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Question

All the 3 sides of a right triangle are integers and one side has a length 11 units. Area of the triangle in square units lies between

A
1 and 100
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B
100 and 200
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C
200 and 300
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D
More than 300
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Solution

The correct option is B More than 300
Case 1 (When 11 is the biggest side)
a2+b2=121
Since a & b are integers
Possible values of a2 & b2 are:1,4,9,16,25,36,49,64,81,100
but none of them sum up to be 121
Hence the case is VOID
Case 2 (Let us assume a is hypotenuse)
b2+112=a2
a2b2=121
(ab)(a+b)=121
Now (ab) and (a+b) are integers as well
Only 2 combinations are possible
11×11 & 121×1
(11×11) is not possible
Hence
a+b=121
a=(121b)
a2b2=121
1212+b2242bb2=121
b=1212121242=60
a=61
Area=60×612>300

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