Step 1: Identify the intended symmetry axes.
For a $4 \times 5$ grid, the two natural symmetry lines are:
(i) a vertical line between columns 2 and 3, and
(ii) a horizontal line through the middle row (row 3).
Step 2: Check which existing blacks already satisfy symmetry.
- Horizontal symmetry pairs: $(r1,c2)\leftrightarrow(r5,c2)$, $(r1,c3)\leftrightarrow(r5,c3)$, $(r2,c1)\leftrightarrow(r4,c1)$, $(r2,c4)\leftrightarrow(r4,c4)$ are already black–black and hence fine.
- The mismatch for horizontal symmetry occurs at $(r4,c3)$ which is black; its mirror across row 3 is $(r2,c3)=d$ (currently white). Thus d must be coloured black.
Step 3: Enforce vertical symmetry (mirror across columns 2 and 3).
- Cell $(r4,c3)$ is black; its vertical mirror is $(r4,c2)=i$ (white). Hence i must be coloured black.
- With d black (from Step 2), its vertical mirror is $(r2,c2)=c$; therefore c must also be black for vertical symmetry.
Step 4: Minimality check.
Colouring c, d, and i makes all horizontal and vertical mirror pairs match. No other cells are required; choosing fewer (e.g., only c,i) fails the horizontal pair $(r2,c3)\leftrightarrow(r4,c3)$, and any extra cells (e.g., f,g) are unnecessary.
\[
\boxed{\text{Colour exactly } c, d, \text{ and } i \text{ to achieve two lines of symmetry (minimum = 3).}}
\]
If a mirror is placed vertically next to the picture given above at the dotted line, then the image of the picture on the left looks like:
Find the correct mirror image for the following problem figure from the alternatives:
The figures, I, II, and III are parts of a sequence. Which one of the following options comes next in the sequence as IV?
For the beam and loading shown in the figure, the second derivative of the deflection curve of the beam at the mid-point of AC is given by \( \frac{\alpha M_0}{8EI} \). The value of \( \alpha \) is ........ (rounded off to the nearest integer).