The diagonal of a rectangular solid is 5√2 dm. If its length and breadth are 5 dm and 4 dm, then find the height of the solid.
Answers
Step-by-step explanation:
Given :-
The diagonal of a rectangular solid is 5√2 dm
It's length and breadth are 5 dm and 4 dm
To find :-
Height of the solid
Solution :-
Given that
Length of the rectangular solid
(l) = 5 dm
Breadth of the rectangular solid
(b) = 4 dm
Let the height of the rectangular solid be h dm
We know that
Diagonal of a rectangular solid (cuboid) is d = √(l²+b²+h²) units
The diagonal of the rectangular solid
=> d = √(5²+4²+h²) dm
=> d = √(25+16+h²) dm
=> d = √(41+h²) dm
According to the given problem
The diagonal of the rectangular solid
= 5√2 dm
=> √(41+h²) = 5√2
On squaring both sides then
=> [√(41+h²) ]² = (5√2)²
=> 41+h² = 25×2
=> 41+h² = 50
=> h² = 50-41
=> h² = 9
=> h = ±√9
=> h = ±3
Therefore, h= 3 dm
Since, height can't be negative.
Answer:-
The height of the rectangular solid is
3 dm
Used formulae:-
→ Diagonal of a rectangular solid (cuboid) is d = √(l²+b²+h²) units
- l = length
- b = breadth
- h = height
Refer the given attachment
