SOLUTION: I have a perfect circle inscribed inside of a perfect square where the sides of the square are tangent to the circle. In one corner of the square is a rectangle that ia 140mm x 70m

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Question 49774: I have a perfect circle inscribed inside of a perfect square where the sides of the square are tangent to the circle. In one corner of the square is a rectangle that ia 140mm x 70mm. The rectangle is perfectly squared in the corner and the bottom corner just touches the outside of the circle.
What is the diameter of the circle?
Answer by venugopalramana(3286) About Me  (Show Source):
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I have a perfect circle inscribed inside of a perfect square where the sides of the square are tangent to the circle. In one corner of the square is a rectangle that ia 140mm x 70mm. The rectangle is perfectly squared in the corner and the bottom corner just touches the outside of the circle.
What is the diameter of the circle?
GOOD PROBLEM.TRY TO TO DRAW A FIGURE AS DESCRIBED BELOW FOR BETTER UNDERSTANDING


LET THE CENTRE OF CIRCULAR PIZZA BE 'C'.THIS IS SAME AS THE CENTRE OF THE BOX.LET THE CARD BE PQRS..LET IT TOUCH THE CIRCULAR PIZZA AT P.JOIN OP
OP IS THE RADIUS=R SAY TO BE FOUND.WE HAVE..
PQ=RS=140....
PS=QR=70...GIVEN
DRAW CT FROM CENTRE C A RADIUS PERPENDICULAR TO RS.
EXTEND QP TO MEET CT AT X.
WE HAVE CT=RADIUS=R=CP
PSTX IS A RECTANGLE.
PS=70=XT
PX=QPX-QP=R-140
CX=CXT-XT=R-70.
NOW IN THE RIGHT ANGLED TRIANGLE CPX,WE HAVE
CP=R=HYPOTENUSE
CX=ONE SIDE=R-70
PX=ANOTHER SIDE=R-140
USING PYTHOGARUS THEOREM ,WE HAVE....HYPOTENUSE^2=SIDE^2+SIDE^2
R^2=(R-70)^2+(R-140)^2
R^2=R^2-140R+4900+R^2-280R+19600
R^2-420R+24500=0
R^2-70R-350R+24500=0
R(R-70)-350(R-70)=0
(R-70)(R-350)=0
R-70=0.....OR....R=70....NOT POSSIBLE AS THEN IT TOUCHES THE LOWER EDGE OVERLAPPING THE PIZZA.
R-350=0......R=350 IS THE ANSWER.