Task B-1.6. The sum of the lengths of the diagonals of a rhombus is $84 \mathrm{~cm}$. If we extend one diagonal by $4 \mathrm{~cm}$ and shorten the other diagonal by $4 \mathrm{~cm}$, the area of the rhombus will increase by $16 \mathrm{~cm}^{2}$. Calculate the length of the side and the lengths of the diagonals of this rhombus.
## Solution.

If the diagonals of the rhombus are $e$ and $f$, the area of the rhombus is

$$
P=\frac{e \cdot f}{2}
$$

After one diagonal is extended and the other is shortened by $4 \, \text{cm}$, we have

$$
\frac{(e+4)(f-4)}{2}=\frac{e \cdot f}{2}+16
$$

Then,

$$
\begin{aligned}
& (e+4)(f-4)=e \cdot f+32 \\
& e f-4 e+4 f-16=e f+32 \\
& -e+f=12
\end{aligned}
$$

which, together with $e+f=84$, gives the system of equations from which we obtain $f=48 \, \text{cm}, e=36 \, \text{cm}$.

From $a^{2}=\left(\frac{e}{2}\right)^{2}+\left(\frac{f}{2}\right)^{2}$, we get $a^{2}=18^{2}+24^{2}=900$, so $a=30 \, \text{cm}$.