The length (in metres) of the longest rod that can he put in a room of dimensions 10 m x 10 m x 5m is
The longest length in the room will be along to its diagonal
hence let's say diagonal of room = a
diagonal of base = b
Height = 5 (Given)
now diagonal of base will be $$\sqrt{100+100}$$ = $$10\sqrt{2}$$
so b = $$10\sqrt{2}$$
now as we know $$a^{2}$$ = $$b^{2}$$ + 25
hence a = 15
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