Geometry Problem #90: Arc vs. Chord Distance on a Semicircular Path

  • Thread starter Thread starter Jameson
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on a geometry problem involving the distance between two points, A and B, on a semicircular path versus a straight chord. Given that the diameter CD measures 180 meters, the problem requires calculating the difference in distance between walking along the arc AB and the chord AB. The correct solutions were provided by members MarkFL, mente oscura, anemone, and eddybob123, showcasing their understanding of basic geometric principles.

PREREQUISITES
  • Understanding of semicircles and their properties
  • Basic knowledge of arc length calculations
  • Familiarity with chord length in geometry
  • Ability to apply the Pythagorean theorem
NEXT STEPS
  • Study the formula for arc length in semicircles
  • Learn how to calculate the length of a chord in a circle
  • Explore the relationship between diameter, radius, and arc length
  • Practice solving similar geometry problems from The Art of Problem Solving (AoPS)
USEFUL FOR

This discussion is beneficial for beginning geometry students, educators teaching geometric concepts, and anyone looking to enhance their problem-solving skills in mathematics.

Jameson
Insights Author
Gold Member
MHB
Messages
4,533
Reaction score
13
This is for beginning geometry students so if you are new to geometry, take a look! :)

Marcia could walk from A to B along arc AB on the semicircular path, or she can walk along
chord AB. Diameter CD has length 180m. How much farther is it to walk along the arc as
opposed to the chord?

View attachment 1772
--------------------
 

Attachments

  • Screen Shot 2013-12-15 at 10.40.42 PM.png
    Screen Shot 2013-12-15 at 10.40.42 PM.png
    2.3 KB · Views: 162
Physics news on Phys.org
This problem came from The Art of Problem Solving (AoPS) website.

Congratulations to the following members for their correct solutions:

1) MarkFL
2) mente oscura
3) anemone
4) eddybob123

Solution (from MarkFL):
Please consider the following diagram:

mimdmx.png


From this we find:

$$\cos(x)=\frac{\frac{r}{2}}{r}=\frac{1}{2}\implies x=\frac{\pi}{3}$$

Hence, arc $AB$ (denoted by $s$) has length:

$$s=r(2x)=\frac{2\pi r}{3}$$

Line segment $\overline{AB}$ may be found from:

$$\sin(x)=\frac{\frac{1}{2}\overline{AB}}{r}$$

$$\overline{AB}=2r\sin(x)=\sqrt{3}r$$

Thus, the difference $\Delta$ between the two paths is:

$$\Delta=s-\overline{AB}=\frac{2\pi r}{3}-\sqrt{3}r=\frac{r}{3}\left(2\pi-3\sqrt{3} \right)$$

Using the given data:

$$r=90\text{ m}$$

we have:

$$\Delta=30\left(2\pi-3\sqrt{3} \right)\text{ m}$$
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 40 ·
2
Replies
40
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
31
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
1
Views
6K