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 $100$
• B. $100$ and $200$
• C. $200$ and $300$
• D. More than $300$

Answer 1

Answer: (D)

More than $300$

Case $1$ (When $11$ is the biggest side)

$a^2+b^2=121$

Since a & b are integers

Possible values of $a^2$ & $b^2$ 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)

$b^2+{11}^2=a^2$

$a^2-b^2=121$

$(a-b)(a+b)=121$

Now $(a-b)$ and $(a+b)$ are integers as well

Only $2$ combinations are possible

$11\times 11$ & $121\times 1$

$(11\times 11)$ is not possible

Hence

$a+b=121$

$a=(121-b)$

$a^2-b^2=121$

${121}^2+b^2-242b-b^2=121$

$b=\frac{{121}^2-121}{242}=60$

$a=61$

Area$=\frac{60\times 61}{2}>300$