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555-605 (Medium)|   Geometry|      
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Bunuel
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A room is 15 meters long, 12 meters wide & 10 meters in height. The lo 12 Sep 2019, 21:30
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C
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71% (01:44) correct 29% (01:33) wrong based on 62 sessions

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A room is 15 meters long, 12 meters wide & 10 meters in height. The longest possible rod which can be placed in the room is

A. 19.13 meters

B. 20.22 meters

C. 21.65 meters

D. 22.34 meters

E. 23.32 meters
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C
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A room is 15 meters long, 12 meters wide & 10 meters in height. The lo 12 Sep 2019, 21:38
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A room is 15 meters long, 12 meters wide & 10 meters in height. The longest possible rod which can be placed in the room is

A. 19.13 meters

B. 20.22 meters

C. 21.65 meters

D. 22.34 meters

E. 23.32 meters

The longest possible rod which can be placed in the room is = sq. root of (15^2+12^2+10^2)
= 21.65 meters C

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A room is 15 meters long, 12 meters wide & 10 meters in height. The lo 12 Sep 2019, 22:02
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The longest rod that can placed in a room which is 15m long, 12m wide, and 10m tall can be gotten by first computing the diagonal of the length and width of the room and and computing the diagonal the computed diagonal forms with the height of the building.

So (12*12+15*15)^0.5 = 369^0.5
Longest possible rod = (369+10*10)^0.5 = (469)^0.5 = 21.65.

The answer is therefore C in my view.
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A room is 15 meters long, 12 meters wide & 10 meters in height. The lo 12 Sep 2019, 22:09
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the longest possible rod which can be placed in the room in a cuboid shape... should be placed at the diagonal
Diagonal length in cuboid = sqrt(15^2+12^2+10^2)= sqrt(469) = 21.65
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A room is 15 meters long, 12 meters wide & 10 meters in height. The lo 12 Sep 2019, 23:12
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Given: length l = 15m, breadth b=12m and height h=10m

Longest rod that can be placed in a room is by placing it diagonally.

Length of diagonal of a cuboid(Length of longest rod)= √(l^2 + b^2 + h^2)

= √(225 + 144 + 100) m

= √(469)m

20^2=400 so √(469) would be a bit greater than 20.

Thus the length of the longest rod would be 21.65 meters (OPTION C)

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A room is 15 meters long, 12 meters wide & 10 meters in height. The lo Updated on: 13 Sep 2019, 21:55
Originally posted by sampriya on 12 Sep 2019, 23:13.
Last edited by sampriya on 13 Sep 2019, 21:55, edited 2 times in total.
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we know that hypotenuse is the longest side in a triangle therefore the hypotenuse of the triangle with the larger sides will be largest

therefore the answer is L= under-root(15^2+12^2+10^2) ---space diagonal formula
=21.6

therefore C is correct
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A room is 15 meters long, 12 meters wide & 10 meters in height. The lo 13 Sep 2019, 01:38
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longest possible rod ; √15^2+12^2+10^2 ; √469 ; 21.65 mtrs
IMO C


A room is 15 meters long, 12 meters wide & 10 meters in height. The longest possible rod which can be placed in the room is

A. 19.13 meters

B. 20.22 meters

C. 21.65 meters

D. 22.34 meters

E. 23.32 meters
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A room is 15 meters long, 12 meters wide & 10 meters in height. The lo 13 Sep 2019, 03:40
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Quote:
A room is 15 meters long, 12 meters wide & 10 meters in height. The longest possible rod which can be placed in the room is

A. 19.13 meters
B. 20.22 meters
C. 21.65 meters
D. 22.34 meters
E. 23.32 meters
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longest possible rod placed in a cuboid is a diagonal through the middle:
\(d^2=l^2+w^2+h^2…d^2=15^2+12^2+10^2…d^2=225+144+100…d^2=469\)
\(d^2=469…(400=20^2,…441=21^2,…484=22^2)…441<d^2=469<484…21<d<22\)

Answer (C)
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A room is 15 meters long, 12 meters wide & 10 meters in height. The lo 13 Sep 2019, 04:48
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IMO C is the answer.
It would be sqrt(15^2 + 12^2 + 10^2)= sqrt(225+144+100)= sqrt(469)
Then POE will come, as the calculation is cumbersome.
21^2=441
22^2=484
Hence C is the answer.
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