Write the formula for the area of a triangle.

\(\displaystyle A=\frac{BH}{2}\)

Substitute the dimensions.

\(\displaystyle A = \frac{x^2\cdot \frac{1}{2}x}{2} = \frac{1}{2}(x^2\cdot \frac{1}{2}x)\)

The answer is:  \(\displaystyle \frac{1}{4}x^3\)

Write the formula for the area of a circle.

\(\displaystyle A=\pi r^2\)

Substitute the radius into the equation.

\(\displaystyle A=\pi (\frac{3}{2})^2 = \frac{9}{4} \pi\)

The area is:  \(\displaystyle \frac{9}{4}\pi\)

The area of a rectangle is:  \(\displaystyle A =LH\)

Substitute the length and height into the formula.

\(\displaystyle A = (x+3)\cdot \frac{1}{4} (x+3)\)

We will move the constant to the front and apply the FOIL method to simplify the binomials.

\(\displaystyle A = \frac{1}{4} (x+3)(x+3) = \frac{1}{4} (x^2+6x+9)\)

Distribute the fraction through all the terms of the trinomial.

The answer is:  \(\displaystyle \frac{1}{4}x^2 + \frac{3}{2}x + \frac{9}{4}\)

Write the formula for the area of a triangle.

\(\displaystyle A= \frac{BH}{2}\)

Substitute the dimensions.

\(\displaystyle A= \frac{(\frac{1}{2}x+\frac{1}{4})(2x+2)}{2}=\frac{1}{2}(\frac{1}{2}x+\frac{1}{4})(2x+2)\)

\(\displaystyle A= (\frac{1}{2}x+\frac{1}{4})(x+1)\)

Use the FOIL method to expand this.

\(\displaystyle A = (\frac{1}{2}x)(x)+ (\frac{1}{2}x)(1)+ (\frac{1}{4})(x)+ (\frac{1}{4})(1)\)

Simplify the terms.

\(\displaystyle A = \frac{1}{2}x^2+\frac{1}{2}x+\frac{1}{4}x+\frac{1}{4}\)

Combine like-terms.

The answer is:  \(\displaystyle \frac{1}{2}x^2+\frac{3}{4}x+\frac{1}{4}\)

Write the formula for the area of a square.

\(\displaystyle A = s^2\)

Substitute the area into the equation.

\(\displaystyle 100x= s^2\)

Square root both sides.

\(\displaystyle \sqrt{100x}= \sqrt{s^2}\)

\(\displaystyle s = 10\sqrt{x}\)

The answer is:  \(\displaystyle 10\sqrt{x}\)

The area of a circle is \(\displaystyle A=\pi r^2\).

Substitute the radius and solve for the area.

\(\displaystyle A=\pi (3x^3)^2 = \pi (3x^3)(3x^3) = 9\pi x^6\)

The answer is:  \(\displaystyle 9\pi x^6\)

Write the formula for the area of a triangle.

\(\displaystyle A =\frac{1}{2}bh\)

Substitute the base and height into the equation.

\(\displaystyle A =\frac{1}{2}(6)(\frac{2}{3}) =2\)

The answer is:  \(\displaystyle 2\)

Divide the diameter by two.  This will be the radius.

\(\displaystyle r=\frac{x}{2}\)

Write the formula for the area of a circle.

\(\displaystyle A=\pi r^2\)

\(\displaystyle A=\pi (\frac{x}{2})^2 = \pi (\frac{x^2}{4})\)

The answer is:  \(\displaystyle \frac{\pi x^2}{4}\)

Quadrilateral \(\displaystyle MBCN\) is a trapezoid, so we need to find the lengths of its bases and its height.

The perimeter of the hexagon is \(\displaystyle 60\), so each side of the hexagon measures one sixth of this, or \(\displaystyle 10\).

