The breadth $ b $ of a room is twice its height and half of its length. Find the length of the longest diagonal of the room.

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**Solution:**

1. Let the length, breadth, and height of the room be $ l, b, h $ respectively.

2. Given:
   - Breadth is twice the height: $ b = 2h $  $ \Rightarrow h = \frac{b}{2} $
   - Breadth is half the length: $ b = \frac{l}{2} $  $ \Rightarrow l = 2b $

3. The longest diagonal $ d $ of a cuboid is given by:
   $$ d = \sqrt{l^2 + b^2 + h^2} $$

4. Substitute $ l = 2b $ and $ h = \frac{b}{2} $:
   $$
   d = \sqrt{(2b)^2 + b^2 + \left(\frac{b}{2}\right)^2} = \sqrt{4b^2 + b^2 + \frac{b^2}{4}}
   $$

5. Simplify inside the square root:
   $$
   d = \sqrt{5b^2 + \frac{b^2}{4}} = \sqrt{\frac{20b^2}{4} + \frac{b^2}{4}} = \sqrt{\frac{21b^2}{4}} = \frac{b\sqrt{21}}{2}
   $$

**Answer:** The length of the longest diagonal is $ \frac{b\sqrt{21}}{2} $.

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