BTGmoderatorLU wrote:Does the rectangle have an area less than 30?
(1) Perimeter = 20.
(2) Diagonal < 10.
The OA is A.
(1) 2L + 2W = 20 --> L + W = 10, max 5, 5; max area 25; 25 < 30. Sufficient.
(2) 1/2 rectangle = right triangle; a^2 + b^2 = c^2, c < 10 (6 - 8 - 10 right triangle); sides less than 6 & 8
given that, does the possibility of the L & W being any combo of #'s < 6&8 make it INS?
Hence, A is the correct answer.
Has anyone another strategic approach to solve this DS question? Regards!
We have to find out whether the rectangle has an area of less than 30.
Say the length and the breadth of the rectangle are a and b, respectively,
Thus we have to determine whether ab < 30.
Let's take each statement one by one.
(1) Perimeter = 20.
=> 2(a + b) = 20
=> a + b = 10
Note that for two numbers (here a and b), whose sum is constant (here 10), their product would be maximum if the numbers are equal.
Thus, the maximum value of a*b is when a = b = 10/2 = 5.
=> the maximum value of the area of the rectangle = a*b = 5*5 = 25 < 30. The answer is yes, the rectangle has an area of less than 30. Sufficient.
(2) Diagonal < 10.
=> a^2 + b^2 < 10^2
=> a^2 + b^2 < 100
Again, the maximum value of (a^2)*(b^2) would be attained when a^2 = b^2.
=> 2a^2 < 100 => a^2 < 50 => Maximum possible value of a = ~7.
Thus, the maximum value of (a^2)*(b^2) = (7^2)*(7^2) = 7^4
=> the maximum value of the area of the rectangle = a*b = sqrt[(a^2)*(b^2)] = sqrt[7^4] = 7^2 = 49 > 30. The answer is no, the rectangle does not have an area of less than 30.
Given a^2 + b^2 < 100, the minimum value of a*b can be too less than 30. Say a = b = 1, then 1^1 + 1^2 < 100. The area of the rectangle = a*b = 1*1 = 1 < 30. The answer is yes, the rectangle has an area of less than 30. Sufficient.
No unique answer. Insufficient.
The correct answer:
A
Hope this helps!
-Jay
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