Because this is a right triangle, we can use the Pythagorean Theorem which says a² + b² = c², or the squares of the two sides of a right triangle must equal the square of the hypotenuse. Here we have a = 5 and b = 8.

a² + b² = c²

5² + 8² = c²

25 + 64 = c²

89 = c²

c = √89

The hypotenuses of the triangles measure as follows:

(a) \(\displaystyle c = \sqrt {20^{2} + 20^{2} } = \sqrt {400 + 400 } = \sqrt {800}\)

(b) \(\displaystyle c = \sqrt {19^{2} + 21^{2} } = \sqrt {361 + 441} = \sqrt {802}\)

\(\displaystyle 800 < 802\), so \(\displaystyle \sqrt {800} < \sqrt {802}\), making (b) the greater quantity

The hypotenuses of the triangles measure as follows:

(a) \(\displaystyle c = \sqrt {10^{2} + 15^{2} } = \sqrt {100 + 225 } = \sqrt {325}\)

(b) \(\displaystyle c = \sqrt {12^{2} + 13^{2} } = \sqrt {144 + 169 } = \sqrt {313}\)

\(\displaystyle 325 > 313\), so \(\displaystyle \sqrt {325} > \sqrt {313}\), making (a) the greater quantity.

The length of the second leg can be calculated using the Pythagorean Theorem. Set \(\displaystyle c = 7 \frac{1}{2} = \frac{15}{2} , a =4 \frac{1}{2} = \frac{9}{2}\):

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

\(\displaystyle b ^{2} = \left ( \frac{15}{2} \right ) ^{2}- \left ( \frac{9}{2} \right )^{2}\textup{}\)

\(\displaystyle b ^{2} = \frac{225}{4} -\frac{81}{4}\)

\(\displaystyle b ^{2} = \frac{144}{4}\)

\(\displaystyle b ^{2} = 36\)

\(\displaystyle b = \sqrt{36} = 6\)

The second leg therefore measures \(\displaystyle 6 \times 12 = 72\) inches.

Since we're dealing with right triangles, we can use the Pythagorean Theorem (\(\displaystyle a^2+b^2=c^2\)). In this formula, a and b are the sides, while c is the hypotenuse. The hypotenuse of a right triangle is the longest side and the side that is opposite the right angle. Now, we can plug into our formula, which looks like this: \(\displaystyle 9^2+12^2=c^2.\) We simplify and get \(\displaystyle 225=c^2\). At this point, isolate c. This means taking the square root of both sides so that your answer is 15in.

By the Pythagorean Theorem, the hypotenuse of the right triangle is 

\(\displaystyle \sqrt{14^{2}+48^{2}} = \sqrt{196+2,304} = \sqrt{2,500} = 50\) inches, making its perimeter

\(\displaystyle 14 + 48 + 50 =112\) inches.

The pentagon in question has sides of length 75% of 112, or 

\(\displaystyle 112 \times 0.75 = 84\).

Since a pentagon has five sides of equal length, each side will have measure

\(\displaystyle 84 \div 5 = 16 \frac{4}{5}\) inches.

One and a half feet are equivalent to \(\displaystyle 12 \times 1 \frac{1}{2} = 18\) inches, so (B) is the greater quantity.

By the Pythagorean Theorem, the distance from B to C is 

\(\displaystyle \sqrt{600^{2} + 800^{2} }\)

\(\displaystyle = \sqrt{360,000 + 640,000 }\)

\(\displaystyle = \sqrt{1,000,000} = 1,000\)  feet

Cary runs 

\(\displaystyle 800 + \frac{1}{2} \times1,000 = 800 + 500 = 1,300\) feet

Since 5,280 feet make a mile, one-fourth of a mile is equal to 

\(\displaystyle 5,280 \div 4 = 1,320\) feet.

(B) is greater

If we let \(\displaystyle c\) be the length of the hypotenuse, then by the Pythagorean theorem,

\(\displaystyle c = \sqrt{(k+2)^{2}+(k-2)^{2}}\)

\(\displaystyle c = \sqrt{(k ^{2}+4k+4) +(k ^{2}-4k+4)}\)

\(\displaystyle c = \sqrt{2k ^{2} +8}\)

The figure referenced is below:

For the sake of simplicity, assume that the square has sides of length 4. The following reasoning is independent of the actual lengths, and the reason for choosing 4 will become apparent in the explanation.

\(\displaystyle Q\) and \(\displaystyle X\) are midpoints of their respective sides, so \(\displaystyle QS = SX = 2\), making \(\displaystyle \overline{QX}\) the hypotenuse of a triangle with legs of length 2 and 2. Therefore,

\(\displaystyle (QX ) ^{2}= (SQ)^{2}+ (SX)^{2} = 2^{2}+ 2^{2} = 4+ 4 = 8\).

Also, \(\displaystyle QA = 2\), and since \(\displaystyle U\) is the midpoint of \(\displaystyle \overline{AQ}\), \(\displaystyle AU = 1\). \(\displaystyle AR = 4\), making \(\displaystyle \overline{UR}\) the hypotenuse of a triangle with legs of length 1 and 4. Therefore, 

\(\displaystyle (UR)^{2} = (AU)^{2}+ (AR)^{2} = 1^{2}+ 4^{2} = 1+ 16 = 17\)

\(\displaystyle (UR)^{2} >(QX ) ^{2}\), so \(\displaystyle UR > QX\)

The altitude of a right triangle from the vertex of its right angle divides the triangle into two smaller triangles each similar to the larger triangle. In particular, 

\(\displaystyle \bigtriangleup BXC \sim \bigtriangleup ABC\).

Their corresponding sides are in proportion, so, setting the ratios of the hypotenuses to the short legs equal to each other,

\(\displaystyle \frac{AC}{BC} = \frac{BC}{CX}\)

\(\displaystyle \frac{AC}{13} = \frac{13}{5}\)

\(\displaystyle \frac{AC}{13} \cdot 13 = \frac{13}{5} \cdot 13\)

\(\displaystyle AC = \frac{169}{5} = 33 \frac{4}{5 }\)