The altitude of a right triangle to its hypotenuse divides the triangle into two smaller trangles similar to each other and to the large triangle. 

Therefore, 

\(\displaystyle \Delta ABC \sim \Delta ADB\)

and, consequently, 

\(\displaystyle \frac{DB}{BC} = \frac{AB}{AC}\),

or, equivalently,

\(\displaystyle DB = \frac{AB \cdot BC}{AC}\)

\(\displaystyle AC = \sqrt{ n^{2}+36}\) by the Pythagorean Theorem, so

\(\displaystyle DB = \frac{6n }{\sqrt{ n^{2}+36}}\).

The height of a right triangle from the vertex of its right angle is the geometric mean - in this case, the square root of the product - of the lengths of the two segments of the hypotenuse that it forms. Therefore, 

\(\displaystyle BD = \sqrt{(AD)(CD)} = \sqrt{6n}\)

The two triangles formed by an altitude from the vertex of a right triangle are similar to each other and the large triangle, so all three are 30-60-90 triangles. Take advantage of this, applying it twice.

Looking at \(\displaystyle \Delta ABC\). By the 30-60-90 Theorem, the shorter leg of a hypotentuse measures half that of the hypotenuse. \(\displaystyle AB = \frac{1}{2} AC = \frac{1}{2} \cdot 10 = 5\). 

Now, look at \(\displaystyle \Delta ADB\). By the same theorem, 

\(\displaystyle AD = \frac{1}{2} AB = \frac{1}{2} \cdot 5 = \frac{5}{2}\)

and 

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

The three roads form the legs and the hypotenuse of a right triangle with legs 5 and 12 miles; by the Pythagorean Theorem, the hypotenuse is

\(\displaystyle \sqrt{5^{2}+12^{2}} = \sqrt{25+144} = \sqrt {169}= 13\) miles. 

To find the length of the shortest possible road, which must be perpendicular to Highway 2, it should be observed that this road serves as an altitude from the vertex of the triangle to the hypotenuse.

The area of a right triangle can be calculated two ways - by taking half the product of the lengths of the legs or by taking half the product of the length of the altitude (height) and that of the hypotenuse. Setting \(\displaystyle h\) as the height, we can solve for \(\displaystyle h\) in the equation:

\(\displaystyle \frac{1}{2} \cdot 13 \cdot h = \frac{1}{2} \cdot 5 \cdot 12\)

\(\displaystyle 6.5 h = 30\)

\(\displaystyle h = 30 \div 6.5 \approx 4.6\) miles.

This is 46 tenths of a mile, so the approximate cost of the road in dollars will be

\(\displaystyle 46 \times 1,500 = 69,000\)

Among the five choices, $70,000 comes closest.

The sum of the two given angles is 90 degrees, which means that the third angle should be a right angle (90 degrees). We also know that two of the angles are equal. Therefore, the triangle is right and isosceles.

In a right triangle, there is one right angle of 90 degrees, while the two acute angles add up to 90 degrees. 

Given that the two acute angles are equal to 5x and 25x, the value of x can be calculated with the equation below:

\(\displaystyle 5x+25x=90\)

\(\displaystyle 30x=90\)

\(\displaystyle x=3\)

This triangle has two angles of 45 and 90 degrees, so the third angle must measure 45 degrees; this is therefore an isosceles right triangle.

By the Pythagorean Theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Let \(\displaystyle c=\) hypotenuse and \(\displaystyle s=\) side length.

\(\displaystyle c^2=s^2+s^2\Rightarrow c^2=2s^2\Rightarrow 8^2=2s^2\Rightarrow s^2=32\)

\(\displaystyle \Rightarrow s=\sqrt{32}=\sqrt{16\times 2}=4\sqrt{2}\ cm\)

We can then plug this side length into the formula for area.

\(\displaystyle area=\frac{s^2}{2}=\frac{(4\sqrt{2})^2}{2}=\frac{32}{2}=16\ cm^2\)

The hypotenuse of the triangle can be calculated using the Pythagorean Theorem. Set \(\displaystyle a = 2 \frac{2}{5} = \frac{12}{5} , b=3 \frac{1}{5}= \frac{16}{5}\):

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

\(\displaystyle c ^{2} = \left ( \frac{12}{5} \right ) ^{2} + \left ( \frac{16}{5} \right ) ^{2}\)

\(\displaystyle c ^{2} = \frac{144}{25} + \frac{256}{25}\)

\(\displaystyle c ^{2} = \frac{400}{25} = 16\)

\(\displaystyle c = \sqrt{16} = 4\)

Add the three sidelengths:

\(\displaystyle P =2 \frac{2}{5} + 3 \frac{1}{5} + 4 = 9 \frac{3}{5}\)

The altitude of a right triangle from the vertex of its right angle - which, here, is \(\displaystyle \overline{BX}\) - divides the triangle into two triangles similar to each other. The ratio of the hypotenuse of the white triangle to that of the gray triangle (which are corresponding sides) is 

\(\displaystyle \frac{BC}{AB} = \frac{40}{30} =\frac{4}{3}\) ,

making this the similarity ratio. The ratio of the areas of two similar triangles is the square of their similarity ratio, which here is

\(\displaystyle \left ( \frac{4}{3} \right )^{2} = \frac{4^{2}}{3^{2}} = \frac{16}{9}\), or \(\displaystyle 16: 9\).

To find the area of a right triangle, we will use the following formula:

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

where b is the base and h is the height of the triangle. 

We know the base of the triangle is 7cm.  We know the height of the triangle is 20cm.  Knowing this, we can substitute into the formula.  We get

\(\displaystyle A = \frac{1}{2} \cdot 7\text{cm} \cdot 20\text{cm}\)

\(\displaystyle A = 7\text{cm} \cdot 10\text{cm}\)

\(\displaystyle A = 70\text{cm}^2\)