The figure shows a quarter circle ABC and a rectangle CDEF. The length of AC is 10 cm. The sum of the lengths of EF and ED is 14 cm. (Take $\pi = 3.14$)

1. Identify the radius of the quarter circle

The length of $AC$ is given as $10\text{ cm}$. Since $ABC$ is a quarter circle with centre $A$, the radius $r = AC = 10\text{ cm}$.

2. Calculate the area of the quarter circle

$\text{Area} = \frac{1}{4} \times \pi \times r^2$
$\text{Area} = \frac{1}{4} \times 3.14 \times 10^2$
$\text{Area} = \frac{1}{4} \times 3.14 \times 100$
$\text{Area} = \frac{1}{4} \times 314 = 78.5\text{ cm}^2$

1. Find the length of the diagonal DF and the arc AB.

In rectangle $CDEF$, the diagonals are equal in length, so $DF = CE$. Since $E$ lies on the arc of the quarter circle with center $C$, $CE$ is a radius, meaning $DF = 10$ cm.
The length of arc $AB = \frac{1}{4} \times 2 \times 3.14 \times 10 = 15.7$ cm.

2. Find the sum of lengths DA and BF, and calculate the total perimeter.

Since $CA = CB = 10$ cm, we have $DA = 10 - CD$ and $BF = 10 - CF$. Thus, the sum of their lengths is $DA + BF = 20 - (CD + CF)$.
In rectangle $CDEF$, opposite sides are equal, so $CD = EF$ and $CF = ED$. Given $EF + ED = 14$ cm, we have $CD + CF = 14$ cm.
Therefore, $DA + BF = 20 - 14 = 6$ cm.
The perimeter of the shaded part $DABF = \text{arc } AB + DF + (DA + BF) = 15.7 + 10 + 6 = 31.7$ cm.