The diagram shows a pattern of shaded squares on a \(6\) by \(6\) grid. The bottom left corner of the grid is at the origin and each square has side \(1\) u...

Assessment: Mathematics 0580 | Paper 1 Mock 01 | Non-calculator (Core) Subject: Mathematics - 0580

Question 1 Report

The diagram shows a pattern of shaded squares on a \(6\) by \(6\) grid. The bottom left corner of the grid is at the origin and each square has side \(1\) unit.

(a) Write down the number of shaded squares. [1]

(b) Write down the number of lines of symmetry of the pattern. [1]

(c) Write down the order of rotational symmetry of the pattern. [1]

(d) Write down the equations of the two lines of symmetry that are parallel to the axes. [2]

(e) Write down the coordinates of the centre of rotation. [1]

(f) Write down the smallest angle of rotation, in degrees, that maps the pattern onto itself. [1]

Answer Details

The grid is \(6\) by \(6\) with its bottom left corner at the origin and squares of side \(1\) unit, so it stretches from \(0\) to \(6\) in each direction and its centre is at \((3,\,3)\). All the later parts follow from that centre.

(a) Counting the shaded squares in the pattern gives \(16\) [B1]. Count them row by row and add the row totals rather than trying to see them all at once.

(b) Testing the four candidate mirror lines, the vertical centre line, the horizontal centre line and the two diagonals, all four reflect shaded squares onto shaded squares, giving \(4\) lines of symmetry [B1].

(c) Turning the pattern about the centre, each quarter turn reproduces it, so the order of rotational symmetry is \(4\) [B1].

(d) The two mirror lines parallel to the axes are the centre lines of the grid: \(x=3\) [B1] and \(y=3\) [B1].

(e) The centre of rotation is where all the lines of symmetry cross, at \((3,\,3)\) [B1].

(f) The smallest angle of rotation is \(360\div 4=90\) degrees [B1].

The self-check across parts (c), (e) and (f) is that the order, the centre and the smallest angle must agree: order \(4\) forces the angle \(\frac{360^\circ}{4}=90^\circ\), and the centre of rotation must be the common point of the mirror lines found in (d).

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