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High School Math : Plane Geometry

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Example Question #131 : High School Math

What is the perimeter of a regular hendecagon with a side length of 32?

Possible Answers:

334

320

384

352

Correct answer:

352

Explanation:

To find the perimeter of a regular hendecagon you must first know the number of sides in a hendecagon is 11.

When you know the number of sides of a regular polygon to find the perimeter you must multiply the side length by the number of sides.

In this case it is 11*32=352 .

The answer for the perimeter is 352 .

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Example Question #1 : Other Polygons

All segments of the polygon meet at right angles (90 degrees). The length of segment $\overline{AB}$ is 10. The length of segment $\overline{BC}$ is 8. The length of segment $\overline{DE}$ is 3. The length of segment $\overline{GH}$ is 2.

Find the perimeter of the polygon.

Possible Answers:

$\dpi{100} \small 46$

$\dpi{100} \small 44$

$\dpi{100} \small 40$

$\dpi{100} \small 48$

$\dpi{100} \small 42$

Correct answer:

$\dpi{100} \small 46$

Explanation:

The perimeter of the polygon is 46. Think of this polygon as a rectangle with two of its corners "flipped" inwards. This "flipping" changes the area of the rectangle, but not its perimeter; therefore, the top and bottom sides of the original rectangle would be 12 units long $\dpi{100} \small (10+2=12)$ . The left and right sides would be 11 units long $\dpi{100} \small (8+3=11)$ . Adding all four sides, we find that the perimeter of the recangle (and therefore, of this polygon) is 46.

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Example Question #2 : Geometry

What is the perimeter of a regular nonagon with a side length of ?

Possible Answers:

150

120

105

135

Correct answer:

135

Explanation:

To find the perimeter of a regular polygon, we take the length of each side, , and multiply it by the number of sides, .

P=s*l

In a nonagon the number of sides is , and in this example the side length is .

P=9*15=135

The perimeter is 135 .

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Example Question #3 : Geometry

Find the perimeter of the following octagon:

Possible Answers:

$56\ m$

$86\ m$

$66\ m$

$96\ m$

$76\ m$

Correct answer:

$96\ m$

Explanation:

The formula for the perimeter of an octagon is P = 8(side) .

Plugging in our values, we get:

$P = 8(12\ m)$

$P = 96\ m$

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Example Question #4 : Geometry

What is the area of a regular heptagon with an apothem of 4 and a side length of 6?

Possible Answers:

Correct answer:

Explanation:

What is the area of a regular heptagon with an apothem of 4 and a side length of 6?

To find the area of any polygon with the side length and the apothem we must know the equation for the area of a polygon which is

First, we must calculate the perimeter using the side length.

To find the perimeter of a regular polygon we take the length of each side and multiply it by the number of sides.

In a heptagon the number of sides is 7 and in this example the side length is 6 so

The perimeter is .

Then we plug in the numbers for the apothem and perimeter into the equation yielding

We then multiply giving us the area of .

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Example Question #5 : Geometry

What is the area of a regular decagon with an apothem of 15 and a side length of 25?

Possible Answers:

1875

2750

375

3750

Correct answer:

1875

Explanation:

To find the area of any polygon with the side length and the apothem we must know the equation for the area of a polygon which is

First, we must calculate the perimeter using the side length.

To find the perimeter of a regular polygon we take the length of each side and multiply it by the number of sides.

In a decagon the number of sides is 10 and in this example the side length is 25 so 25*10=250

The perimeter is 250 .

Then we plug in the numbers for the apothem and perimeter into the equation yielding $Area\:of\:Polygon=\frac{1}{2}(250)15$

We then multiply giving us the area of 1875 .

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Example Question #6 : Geometry

What is the area of a regular heptagon with an apothem of and a side length of ?

Possible Answers:

112

224

Correct answer:

112

Explanation:

To find the area of any polygon with the side length and the apothem we must know the equation for the area of a polygon which is

We must then calculate the perimeter using the side length.

To find the perimeter of a regular polygon we take the length of each side and multiply it by the number of sides .

In a heptagon the number of sides is and in this example the side length is so 8*7=56

The perimeter is 56.

Then we plug in the numbers for the apothem and perimeter into the equation yielding $Area=(\frac{1}{2})(56)(4)$

We then multiply giving us the area of 112 .

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Example Question #7 : Geometry

Find the area of the shaded region:

Screen_shot_2014-02-27_at_6.53.35_pm

Possible Answers:

$100m^2 + 25\pi m^2$

$100m^2 - 25\pi m^2$

$100\pi m^2 - 25m^2$

$75\pi m^2$

75m^2

Correct answer:

$100m^2 - 25\pi m^2$

Explanation:

To find the area of the shaded region, you must subtract the area of the circle from the area of the square.

The formula for the shaded area is:

$A=A_{square}-A_{circle}$

$A = (s)^2 - \pi(r)^2$ ,

where is the side of the square and is the radius of the circle.

Plugging in our values, we get:

$A = (10m)^2 - \pi(5m)^2$

$A = 100m^2 - 25\pi m^2$

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Example Question #8 : Geometry

Find the area of the shaded region:

Screen_shot_2014-03-01_at_9.02.03_pm

Possible Answers:

$6\pi - 9cm^2$

$9\pi - 18cm^2$

$9\pi - 9cm^2$

$9\pi - 24cm^2$

$6\pi - 18cm^2$

Correct answer:

$9\pi - 18cm^2$

Explanation:

To find the area of the shaded region, you need to subtract the area of the triangle from the area of the sector:

$A = A_{sector} - A_{triangle}$

$A = \pi r^2 (part) - \frac{1}{2} (b)(h)$

Where is the radius of the circle, part is the fraction of the circle, is the base of the triangle, and is the height of the triangle

Plugging in our values, we get

$A = \pi (6cm^2) (\frac{90}{360}) - \frac{1}{2} (6cm)(6cm)$

$A = 9\pi - 18cm^2$

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Example Question #9 : Geometry

Find the area of the shaded region:

Screen_shot_2014-03-01_at_9.04.49_pm

Possible Answers:

$\frac{25\pi }{6}- \frac{25\sqrt{2}}{4} cm^2$

$\frac{25\pi }{4}- \frac{25\sqrt{3}}{6} cm^2$

$\frac{25\pi }{4}- \frac{25\sqrt{3}}{4} cm^2$

$\frac{25\pi }{6}- \frac{25\sqrt{3}}{6} cm^2$

$\frac{25\pi }{6}- \frac{25\sqrt{3}}{4} cm^2$

Correct answer:

$\frac{25\pi }{6}- \frac{25\sqrt{3}}{4} cm^2$

Explanation:

To find the area of the shaded region, you need to subtract the area of the equilateral triangle from the area of the sector:

$A = A_{sector} - A_{equilateral triangle}$

$A = \pi r^2 (part) - \frac{s^2 \sqrt{3}}{4}$

Where is the radius of the circle, part is the fraction of the circle, and is the side of the triangle

Plugging in our values, we get

$A = \pi (5cm^2) \frac{60}{360} - \frac{(5cm)^2\sqrt{3}}{4}$

$A=\frac{25\pi }{6}- \frac{25\sqrt{3}}{4} cm^2$

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High School Math : Plane Geometry

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #131 : High School Math

Example Question #1 : Other Polygons

Example Question #2 : Geometry

Example Question #3 : Geometry

Example Question #4 : Geometry

Example Question #5 : Geometry

Example Question #6 : Geometry

Example Question #7 : Geometry

Example Question #8 : Geometry

Example Question #9 : Geometry

All High School Math Resources

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