Square and triangle
Can you find the area of the yellow square?
Problem
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Find the area of the yellow square in this diagram.
This problem is taken from the World Mathematics Championships
Student Solutions
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The sides of the large right-angled triangle are both 6 cm, so its angles are 90$^\circ$, 45$^\circ$, and 45$^\circ$.
The angles in the square are all right angles, so the smaller triangles around the square are also right-angled isosceles triangles.
This means that the angles shaded green on the diagram are all 45$^\circ$, and the lengths labelled $k$ cm are all equal to the side length of the square.
This is important for all of the three methods shown below:
Using congruent triangles
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There are 9 of these triangles altogether, and 4 of them fit into the square, so the area of the square is $\frac49$ of the area of the whole triangle.
The area of the whole triangle is $\frac12\times6\times6 = 18$ cm$^2$. So the area of the square is $\frac49$ of $18$ cm$^2$, which is $8$ cm$^2$.
Using scale factors
The diagonal side of the whole triangle is equal to 3$k$ cm, and the diagonal side of the small triangle in its corner (to the bottom left of the square) is $k$ cm. So the scale factor from the small triangle to the whole triangle is 3.
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This means that the sides of the small triangle are 2 cm, since 2 cm $\times$ 3 = 6 cm.
So the area of the small triangle is 2$\times$2$\div$2 = 2 cm$^2$.
The area of the whole triangle is 6$\times$6$\div$2 = 18 cm$^2$.
So the area of the square and the two triangles on either side of it is 18 $-$ 2 = 16 cm$^2$. Each of these two triangles is congruent to half of the square, so the square occupies half of this area. So the area of the square is 8 cm$^2$.
Using Pythagoras' Theorem on the whole triangle
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The hypotenuse of the triangle is equal to $3k$, and can also be found using Pythagoras' Theorem: $$6^2+6^2=(3k)^2\\\Rightarrow 72=9k^2\\
\Rightarrow 8=k^2$$ But $k^2$ is the area of the square. So the area of the square is $8$ cm$^2$.