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Geometry Difficulty 2.8 Junior Find the answer Canada

In right-angled, isosceles triangle $FGH$ , $FH=\sqrt{8}$ . Arc $FH$ is part of the circumference of a circle with centre $G$ and radius $GH$ , as shown.

The area of the shaded region is

Pick one

Solution

We use labels, $m$ and $n$ , in the fourth row of the grid, as shown. [[IMAGE0]] Then, $10,m,36,n$ are four terms of an arithmetic sequence. Since 10 and 36 are two terms apart in this sequence, and their difference is $36-10=26$ , the constant added to one term to obtain the next term in the fourth row is $\frac{26}{2}$ or 13. That is, $m=10+13=23$ , and $n=36+13=49$ . (We confirm that the terms $10,23,36,49$ do form an arithmetic sequence.) In the fourth column, 25 and $n$ (which equals 49) are two terms apart in this sequence, and their difference is $49-25=24$ . Thus, the constant added to one term to obtain the next term in the fourth column is $\frac{24}{2}$ or 12. That is, $x=25+12=37$ (or $x=49-12=37$ ).
The completed grid is as shown.

Figure for this problem

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.