#1

Hey, was wondering if anyone on the Pit is smart enough to figure out these questions:

(NO calculators allowed, the questions must be done by hand and exact values)

A basketball with a radius of 20 centimeters sits in the corner of a room, touching the floor and two walls simultaneously. A softball sits on the floor between the basketball and the corner. The softball touches the floor, the two walls, and the basketball. The radius of the softball, in centimeters is?

Answer I got is: 20[(sqrt2)-1] / [(sqrt2)+1] but all the multiple choice answers have (sqrt3) in them

Second question is: A square of side 13 is inscribed inside a square of side 17 as shown in the diagram, The

Here's the diagram, no clue how to practically figure this out without a calculator though:

(NO calculators allowed, the questions must be done by hand and exact values)

A basketball with a radius of 20 centimeters sits in the corner of a room, touching the floor and two walls simultaneously. A softball sits on the floor between the basketball and the corner. The softball touches the floor, the two walls, and the basketball. The radius of the softball, in centimeters is?

Answer I got is: 20[(sqrt2)-1] / [(sqrt2)+1] but all the multiple choice answers have (sqrt3) in them

Second question is: A square of side 13 is inscribed inside a square of side 17 as shown in the diagram, The

**greatest distance**between a vertex of the inner square and a vertex of the outer square is?Here's the diagram, no clue how to practically figure this out without a calculator though:

#2

for number 2... ok. the whole side is 17. look at it as a right triangle. you know that the two legs of any right triangle (the corners) add up to 17 and have a hypoteneuse of 13. the greatest distance would be 12. 144 + 25 = 169. do you get it? or is this just really confusing?

#3

I really don't have the kind of attention span at almost nine o'clock.

Sorry.

Sorry.

#4

Is the second one something to do with the midpoint of each vertexx?

vertex are corners?!

vertex are corners?!

*Last edited by let__it__bleed at May 28, 2008,*

#5

for number 2... ok. the whole side is 17. look at it as a right triangle. you know that the two legs of any right triangle (the corners) add up to 17 and have a hypoteneuse of 13. the greatest distance would be 12. 144 + 25 = 169. do you get it? or is this just really confusing?

I understand what you're saying but I don't believe those are the numbers I need to use, it wants to know the greatest distance between the farthest vertexes of each square, so like this distance I'm going to draw here with a red line:

The possible answers it lists are:

A. 17

B. (sqrt 314)

C. (sqrt 433)

D. 22

E. (sqrt 458)

#6

edit: I need paper...will reply later. The solving for x one below me is correct.

*Last edited by Fast_Fingers at May 29, 2008,*

#7

I understand what you're saying but I don't believe those are the numbers I need to use, it wants to know the greatest distance between the farthest vertexes of each square, so like this distance I'm going to draw here with a red line:

The possible answers it lists are:

A. 17

B. (sqrt 314)

C. (sqrt 433)

D. 22

E. (sqrt 458)

then it's root 433? because... the side is 17, and the one part of the side is 12, the longest side, I mean, so 17 squared + 12 squared = 433? and then the root of that... is root 433.

#8

For the second question, the exact answer would be (sqrt433). Just use algebra and phytagorean theorem. The equation would be 13^2=(17-x)^2+x^2 then solve for x. hope this helps.

EDIT: I forgot to mention that x here is the shorter leg of one side of the triangle which is 5. that means the longer leg is 12. after solving for this one use the phytagorean theorem to solve for the distance of the greatest vertex: sqrt433=sqrt((5+12)^2+12^2).

EDIT: I forgot to mention that x here is the shorter leg of one side of the triangle which is 5. that means the longer leg is 12. after solving for this one use the phytagorean theorem to solve for the distance of the greatest vertex: sqrt433=sqrt((5+12)^2+12^2).

*Last edited by deadpoolxs at May 29, 2008,*

#9

With the balls, I'm guessing you'll end up with triangles of 30,60,90, which the hypotenuse is twice one side. Still, I did end up with a rt. 2 answer like yourself.

http://www.krellinst.org/UCES/archive/resources/trig/node20.html

http://www.krellinst.org/UCES/archive/resources/trig/node20.html

#10

Alright I got the second one, thanks guys =D

The problem I was having with it was solving it for x without using a calculator hehe but I finally got it as 5.

