The square root of 1,000,000 is 1000; therefore, there will be 1000 perfect squares:
12 = 1
22 = 4
32 = 9
9982 = 996,004
9992 = 998,001
10002 = 1,000,000
Now, we have to cross-check that with a list of perfect cubes.
Since the cube root of 1,000,000 is 100, there will be 100 numbers on this list.
13 = 1 Cross this one off
23 = 8
33 = 27
43 = 64 Cross this one off
983 = 941,192
993 = 970,299
1003 = 1,000,000 Cross this one off
So, there will be a total of 1000 from the list of perfect squares plus all those on the list of perfect cubes that
aren't crossed off.
Then, subtract that total from 1,000,000.
Since angle(CAB) = 45o and angle(CAD) = 30o, angle(DAB) = 15o.
Since angle(ADB) = 150o and angle(DAB) = 15o, angle(DBA) = 15o.
Therefore, AD = DB and since AD = 2, DB = 2.
We can find the value of AB by using the Law of Cosines:
AB2 = AD2 + BD2 - 2·AD·BD·cos(ADB) ---> AB2 = 22 + 22 - 2·2·2·cos(150o)
---> AB2 = 4 + 4 - 8 · -sqrt(3)/2 ---> AB2 = 8 + 4sqrt(3)
---> AB = sqrt( 8 + 4sqrt(3) )
Since triangle(ABC) is an isosceles right triangle, BC = AB / sqrt(2)
BC = sqrt( 8 + 4sqrt(3) ) / sqrt(2)
BC = sqrt( 4 + 2sqrt(3) )
Look at triangle(SPR):
-- MX is a midline of this triangle, so its length is one-half the length of the base, SR:
MX = 10
Look at triangle(RSQ)
-- YN is a midline of this triangle, so its length is one-half the length of the base, SR:
YN = 10
Now look at MN
-- MN = MY + XY + YN ---> MN = 10 + 7 + 10 = 27
To find PQ:
-- MN must have a value half-ways between SR and PQ.
-- MN = 27, SR = 20 ---> PQ = 34
#1) Draw a line segment from O to the point of tangency of te quarter-circle (call that point T).
The center of the circle will be on OT.
Call the cdnter of the circle C.
Drop perpendiculars from C to AO (call that point X) and from C to OB (call tht point Y).
The distances from C to T, from C to X, and from C to Y are all equal and they are radii of
the smaller circle.
Let the radius of the smaller circle be x.
All the distances: XO, YO, CX, CY, and OT are equal to x.
Since OXCY is a square with side = x, the distance from O to C = x·sqrt(2).
The distance from OT = OC + CT
OT = x·sqrt(2) + x = x(sqrt(2) + 1)
But, OT is a radius of the quarter-circle = 5.
5 = x(sqrt(2) + 1)
x = 5 / (sqrt(2) + 1)
#2) Drop a perpendicular from X to YZ. Call this point O.
O will be the midpoint of YZ.
Since YZ = 40, OZ = 20.
Use the Pythagorean Theorem on right trianle(XOZ)
---> XO2 + OZ2 = XZ2 ---> XO2 + 202 = 252 ---> XO = 15
Draw a perpendicular from O to XZ. Call the point of intersection P.
Since this is perpendicular to XZ, this is the point of tangency to the semicircle.
Consider triangle(OPZ); this right triangle is similar to triangle((XOZ) by AA.
Therefore: OP / OZ = XO / XZ.
OP / 20 = 15 / 25 ---> OP = 20 · 15 / 25 ---> OP = 12.
OP is the radius of the semicircle; now use the area formula:
area semicircle = ½ · pi · radius2 to find the area of the semicircle.
The diagonals of a rhombus are perpendicular to each other.
If we draw the rhombus and its two diagonals, we get four congruent right triangles.
Looking at one of these right triangles, we can see that the hypotenuse is 169 (the side of the rhombus) and one of its sides is 65 (one-half of the one diagonal0.
I'm going to use the Pythagorean Theorem to find the other half-diagonal:
c2 = a2 + b2 ---> 1692 = 652 + b2 ---> b = 156 ---> so the other diagonal is 312.
Now that I know the length of both diagonals, I can find the area of the rhombus:
A = ½·diagonal1·diagonal2 ---> A = ½·130·312 = 20280
Since all four triangles are congruent, each triangle has an area of 20280 / 4 = 5070.
I'm going to use this area to find the radius of the inscribed circle.
The center of the circle is the center of the rhombus.
The radius of the circle is the height of each of the triangles (the base will be a side of the rhombus).
A = ½·base·height ---> 5070 = ½·169·height ---> height = 60
The radius of the circle is 60.
Find the area of the circle and subtract from the area of the rhombus.
You can win in any of four ways:
1) Getting two of the three winning white balls and the red SuperBall.
Two of the three wining white balls and one out of seven losing white balls:
3C2·7C1 / 10C3 = 3·7/120 = 7/40
Combine this with getting the red SuperBall = 1/10 ---> 7/400
2) Getting two of the three white balls and not getting the red SuperBall:
The 7/40 from above combined with not getting the red SuperBall = 9/10 ---> 63/400
3) Getting all three of the white balls and the red SuperBall:
Three of three winning white balls and none of the seven losing white balls:
3C3·7C0 / 10C3 = 1·1/120 = 1/120
Combine this with getting the red SuperBall = 1/10 ---> 1/200
4) Getting all three of the white balls and not getting the SuperBall:
The 1/120 from above combined with not getting the red SuperBall = 9/10 ---> 3/400
To get the final answer, you must add the individual answers from above.