Help needed with geometry based math problem.

[Satish555]

can anyone help me with this one? it looks a bit though.

Here's a rough sketch of the problem which i made in photoshop:

ill still type the details though. first there is a big square. then a circle inside with touches the edges of the square. then there is a little rectangle on the top right which has an edge (right bottom) touching the circle.and the dimensions of that little rectangle are 1x2. you need to find the radius of the circle. note the diagram is not drawn to scale.

Please Explain in detail and also if u can give a visual demonstration like i did. i need to submit this in 2 days and its worth good marks so please help.

 

[Adarsh]

Visualising it in my head, it looks like the radius is 5 cm. I'm not sure. I'll work on it and tell you.

EDIT: Gee this is annoying me. I'm pretty sure the radius is 5cm, but I'm not able to prove it.

 

[Harrypotter_fan]

hey the radius is 18.95 approx.
(the solution is given by my sister,who has used some simple geometry and logarithmic tables and found it)

p.s. the proof is there but very tedious to type it.

btw adarsh how did you get the radius to be 5?

 

[Adarsh]

Mine was nothing more than an educated guess, tbh! 18.95... doesn't seem right, but if you say so it is then I can't argue.

 

[Harrypotter_fan]

why do you say so?
any solid reasoning behind it?

 

[Adarsh]

I've nearly given up. Either this uses techniques beyond the realms of an A level student or this is one of those questions where you either get it straight away or you don't.

YES!! I've got it. Ridiculously easy!! A bit of proof for my solution. If X = radius, you can see that the lines (x-1) and (x-2) and x form a right angled triangle. This gives (x-1)^2 + (x-2)^2= x^2. You can either factorise it and solve it or use common knowledge of the Pythagorean triple 5, 4 ,3. Therefore, x = 5 !!

 

[andrew_nixon]

Adarsh, that equation you give simplifies to x^2 = -5, which is impossible.

 

[Adarsh]

Does it really ? Please correct me if I am wrong. The equation in itself does NOT simplify to x^2 = -5. It simplifies to (x-5)(x-1)=0. Therefore, the 2 solutions are 5 and 1. And since 1 can't be the right answer as (x-1) will be 0 which is impossible, 5 is the only right answer. If you get 3^2 + 4^2 = x^2, then hypothetically -5 is possible but surely it isn't in this case.

Or is this another example of Andrew Nixon's pedantry ?

 

[Satish555]

i have something else here. and i cant understand adarsh reasoning adarsh could u explain a bit better? i am not that advanced and im only 15. so what i have is this.

first cut the rectangle into two little squares so one has a dimension of 1x1. and the diagonal of that square will be sq.rt. of 2. and then draw a triangle, from the center of circle to the little square measuring 1x1 which has a diagonal of sqrt 2. so the two sides of the triangle are equal but the third side is equal to one of the other sides + sqrt 2.
2. You have a triangle with sides - two equal sides= R & hypotenuse is R+(sqrt2)
3. Using pythogoras for this triangle gives -
R? +R?=(R +(sqrt2))?
R? +R?= R? +2xRx(sqrt2) +(sqrt2)?
2R? = R? +2xRx(sqrt2) + 2
2R? - R? = 2xRx(sqrt2) + 2
R? = 2xRx (sqrt2) +2
R? -2xRxsqrt2 - 2 =0
(sqrt2 = 1.414)
R? - 2.83R -2 =0
then solve this quadratic equation using quadratic formula.
R = 3.415cm?

and one more thing i cant understand the (x-1) and (x-2) which points are these. i think ill need i diagram to understand.

 

[Harrypotter_fan]

hey here is something else......
using simple geometry
triangle ABC is similar to triangle ADE
therefore AB/AD = BC/DE = AC/AE
considering ratios BC/DE = AC/AE
BC= 2 ; DE= r ; AC=sqrt 5 (by pythagoras theoram) ; AE=sqrt5+r
so by solving equations the value of r=18.95
p.s. sqrt5=2.236 ......(my log book says so.)

if there is some problem in the logic please correct me

p.s. posted by my sis.

 

[Adarsh]

Yes there is a problem with that. You guys are not getting it!!! Satish, here's my solution for you.My solution is right.

Shreyas - ACE is NOT a straight line. The diagram is misleading. Anyway here's my proof again, with a pic.

P.S: I'm 15 too....

 

[blackleopard92]

adarsh is right on this one, it's 5 cm only

 

[Kshitiz_Indian]

You're a genius adarsh. but one thing - the guys says the rectangle's dimensions are 1 X 2, not of the square it makes on the bigger square. And therefore we cant take x - 1 ?

Your thery is perfect but i think are we missing something ? because the figure says that 1 X 2 is the dimensions of the square being formed by the rectangle. But the guys says that 1 X 2 is the rectangle ? :S

 

[Adarsh]

Oh no. You've misread the question. The rectangle is completely inside the big square, and forms a rectangle inside. As I said, the diagram is a bit misleading.

 

[Kshitiz_Indian]

Ok then its the correct answer ! Well done