A diagonal of a square cuts the square into two isosceles right triangles, of which the diagonal is the common hypotenuse. Therefore, each figure is the hypotenuse of an isosceles right triangle with legs 10 inches, making them equal in length.

The diagonal of a square has length $\displaystyle \sqrt{2}$, or about 1.414, times the length of a side, which here is 600 feet; this makes the diagonal path about

$\displaystyle 600 \times 1.414 \approx 848$ feet long.

Les runs around the square track twice, meaning that he runs the length of one side eight times; he also runs the length of the diagonal twice, This is a total of about

$\displaystyle 600 \times 8 + 848 \times 2 = 4,800 + 1,696 = 6,496$ feet.

Divide by 5,280 to convert to miles:

$\displaystyle 6,496 \div 5,280 \approx 1.23$

Of the given responses, $\displaystyle 1\frac{1}{4}$ miles comes closest to the correct distance.

The diagonal of a square has length $\displaystyle \sqrt{2}$, or about 1.414, times the length of a side, which here is 700 feet; this makes the diagonal path about

$\displaystyle 700 \times 1.414 \approx 990$ feet long.

We will call one complete circuit one running of the diagonal, which is 990 feet long, and one running around the square; the completion of one complete circuit amounts to running a distance of 

$\displaystyle 990 + 4 \times 700 = 990 + 2,800 = 3,790$ feet.

Ellen seeks to run three miles, or 

$\displaystyle 5,280 \times 3 = 15, 840$ feet, which, divided by 3,790 feet, is about:

$\displaystyle 15, 840 \div 3,790 \approx 4.17$,

or four complete circuits and 0.17 of a fifth.

After four complete circuits, Ellen is backat Point A. She has yet to run

$\displaystyle 3,790 \times 0.17 = 629$ feet. 

She will now run along the diagonal from Point A to Point C, but since the diagonal has length 990 feet, which is greater than 629 feet, she will finish running three miles when she is on this diagonal path.

The diagonal of a square has length $\displaystyle \sqrt{2}$, or about 1.414, times the length of a side, which here is 600 feet; this makes the diagonal path about

$\displaystyle 600 \times 1.414 \approx 848$ feet long.

Kenny runs along this path twice, and he runs along the entire perimeter of the square path, so his run is about

$\displaystyle 848 \times 2 + 600 \times 4 = 1,696 +2,400 = 4,096$ feet. Since one mile is equal to 5,280 feet, the greater quantity is (b).

The sidelength of a square is the square root of its area and one-fourth of its perimeter, so:

(a) A square with area 400 square inches has sidelength $\displaystyle \sqrt{400} = 20$ inches.

(b) A square with perimeter 80 inches has sidelength $\displaystyle 80 \div 4 = 20$ inches.

The two quantities are equal.

The sidelength of a square is the square root of its area and one-fourth of its perimeter, so:

(a) A square with area $\displaystyle 225$ square inches has sidelength $\displaystyle \sqrt{225} = 15$ inches.

(b) A square with perimeter $\displaystyle 50$ inches has sidelength $\displaystyle 50 \div 4 = 12.5$ inches.

(a) is the greater quantity.

Divide the perimeter of a square by 4 to get its sidelength:

$\displaystyle \frac{2 ^{x} }{4} = \frac{2 ^{x} }{2^{2}} = 2^{x-2}$

The perimeter of a rectangle is twice the sum of its length and its width:

$\displaystyle P = 2 (2x + x) = 2 (3x) = 6x$

Since the height of the trapezoid in the figure is $\displaystyle x$, both of its legs must have length greater than or equal to $\displaystyle x$. But for a leg to be of length$\displaystyle x$, it must be perpendicular to the bases. Since perpendicularity of both legs would make the trapezoid a rectangle - which it cannot be - it follows that both legs cannot be of length $\displaystyle x$. Therefore, the perimeter of the trapezoid is:

$\displaystyle P > x + 3x + x + x = 6x$

The perimeter of the trapezoid must be greater than that of the rectangle.

The length of the midsegment of a trapezoid is half sum of the lengths of the bases, so

$\displaystyle XY = \frac{1}{2} (12 + 40) = \frac{1}{2} \cdot 52 = 26$.

Also, by definition, since Trapezoid $\displaystyle TRAP$ is isosceles, $\displaystyle TP = RA$. The midsegment divides both legs of Trapezoid $\displaystyle TRAP$ into congruent segments; combining these facts:

$\displaystyle TX = XP = \frac{1}{2} TP = \frac{1}{2} RA = RY = YA$

$\displaystyle TX = \frac{1}{2} XY = \frac{1}{2} \cdot 26 = 13$.

$\displaystyle XP = YA= TX = 13$, so the perimeter of Trapezoid $\displaystyle XYAP$ is 

$\displaystyle XP+XY + YA + AP = 13+26+13+40 = 92$.

The midsegment of a trapezoid bisects both of its legs, so 

$\displaystyle TX = XP$  and  $\displaystyle RY = YA$.

For reasons that will be apparent later, we will set

$\displaystyle z = TX + RY = X P + YA$

Also, the length of the midsegment is half sum of the lengths of the bases:

$\displaystyle XY = \frac{1}{2} (12 + 36) = \frac{1}{2} \cdot 48= 24$.

The perimeter of  Trapezoid $\displaystyle TRYX$ is 

$\displaystyle TR + RY + YX + TX = TR + YX +( TX + RY) = 12+24+ z = 36 + z$

Twice this is 

$\displaystyle 2 (36 + z) = 72 + 2z$

The  perimeter of  Trapezoid $\displaystyle XYAP$ is

$\displaystyle XY + YA + AP + XP = XY + AP + (XP+ YA) = 24 + 36 + z = 60 + z$

$\displaystyle 72 > 60$ and $\displaystyle 2z > z$, so $\displaystyle 72 + 2z > 60+ z$, making (a) the greater quantity.