In the diagram above, the four triangles ABE, CBE, ADE, and CDE are al

Each of the 4 equal triangles is such that the bigger radius forms one the sides, the smaller radius forms the other side and the length of hypotenuse is given (CD = 5).

Statement 1

AE =3, AB = CD = 5, Therefore BE is 4 (3, 4, 5 is a Pythagorean pair. Triangle ABE is a right angled \(/triangle\))

The bigger radius is known. Area of bigger circle is known.
Smaller radius is known. Area of smaller circle is known.
We can calculate the area of the space between two circles

Sufficient.

Statement 2
This information is already known and does not give any additional information. Further, we cannot calculate the radius of the two circles.

Hence not sufficient

A is the answer.

Hi robu ,
Triangles are all equal means they are congruent .
If two or more triangles are congruent, then all of their corresponding angles and sides are congruent as well.
Here Since AE= 3 , EC= 3
Also, AB = BC = CD = AD = 5
and \(\angle\) BEA = \(\angle\) BEC = \(\angle\) CED = \(\angle\) DEA = 90

Hope this helps!!

If the angle is 90 degree in statement 2. Isnt it obvious that the other two sides would be 3 & 4 as the 3,4,5 are pythagorean triplets. Bunuel

Two points:

1. It can be inferred that angle BEC equals 90 degrees from the stem of the problem itself. In triangles ABE, CBE, ADE, and CDE, all angles are equal, and since the sum of the angles' measures at vertex E is 360 degrees, each angle at E must be 90 degrees. Therefore, when you assert that statement (2) is sufficient, you are actually suggesting that the problem's stem itself provides enough information to answer the question, implying that the question is flawed.


2. Just because CD, the hypotenuse in triangle CDE, has a length of 5, it doesn't necessarily mean that the sides of CDE form a Pythagorean triplet. We are not given that the side lengths must be integers. In other words, EC^2 + ED^2 = 5^2 has infinitely many solutions for EC and ED, with only one of them being EC = 3 and ED = 4. For instance, consider EC = 1 and ED = √24; EC = 2 and ED = √21; EC = 1/2 and ED = √24.75, and so on. If we were given that the lengths of EC and ED are integers, then a hypotenuse of 5 would imply that EC and ED are 3 and 4. Otherwise, knowing the length of the hypotenuse alone is insufficient to determine the other two sides.

Hope it's clear.