Construct the diameters of the hexagon, which meet at center \(\displaystyle O\); construct the apothem from \(\displaystyle O\) to \(\displaystyle \overline{BC}\), with point of intersection \(\displaystyle P\). The diagram is below:

The six triangles formed by the diameters are equilateral, so \(\displaystyle AO = OD = BC = 10\), and \(\displaystyle AD = AO + OD = 10+10= 20\). Quadrilateral \(\displaystyle ABCD\) is a trapezoid with bases of length 10 and 20. Since \(\displaystyle \overline{MN}\) has its endpoints at the midpoints of the legs of Trapezoid \(\displaystyle ABCD\), it follows that \(\displaystyle \overline{MN}\) is a midsegment, and has as its length \(\displaystyle MN = \frac{1}{2} (AD + BC) = \frac{1}{2} (20+ 10 ) = 15\).

The trapezoid has bases of length \(\displaystyle 15\) and \(\displaystyle 20\); we now need to find its height. This is the measure of \(\displaystyle \overline{PQ}\), which is half the length of apothem \(\displaystyle \overline{OP}\). \(\displaystyle OP\) is the height of an equilateral triangle \(\displaystyle \bigtriangleup BCO\) and, consequently, the long leg of a right triangle \(\displaystyle \bigtriangleup BCO\). By the 30-60-90 Theorem, 

\(\displaystyle OP = BP \cdot \sqrt{3} = \frac{1}{2} \cdot BC \cdot \sqrt{3} = \frac{1}{2} \cdot 10 \cdot \sqrt{3}=5\sqrt{3}\).

\(\displaystyle PQ = \frac{1}{2} \cdot OP = \frac{1}{2} \cdot 5 \sqrt{3} = \frac{5}{2} \sqrt{3}\) 

The area of a trapezoid of height \(\displaystyle h\) and base lengths \(\displaystyle B\) and \(\displaystyle b\) is

\(\displaystyle A = \frac{1}{2} (B+b) h\);

Setting \(\displaystyle h =PQ = \frac{5}{2} \sqrt{3} , B = MN = 15, b = BC = 10\):

\(\displaystyle A = \frac{1}{2} (B+b) h\)

\(\displaystyle = \frac{1}{2} \cdot (15 + 10 ) \cdot \frac{5}{2} \sqrt{3}\)

\(\displaystyle = \frac{25}{2} \cdot \frac{5}{2} \sqrt{3}\)  

\(\displaystyle = \frac{125}{4} \sqrt{3}\)

In a regular pentagon, called Pentagon \(\displaystyle ABCDE\), construct the five perpendicular segments from each vertex to its opposite side, as shown below:

The segments divide the pentagon into ten congruent triangles. 

In particular, examine \(\displaystyle \bigtriangleup OXD\). \(\displaystyle \overline{OD}\), a radius of the pentagon, bisects \(\displaystyle \angle EDC\), which, as the interior angle of a regular pentagon, has measure \(\displaystyle 108 ^{\circ }\); therefore, \(\displaystyle m \angle ODX = 54 ^{\circ }\). \(\displaystyle \overline{OX}\) is an apothem and therefore bisects \(\displaystyle \overline{CD}\); since the pentagon has perimeter 50, \(\displaystyle \overline{CD}\) has length one fifth of this, or 10, and \(\displaystyle XD= 5\). 

Using trigonometry,

\(\displaystyle \tan \angle ODX = \frac{OX}{XD }\), 

or, substituting,

\(\displaystyle \frac{OX}{5} = \tan 54 ^{\circ }\)

Solving for \(\displaystyle OX\):

\(\displaystyle 5 \cdot \frac{OX}{5} =5 \cdot \tan 54 ^{\circ }\)

\(\displaystyle OX \approx 5 \cdot (1.3764 )\)

\(\displaystyle OX \approx 6.8819\)

The area of this triangle is half the product of the lengths of legs \(\displaystyle \overline{XD}\) and \(\displaystyle \overline{OX}\):

\(\displaystyle A \approx \frac{1}{2} \cdot 5 \cdot 6.8819 \approx 17.2048\)

Since the pentagon comprises ten triangles of this area, multiply:

\(\displaystyle 17.2048 \cdot 10 = 172 .048\)

To the nearest whole, this is 172.