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Puzzle #18

Two squares, one is small and the other is big. One of the sides of the big square starts in the top right vertex of the small square and finishes at the left side of the small square. The distance between the bottom left vertex of the small square and the left vertex of the big square is labeled/labelled one. The horizontal distance between the left and the right vertices of the big square is represented by a double arrow and is labeled/labelled nine.

The diagram shows two squares. The rightmost vertex of the big square is aligned with the base of the small square.

What is the area of the big square?

This puzzle was published in the Mathematical Association’s magazine Mathematics in School.

Solution

Let xx and yy be the sides of the small and the big squares, respectively.

We can label our diagram as follows:

Two squares, one is small and the other is big. The side length of the first one is labeled/labelled x and the second one is labeled/labelled y. One of the sides of the big square starts in the top right vertex of the small square and finishes at the left side of the small square. The distance between the bottom left vertex of the small square and the left vertex of the big square is labeled/labelled one. The rest of the small's square left side is labeled/labelled x minus one. The horizontal distance between the left and the right vertices of the big square is represented by a double arrow and is labeled/labelled nine. A right(-angled) triangle with hypotenuse the top right side of the big square, labeled/labelled y, and the other two short sides, one horizontal and the other vertical, are labeled/labelled nine minus x and x, respectively.

Using the Pythagorean/Pythagoras’ theorem in the leftmost right(-angled) triangle:

y2=x2+(x−1)2y^2=x^2+\left(x-1\right)^2
y2=2x2−2x+1(1)y^2=2x^2-2x+1\;\;\;\left(1\right)

Using the Pythagorean/Pythagoras’ theorem again but now in the rightmost right(-angled) triangle:

y2=x2+(9−x)2y^2=x^2+\left(9-x\right)^2
y2=2x2−18x+81(2)y^2=2x^2-18x+81\;\;\;\left(2\right)

Equaling/equalling (1) and (2) we get

2x2−2x+1=2x2−18x+812x^2-2x+1=2x^2-18x+81
16x=8016x=80
x=5x=5

Substituting the value of xx in (1), we have

y2=2(5)2−2(5)+1y^2=2\left(5\right)^2-2\left(5\right)+1
y2=50−10+1y^2=50-10+1
y2=41y^2=41

So, the area of the big square is 41.

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