Question

find the lenght of the longest pole that can be placed in a room 12 m long , 8 m broad and 9 m high .​

Answer 1

Answer:

To put longest pole in a room , it should be placed diagonally .

It should be placed along the diagonal (d) of room as seen in the picture below.

Let a= 12 m , b= 8 m , c= 9 m.

As we know , for longest pole it should be placed diagonally along diagonal (d) so, its length should equals to the length of diagonal of the room.

Length of pole = Length of diagonal (d)

Now,

By applying Pythagoras theorem , in the right triangle given in figure whose base is d1 and hypotenuse is d and height is same as height of room which is given to us as c = 9 m.

d(square) = 9(square) + d1(square). [equation 1]

Now first we need to find d1 ,

Again apply Pythagoras theorem , on the right triangle whose base is a = 12 m ,hypotenuse is d1 and height is b = 8 m.

d1(square) = b(square) + a(square)

d1(square) = 8(square) + 12(square)

d1(square) = 64 + 144 + 208

Hence , d1 = sq. root(208) = 14.422 m

Now putting this value of d1 in [equation 1],

d(square) = 9(square) + 14.422(square)

d(square) = 81 + 208 = 289

Now, d = sq. root(289) = 17 m

Hence , the longest pole which can be placed in the room is of 17 m in length.

Hope It Helps

Thanks .

Answer 2

Answer:

Given:

Length of the room = l= 12 m

Breadth of the room = b= 8 m

Height of the room = h= 9 m

To find:

The length of the longest pole to be placed in the given room

Proof:

Length of the longest pole that can be placed in the given room = Length of the diagonal of the cuboidal room

Diagonal = $\sqrt{ (l)^2 + (h)^2 + (b)^2}$

$= \sqrt{12^2 + 9^2 + 8^2 } \\= \sqrt{144 + 81 + 64 } \\= \sqrt{289} \\= 17 m$

Hence, the longest pole that can be placed in a cuboidal room of the given dimensions is of length = 17 m