Question

if the diagonal of a square is increased to √2 times , then its area is increased/decreased to how many times

Answer 1

Let's say we have a square of length x m.
Now,
Divide the square into two equal triangles. i.e. draw a line from a corner to its corresponding opposite corner.

Hence, this line divides the square into two equal parts.
Choose one part and apply the Pythagoras Theorem and find out the hypotenuse i.e. the diagonal.
Assume Hypotenuse = Diagonal = c
Let c = 1. (because it is easy to compare)
Assume Sides = a,b
Also a = b.

Therefore,
c² = a² + b²
⇒c² = a² + a² (As a = b, a² = b². Replacing a² with b² will not change the answer.)
⇒ 1 = 2a²
⇒ a² = 1/2
Therefore a = 1/√2 m
So, side = 1/√2 m
Then, Area = a² = (1/√2) * (1/√2)
Therefore Area = 1/2 m²


Now,
Final diagonal length = √2 * initial diagonal length
Therefore, final diagonal length = √2*1 = √2 m
Again, c² = a² + a²
c² = 2a²
⇒(√2)² = 2*a
⇒ 2 = 2a
Therefore a = 1.
Therefore side = 1 m
So, Area = 1*1 = 1 m²

Comparing,
 
Initial area = 1/2 m²
Final area = 1 m²

Final / Initial = Time increased
1/(1/2) = 2 times.

Therefore,
If the diagonal of a square is increased to √2 times , then its area is increased by 2 times.

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