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# Four or Five

**Answer**: 1620 square centimetres

**Using the options given**

450 $\div$ 9 = 50

50 = length $\times$ width, where length : width = 5 : 4

50 = 5 $\times$ 10

No closer factor pairs $\therefore$ no factor pairs in the ratio 5 : 4

1260 $\div$ 9 = 140

140 = 2 $\times$ 70

= 4 $\times$ 35

= 10 $\times$ 15

No closer factor pairs $\therefore$ no factor pairs in the ratio 5 : 4

1620 $\div$ 9 = 180

180 = 10 $\times$ 18

= 12 $\times$ 15 correct ratio

So 1620 is possible

**Using $x$**

The ratio length:breadth of the smaller rectangles is $5:4$. Let the length and breadth of these rectangles be $5x \; \text{cm}$ and $4x\; \text{cm}$ respectively. The area of the large rectangle, in cm$^2$, is $9 \times 20x^2 = 180x^2$ and the only one of the alternatives which is a product of $180$ and a perfect square is $1620$, which corresponds to $x = 3$.

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Age 11 to 14

ShortChallenge Level

- Problem
- Solutions

450 $\div$ 9 = 50

50 = length $\times$ width, where length : width = 5 : 4

50 = 5 $\times$ 10

No closer factor pairs $\therefore$ no factor pairs in the ratio 5 : 4

1260 $\div$ 9 = 140

140 = 2 $\times$ 70

= 4 $\times$ 35

= 10 $\times$ 15

No closer factor pairs $\therefore$ no factor pairs in the ratio 5 : 4

1620 $\div$ 9 = 180

180 = 10 $\times$ 18

= 12 $\times$ 15 correct ratio

So 1620 is possible

The ratio length:breadth of the smaller rectangles is $5:4$. Let the length and breadth of these rectangles be $5x \; \text{cm}$ and $4x\; \text{cm}$ respectively. The area of the large rectangle, in cm$^2$, is $9 \times 20x^2 = 180x^2$ and the only one of the alternatives which is a product of $180$ and a perfect square is $1620$, which corresponds to $x = 3$.

This problem is taken from the UKMT Mathematical Challenges.

You can find more short problems, arranged by curriculum topic, in our short problems collection.

Seven small rectangular pictures have one inch wide frames. The frames are removed and the pictures are fitted together like a jigsaw to make a rectangle of length 12 inches. Find the dimensions of the pictures.

It's easy to work out the areas of most squares that we meet, but what if they were tilted?

Read about David Hilbert who proved that any polygon could be cut up into a certain number of pieces that could be put back together to form any other polygon of equal area.