2
$\begingroup$

Problem: The length of arc of circle of radius $r$ which subtends an angle $\theta$ radians at the centre of the circle $r\theta$. The area of the sector formed by this arc and radii to either end of the arc is $\frac{1}{2}r^2\theta$. Find the angle $\theta$ if the magnitude of the arc length is $10\pi $ $cm$ and the area is $0.740\pi$ $cm^2$. (Source: Engineering Maths by Leslie Mustoe, Chapter 5, Exercise 5.1 on page 88.)

Result: $\frac{5\pi}{4}$ (According to the book.)

Question: My solution attempt yielded a very different number, and shame on me, but I do not understand what is wrong, in this seemingly simple calculation. So, could you please point out what is wrong in the below solution? (Please, don't tell me the right solution, as I want to arrive to it myself. I just want to understand where I'm making a mistake.)

Attempt:

$r\theta=10\pi$ $cm$ (1)

$\frac{1}{2}r^2\theta=0.74 \pi$ $cm^2$ (2)

Dividing (2) by (1):

$\frac{1}{2}r = \frac {0.74}{10}$ $cm$

$r = \frac{0.74}{5} $ $cm$ = $0.148$ $cm$.

From (1):

$\theta = \frac{10\pi}{r} = \frac{10\pi}{0.148} $

$\theta = 212.27 = 67.57 \pi$

Similar questions

Arc length and area of sector

--> This is basically the proof for a slightly different (degree-based) version of (2).

Find the Radius, Central Angle, and Arc Length Given Sector Area and Perimeter --> The answer gives me (1) and (2).

Central angle of a circular sector from area and arc length --> This is the same problem I have, with different numbers. Strange thing is, if I try to do it, I get the same (right) result as the OP.

$\endgroup$
  • $\begingroup$ Your solution seems correct to me. I put the values in a calculator and they match up as well. I am inclined to believe the book's answer is wrong... +1 for a very well asked question anyway. $\endgroup$ – Gaurang Tandon Nov 12 '17 at 15:48
  • $\begingroup$ Unfortunately, even in math books, sometimes answers are wrong. +1 for anyone who finds such a mistake. Then of course, math community can set it all straight. $\endgroup$ – imranfat Nov 12 '17 at 15:57
  • $\begingroup$ There is something wrong with the question. Obviously, $\theta$ cannot exceed $2 \pi$. I believe the area is $40 \pi$ instead of $0.7~40\pi$. $\endgroup$ – Math Lover Nov 12 '17 at 15:58
  • $\begingroup$ It seems that the real problem is the arc length is given as $10\pi$ cm while the area is $0.740\text{ cm}^2$. Perhaps the area should be in $\text{m}^2$? Currently you have an arc length of about a foot and an area less than the size of a thumbnail. $\endgroup$ – Dan Robertson Nov 12 '17 at 16:01
  • $\begingroup$ Small nitpick: when you work out $\thera,$ you divide $10\pi$ cm by 0.148 cm so the result should be dimensionless (cm/cm=1). You do seem to change this in the next line $\endgroup$ – Dan Robertson Nov 12 '17 at 16:03
1
$\begingroup$

I think that there are no doubts. Data are wrong. $\theta $ cannot be more than $2\pi$ and the picture below is the only possibility to "save" the exercise, otherwise is meaningless: a sector having perimeter $31$ cm and more CANNOT have an area of $2.3$ cm$^2$

Hope this is useful

$$...$$

enter image description here

$\endgroup$
  • $\begingroup$ Thanks for pointing it out, this makes sense. However, the calculations still don't match for me. As per (1), I have $14.8\theta = 10\pi$, thus $\theta=\frac{10\pi}{14.8} = \frac{25\pi}{37}$. I get the point, though. $\endgroup$ – Attilio Nov 19 '17 at 15:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.