Question
A square has two of its vertices on a circle and the other two are on the tangent to the circle. Find the area of the square if the radius of the circle is 5 units. 
Answer: Option A
:
A
Area of the square: (x + 5)2
So the answer has to be a perfect square: 36 or 64
If area is 36 then x = 1; so EF = 6, OE = 1, OA = 5 and EA = 4
Angle OFB = ABF = BAE = 90, so angle OEA = 90
OE, EA and OA do not form Pythagoras triplet.
If area is 64 then x = 3, EF = 8, EA = 4, OA = 5
Shortcut 1
Reverse Gear approach-With a radius of 5, we will get a square of side 5%undefined2. area will be 50, but we know the side has to be slightly greater than 5%undefined2. hence, choose an option which is slightly greater than 50; i.e. 64
Shortcut 2
Graphical division:- The square gets divided into 8 parts. 4 parts are each right angled triangles of sides 3,4,5 and hence with area 6. 4 of them in total have an area of 24. The other 4 are triangles of the form 4, 5, √41 and whose areas are 10 each. Total area = (6+10)*4 = 64.
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:
A
Area of the square: (x + 5)2
So the answer has to be a perfect square: 36 or 64
If area is 36 then x = 1; so EF = 6, OE = 1, OA = 5 and EA = 4
Angle OFB = ABF = BAE = 90, so angle OEA = 90
OE, EA and OA do not form Pythagoras triplet.
If area is 64 then x = 3, EF = 8, EA = 4, OA = 5
Shortcut 1
Reverse Gear approach-With a radius of 5, we will get a square of side 5%undefined2. area will be 50, but we know the side has to be slightly greater than 5%undefined2. hence, choose an option which is slightly greater than 50; i.e. 64
Shortcut 2
Graphical division:- The square gets divided into 8 parts. 4 parts are each right angled triangles of sides 3,4,5 and hence with area 6. 4 of them in total have an area of 24. The other 4 are triangles of the form 4, 5, √41 and whose areas are 10 each. Total area = (6+10)*4 = 64.
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