The rectangular mirror has area greater than 10 square cm

Question

Does the rectangular mirror have an area greater than  10 cm^2  ?

1) The perimeter of the mirror is 24 cm.

2) The diagonal of the mirror is less than 11 cm.

mikemcgarry · Answer

Dear samichange , This is a great question! I'm happy to respond! :-)

Statement #1: Since a rectangle is a square, we could have a 6 x 6 square, that has an area of 36, more than 10. If we made this long and skinny, an 11 by 1 rectangle, that still has an area of 11, more than 10. But if we make it really skinny, 11.9 by 0.1, then it has an area of 1.19, much less than 10. This statement, alone and by itself, is not sufficient . Statement #2: The 6 x 6 square discussed in the first statement has a diagonal of 6sqrt(2) , which is approximately 6*(1.4) = 8.4, which is less than 11. Again, this square has an area of 36, more than 10. Alternately, we could have a really tiny rectangle, 1 x 2, than has a diagonal much less than 11 and an area much less than 10. We could go either way. This statement, alone and by itself, is not sufficient . Combined. Now, perimeter is fixed at 24, and the diagonal must be less than 11. Clearly, the square and rectangles in which length and width are close will have areas more than 10. As the rectangles of perimeter 24 get skinnier, the diagonals get longer and longer, and the areas go down. Let's go back to the 11 x 1 triangle, which has an area of 11, greater than 10. This clearly has a diagonal greater than 11, so it wouldn't could under the combined statements. All the rectangles that are "less skinny", with a length/width ratio < 11, will have areas more than this, more than 11. Some of those rectangles will have diagonals less than 11, and so will count under the combined statements. On the other hand, if we look at the rectangles that are "more skinny" than the 11 x 1 rectangle, with a length/width ratio > 11, then many of those will have areas less than 10, but all of those, like the 11 x 1 rectangle, will have diagonals larger than 11, so none of those would be included among the combined statements. The only rectangles that would count are some subset of the rectangles that are "less skinny" than the 11 x 1, and we definitively can say that all of these have an area greater than 10. Combined, the statements are sufficient. Does all this make sense? Mike :-)

mbadude2017 · Answer

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.

KarishmaB · Answer

Here is how I thought about it:

Question: Does the rectangle have an area >  10 cm^2  ?

Statement 1: Perimeter 24 means the sum of the length of two sides is 12.
So maximum area will be obtained when the length of the two sides are equal i.e. 6 each. Area = 36
Minimum area will be obtained when the length of the two sides are as far apart as possible i.e. almost 12 and slightly more than 0. The area is slightly more than 0 in this case. 
Hence area will range from slightly more than 0 to 36. Is it more than 10, we cannot say. Not sufficient.

Statement 2: Diagonal could be 1 in which case each side will be less than 1 and area will be less than 1 too. 
The diagonal could be almost 11. The sides could be 1 and 'slightly less than diagonal'. So area in this case would be almost 11. 
Not sufficient.

Using both, the sum of sides is 12 and diagonal is 11 so each side is less than 11. To get the minimum area, we need the sides to be as apart as possible. So one side would be slightly less than 11 and the other slightly more than 1. In this case, the area would be about 11. This is the minimum area. Hence area will be more than 10. 
Sufficient.

Answer (C)

mbadude2017 · Answer

Clarifying - when I put 1 & 2 sufficient I meant both are required to be sufficient and the answer is C.

locklock · Answer

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.

Sanjay7392 · Answer

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.

bumpbot · Answer

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The rectangular mirror has area greater than 10 square cm

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