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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 $x = 3$.
Four or Five
Age 11 to 14
ShortChallenge Level 
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 $x = 3$.
This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.
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