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Editorial
Once you've found the side length of the triangle, you are done. Here all I can help you by telling,
$\cos 15$° = $\frac{\sqrt{2}+\sqrt{2}\times\sqrt{3}}{4}$
$AEF$ is an equilateral triangle inscribed in square $ABCD.$ If the square has an area of $968$, can you find the area of triangle $AEF?$ (We know the answer is not perfect. Just write the integer part of your solution)
Once you've found the side length of the triangle, you are done. Here all I can help you by telling,
$\cos 15$° = $\frac{\sqrt{2}+\sqrt{2}\times\sqrt{3}}{4}$