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A rectangular box P is inscribed in a sphere of radius r. The surface

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A rectangular box P is inscribed in a sphere of radius r. The surface  [#permalink]

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New post 24 Mar 2019, 21:31
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A
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E

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50% (02:56) correct 50% (04:03) wrong based on 4 sessions

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Re: A rectangular box P is inscribed in a sphere of radius r. The surface  [#permalink]

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New post 24 Mar 2019, 22:15
1
Let l be length, b be breadth and h be height of the box
Surface area is given by = 2(lb+bh+hl) = 384

Also, ata
4(l+b+h)=112
L+b+h= 28
Radius of the sphere={rt(l^2+b^2+h^2)}/2

l^2+b^2+h^2= (l+b+h)^2-2(lb+bh+hl)
= (28)^2-384=784-384=400

So radius =20/2=10

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Re: A rectangular box P is inscribed in a sphere of radius r. The surface  [#permalink]

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New post 24 Mar 2019, 23:07
Bunuel wrote:
A rectangular box P is inscribed in a sphere of radius r. The surface area of P is 384, and the sum of the lengths of its 12 edges is 112. What is r?

(A) 8
(B) 10
(C) 12
(D) 14
(E) 16


SA of box ; 2*(lb+bh+lh)= 384
sum of lengths = 4l+4w+4h=112 ; l+w+h = 112/4 = 28

digonal of the box = diameter of the sphere = √l^2 + w^2 +h^2

(l + w + h)^2 - (2lw + 2lh + 2wh) = l^2 + w^2 + h^2 = 784 - 384 = 400
so
√400= 20 ; radius = 10
IMO B
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Re: A rectangular box P is inscribed in a sphere of radius r. The surface   [#permalink] 24 Mar 2019, 23:07
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A rectangular box P is inscribed in a sphere of radius r. The surface

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