Parallelogram 7 Year 10 31 Oct 2019Skyscrapers

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Noun: Parallelogram Pronunciation: /ˌparəˈlɛləɡram/

  1. a portmanteaux word combining parallel and telegram. A message sent each week by the Parallel Project to bright young mathematicians.

These challenges are a random walk through the mysteries of mathematics, most of which will be nothing to do with what you are doing at the moment in your classroom. Be prepared to encounter all sorts of weird ideas, including a few questions that appear to have nothing to do with mathematics at all.

  • 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. Skyscrapers

No mathematics in this clip, but just extraordinary footage of the skyscraper acrobats who built New York’s skyline.

2 marks

1.1. The Empire State Building is 381m tall, it has 103 storeys and it was constructed in 410 days.

Imagine you had a stack of one penny coins that reached from the ground to the top of the Empire State Building. Could you fit all those pennies inside your classroom?

  • Yes
  • No

One approach is to say that a penny is about 1 mm in thickness. So, your stack consists of 381,000 pennies. Let’s round it up to 400,000 pennies.

Your classroom ceiling is at least 2.5 m high, so you can create 160 stacks of 2,500 pennies, which is a grid of 16 × 10 stacks, each one 2,500 pennies tall.

So you can easily fit them into a classroom.

Another approach is to say that the pennies are roughly as tall as 100 storeys, which is roughly the number of storeys in the Empire State Building. So, we can break the total stack of pennies into 100 stacks, each one about the height of one storey. So, then the question changes to: can you fit 100 stacks of pennies (each one going from floor to ceiling)? And the answer is YES.

2. Approximating square roots

Here is a video all about estimating square roots with a high degree of accuracy without using a calculator.

2 marks

2.1 Use this approach to work out the square root of 10 by hand accurate to one decimal place.

(Of course, you can cheat and use a calculator, but what’s the point? This is not the world’s most important test, but it is a chance for you to try and stretch your brain.)

  • 3.0
  • 3.1
  • 3.2
  • 3.3
  • 3.4
3 marks

2.2 Use this approach to work out the square root of 66 by hand accurate to two decimal places.

  • 7.92
  • 8.12
  • 8.32
  • 8.52
  • 8.72

3. Intermediate Maths Challenge Problem (UKMT)

3 marks

3.1 One third of the animals in Jacob’s flock are goats, the rest are sheep. There are twelve more sheep than goats.

How many animals are there altogether in Jacob’s flock?

  • 12
  • 24
  • 36
  • 48
  • 60

One third of the animals in Jacob’s flock are goats. Therefore two thirds of these animals are sheep. Therefore, the 12 more sheep than goats amount to one third of the flock. Since 12 animals make up one third of the flock, there are 3 × 12 = 36 animals in Jacob’s flock.

4. Intermediate Maths Challenge Problem (UKMT)

4 marks

4.1 The circle has radius 1 cm. Two vertices of the square lie on the circle. One edge of the square goes through the centre of the circle, as shown.

What is the area of the square, in cm2?

  • 45
  • π5
  • 1
  • π4
  • 54
Show Hint (–1 mark)
–1 mark

Let the vertices of the square be P, Q, R and S, as shown in the diagram. Let O be the centre of the circle. We also suppose that the length of the sides of the square is s cm. It follows that the area of the square is s2 cm2.

Let the vertices of the square be P, Q, R and S, as shown in the diagram. Let O be the centre of the circle. We also suppose that the length of the sides of the square is s cm. It follows that the area of the square is s2 cm2.

We consider the triangles POQ and SOR. In these triangles OPQ=OSR=90°, as they are both angles in the square; OQ=OR, as they are radii of the same circle; and PQ=SR, as they are sides of the same square.

It follows that the triangles POQ and SOR are congruent, by the RHS congruence condition. [RHS stands for Right-angle, Hypotenuse, Side.] It follows thatOP=OS=12s cm. Therefore, by applying Pythagoras’ Theorem to the triangle SOR, we have

s2+12s2=12,

that is,

s2+14s2=1,

and hence

54s2=1,

from which it follows that

s2=45.

Therefore the area of the square is 45 cm2.

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.