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Bunuel
For a rectangle whose diagonal is 10cm. Which of the following cannot be its area?

A. 8
B. 12
C. 20
D. 45
E. 54

Are You Up For the Challenge: 700 Level Questions

Since there are no restrictions on the sides on being integer, we can have all the areas possible except when they are not in range.

Minimum
We can have one side almost equal to the diagonal, that is say 9.999999, the. The other side will be 0.000002.
The area here can be just above 0.
So, all options are possible

Maximum
The area of a right angled triangle with given diagonal is maximum when it is an isosceles triangle.
Thus sides become \(5\sqrt{2}\) each.
Area of triangle = \(\frac{1}{2}*5\sqrt{2}*5\sqrt{2}=25\)
Area of rectangle = 2*25 = 50.
54 is not possible


E
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jackoftrades
\(a^2 + b^2 = c^2\)
\(a^2 + b^2 = 10^2\)
\(a^2 + b^2 = 100\)

\((a+b)^2 = a^2 + 2ab + b^2\)

\((a+b)^2 = 100 - 2ab\)

Plug in answers for ab (the area of rectangle).
We also know it has to be one of the extremes (8 or 54).

Let's test them:
\((a+b)^2 = 100- 2(8) = 84\) (this is a possibility)
\((a+b)^2 = 100 - 2(54) = -8 \) (this can't work)

Therefore: E

what is the reason for using (a+b)^2
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