Solution: Australian Mathematics Competition 2022 – Upper Primary, Problem 16

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Let's discuss a problem from AMC 2022 - Upper Primary - 16 on Geometry.


(A) 12 cm (B) 14 cm (C) 20 cm (D) 24 cm (E) 30 cm


In this problem, it is given that the big square has a perimeter of \(32\)cms. We know how to find the perimeter of a square.
The perimeter is nothing but the length of the boundary. So \(4a\) is the perimeter of a square. Thus \(4a = 32\),

so, \(a = \frac {32}{4} = 8\).

Thus the side length of the bigger square is = \(8\)cms.

The small square has the area as \(9\) sq cms.

The area of a square is = \( a^2\). Thus \(a^2 = 9, a = 3\).

Thus the side length of the smallest square is 3 cms.

Thus the medium square has a side length of = \(8-3\) cm = \(5\)cm.

Thus the perimeter of the medium square is = \(4a = 4 \times 5 = 20\) cm.

What is AMC (Australian Mathematics Competition)?

The Australian Mathematics Competition (AMC) is one of the largest and oldest annual mathematics competitions in Australia, aimed at fostering interest and excellence in mathematics among students

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