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SAT Math : How to find the perimeter of a hexagon

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Example Question #1 : How To Find The Perimeter Of A Hexagon

Hexagon

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?

Possible Answers:

Point C

Point B

Point F

Point E

Point D

Correct answer:

Point E

Explanation:

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 = rt

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

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Example Question #2 : Hexagons

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

Possible Answers:

150

$60 \sqrt{10}$

$135 \sqrt{2}$

$120 \sqrt{3}$

None of these

Correct answer:

$120 \sqrt{3}$

Explanation:

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.

Hexagon 3

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

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

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SAT Math : How to find the perimeter of a hexagon

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Example Question #1 : How To Find The Perimeter Of A Hexagon

Example Question #2 : Hexagons

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