(a) A 2-digit number is such that the product of the digits is 14. When 45 is added to the number, the digits are reversed. Find the number.

Let the two digit number be \(\displaystyle 10 x+y\)

\[\begin{aligned} & x y=14 \ldots \ldots \text { (i) } \\ & 10 x+y+45=10 y+x \\ & y-x=5 \ldots . . \text { (ii) } \end{aligned} \]

from (i) and (ii)

\[\begin{aligned} & \frac{14}{x}-x=5 \\ & x^{2}+5 x-14=0 \\ & (x+7)(x-2)=0 \\ & x \neq-7 \\ & x=2 \\ & \text { Number }=27 \end{aligned} \]

Let the side of first square \(\displaystyle =x \mathrm{~cm}\)

\[\begin{aligned} & \therefore \text { Side of second square }=(x+4) \mathrm{cm} \\ & \quad x^{2}+(x+4)^{2}=400 \\ & x^{2}+4 x+192=0 \\ & (\mathrm{x}+16)(\mathrm{x}-12)=0 \\ & \mathrm{x} \neq-16 \\ & \mathrm{x}=12 \end{aligned} \]

Side of square \(\displaystyle =12 \mathrm{~cm}\) and $\displaystyle 16$ cm