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Question
For the following statement, state whether true (T) or false (F):
The sum of the squares on the sides of a rhombus is equal to the sum of the squares on its diagonals.
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Solution
This statement is true.
Explanation:
Suppose ABCD is a rhombus having AC and BD its diagonals.
Since, the diagonals of a rhombus perpendicular bisect each other.
Hence, AOC is a right angle triangle
In right triangle AOC
By using Pythagoras theorem, we have
`AB^2=((AC)/2)^2+((BD)/2)^2`
[∴ Diagonals of a rhombus perpendicularly bisect each other]
⇒ `AB^2=((AC)/2)^2+((BD)/2)^2`
⇒ `4AB^2=AC^2+BD^2`
⇒ `AB^2+AB^2+AB^2+AB^2=AC^2+BD^2`
⇒ `AB^2+BC^2+CD^2+DA^2=AC^2+BD^2` [โต๐ด๐๐ ๐ ๐๐๐๐ ๐๐ ๐ ๐โ๐๐๐๐ข๐ ๐๐๐ ๐๐๐ข๐๐]
