A triangle has sides 25, 39, 34 units. If the area of a square exceeds the area of this triangle by 21 units, then the side of the square is:

Answer: Option B

$$\begin{array}{rl}& s=\frac{25+34+39}{2}=49\\ & \text{Area of }\Delta ABC=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\\ & =\sqrt{49\times 15\times 10\times 24}\\ & =\sqrt{s^2\times 3\times 2\times 2^3\times 3}\\ & =7\times 5\times 3\times 4\\ & =420\end{array}$$
So, area of square is 21 more than area of Δ = (i.e.)
= 420 + 21 = 441 square unit
$$\begin{array}{rl}& {\text{A}}^2=441\\ & \text{A}={\left[{\left(21\right)}^2\right]}^{\frac{1}{2}}=21\end{array}$$