Math Is Fun Forum

Ok, I redid my calculations and got the same answer as you. Here is my work:

Hi Bob, for number 1, I used the equation

Hi bob for number 2, don't we need to know the arc length of AC? Since the full arc length of a circle is pi*EF we divide that by half and then subtract 80+35 but that's getting me absolutely nowhere

Also for number 3 I still don't get it

Ok I will clean up confusion, the sector is 32π

In the figure with four circles below, let A_1 be the area of the smallest circle, let A_2 be the area of the region inside the second-smallest circle but outside the smallest circle, and so on. If A_1 : A_2 : A_3 : A_4 = 1 : 2 : 3 : 4, then find the ratio of the largest radius to the smallest radius. Answer: sqrt{10}: 1

Let A, B, C, be three points on a line such that AB = 2 and BC = 4. Semicircles are drawn with diameters \overline{AB}, \overline{AC}, and \overline{BC}. Find the area of the shaded region. Answer: 2π

A sector of a circle has a central angle of 80 degrees. If the area of the sector is 32π, what is the radius of the circle? Answer: 12

Bob your hints are really nice
Please confirm my answers, sorry for the confusion thanks

Hi can I have a solution aswell?

I am getting 2pi for number 2

Wait is the question really trivial or am I missing something

Derrick: 0
Ari: 0.50
Wesley: 2.10

Question 18 doesn't make sense to me...are they side lengths? Anyway its too late here, when I get up tomorrow morning ill do some of the problems

Thanks bob, Ill get back to you with the answers for the 2 problems
I am getting 86 and 24, can you check that they are right? Thanks

Hi, sorry I was very sick with a fever for the last couple of days. Here is the solution for Q3:

Let the line through P parallel to \overline{AB} intersect \overline{BC} and \overline{AD} at T and V, respectively.  Let the line through P/parallel to \overline{AD} intersect \overline{AB} and \overline{CD} at S and U, respectively. Then by Pythagoras on right triangles PAV, PBT, PCT, and PDV, PA^2 = AV^2 + PV^2,  PB^2 = BT^2 + PT^2,  PC^2 = CT^2 + PT^2,  PD^2 = DV^2 + PV^2, so PA^2 + PC^2 = AV^2 + CT^2 + PV^2 + PT^2 and PB^2 + PD^2 = BT^2 + DV^2 + PV^2 + PT^2. But quadrilaterals ABTV and CDVT are rectangles, so AV = BT and CT = DV. Hence, PA^2 + PC^2 = PB^2 + PD^2. Substituting the given information, we get 1^2 + 8^2 = 7^2 + PD^2, so PD^2 = 1^2 + 8^2 - 7^2 = 16, which means PD ={4}.

Someone please put math tags in for me and sorry about the diagram being too big