Let's go through each part of the question step by step.

Part (a)

Explanation for why triangle $DEC$ is equilateral:

Since $ABCD$ is a square with a side length of $62$ mm:

1. The sides $AD$, $DC$, $AB$, and $BC$ are all equal, so each has a length of $62$ mm.
2. Arc $AC$ is drawn with center $D$, meaning the radius of this arc is $DA = DC = 62$ mm.
3. Similarly, arc $BD$ is drawn with center $C$, making $CB = CD = 62$ mm.

In triangle $DEC$:

• $DE = DC = CE = 62$ mm, so all three sides are equal.

Since all sides of $DEC$ are equal, triangle $DEC$ is an equilateral triangle.

Part (b)

Calculating the perimeter of the shaded shape $ABE$:

The shaded shape $ABE$ consists of:

1. Line segment $AB$.
2. Arc $AE$ (a part of arc $AC$).
3. Arc $BE$ (a part of arc $BD$).

Given:

• $AB = 62$ mm (since it's a side of the square).
• Both arcs $AE$ and $BE$ subtend an angle of $60^\circ$ (since they each span one side of the equilateral triangle $DEC$).

To find the arc lengths:

1. Arc $AE$:

   • Arc $AE$ is $\frac{60^\circ}{360^\circ} \times$ the circumference of the circle centered at $D$ with radius $62$ mm.
   • Circumference of the circle = $2 \pi \times 62 = 124 \pi$.
   • Arc $AE = \frac{60}{360} \times 124 \pi = \frac{1}{6} \times 124 \pi = \frac{124 \pi}{6}$.

2. Arc $BE$:

   • Similarly, arc $BE = \frac{124 \pi}{6}$.

Now, adding up these lengths for the perimeter of $ABE$: $\text{Perimeter of } ABE = AB + \text{Arc } AE + \text{Arc } BE$ $= 62 + \frac{124 \pi}{6} + \frac{124 \pi}{6}$ $= 62 + \frac{248 \pi}{6}$ $= 62 + \frac{124 \pi}{3}$

Using $\pi \approx 3.1416$: $\text{Perimeter} \approx 62 + \frac{124 \times 3.1416}{3}$ $\approx 62 + 129.9467$ $\approx 191.9 \text{ mm (to 1 decimal place)}$

Final Answer:

• (a) Triangle $DEC$ is equilateral because all three sides are equal in length (62 mm).
• (b) The perimeter of the shaded shape $ABE$ is approximately 191.9 mm.

Would you like further details or clarification on any part?

Related Questions:

1. How would the perimeter change if the side length of the square was different?
2. What would be the area of the shaded region $ABE$?
3. How would you calculate the perimeter if $ABE$ included a different arc length?
4. Can you find the area of triangle $DEC$?
5. What if $AC$ and $BD$ were semicircles instead of arcs of $60^\circ$?

Tip:

When dealing with arc lengths, always make sure to convert angles to fractions of $360^\circ$ to find their respective portions of the circle’s circumference.