Problem Solving

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Could you please help me in solving this?

thanks
Andrew

Is the answer 4-(2*sqrt(2)). ?

I will explain if i am correct.

yes it should be,
please explain in details....

cheers
Andrew

I will try my best to explain.

Name the square as ABCD where AC represents the diagonal of the square.

Name the centers of the circles as O1 (of the circle near to vertex A) and O2 (of the circle near to vertex C)

So the diagonal line is AO1O2C ( the length = 4sqrt(2) ). Let the radius of the circles be r.

To determine r, we should know the distance between A and O1 and the distance between O2 and C. The distance between O1 and O2 is 2r.

To determine distance between A and O1
------------------------------------------------

Draw a line between A and O1 and a line between O1 and the tangential point P in AB. Drawing a line between O1 and the tangential point P in AB creates a right angle ( If i remember correctly ). The distance of the line O1P is again r.

Apply hypotenuse theorem here. AO1^2 = r^2 + r^2 = 2*r^2
=> AO1 = r*sqrt(2)

Similarly O2C = r*sqrt(2)

AC = AO1 + O1O2 + O2C
= r*sqrt(2) + 2r + r*sqrt(2)
4sqrt(2) = 2r*sqrt(2) + 2r
2sqrt(2) = r(1+ sqrt(2))

Solving this, we get r = 4-(2*sqrt(2))

HTH.

My attempt!!

Square side = 4
diagonal = 4sqrt(2)

this squares diagonal is sum to two diagonals of circle
so 2d = 4sqrt(2)
d = 2sqrt(2)

and radius = sqrt(2)

stop@800 wrote:My attempt!!

Square side = 4
diagonal = 4sqrt(2)

this squares diagonal is sum to two diagonals of circle
so 2d = 4sqrt(2)
d = 2sqrt(2)

and radius = sqrt(2)

Good approach to be able to select the ans from amongst the choices, but this is not exactly the radius, and the actual radius would be slightly smaller than this, cos the diagonal of the square is not exactly equal to 4r, but extends a bit on both sides beyond the circles.

However, given that thr would be 5 choiced to select from, this should be pretty good to estimate and guess the right answer!

yes

please post the answer choices?

Mani_mba wrote: 2sqrt(2) = r(1+ sqrt(2))

Solving this, we get r = 4-(2*sqrt(2))

I am stuck with this thing

Anybody. please elaborate

anybody can help?

2sqrt(2) = r(1+sqrt(2))
=> r = 2*sqrt(2) / (1+sqrt(2))

Multiply (1-sqrt(2)) on both Numerator and Denominator

=> r = 2sqrt(2) * (1-sqrt(2)) / (1+sqrt(2)) (1-sqrt(2))
= (2sqrt(2) - 4) / (-1)
= 4-2sqrt(2)

HTH.

Mani_mba wrote:I will try my best to explain.

Name the square as ABCD where AC represents the diagonal of the square.

Name the centers of the circles as O1 (of the circle near to vertex A) and O2 (of the circle near to vertex C)

So the diagonal line is AO1O2C ( the length = 4sqrt(2) ). Let the radius of the circles be r.

To determine r, we should know the distance between A and O1 and the distance between O2 and C. The distance between O1 and O2 is 2r.

To determine distance between A and O1
------------------------------------------------

Draw a line between A and O1 and a line between O1 and the tangential point P in AB. Drawing a line between O1 and the tangential point P in AB creates a right angle ( If i remember correctly ). The distance of the line O1P is again r.

Apply hypotenuse theorem here. AO1^2 = r^2 + r^2 = 2*r^2
=> AO1 = r*sqrt(2)

Similarly O2C = r*sqrt(2)

AC = AO1 + O1O2 + O2C
= r*sqrt(2) + 2r + r*sqrt(2)
4sqrt(2) = 2r*sqrt(2) + 2r
2sqrt(2) = r(1+ sqrt(2))

Solving this, we get r = 4-(2*sqrt(2))

HTH.

Hi Mani_mba,
I dint quite understand how the length of AP = r ??
Can u pls elaborate??

thanks
-V

I just knew AP should be r by intuition. I will try to explain mathematically:

Let us name the tangential point in the side AD as P1.

As i said already, Angle O1PA = 90 and Angle A = 90. Hence the lines DA and O1P must be parallel.

Applying same for the tangential point P1, we get
Angle O1P1A=90 and it is known that Angle A = 90. Hence the lines O1P1 and PA must be parallel.

=> APO1P1 looks like a square and all the sides measure the same. Hence AP = PO1 = O1P1 = P1A = r.

HTH.