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:

Solution (By Examveda Team)

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