A 6-foot-diameter disk and an 18-foot-diameter disk sit side by side as shown in the figure below. Determine?

Question

A 6-foot-diameter disk and an 18-foot-diameter disk sit side by side as shown in the figure below. Determine the shortest length of rope that will go around both disks.

http://i117.photobucket.com/albums/o77/florpz/2.jpg

Oh and if possible, can someone help me with this as well?

Given a circle O with a diameter of 10 cm and equilateral triangle ABC as shown in the diagram, find the area of the shaded region.

http://i117.photobucket.com/albums/o77/florpz/10.jpg

Answer 1

2) Let's start with the second question.

Given a circle O with a diameter of 10 cm and equilateral triangle ABC as shown in the diagram, find the area of the shaded region.

Let's add two points. D is the point on CA where CA intersects circle O. E is the point on CB where CB intersects circle O.

Angle C = (1/2)(Arc ADEB - Arc DE)
π/3 = (1/2)(π - Arc DE)
2π/3 = π - Arc DE
Arc DE = π/3

Angle A = (1/2)(Arc DEB)
π/3 = (1/2)(Arc DEB)
2π/3 = Arc DEB
Arc DEB = 2π/3

Arc EB = (Arc DEB) - (Arc DE) = 2π/3 - π/3 = π/3

By symmetry

Arc AD = Arc EB = π/3

Now the area K, of the pie shaped region of the circle AOD is:

K = [(Arc AD/(2π)]*πr² = [(π/3)/(2π)]*πr²
K = (1/6)πr²

But the shaded area exludes the equilateral triangle AOD. THe area of triangle AOD = [(√3)/4]r²

So the area S, of the shaded area is:

S = (1/6)πr² - [(√3)/4]r²

However there are two of them and they are symmetric. So the total shaded area is:

2S = 2{(1/6)πr² - [(√3)/4]r²} = (1/3)πr² - [(√3)/2]r²
2S = [(1/3)π - (√3)/2]r²

But r = d/2 = 10 cm/2 = 5 cm

2S = [π/3 - (√3)/2](5²)
2S = 25[π/3 - (√3)/2] cm²
2S = 4.5293037 cm²

Answer 2

Hope you can do the diagram stuff yourself because I don't know how to show it here.
Join the centres of the circles (24 units long), then draw the radius from each centre to the top point where the rope touches each circle. Now you have a trapezium with a straight bit of rope for one side, two sides at rightangles to it (6 units and 12 units) and the other side 24 units long.

Complete a rectangle by a line from the centre of the small circle, parallel to the rope bit, to meet the radius of the large circle.

Now our trapezium is a rectangle plus a rightangled triangle, and the triangle has hypotenuse 24 and one side 12 (it's 18 - 6).

Do you know enough trigonometry to realise this triangle must be half an equilateral triangle? Its acute angles are 30 and 60 deg, and its other side is 12Sqrt3.

SORRY! You said diameter, and I've been doing it as if it's radius. Halve all lengths, so the straight bit of rope is 6*sqrt3 units.

OK. So the two straight bits of rope total 2*6*sqrt3 units.

The bit wrapped aroung the big circle subtends 240 deg at the centre, so it's 2/3*18*pi. On the small one it's only 120 deg, so it's 1/3*6*pi. Adding all this up gives

14*pi + 12*sqrt3.

Love to do the other one but have to go soon. Sorry.

Answer 3

The answer to the second question is....

There is a theory in Geometry that the arc subtended (cut) by an interior angle in a circle is twice the number of degrees as the angle is.

Thus angle A and angle B (each being 60-degrees since it is an equilateral triangle) each cut off a 120-degree arc of the circle.

That means the remaining arc (around each shaded part) is 60-degrees, or 1/6th of an entire circle. Since the circle is r=5 (half of the diameter), the whole circle is pi*r^2, or 25pi. We need 1/6th of it, or 25pi/6.

We need to subtract out the small traingle out of the 60-degree circle wedge. Since it is a 60-degree wedge, and we already know A or B is 60, that means the 3rd angle must be 60, so the small triangle is also an equilateral triangle with side = 5. Area of any equilateral triangle is (side^2)*sqrt(3)/4. This example is then 25*sqrt(3)/4.

Total area of 1 shaded piece = 25pi/6-25sqrt(3)/4. We need both, so the answer is 25pi/3-25sqrt(3)/2.

Answer 4

For the first problem, the rope is tangent to both circles, hence perpendicular to the point where they first touch the circle, and a radius. Since the radius of the smaller circle is 3 feet, and the radius of the larger circle is 9 feet, we can draw a diameter through both circles that complete a quadrilateral, with two right angles and the parallel sides have length 3 feet and 9 feet. The diagonal (longest) side of the quadrilateral has length 12 feet. You can solve for the base using the pythagorean theorem. The angle the radii make with the common diameters is 60 degrees. So the rope goes around a third of the smaller circle, and two thirds of the larger circle.

For the second problem, the triangle intersects the circle at two points. Connect these two points to the center, and calculate the area of each of these smaller equilateral triangles. Add these areas together and subtract from one third of the area of the circle.

Answer 5

mark the center D, so AD is 5cm, mark the piont where AB cuts the circle E, forming an other triangle ADE, this triangle will be isoceles, since 2 of its sides ar both radios of a cicle, so since tri ABC is equilateral than angle BAC is 60 ans since tri ADE is isoceles then ang AED is also 60, so angle EDA is 60 too. then u find the area bye the equation x/360(pie R^2), x being the angle ADEand R the radious. the u find the aria of triangle ADE by the formula 1/2 base * height , and u subtract the area of the triangle from the area of the sector, then repeat the same procedure for the other part

Answer 6

in case you draw 3 radii in each circle so as that they go from a) the midsection to the tangent factor of the rope on one factor b) the midsection to the tangent factor of the rope on the different factor c) the factor of intersection of the two circles then make a rectangle with one factor the piece of rope between the circles, one factor the smaller radius (3), in spite of this to a piont 3 from the sting of the bigger circle, you will additionally see an exact triangle with leg 6 and hypotenuse 12. to that end the precious attitude is 60 tiers and the long leg 6 sqrt 3, that's additionally the area of rope between the two circles. This area is on the different factor additionally. The area of the great circle with a rope wrapped around that's to that end 2/3 of that circle, and the area around the toddler is a million/3 of that circle. discover their circumferences, then get the fractional factors, upload the two 6 sqrt 3 factors and you have it As to your different, the shaded area is termed a phase. To get its section, draw yet another radius to the different end of each phase. This makes little 5x5x5 equilateral triangles. Subtract the portion of one of those from a million/6 of the portion of the circle to get the portion of each phase, then double it when you consider that there are 2 of them.