How to find the perimeter of a hexagon - SAT Math

Card 0 of 2

Question

Archimedes High School has an unusual track in that it is shaped like a regular hexagon, as above. Each side of the hexagon measures 264 feet.

Alvin runs at a steady speed of seven miles an hour for twelve minutes, starting at point A and working his way clockwise. When he is finished, which of the following points is he closest to?

Answer

Alvin runs at a rate of seven miles an hour for twelve minutes, or $\frac{12}{60} = \frac{1}{5}$ hours. The distance he runs is equal to his rate multiplied by his time, so, setting $r = 7 , t = \frac{1}{5}$ in this formula:

$d = 7 \cdot \frac{1}{5} = \frac{7}{5}$ miles.

One mile comprises 5,280 feet, so this is equal to

$\frac{7}{5} \cdot 5,280 = 7,392$ feet

Since each side of the track measures 264 feet, this means that Alvin runs

$7,392 \div 264 = 28$ sidelengths.

$28 \div 6 = 4 \textrm{ R }4$ ,

which means that Alvin runs around the track four complete times, plus four more sides of the track. Alvin stops when he is at Point E.

Compare your answer with the correct one above

Question

A circle with circumference $60 \pi$ is inscribed in a regular hexagon. Give the perimeter of the hexagon.

Answer

Below is the figure referenced; note that the hexagon is divided by its diameters, and that an apothem—a perpendicular bisector from the center to one side—has been drawn.

The circle has circumference $C = 60 \pi$ ; its radius, which coincides with the apothem of the hexagon, is the circumference divided by $2\pi$ :

$r = \frac{C}{2\pi} = \frac{60 \pi }{2\pi} = 30$

The hexagon is divided into six equilateral triangles. One, $\bigtriangleup AOB$ , is divided by an apothem of the hexagon $\overline{OM}$ - a radius of the circle - into two 30-60-90 triangles, one of which is $\bigtriangleup AMO$ . Since $\overline{OM}$ has length 30, and it is a long leg of $\bigtriangleup AMO$ , then short leg $\overline{AM}$ has length

$AM = \frac{OM}{\sqrt{3}} = \frac{30}{\sqrt{3}} = \frac{30 \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} =\frac{30 \sqrt{3}}{3} = 10 \sqrt{3}$

is the midpoint of $\overline{AB}$ , one of the six congruent sides of the hexagon, so

$AB = 2 \cdot AM = 2 \cdot 10 \sqrt{3} = 20\sqrt{3}$ ;

this makes the perimeter of the hexagon six times this, or

$P = 6 \cdot 20 \sqrt{3} = 120 \sqrt{3}$ .

Compare your answer with the correct one above

Tap the card to reveal the answer