In the figure below, find the area of the non-shaded region. Leave your answer in terms of r. To receive full?

Question

http://i40.photobucket.com/albums/e215/xxs3xp1stolxx/2.jpg

Answer 1

The length of the out side rectangle =

diamete + r + r = 2r + 2r = 4r

width of rectangle = diameter of circle = 2r

area of rectangle = 4r*(2r) = 8r^2

area of shaded portion= area of circle + 2 area of semi circle)

= 2(area of circle) = 2 pi r^2

area of unshaded portion

= area of rectangle - shaded portion

= 8r^2 - 2pi r^2

r^2( 8 - 2pi)

1.72 r^2 sq. units

Answer 2

wish you're able to do the diagram stuff your self because of the fact i do no longer understand a thank you to coach it right here. connect the centres of the circles (24 gadgets long), then draw the radius from each and each centre to the appropriate component the place the rope touches each and each circle. Now you have a trapezium with a right away little bit of rope for one area, 2 factors at rightangles to it (6 gadgets and 12 gadgets) and the different area 24 gadgets long. complete a rectangle via a line from the centre of the small circle, parallel to the rope bit, to fulfill the radius of the great circle. Now our trapezium is a rectangle plus a rightangled triangle, and the triangle has hypotenuse 24 and one area 12 (it rather is 18 - 6). do you recognize sufficient trigonometry to grasp this triangle must be 0.5 an equilateral triangle? Its acute angles are 30 and 60 deg, and its different area is 12Sqrt3. SORRY! You pronounced diameter, and that i've got been doing it as though it rather is radius. Halve all lengths, so the right away little bit of rope is 6*sqrt3 gadgets. ok. So the two right away bits of rope entire 2*6*sqrt3 gadgets. The bit wrapped aroung the great circle subtends 240 deg on the centre, so it rather is 2/3*18*pi. on the small one it rather is only one hundred twenty deg, so it rather is a million/3*6*pi. including all this up supplies 14*pi + 12*sqrt3. decide for to do the different one yet ought to bypass quickly. Sorry.

Answer 3

If you know how to find the area of a rectangle simple take the area of the rectangle then subtract the area of the two circles (2 full circles because you have 1 full and 2 halves which means you have 2 full circles). Once you have those equations set up solve for r. If you don't know what the equations for the area of a rectagle or the area of a circle are, check your text book.

Answer 4

Shaded area = 2 π r ²
Area of rectangle = 4 r x 2 r = 8 r ²
Non shaded area
= 8 r ² - 2 π r ²
= (8 - 2 π) r ²
= 2 (4 - π) r ²
= 1.72 r ²

Answer 5

Since all the circles are identical (same radii), then their areas are the same, i.e. total area of the 2 semi-circles + the complete circle is 2#r2..(2 pi r squared)
Also, the length of the square = the combined radii of all the circles = 4r
Also, the width of the rectangle = the diameter of any circle = 2r
therefore, the area of the unshaded region is given by, Area of rectangle, (4r).(2r) - total Area of circles, 2(*r2)
= 8r2 - 2*r2
= 2r2 (4 - *)

Answer 6

so you basically want the area of the box minus the area of a half circle+a circle+a half circle (which is 2 circles).

so what is the length of the box? it is 4 radiuses (4r).
what is the height of the box? the diameter which is 2 radiuses (2r).

so the area of that box is (4r)(2r)=8r^2

now let's find the area of one circle.
it's pir^2 right?

so since we need to subtract the area of two circles we just multiply the area of one circle by 2.

so the area of the two circles is just 2pir^2.

so the area of the box minus the area of the two circles is:
8r^2-2pir^2

you can factor out 2r^2 from both terms and get
2r^2(4-pi)

Answer 7

The entire box is 4r across by 2r tall. (4r)(2r) = 8r².
The shaded portion is two full circles: 2(πr²).
The unshaded region is the difference between the two.
A = 8r² - 2πr² = 2r²(4 - π).

Answer 8

The first 2 answers are both right :)

Answer 9

1.72r^2