Parallelogram 17 • Level 3 • 28 Dec 2023Chrismaths - Part 4

We first convert the fraction into a decimal. We multiply both the numerator (the top) and the denominator (the bottom) by 4. This gives:

125=4100.

Therefore, expressed as a decimal, 125 is 0.04.

Hence 125+0.25=0.04+0.25=0.29.

To fold a figure exactly in half we need to fold it along a line of symmetry.

A square has four lines of symmetry, but these are of just two types. There are two lines of symmetry joining the midpoints of opposite edges, and two lines of symmetry joining opposite vertices (see figure 1).

When a square is folded along a line of symmetry joining the midpoints of edges, such as l, the resulting shape is a rectangle whose longer sides are twice as long as the shorter sides (see figure 2). When it is folded along a line of symmetry joining opposite vertices, such as m, the resulting shape is an isosceles right-angled triangle (see figure 3).

The rectangle shown in figure 2 has two lines of symmetry, n and p, joining midpoints of opposite edges. When the rectangle is folded along n, the resulting shape is a square with half the side length of the original square. This is the shape of option A.

When the rectangle is folded along p, the resulting shape is a rectangle whose longer sides are four times as long as the shorter sides. This is the shape of option B.

The isosceles right-angled triangle shown in figure 3 has one line of symmetry, labelled q. When the triangle is folded along this line the resulting shape is a smaller isosceles right-angled triangle. This is the shape of both options C and E.

We therefore see that the shapes of options A, B, C and E can be achieved. Since we have covered all the possible ways of folding the square in half and in half again, it follows that the shape that could not be achieved is D.

We find the value of each expression to see which of them are squares. After we have found the value of the one expression, we can find the value of the next expression by adding the extra term to the sum we have already calculated. In this way, we obtain:

• 1³ + 2³ = 1 + 8 = 9 = 3².
• 1³ + 2³ + 3³ = 9 + 3³ = 9 + 27 = 36 = 6².
• 1³ + 2³ + 3³ + 4³ = 36 + 4³ = 36 + 64 = 100 = 10².
• 1³ + 2³ + 3³ + 4³ + 5³ = 100 + 5³ = 100 + 125 = 225 = 15².

We now see that all four of the expressions are perfect squares.

Note The first cube 1³ is, of course, also a square since 1³ = 1². We see from the above calculations that the sum of the first two cubes is square, and so also are the sums of the first three cubes, the first four cubes and the first five cubes. Just from these five examples we cannot deduce that the sum of the first n cubes is always a square, but it looks as though it might be true.

You should not be satisfied until you have found either a proof that this is always true, or a counterexample to show that it is false for at least one positive integer n.

In Brahmagupta’s formula, we have a=4,b=5,c=7 and d=10.

The total length of the perimeter, in cm, is 4 + 5 + 7 + 10 = 26. Therefore s=26÷2=13.

It follows that:

• s − a = 13 − 4 = 9,
• s − b = 13 − 5 = 8,
• s − c = 13 − 7 = 6, and
• s − d = 13 − 10 = 3.

Hence: s−as−bs−cs−d=9×8×6×3=9×144.

Therefore:

Therefore the area of the cyclic quadrilateral is 36 cm².

The board has 25 squares. Each domino covers two squares. So, however many dominoes are placed on the board, the total number of squares that are covered will be even. So the number of uncovered squares will be odd.

In particular, it is not possible to have 8 uncovered squares. The diagram shows that it is possible for Beatrice to place 9 dominoes on the board in such a way that she cannot place another domino. This leaves 7 uncovered squares.

As this is the largest odd number given as one of the options, we conclude that D is the correct option.