(39-20)(39-26)(39-32)}\)

\(\displaystyle = \sqrt{39(19)(13)(7)}\)

\(\displaystyle = \sqrt{67,431 }\)

\(\displaystyle \approx 259.7\)

To the nearest whole, this is 260.