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:

Given, length of sides of triangle

a=25, b=39, c=34

Semiperimeter of triangle(s) = $$\frac{a+b+c}{2}$$ = $$\frac{25+39+34}{2}$$ = 49

Area of triangle = $$\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$$

$$=\sqrt{49\left(49-25\right)\left(49-39\right)\left(49-34\right)}$$

$$=\sqrt{49\times24\times10\times15}$$

$$=\sqrt{49\times4\times6\times2\times5\times3\times5}$$

$$=\sqrt{49\times4\times6\times6\times5\times5}$$

$$=7\times2\times6\times5=420$$

Area of Square = Area of triangle + 21 = 420+21 = 441

$$\therefore\ $$Side of Square = $$ \sqrt{441}=21$$ units

Hence, the correct answer is Option D