Parallelogram 43 Level 4 26 Jun 2025The Wright Stuff

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Noun: Parallelogram Pronunciation: /ˌparəˈlɛləɡram/

  1. a portmanteau word combining parallel and telegram. A message sent each week by the Parallel Project to bright young mathematicians.
  • Tackle each Parallelogram in one go. Don’t get distracted.
  • Finish by midnight on Sunday if your whole class is doing parallelograms.
  • Your score & answer sheet will appear immediately after you hit SUBMIT.
  • Don’t worry if you score less than 50%, because it means you will learn something new when you check the solutions.

1. Inspirational teachers

Arthur Cayley is one of the great British mathematicians of the Victorian era.

His father wanted him to become a merchant join the family business, but Cayley’s teacher at King’s College School persuaded the father that the boy had a tremendous mathematical talent which should be encouraged, not wasted.

As a result, the young Cayley went to Cambridge University and flourished.

I am mentioning this, because teachers can have a tremendous influence on our lives.

Here is a video about England Arsenal footballer Ian Wright meeting the teacher who inspired him and who changed his life.

It’s a really emotional clip. I have watched it several times, and it still moves me.

(If you have problems watching the video, right click to open it in a new window)

And here is a video of cosmologist Stephen Hawking recalling the teacher who changed his life.

(If you have problems watching the video, right click to open it in a new window)

2 marks

1.1 How did Hawking’s teacher describe mathematics?

  • The blueprint of the universe
  • The heart of the universe
  • The imprint of the universe
  • The manuscript of the universe
  • The child of the universe
  • (Not answered)

2.

4 marks

2.1 What is the area of the shaded region in the rectangle?

  • 21 cm2
  • 22 cm2
  • 23 cm2
  • 24 cm2
  • more information needed
  • (Not answered)
Show Hint (–1 mark)
–1 mark

This question is not that hard to solve algebraically or by drawing some lines and seeing some symmetries. However, I decided to take an extreme version of squeezing one of the triangles and expanding the other, so that we have the following, which has an area which is easy to calculate. This means that the answer is either the shaded area or “more information needed”.

We divide the rectangle into four smaller rectangles by the vertical dashed lines as shown in the figure. It is then evident that the shaded area is the sum of exactly half the area of each of the smaller rectangles.

Therefore the total shaded area is half the total area of the original rectangle. So the shaded area is 123×14 cm2 = 21 cm2.

We could also solve this question algebraically. We let the top edges of the two shaded triangles have lengths a cm and b cm, respectively, so that a+b=14. The two shaded triangles have areas 12a×3 cm2 =32a cm2 and 12b×3 cm2 =32b cm2, respectively. Hence their total area is

32a+32b cm2 =32a+b cm2 =32×14 cm2 = 21 cm2.

3.

4 marks

3.1 In a sequence, each term after the first two terms is the mean of all the terms which come before that term. The first term is 8 and the tenth term is 26.

What is the second term?

  • 17
  • 18
  • 44
  • 52
  • 68
  • (Not answered)

Here it is actually a little easier to see what is going on if we think about the general case of a sequence in which every term after the second is the mean of all the terms that come before it.

Consider a sequence generated by this rule whose first two terms are a and b. The mean of these terms is a+b2, and so the first three terms are

a,b,a+b2.

The sum of these three terms is the same as the sum of the three terms

a+b2,a+b2,a+b2

and it follows that the mean of the first three terms is also (a + b)/2. So the first four terms of the sequence are

a,b,a+b2,a+b2.

These have the same sum as the four terms

a+b2,a+b2,a+b2,a+b2.

It follows that the mean of the first four terms is also a+b2 and that therefore this is the fifth term of the sequence, and so on. We thus see that each term of the sequence after the first two is equal to the mean of the first two terms.

In the sequence we are given in the question the first term is 8 and the tenth term is 26. It follows that 26 is the mean of the first two terms. Hence their sum is 2 × 26 = 52. Since the first term is 8, we deduce that the second term is 52 − 8 = 44.

4.

4 marks

4.1 The figure shows four smaller squares in the corners of a large square.

The smaller squares have sides of length 1 cm, 2 cm, 3 cm and 4 cm (in anticlockwise order) and the sides of the large square have length 11 cm.

