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There is a perfect sphere of diameter 40 cms. resting up
against a perfectly straight wall and a perfectly straight
floor i.e. the wall and the floor make a perfect right angle.

Can a perfect sphere of diameter 7 cms. pass through the
space between the big sphere, the wall and the floor?
Support your answer with valid arguments. Don't submit just
"Yes" or "No".

Answers were Sorted based on User's Feedback



There is a perfect sphere of diameter 40 cms. resting up against a perfectly straight wall and a pe..

Answer / munna

The answer is yes.
Consider a 2-D diagram,
given that the diameter of the sphere is 40 cms.
so radius = 20 cms.
c0nsider the points that touch the floor and the wall and
the center of the sphere and the intersection of the floor
and wall.This forms a square of length of side 20 cms.
so,
the distance left for any sphere of unknown diameter to
pass through be x.
x=sqrt(800)-20=8.284... cms.
so a sphere of radius 7 cms passes.

Is This Answer Correct ?    2 Yes 0 No

There is a perfect sphere of diameter 40 cms. resting up against a perfectly straight wall and a pe..

Answer / raj kiran

consider 2-d fig for simplicity.. i.e two lines at right
angles and a circle with its circumference touching the
lines(walls).. now the line joining the center of the circle
n the point of contact of the circle wit the walls forms a
square of side 40 cm..

now join the center of the circle n the meeting point of the
wall, this forms the diagonal of the square..
by right angle theorem, we get the diagonal as,
(40*1.414)=56.56..

so the remaining space left ( corner of the wall to the
circumference of th circle) is 16.6cm...

through which a 7cm ball can pass through..

Is This Answer Correct ?    1 Yes 0 No

There is a perfect sphere of diameter 40 cms. resting up against a perfectly straight wall and a pe..

Answer / guest

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For the sake of simplicity, consider two-dimension i.e view
sphere as a two dimensional circle with diameter 40 cms.

From Figure I, (40 cms diameter sphere)

OC2 = OD2 + CD2

OC2 = 202 + 202

OC = 28.28427 cms

Also, X is the closest point to origin O on the sphere.

CX = 20 cms (radius)

OX = OC - CX

OX = 28.28427 - 20

OX = 8.28427 cms

From Figure II, (7 cms diameter sphere)

OP2 = OQ2 + PQ2

OP2 = (3.5)2 + (3.5)2

OP = 4.94974 cms

Also, Y is the farthest point to origin O on the sphere.

PY = 3.5 cms (radius)

OY = OP + PY

OY = 4.94974 + 3.5

OY = 8.44974 cms

Now, as OY > OX i.e. smaller sphere requires more space than
the space available. Hence, smaller sphere of 7 cms diameter
can not pass through the space between the big sphere, the
wall and the floor.

The puzzle can be solved by another method.

Draw a line tangent to the big sphere at the point X such
that X is the closest point to the origin O on sphere. The
tanget will cut X and Y axes at A and B respectively such
that OA=OB. [See Fig III] From above, OX=8.28427 cms.

From the right angle triangle OAB, we can deduct that

OA = OB = 11.71572 cms

AB = 16.56854 cms

Now, the diameter of the inscribed circle of right angle
triangle is given by d = a + b - c where a <= b < c

The maximum possible diameter of the circle which can pass
through the space between the big sphere, the wall and the
floor is

= OA + OB - AB

= 11.71572 + 11.71572 - 16.56854

= 6.86291 cms

Hence, the sphere with 7 cms diameter can not pass through
the space between the big sphere, the wall and the floor.

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