Parallelogram 28 • Level 4 • 13 Mar 2025Venn diagrams

Noun: Parallelogram Pronunciation: /ˌparəˈlɛləɡram/

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. Venn diagrams

This is my all time favourite Venn diagram. What? Doesn’t everyone have a favourite Venn Diagram? Anyway... If you don’t yet have a favourite Venn diagram, then you can borrow this one for the time being. Can you work out the label for the two circles? You can find my answer below.

I guess that the left set contains “creatures with flat tails playing guitars”, while the right set contains “creatures with bills playing keyboards”. Hence, the animal in the overlap is a platypus playing a keytar, because it is a creature with a bill AND a flat tail, which is playing a guitar AND a keyboard.

2.

3 marks

2.1 ABCDE is a regular pentagon and BCF is an equilateral triangle such that F is inside ABCDE. What is the size of ∠FAB?

48°
63°
66°
69°
72°
(Not answered)

Show Hint (–1 mark)

1 mark

AB=FB, so the triangle ABF is isosceles, and hence ∠FAB=∠BFA.

Since the pentagon ABCDE is regular, BC=AB. Since the triangle BCF is equilateral, BC=FB. Therefore, AB=FB. So the triangle ABF is isosceles, and hence ∠FAB=∠BFA.

∠ABC=108°, as it is the interior angle of a regular pentagon. As the triangle BCF is equilateral, ∠FBC=60°. Therefore ∠ABF=∠ABC−∠FBC=108°−60°=48°.

So, from triangle ABF, we can deduce that ∠FAB+∠BFA=180°−48°=132°.

Since ∠FAB=∠BFA, it follows that ∠FAB=12132°=66°.

3.

3 marks

3.1 The sum:

one+four=seventy

becomes correct if we replace each word by the number of letters in it to give 3 + 4 = 7.

Using the same convention, which of these words could be substituted for x to make the following sum true?

three+five=x

eight
nine
twelve
seventeen
eighteen
(Not answered)

If we replace ‘three’ and ‘five’ by the number of letters in these words, we obtain the equation:

5+4=x.

Hence x=9. So we need to replace x by a word with 9 letters in it. Of the options we are given ‘seventeen’ is the only word with 9 letters in it.

4.

5 marks

4.1 The diagrams show squares placed inside two identical semicircles. In the lower diagram the two squares are identical.

What is the ratio of the areas of the two shaded regions?

1 : 2
2 : 3
3 : 4
4 : 5
5 : 6
(Not answered)

Show Hint (–1 mark)

1 mark

Choose units so that the two semicircles have radius 1. We let x be the side length of the square in the first diagram, and y be the side lengths of the squares in the second diagram.

We choose units so that the two semicircles have radius 1. We let x be the side length of the square in the first diagram, and y be the side lengths of the squares in the second diagram.

By Pythagoras’ Theorem, we have x2+12x2=1 and y2+y2=1. That is 54x2=1 and 2y2=1. Therefore x2=45 and y2=12.

The shaded region in the top diagram consists of a square of side length x and hence of area x2, that is, 45. The shaded region in the lower diagram consists of two squares each of side length y, and hence it has area 2y2, that is, 1. Hence the ratio of the two shaded regions is 45:1=4:5.

5.

5 marks

5.1 The diagram shows a shaded shape bounded by circular arcs with the same radius. The centres of three arcs are the vertices of an equilateral triangle; the other three centres are the midpoints of the sides of the triangle.

The sides of the triangle have length 2.

What is the difference between the area of the shaded shape and the area of the triangle?

π6
π4
π3
π2
π
(Not answered)

Show Hint (–1 mark)

1 mark

One way to think about this problem is that the difference between the area of the shaded shape and the area of the triangle is:

Shaded areas – Triangle area
(Shaded areas outside Triangle + shaded areas inside Triangle) – Triangle area
(Shaded areas outside Triangle + [Triangle area – 3 semicircles]) – Triangle area
Shaded areas outside Triangle + Triangle area – 3 semicircles – Triangle area
Shaded areas outside Triangle – 3 semicircles

One way to think about this problem is that the difference between the area of the shaded shape and the area of the triangle is:

Shaded areas – Triangle area
(Shaded areas outside Triangle + shaded areas inside Triangle) – Triangle area
(Shaded areas outside Triangle + [Triangle area – 3 semicircles]) – Triangle area
Shaded areas outside Triangle + Triangle area – 3 semicircles – Triangle area
Shaded areas outside Triangle – 3 semicircles

Since the sides of the triangle have length 2, each of the circular arcs has radius 12. The difference between the area of the shaded shape and that of the triangle, is the difference between the area of the 3 shaded sectors of circles outside the triangle, and that of the 3 unshaded semicircles within the triangle.

The angles of the equilateral triangle are 60° which is 16th of a complete revolution. So the areas of the shaded sectors of circles outside the triangle are 56ths of the total areas of these circles, each of which has radius 12.

So the area of these sectors is 3×56π122=58π. The area of the 3 unshaded semicircles inside the triangle is 3×12π122=38π. The difference between these is 58π−38π=14π. Hence, the the difference between the area of the shaded shape and the area of the triangle is π4.

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.