Coordinate Geometry questions (1 Viewer)

[velox]

Not sure how to answer this question:



And not sure how to answer this one either:



And last one:



Thanks.

 

[shafqat]

Last one:
area = pi.r^2
So area = pi.r^2 + piR^2
= pi.r^2 + pi(20-r)^2 (given)
= answer
Find dS/dr and put = 0. Then r = 10. Show that this is a max from double derivative. So max area when r= 10, R = 20 - 10 = 10. ie they are both the same.

 

[velox]

Thanks Shafqat. Damn, can't believe i missed that.

 

[shafqat]

For the first one, triangles similar: AA, corresponding angles, common angle
So FB/DC = EB/EC (matching sides of similar triangles)
x/y = 3/(3+5) = 3/8

 

[sarabeara]

If i had more time i would spend it doing these questions, but i am currently doing an assessment so I only got out the first part of the last question for you.

"the cirlces shown are such that the sum of their radii is always 20 cm, ie
R + r = 20."

a) Prove that if S is the sum of their areas then S= 2pi(rsq - 20r + 200).

solution:

Area of smaller circle
= pi x rsq (pi times r squared)

Area of larger circle
= pi x Rsq (pi times R squared)

Sum of areas
= (pi x rsq) + (pi x Rsq)
= pi (rsq + Rsq)
= pi [(r + R)sq - 2rR]
= pi [400 - 2rR]
= 2pi (200 - rR)
as R + r = 20, then R = 20 - r
therefore
= 2pi [200 - r(20-r)]
= 2pi [200 - 20r + rsq]
= 2pi [rsq - 20r + 200]

Hope that helps a little bit.

 

[velox]

thanks sarabeara. Shaf i tried the second part of the question but it still doesnt work out.

 

[shafqat]

Are you talking about the circles question?
ds/dr = 4pi.r - 40pi = 0
r = 40pi/4 = 10

 

[velox]

aren't we supposed to be looking for a minimum not max since it wants u to show that the area is 'least' when the radii are equal?

 

[shafqat]

sorry i meant to say min. it works out: double derivative is 4pi, and remember that if its positive, its a min.

 

[velox]

hahah cool, yeh i was confused at why you were saying that its max. thanks.

 

[velox]

What about the second part of this one?

 

[velox]

thanks got em all out in the end.