A polygon with \(8\) sides is called an octagon.

A quadrilateral has \(4\) sides.

All four sides are the same length and the opposite sides are parallel, so it is a rhombus. It is not a square, because none of the four corners is a right angle.

Opposite sides equal and parallel narrows it to a parallelogram, rhombus, rectangle or square. The last three all have at least one line of symmetry, so it is a parallelogram.

A regular polygon has one line of symmetry for every side it has.

The shape has \(10\) sides, so it has \(10\) lines of symmetry.

With an odd number of sides each line runs from a corner to the middle of the opposite side; with an even number, half of them join opposite corners and half join the middles of opposite sides.

Turning the shape through a full circle brings it back onto itself once for every side.

\(360 \div 10\) is the smallest turn that works, and it fits into \(360^\circ\) exactly \(10\) times, so the order of rotational symmetry is \(10\).

For a regular polygon the order of rotational symmetry and the number of lines of symmetry are always the same number. That is only true because it is regular — a rectangle has order 2 and two lines, but a parallelogram has order 2 and no lines at all.

An isosceles trapezium has one line of symmetry, the line through the middles of the two parallel sides. A diagonal is not one: the two parallel sides are different lengths, so folding along a diagonal cannot match them.

The isosceles trapezium has one line of symmetry.

Count how many times the shape fits onto itself in one complete turn, including the turn that brings it all the way back to the start.

No turn short of a whole one brings an isosceles trapezium back onto itself.

For the isosceles trapezium the order of rotational symmetry is \(1\).

Parallel sides are the ones that never meet, however far they are extended. \(DE\) runs along the bottom of the shape and \(GF\) runs along the top, and the two stay the same distance apart, so \(GF\) is parallel to \(DE\).

\(EF\) slopes, so it is not parallel to \(DE\); \(DG\) meets \(DE\) at \(D\), so it certainly is not.

Perpendicular means "at right angles to". The square marked at \(D\) shows that \(DG\) meets \(DE\) at \(90^\circ\), so \(DG\) is perpendicular to \(DE\).

\(EF\) is the only other side that meets \(DE\), and there is no square marked at \(E\), so \(EF\) is not perpendicular to it.

\(DEFG\) has exactly one pair of parallel sides, \(DE\) and \(GF\), so it is a trapezium. Because two of its angles are right angles it is sometimes called a right-angled trapezium, but "trapezium" is the name being asked for.

Put the right angle of the drawn triangle at the origin, so its corners are \((0, 0)\), \((4, 0)\) and \((0, 7)\).

Rotate a copy half a turn about the middle of the vertical side, which has length \(7\) squares. Its corners are \((0, 7)\), \((0, 0)\) and \((-4, 7)\).

The outline is \((0, 0)\), \((4, 0)\), \((0, 7)\), \((-4, 7)\).

The bottom side and the top side are both \(4\) squares long and horizontal, and the two sloping sides are both hypotenuses, so opposite sides are equal and parallel: a parallelogram.

Two copies joined along their hypotenuse, rotated half a turn, make the \(4\) by \(7\) rectangle \((0, 0)\), \((4, 0)\), \((4, 7)\), \((0, 7)\), which is also a parallelogram.

Start with the two triangles from part (a), with corners \((0, 0)\), \((4, 0)\), \((0, 7)\) and \((0, 0)\), \((0, 7)\), \((-4, 7)\).

Add a third copy along the sloping side from \((0, 0)\) to \((-4, 7)\): its corners are \((-4, 0)\), \((0, 0)\) and \((-4, 7)\).

The outline is \((-4, 0)\), \((4, 0)\), \((0, 7)\), \((-4, 7)\).

The bottom side is \(4 + 4 = 8\) squares long and the top side is \(4\) squares long. Both are horizontal, so they are parallel, and they are different lengths. The other two sides are not parallel, so the shape has exactly one pair of parallel sides: a trapezium.

Place the four right angles together at the origin. Reflecting the drawn triangle in the vertical side, then in the horizontal side, gives triangles with corners:

\((0, 0)\), \((4, 0)\), \((0, 7)\)

\((0, 0)\), \((0, 7)\), \((-4, 0)\)

\((0, 0)\), \((-4, 0)\), \((0, -7)\)

\((0, 0)\), \((0, -7)\), \((4, 0)\)

The outline is \((4, 0)\), \((0, 7)\), \((-4, 0)\), \((0, -7)\).

Every side of this outline is the hypotenuse of a copy, with length \(\sqrt{ 65 }\) squares, so all four sides are equal and it is a rhombus. It is not a square because the diagonals \(8\) and \(14\) squares long are different.

Diagonals that bisect each other at right angles make all four sides equal. If the diagonals were also equal in length the shape would be a square, so the shape is a rhombus.

The quadrilateral is a rhombus.

The grid has \(5\) rows of \(8\) squares, so it has \(5 \times 8 = 40\) squares. One quarter of them is \(40 \div 4 = 10\) squares.

Every shaded square needs its mirror images in both middle lines, so the squares are shaded in groups of four. A square on the middle row is its own reflection in the horizontal line, so those squares are shaded in pairs.

The grid is not square, so the two middle lines are its only possible lines of symmetry.

The model answer shows one correct shading of \(10\) squares. Any other shading that meets both conditions is also correct.

\(ABC = 90^\circ\), so \(BC\) is perpendicular to \(BA\). Count 2 cm up from \(B\) to plot \(C(4, 5)\).

\(CD\) is parallel to \(BA\) and the direction from \(B\) to \(A\) is right. Count 3 cm right from \(C\) to plot \(D(7, 5)\). Going the other way would make \(AD\) cross \(BC\).

Joining \(A\) to \(B\), \(B\) to \(C\), \(C\) to \(D\) and \(D\) to \(A\) makes a right trapezium.

Counting 2 cm down from \(B\) instead is also correct, giving \(C(4, 1)\) and \(D(7, 1)\).