If segments are constructed in which the endpoints form the midpoints of the sides of a triangle, then four triangles, congruent to each other and similar to the larger triangle, are formed. Therefore, one of these triangles - specifically, \(\displaystyle \Delta DEF\) - would have one-fourth the area of \(\displaystyle \Delta ABC\). This means \(\displaystyle \Delta ABC\) has more than twice the area of \(\displaystyle \Delta DEF\).

Note that the fact that the triangle is equilateral is irrelevant.

By definition, an equilateral triangle has three sides of equal length. We can identify the equilateral triangle by converting the given sidelengths to the same units and comparing them.

We can eliminate the following by showing that at least two sidelengths differ.

\(\displaystyle 2 \textrm{ yd, } 84 \textrm{ in, }7\textrm{ ft}\)

2 yards = \(\displaystyle 2 \times 3 = 6\) feet.

Two sides have lengths 6 feet and 7 feet, so we can eliminate this choice.

\(\displaystyle 1\frac{1}{2} \textrm{ yd, } 50 \textrm{ in, }4\textrm{ ft}\)

4 feet = \(\displaystyle 4 \times 12 = 48\) inches

Two sides have lengths 48 inches and 50 inches, so we can eliminate this choice.

\(\displaystyle 1\frac{1}{3} \textrm{ yd, } 48 \textrm{ in, }5\textrm{ ft}\)

5 feet = \(\displaystyle 5 \times 12 = 60\) inches

Two sides have lengths 48 inches and 60 inches, so we can eliminate this choice.

\(\displaystyle 1\frac{2}{3} \textrm{ yd, } 48 \textrm{ in, }4\textrm{ ft}\)

\(\displaystyle 1\frac{2}{3}\) yards = \(\displaystyle 1\frac{2}{3} \times 3 = 5\) feet

Two sides have lengths 4 feet and 5 feet, so we can eliminate this choice.

\(\displaystyle \frac{2}{3} \textrm{ yd, } 24 \textrm{ in, }2\textrm{ ft}\)

\(\displaystyle \frac{2}{3}\) yards = \(\displaystyle \frac{2}{3} \times3 = 2\) feet = \(\displaystyle 2 \times 12 = 24\) inches 

All three sides have the same length, making this the triangle equilateral. This choice is correct.

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,

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

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

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