Xavier Aptitude Test 2012 Paper

Question : 81 of 85

Marks: +1, -0

Carpenter Rajesh has a circular piece of plywood of diameter 30 feet. He has cut out two disks of diameter 20 feet and 10 feet. What is the diameter of the largest disk that can be cut out from the remaining portion of the plywood piece?

> 8.00 feet and ≤ 8.20 feet
> 8.21 feet and ≤ 8.40 feet
> 8.41 feet and ≤ 8.60 feet
> 8.61 feet and ≤ 8.80 feet
> 8.81 feet and ≤ 9.00 feet

Validate

Solution:

\ {C}_{1}, \ {C}_{2}, \ {C}_{3}, \ {C}_{4}

be the circle as shown having centre O ,

\ {A}, \ {B}

and

{C}

respectively.

\ {C}_{2}

and

\ {C}_{3}

meet at

\ {D}

.
Let

r

be the radius of circle

\ {C}_{4}

having centre

{C} . \ {O}, {A}, \ {B} \ {be}

the centre of circle

\ {C}_{1}, \ {C}_{2}

and

\ {C}_{3}

respectively.
Now given radius of

\ {C}_{2}

and

\ {C}_{3}

are 10

{cm}

and 5

\ {cm}

respectively and radius of

\ {C}_{1}

is 15

\ {cm} .

∴ \ {AC}=10+r, \ {BC}=5+r, \ {OC}=15-r

Let

{DC}=x

\∴ \ {AO}=\ {OD}

so we can use Apollonius theorem in

\Δ{ADC}

\ {AC}^{2}+\ {DC}^{2}=2 (\ {OC}^{2}+ {OD}^{2}\ )

(10+r)^{2}+x^{2}=2 [5^{2}+(15-r)^{2}\ ]

Similarly, for

\Δ {BCO}

\ {BD}=\ {OD}

\ {OC}^{2}+\ {BC}^{2}=2 [\ {CD}^{2}+ {BD}^{2}\ ]

(15-r)^{2}+(5+r)^{2}=2\ [x^{2}+5^{2}\ ]

Eliminating

x

and solving for

r

r= {60}/{14}= {30}/{7}

2 r=

Diameter

=\ {60}/{7}=8.57

Hence

(\ {c})

is the correct option

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