+0  
 
+1
575
6
avatar+124 

1)Let A, B, and C, be three points on a line such that AB = 2 and BC = 4. Semicircles are drawn with diameters AB, AC, and BC. Find the area of the shaded region.

Picture: https://latex.artofproblemsolving.com/2/f/b/2fba9835516db432c9d4f66df0c10511ce9639d6.png

 

2)In the figure with four circles below, let A1 be the area of the smallest circle, let A2 be the area of the region inside the second-smallest circle but outside the smallest circle, and so on. If A1 = A2 = A3 = A4 then find the ratio of the largest radius to the smallest radius.

Picture: https://latex.artofproblemsolving.com/a/a/5/aa50298b475b48eacab1e9534f5676db52b384d3.png

 

3)A rectangular building is 4 meters by 6 meters. A dog is tied to a rope that is 10 meters long, and the other end is tied to the midpoint of one of the long sides of the building. Find the total area of the region that the dog can reach (not including the inside of the building), in square meters.

 

4)Minor arc AB in the diagram is 30 degrees and the radius of the circle is 6. Find the area of circular segment AB.

Picture: https://latex.artofproblemsolving.com/a/d/f/adfa4bc812b92a40fdd8421107c29286dbc037fd.png

 

5)Given that AB = 2, CD = 4 and that AB and CD are diameters of the respective semicircles shown, find the area of the shaded region.

Picture: https://latex.artofproblemsolving.com/d/c/2/dc234df3c6209415ab27e426df0878a3cb78acec.png

 

6)A, B and C are three points on a plane such that AB = BC = 1 and CA = sqrt 3. Draw three circles whose diameters are AB, BC and CA respectively. The area of the region included in all three circles (the grey region below) is (a/b)pi + (sqrt c /d) in simplest form, so a, b, c, d are positive integers, a and b are relatively prime, and c is not divisible by the square of a prime. Enter the values of a, b, c, d in order, separating by commas.

Picture: https://latex.artofproblemsolving.com/9/f/e/9fe0a447f9c035e61fc690f618e76a639d7fdd2c.png

FiestyGeco  Feb 11, 2018
 #1
avatar
-1

Did you deliberately misspell your username? "FiestyGeco" or "FeistyGeco"??!!!. Just for fun!!.

Guest Feb 11, 2018
 #2
avatar+92816 
+2

1)Let A, B, and C, be three points on a line such that AB = 2 and BC = 4. Semicircles are drawn with diameters AB, AC, and BC. Find the area of the shaded region.

 

BC  has a diameter of 4  so the radius is 2

AB  has a diameter of 2  so the radius  = 1

And AC has a diameter of 6  so the radius  = 3

 

The shaded region  is jusi

 

Area of semi-circle AC   -   area of semi-circle BC  - area of semi-circle AB

 

So we have

 

(1/2) pi  ( 3^2 - 2^2  - 1^2)  =

 

(1/2) pi  ( 9 - 4 - 1 )  =

 

(1/2) pi (4)  =

 

2pi  units^2

 

 

cool cool cool

CPhill  Feb 11, 2018
 #3
avatar+92816 
+1

4. 4)Minor arc AB in the diagram is 30 degrees and the radius of the circle is 6. Find the area of circular segment AB.

 

The area is

 

pi (6)^2  (30/ 360)  - (1/2)6^2sin (30)  =

 

36 pi (1/12)  - (1/2) * 36 * (1/2) =

 

3pi  -  9  =

 

0.4248  units^2

 

 

cool cool cool

CPhill  Feb 12, 2018
 #4
avatar+92816 
0

(3)  was answered here :

 

https://web2.0calc.com/questions/pls-help-need-by-tmrow

 

 

cool cool  cool

CPhill  Feb 12, 2018
 #5
avatar+92816 
+1

5)Given that AB = 2, CD = 4 and that AB and CD are diameters of the respective semicircles shown, find the area of the shaded region.

 

This one is a little difficult

 

Call the mid-point of CD,, M..... connect MA  and MB

 

Triangle  MAB will be equilateral with sides = 2  and will have and area of  

 

(1/2)(2^2) sin (60)   =   √3  units^2  (1)

 

And the area of sector  AMB  is    pi (2)^2 (60/360)  =  (2/3) pi  units ^2   (2)

 

So... the area between chord AB and the circle's edge is just   (2) - (1)   =

 

[ (2/3)pi  - √3 ] units^2      (3)

 

And the area of the small semi-circle is  (1/2) pi ( 1)^2  =  pi / 2    (4)

 

So the shaded area is just  (4)  - (3)   =

 

pi/ 2  -  [ 2/3 pi - √3 ]  =

 

pi/2 + √3  - (2/3) pi   =  

 

[ √3  - pi/6 ]  units^2  ≈   1.208  units^2

 

 

 

cool cool cool

CPhill  Feb 12, 2018
 #6
avatar+92816 
+1

2)In the figure with four circles below, let A1 be the area of the smallest circle, let A2 be the area of the region inside the second-smallest circle but outside the smallest circle, and so on. If A1 = A2 = A3 = A4 then find the ratio of the largest radius to the smallest radius.

 

We can imagine these circles to be concentric..the area between them would still be the same

 

Without a loss of generality....let the smallest radius   = 1

 

And let the next largest radius  = b

 

And we don't need to worry abut "pi" ...  it wll  "cancel"  in the compuations

 

The  Area  of the second smallest circle -  Area of smallest circle = Area of smallest circle

 

b^2  -  1^2  =  1^2

b^2   =  1^2 + 1^2

b^2  = 2

b =  √2

 

Call the radius of the next-to-largest circle  "c"  ..... so...

 

Area of this circle - Area of circle with radius "b"  =  Area of smallest circle

 

c^2  -  2  =  1

c^2 =  3

c^2  =   3

c = √3

 

A call the radius of largest circle  = "d"

 

Area of this circle - Area of circle with radius "c"   =  Area of smallest circle

 

d^2  - 3  =  1

d^2  =  4

d  = 2

 

So.... the ratio of the largest circle's radus to the smallest is just   

 

2 / 1    =   2

 

 

cool cool cool

CPhill  Feb 12, 2018

18 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.