solve it please.?

Question

there are two circular wheels of radius 8 cm and 4 cm and their centres are 20 cm apart. what will be the minimum length of belt required to cover the bothe wheen(like in the bicycle chain)

Answer 1

I saw this last night but was too sleepy to work on it. Now, it appears to me that Raj got the right answer for the wrong reason, while Kevin had the right method but probably missed a sign somewhere. His answer is wrong.

We're going to use the symmetry in this problem; we'll solve the top half, then double it to get the answer.

Draw a long horizontal line. At the left side, draw a half-circle of radius 8 centered at Q, and at the right side, a half-circle of radius 4 centered at T. The distance QT is 20 cm.

Also on that line, points P and R are the left and right endpoints of the diameter through Q, and points S and U are the left and right endpoints of the diameter through T.

Left to right along that horizontal line, we have points P, Q, R, S, T, and U.

Now draw a slanted line tangent to both half-circles, intersecting circles Q and T at A and B respectively. Draw the radii AQ and BT. Note that AQ and BT are parallel, and that angles QAB and TBA are right angles.

Locate point C at the midpoint of radius AQ, and draw TC, which is parallel to and congruent with AB. QC=4, and triangle QCT is a right triangle.

By the Pythagorean Theorem, CT^2 = 20^2 - 4^2 = 384. Taking square roots, AB = CT = sqrt(384) = 8 sqrt(6). Just about everybody got that result.

Now look at angle AQT (equals angle CQT), which we'll call angle x. tan x = CT/CQ = (8 sqrt 6)/4 = 2 sqrt(6), so x = arctan (2 sqrt 6) = 1.3694 radians.

Because of parallel lines, angle UTB = x (corresponding angles). Also, angle PQA = pi - x (using radian measure).

Arc AP = 8 times angle PQA = 8(pi - x). Arc UB = 4 times angle UTB = 4x.

Now we just add to get the top half of the belt. If L is the total length (top and bottom),

L/2 = AB + arc AP + arc UB

L/2 = 8 sqrt(6) + 8(pi - x) + 4x = 8 sqrt(6) + 8 pi - 4x

L = 16 [sqrt(6) + pi - x/2] where x = arctan (2 sqrt 6) = 1.3694 radians

L = 78.50 cm.

This is the same answer that Raj got, but it was Kevin who knew how to do it.

Answer 2

Let points A and B be the first and last(or sort of "endpoints") tangent of the belt and X the center of the 4-cm circle and C and D for the 8 cm belt (center Y). Assuming that the wheels lie on the same plane, let A and D, and B and C be on the same side.
It is better if you draw my figure.
thus, AD = BC.

Draw 2 quadrilaterals BCYX and ADYX. Thus BCYX is congruent to ADYX. Since some are tangent to the circle, ADY, DAX, BCY and CBX are right angles. given the 3 sides lengths, AD and BC measures sqrt(20² - (8 - 4)²) by the Pythagorean theorem and some cancellations. thus,
AD = BC = sqrt(400 - 16) = sqrt(384) = 8sqrt6 cm

To get the percentage of the circumference of the circles, get the angles AXB and the reflex angle CYD.
XYC measures arccos [(8 - 4)/20] = arccos 0.2
CYD measures 2 arccos 0.2
The reflex angle CYD is 360 degrees - 2 arccos 0.2

Since BX and CY are parallel,
BXY and XYC are supp. and
BXY measures 180 - XYC = 180 - arccos 0.2
reflex BXA measures 360 - 2 arccos 0.2
AXB measures 360 - (360 - 2 arccos 0.2 = 2 arccos 0.2

The circumference of the 4 cm circle in which the belt rests (minorarcAB):
2 arccos 0.2/360 degrees (2 · pi · 4 cm) = 2 pi arccos 0.2/45 cm

The circumference of the 8 cm circle in which the belt rests(majorarcCD)
(360 degrees - 2 arccos 0.2)/360 degrees(2 · pi · 8 cm)
= 1 - 4 pi arccos 0.2/45 cm

Thus, the length of the belt (let it be L) is:
L = AD + BC + minorarcAB + majorarcCD
L = 8 sqrt 6 cm + 8 sqrt 6 cm + 2 pi arccos 0.2/45 cm + 1 - 4 pi arccos 0.2/45 cm
L = (16 sqrt 6 + 1 - 2 pi arccos 0.2/45) cm

therefore, the length of the belt is:
L = (16 sqrt 6 - 2/45 pi arccos 0.2 + 1) cm
and
L ≈ 51.1473 cm

^_^

Answer 3

Firstly, we can see that only half of each wheel is effectively in contact with the belt at a given time, therefore, the sum of half the circumference is the length of the belt on the curved area.This gives us 8.pi+ 4.pi= 12.pi cm.

Now looking at the length of belt suspended in the air, we notice that it is NOT HORIZONTAL but at an angle. Since the edge of the bigger wheel is 4 cm higher thanthat of the smaller one, we find the length of the suspended belt by using the pythagoras formula i.e. a^2+b^2=c^2, where 20 cm is a and 4 cm is b. Therefore our answer is 20.4cm. But this is for one side, for both, we simply multiply by 2, i.e. 40.8 cm.

Therefore the total is 40.8+ 12.pi = 78.5 cm!!!

Answer 4

=2*pi*8-4*asin(8/(2*20)*pi/180
+sqrt((2*20)^2-8^2)
88.65188666 Cm

Answer 5

you must take half of circumfrence for both the circles... and then add 20.

Answer 6

77.68