chetan2u
nahid78
If the side lengths of rectangle ABCO are integers, What is the value of a+b?
Statement I tells us that Area is 35.
Since the sides are integers, the sides could be 7*5 or 1*35.
Possible answers: 1+35 or 5+7
Insufficient
Statement II tells us that a and b are integers.
OA is dependent on (a,b) but nothing on values of sides.
Insufficient
Combined
What it tells us combined is that if we drop a perpendicular on x axis from A, say at D, then AOD is a right angled triangle with length of all sides integers.
Now,
Area of rectangle is 35, which can be 7*5, so OA can be 5 and triangle AOD can be 3:4:5
Or
Area can be 1*35, so OA can be 35 and triangle AOD can be 3*7:4*7:5*7, that is 21:28:35
Both 1+35 or 7+5 possible.
Insufficient
E
Please guide where I am wrong as I believe when taken together (a) + (b) should be sufficient:
On the basis of above discussion, we know that the length of OA has to be either 1, 35, 5, or 7
We do know that "a" and "b" both have to be integers and positive integers based on the picture given.
so Length of OA= a^2 + b^2 where "a" and "b" both have to be positive integers.
from 1 to 35 there are only 5 perfect squares that are 1, 4, 9, 16, and 25
Length of OA has to be sum of 2 perfect squares from these, therefore
Length of OA cannot be equal to 1 (1+ anything will greater than 1)
Length of OA cannot be equal to 35 (the closest we can get is 9+25 which is 34)
Length of OA cannot be equal to 7 (none of the pairs of perfect squares satisfy the equation)
The only possible value for Length of OA is 5 as 2^2+1^2 is equal to 5.
So there is only two possible situations:
a=2, b=1 i.e. a+b =3
or
a=1, b=2 i.e. a+b= 3
i.e. a +b =3 only in any case.
So both conditions combined should be sufficient.