Now just to figure out what I could have done wrong on the first question. I got the (sqrt2) from making right triangles with the radius as each side and knowing that for a 45 degree angle if the sides are 1 and 1 the hypotenuse is (sqrt2). I don't see where they could have even come up with a (sqrt3) in their answers.

The problem I was having with it was solving it for x without using a calculator hehe but I finally got it as 5.

Now just to figure out what I could have done wrong on the first question. I got the (sqrt2) from making right triangles with the radius as each side and knowing that for a 45 degree angle if the sides are 1 and 1 the hypotenuse is (sqrt2). I don't see where they could have even come up with a (sqrt3) in their answers.

#11

The rt. 3 probably suggests a 30-60-90 triangle (where the sides are x, x(sqrt3), and 2x) but I don't know where one would come from...I got a rt. 2 answer too.

#12

Hey, was wondering if anyone on the Pit is smart enough to figure out these questions:

(NO calculators allowed, the questions must be done by hand and exact values)

A basketball with a radius of 20 centimeters sits in the corner of a room, touching the floor and two walls simultaneously. A softball sits on the floor between the basketball and the corner. The softball touches the floor, the two walls, and the basketball. The radius of the softball, in centimeters is?

Answer I got is: 20[(sqrt2)-1] / [(sqrt2)+1] but all the multiple choice answers have (sqrt3) in them

I did it two kind of complicated ways but it simplified to 87/30. We never learnt geometry to this depth =/

The first way i did it took a couple pages. I let r be the softball radius and x and y be the lengths along the wall where the softball touches. I also found the distance from the corner to the point at which the softball touches the b-ball. It was 40/sqrt2. That didn't really help because the angle in the corner is then 3pi/8 which i couldn't be bothered using trig identities to simplify. Then i did a massive addition thing with the total area. This part was excessively annoying. I kept splitting it up into smaller segments and using mainly 0.5r^2.theta and the sine 0.5ab.sin(C) rule found the tiny mofo areas. Added em up (my manipulations pretty good so adding up the many r's and pi's was ok), took away from 400, then the old pi.r^2. 87/30.

That felt kinda tedious though so i decided to model them as functions on x and y axes. Fun, but i seriously doubt they'd expect you to do it. Anyway, the second to last line had sqrt3's in it but it was [r/(sqrt3)]/sqrt3 = (9/10)+(1/15) which just simplified to 87/30.

How far off am i?

(btw what maths is this? school/uni?)

#13

The ball one, there was a similar question in my calculus exam today. Are people familiar with complex numbers or vectors? Pretend the wall and floor are axis, then the basketball is the circle at centre (20,20) with radius 20. Therefore the distance from the corner to the point where the softball touches the basketball is 20*root2 -20, so this co ordinate is:

(40-20*root2,40-20*root2)

If z or (x,y) is the centre of the softball, then the distance from z to this point is equal to the distance from z to each of the axis,

ie |z - (40-20*root2,40-20*root2)| = |z - (x,0)| = |z - (0,y)|

we also know x = y,

I'm fairly sure this is solvable, but I cbf doing it. Hey, I had a three hour calc exam today!

edit: I just realised this would only hold for if the ball was against the parallel bit between two walls. For in the corner, you'd have to use 3d vectors. My bad.

I did some working on paper:

http://img102.imageshack.us/my.php?image=save0008el6.jpg

After you find the centre, just find the distance from the centre to one of the points you know and you have the radius.

(40-20*root2,40-20*root2)

If z or (x,y) is the centre of the softball, then the distance from z to this point is equal to the distance from z to each of the axis,

ie |z - (40-20*root2,40-20*root2)| = |z - (x,0)| = |z - (0,y)|

we also know x = y,

I'm fairly sure this is solvable, but I cbf doing it. Hey, I had a three hour calc exam today!

edit: I just realised this would only hold for if the ball was against the parallel bit between two walls. For in the corner, you'd have to use 3d vectors. My bad.

I did some working on paper:

http://img102.imageshack.us/my.php?image=save0008el6.jpg

After you find the centre, just find the distance from the centre to one of the points you know and you have the radius.

*Last edited by shigidab0p at May 29, 2008,*

#14

Number 2 Is 4