Help needed with geometry based math problem.

[AbBh]

yeah Adarsh your answer seems right to me although i think there is nothing wrong with my answer.

 

[Adarsh]

I think your mistake would have been to assume that the points from the corner of the top left corner of the big triangle, the bottom right corner of the small triangle and the centre of the circle lie in a straight line, which they do not.

 

[AbBh]

i took only one triangle with two sides being the radius and the hypotenuse being the radius+the diagonal of the square having the same area as the rectangle. It seems right to me although i dont know as i was half asleep at that time. I assumed that the radius and the diagonal lie in the same line.

 

[Adarsh]

See that's wrong. The diagonal of the small rectangle is not in line with the diagonal of the big square.

 

[Satish555]

i think i understand but but i dont know how to solve this : (x-1)^2 + (x-2)^2= x^2. can anyone help?

 

[Satish555]

got it. (x-5)(x-1). but then how does this say that x=5? its just factorized as (x-5).

 

[surendar]

i think i understand but but i dont know how to solve this : (x-1)^2 + (x-2)^2= x^2. can anyone help?

Its a quadriatic equation problem mate.

(x-1)^2 + (x-2)^2= x^2.

expanding this,

x^2 - 2x + 1 + x^2 - 4x + 4 = x^2 ( Using (a-b)^2 = a^2 - 2ab + b^2 )

On simplifying,

x^2 - 6x + 5 = 0
i.e, (x-5)(x-1) = 0

Which implies, X = 5,1.

Hope it helps you

 

[AbBh]

See that's wrong. The diagonal of the small rectangle is not in line with the diagonal of the big square.

It's not a rectangle but a square which occupies the same area as the rectangle.So the radius of the circle should be in line with it but if the small square touches the circle.

 

[Adarsh]

You have to expand the brackets. I'll show you:
(x-1)^2 + (x-2)^2= x^2
= x^2 - 2x + 1 + x^2 - 4y + 4 = x^2
= 2x^2 -6y +5 = x^2
= x^2 - 6y + 5 = 0
= (x-5)(x-1) = 0 [factorisation]
Therefore, x= 5 , x = 1.
Since the side (x-1) can't be zero, the value for x has to be 5 cm.

Hope this helps.

It's not a rectangle but a square which occupies the same area as the rectangle.So the radius of the circle should be in line with it but if the small square touches the circle.

Well, I don't know what you've done then. It's a bit confusing to follow through, there must be an error somewhere.

 

[Satish555]

ok i got it now. thnx a lot guys.

 

[Satish555]

but can anyone explain to me the logic behing this: " Since the side (x-1) can't be zero, the value for x has to be 5 cm."

 

[AbBh]

but can anyone explain to me the logic behing this: " Since the side (x-1) can't be zero, the value for x has to be 5 cm."

This equation has two roots 1 and 5.
Now as the radius of the circle cant be 1cm it has to be 5 cm.

 

[surendar]

This equation has two roots 1 and 5.
Now as the radius of the circle cant be 1cm it has to be 5 cm.

Who said its a circle :what

A circle has a equation of the form, x^2+ y^2 = a^2 where a is the radius with coordinates ( x,y ).

The given equation is just a plain quadriatic equation i.e degree 2.

 

[Kshitiz_Indian]

No he means that after solving

(X -1)^2 + (X - 2)^2 = X^2

we get

x - 5 and x - 1 as two factors. you multiply them again and you'll get the same expression again. we can either do it with splitting or by the alpha beta formula. You'll get the two roots, or the two possible values of x as 5 and 1 . Since radius of circle cant be 1 it has to be 5 So question solved

 

[andrew_nixon]

Dont again tell that its a circle becos its definitely not!

Did you even read the original question?