Parallelogram 10 Level 5 7 Nov 2024Paper Mysteries

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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.
  • When you finish, remember to hit the SUBMIT button.
  • Finish by Sunday night if your whole class is doing parallelograms.

IMPORTANT – it does not really matter what score you get, because the main thing is that you think hard about the problems... and then examine the solution sheet to learn from your mistakes.

1. Paper mathematics

2 marks

1.1. Take a sheet of A4 paper. What is the length to the nearest centimetre?

Correct Solution: 30 cm

2 marks

1.2. What is the width to the nearest centimetre of the same piece of A4 paper?

Correct Solution: 21 cm

2 marks

1.3. Divide the length of A4 by the width and give the answer to 1 decimal place.

Correct Solution: 1.4

2 marks

1.4. If you measure the paper exactly, and divide the length by the width, what is the answer?

  • 1.4
  • 1.5
  • 1.45
  • 2
  • 1.22
  • π3
  • (Not answered)

2. More paper mathematics

Take a look at this Numberphile video all about the mathematics behind A4 paper.

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

2 marks

2.1 Pythagoreans should not urinate towards…

  • the sun
  • the moon
  • the ground
  • each other
  • the wind
  • (Not answered)
4 marks

2.2 Is it possible to write 2 as a fraction a/b?

  • Yes
  • No
  • (Not answered)

3. Intermediate Maths Challenge Problem (UKMT)

3 marks

3.1 What is the value of 25+250+2500?

  • 0.111
  • 0.222
  • 0.333
  • 0.444
  • 0.555
  • (Not answered)

We have 2/5 = 0.4. Multiplying by 1/10 is the same as dividing by 10, and therefore has the effect of moving the decimal point one place to the left. Therefore 250=110×25=0.04 and 2500=110×250=0.004. Hence 25+250+2500=0.4+0.04+0.004=0.444.

4. Intermediate Maths Challenge Problem (UKMT)

4 marks

4.1 The sum of the areas of the squares on the sides of a right-angled isosceles triangle is 72 cm2.

What is the area of the triangle?

  • 6 cm2
  • 8 cm2
  • 9 cm2
  • 12 cm2
  • 18 cm2
  • (Not answered)
Show Hint (–2 mark)
2 mark

Let the equal sides of the right-angled isosceles triangle have length x cm.

Then the squares on these sides have area x2 cm2. Therefore, by Pythagoras’ Theorem, the area of the square on the hypotenuse of the triangle is x2+x2 cm2, that is 2x2 cm2. Hence, the sum of the areas of the squares on all three sides of the triangle is x2+x2+2x2 cm2 =4x2 cm2. It follows that 4x2=72 and hence that x2=18.

Let the equal sides of the right-angled isosceles triangle have length x cm. Then the squares on these sides have area x2 cm2. Therefore, by Pythagoras’ Theorem, the area of the square on the hypotenuse of the triangle is x2+x2 cm2, that is 2x2 cm2. Hence, the sum of the areas of the squares on all three sides of the triangle is x2+x2+2x2 cm2 =4x2 cm2. It follows that 4x2=72 and hence that x2=18.

The triangle is half of a square measuring x cm ×x cm. Therefore the triangle has area 12x2 cm2.

It follows that the area of the triangle is 9 cm2.

5. Intermediate Maths Challenge Problem (UKMT)

5 marks

5.1 The tiling pattern shown uses two types of tile, regular hexagons and equilateral triangles, with the length of each side of the equilateral triangles equal to half the length of the sides of each side of the hexagons. A large number of tiles is used to cover a floor.

Which of the following is closest to the fraction of the floor that is shaded black?

  • 18
  • 110
  • 112
  • 113
  • 116
  • (Not answered)
Show Hint (–2 mark)
2 mark

We see that the entire plane can be tessellated by the shape labelled S in the above diagram. This shape is made up of one tile in the shape of a regular hexagon and two tiles in the shape of equilateral triangles.

We see that the entire plane can be tessellated by the shape labelled S in the diagram. This shape is made up of one tile in the shape of a regular hexagon and two tiles in the shape of equilateral triangles.

The diagram shows that, as the side length of the triangular tiles is half that of the hexagonal tiles, the area of the hexagonal tiles is 24 times the area of the triangular tiles.

[This may be seen by counting the number of small triangles, each congruent to the triangular tiles, into which the hexagonal tile is divided by the dashed lines.

Alternatively, we see that the hexagonal tile may be divided into 6 equilateral triangles each with side length twice that of the triangular tiles, and hence each with an area 4 times that of the triangular tiles. It follows that the area of the hexagonal tile is 6 × 4, that is, 24, times that of the triangular tiles.]

Therefore the shape S is made up of 2 black triangles and a hexagon equal in area to 24 of the black triangles. So the total area of S is equal to that of 2 + 24 = 26 of the triangles.

It follows that the fraction of the area of S that is black is 226=113.

A large area of the floor will be covered by a large number of tiles which make up shapes congruent to S, together with a few bits of tiles. Therefore the fraction of the floor that is shaded black will be very close to 113.

There will be more next week, and the week after, and the week after that. So check your email or return to the website on Thursday at 3pm.

In the meantime, you can find out your score, the answers and go through the answer sheet as soon as you hit the SUBMIT button below.

When you see your % score, this will also be your reward score. As you collect more and more points, you will collect more and more badges. Find out more by visiting the Rewards Page after you hit the SUBMIT button.

It is really important that you go through the solution sheet. Seriously important. What you got right is much less important than what you got wrong, because where you went wrong provides you with an opportunity to learn something new.

Cheerio, Simon.