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shashanksagar
IMO A..
a = (l+b)/2
Squaring both sides..
a^2 = ( l^2 + b^2 )/2 + lb
So area of square = some positive number + area of rectangle

So area of square > area of rectangle



Sent from my SM-N910H using Tapatalk

Hi Shashank!

Thanks for the answer. But I have a question.

Why do you say - a^2 = ( l^2 + b^2 )/2 + lb?

Isn't this how it should be - a^2 = ( l^2 + b^2 )/4 + (lb/2)? If this is the case then we cannot say - "Area of square = Some positive number + Area of Rectangle". It would rather be - "Area of square = Some positive number + Area of Rectangle/2". Which is not sufficient.
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Bunuel
Is the area of the rectangle more than the area of the square?

Say the length of a side of the square is x and the lengths of the length and breadth of the rectangle are a and b. So, the question is whether ab > x^2.

(1) The side of the square is half the sum of the length and the breadth of the rectangle --> x = (a+b)/2. The question becomes whether ab > ((a+b)/2)^2 --> is 4ab > a^2 + 2ab + b^2 --> is 0 > (a - b)^2. Since the square of a number cannot be negative, then the answer to this question is NO. Sufficient.

(2) The length and breadth of the rectangle are 15 cm and 10 cm respectively. We don't know x. Not sufficient.

Answer: A.

Hope it's clear.

Perfect Bunuel!!

Though I got a bit lost at - is 0 > (a - b)^2. I understand it now. You just took all the terms to RHS and made the LHS = 0. This is something worth taking note of!

Thank you.
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susheelh
shashanksagar
IMO A..
a = (l+b)/2
Squaring both sides..
a^2 = ( l^2 + b^2 )/2 + lb
So area of square = some positive number + area of rectangle

So area of square > area of rectangle



Sent from my SM-N910H using Tapatalk

Hi Shashank!

Thanks for the answer. But I have a question.

Why do you say - a^2 = ( l^2 + b^2 )/2 + lb?

Isn't this how it should be - a^2 = ( l^2 + b^2 )/4 + (lb/2)? If this is the case then we cannot say - "Area of square = Some positive number + Area of Rectangle". It would rather be - "Area of square = Some positive number + Area of Rectangle/2". Which is not sufficient.
Hi Susheelh,

You are absolutely correct. . Sorry I did the calculations in my head and missed the /4 .. thanks for correcting me.

Sent from my SM-N910H using Tapatalk
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susheelh
Is the area of the rectangle more than the area of the square?

(1) The side of the square is half the sum of the length and the breadth of the rectangle.
(2) The length and breadth of the rectangle are 15 cm and 10 cm respectively.
(1) consider two cases

(a)Rectangle L=0.5; B=0.1
Then Side of sq.= 0.3
(b)Rectangle L=100; B=50
Then Side of sq.= 75

Both the cases area sq.> area rectangle.
suff...

(2) no info on side of square
Insuff..

Ans A
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