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Re: The rectangular mirror has area greater than 10 square cm [#permalink]
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I haven't seen anyone post an algebraic approach yet. I'm posting the below to (1) check my work :) (2) provide an algebraic approach if my method is in fact correct.
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Re: The rectangular mirror has area greater than 10 square cm [#permalink]
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brooklyndude wrote:
I haven't seen anyone post an algebraic approach yet. I'm posting the below to (1) check my work :) (2) provide an algebraic approach if my method is in fact correct.


Clarifying - when I put 1 & 2 sufficient I meant both are required to be sufficient and the answer is C. :)
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The rectangular mirror has area greater than 10 square cm [#permalink]
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Interesting problem.
I had skipped directly to using algebra formula...

What I did.
D= Diagonal

1) Perimeter = 2(A+B) = 24, where A = Length and B = Width.
A+B = 12

2) (A+B)^2 = A^2 + 2AB + B^2
12^2 = (A^2 + B^2) + 2AB
144 = D^2 + 2xArea

Area = 1/2 x (144-D^2)
if D = 11
Area = 1/2 x (144-121)
= 23/2 = 11.5

Therefore, at max D of 11, Area is 11.5
Any lower D will give a larger Area.
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Re: The rectangular mirror has area greater than 10 square cm [#permalink]
A-->Perimeter is 24..therefore L+b= 12,
should be true if L= 10, b=2 as area is 20 cm.sq
should be false if L= 11.9 , b= 0.1 as area is 1.19 cm.sq

So A is insufficient

B--> Diagonal is less than 11 - meaning - \sqrt{(L^2 +b^2)} is less than 121.
Clearly insufficient for the same reason as A

Considering both A&B:
considering 11 as diagonal and L+b=12, L cannot be more than 11 because if its more than 11 diagonal cannot be less than 11.
Even if we consider 11 as L, b must atleast be 1 which means that the area cannot be lesser than 11*1=11.
So C is correct.
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Re: The rectangular mirror has area greater than 10 square cm [#permalink]
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Re: The rectangular mirror has area greater than 10 square cm [#permalink]
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