What is the area of the shaded quadrilateral?

  • 35 cm2
  • 36 cm2
  • 37 cm2
  • 38 cm2
  • 39 cm2
  • (Not answered)

We work out the area that is not shaded and subtract this area from that of the large square.

The area that is not shaded is made up of the four corner squares and four trapeziums.

The total area of the squares is, in cm2,

12+22+32+42=1+4+9+16=30.

We now use the fact that the area of a trapezium is given by the formula

12a+bh,

where a and b are the lengths of the parallel sides, and h is their distance apart.

The values of a, b and h for the trapeziums P, Q, R and S, as labelled in the diagram, can be calculated from the side lengths of the squares, and are shown in the diagram. We therefore have that

area of P=121+2×8=12,
area of Q=122+3×6=15,
area of R=123+4×4=14, and
area of S=124+1×6=15.

It follows that the total area of the trapeziums is

12+15+14+15=56.

The large square has area 112 cm2 = 121 cm2. Therefore the area of the shaded quadrilateral is, in cm2, 1213056=35.

5.

5 marks

5.1 In square RSTU a quarter-circle arc with centre S is drawn from T to R.

A point P on this arc is 1 unit from TU and 8 units from RU.

What is the length of the side of square RSTU?

Show Hint (–1 mark)
–1 mark

This is a tough geometry question. So here is an unusual hint – the answer is a vowel.

  • A: 9
  • B: 10
  • C: 11
  • D: 12
  • E: 13
  • (Not answered)

Suppose that x is the length of the sides of the square RSTU. We need to find the value of x.

We let K be the point where the perpendicular from P to RS meets RS, and L be the point where the perpendicular from P to ST meets ST, as shown in the figure. Then PK has length x1, and so also does LS. Also, PL has length x8.

Because PS is a radius of the quarter circle, its length is the same as that of the sides of the square, namely, x.

Therefore SLP is a right-angled triangle in which the hypotenuse has length x, and the other two sides have lengths x1 and x8. Hence, by Pythagoras’ Theorem,

x12+x82=x2.

Multiplying out the squares, we get

x22x+1+x216x+64=x2,

which we can rearrange as

x218x+65=0.

We can factorize the quadratic on the left-hand side of this equation to give the equivalent equation

x5x13=0,

which has the solutions x=5 and x=13. Since x8 is the length of PL and therefore cannot be negative, x5. Therefore x=13.

6.

5 marks

6.1 A point is marked one quarter of the way along each side of a triangle, as shown.

What fraction of the area of the triangle is shaded?

  • 716
  • 12
  • 916
  • 58
  • 1116
  • (Not answered)

We use the notation XYZ for the area of a triangle XYZ.

We let P, Q and R be the vertices of the triangle, and let S, T and U be the points one quarter of the way along each side, as shown in the figure.

We let K and L be the points where the perpendiculars from P and T, respectively, meet QR.

The area of a triangle is half the base multiplied by the height of the triangle. Therefore

PQR=12QR×PK.

We now obtain an expression for TSR.

Because S is one quarter of the way along QR, SR=34QR.

In the triangles PKR and TLR, the angles at K and L are both right angles, and the angle at R is common to both triangles. It follows that these triangles are similar. Therefore

TLPK=TRPR=14,

and hence

TL=14PK.

It follows that

STR=12SR×TL=1234QR×14PK=31612QR×PK=316PQR.

Similarly,

TPU=316PQR.

It follows that the fraction of the area of triangle PQR that is shaded is

1316316=58.

Before you hit the SUBMIT button, here are some quick reminders:

  • You will receive your score immediately, and collect your reward points.
  • You might earn a new badge... if not, then maybe next week.
  • Make sure you go through the solution sheet – it is massively important.
  • A score of less than 50% is ok – it means you can learn lots from your mistakes.
  • The next Parallelogram is next week, at 3pm on Thursday.
  • Finally, if you missed any earlier Parallelograms, make sure you go back and complete them. You can still earn reward points and badges by completing missed Parallelograms.

Cheerio, Simon.