This problem is at around grade 5 to 6 level from Singapore's primary school mathematics curriculum. It is challenging because it seems to require the use of Pythagoras' Theorem. Like: you do not know the length of a side of the square, so you let it be x and the other side is also x. Form the equation, solve it and voilà! You can get the answer!
However, pupils do not learn Pythagoras' Theorem in primary school. It looks like you have to fight the problem with with your hands tied behind your back. Actually, ...
... not quite. There is a trick you can use. In fact examiners have been setting questions like these so much that it has this type of question has become quite common and, maybe, predictable. OK, but what is the trick?
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Try to understand the problem. you are given that the radius is 14 m, and that happens to be equal to diagonal of the square (coloured in yellow).
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Can you think of an equivalent problem?
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Is there something you can do to the area that would leave the area unchanged? (use your imagination)
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How about shifting things around?
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Got it?
If you did, congratulations!
If you didn't, no worries. It's just a learning experience.
Here is my solution:-
You can move the triangle WYX to a new location, forming triangle WZV. Now notice that you get one big triangle VWY (coloured in yellow) with a right-angled triangle at W, with "base" WY = 14 m and
"height" VW = 14 m. Tilt your head around (or turn your tablet or mobile phone around, with portrait/landscape mode locked) if that helps. The area is the same as before, because we are just moving things around. Now we can easily see that the area is 1/2 × 14 × 14 = 98 m².
Ta da!
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Arithmetic Progression in a triangle
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@tradersharpe
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