No need of any jugglery - 2

Geometry Level 4

Given that A B C D ABCD is a square of side length 6 6 . E E is a point on the circumference of the circle with A B AB as diameter such that C B E \angle CBE equals to 30 { 30 }^{ \circ } . The areas of Δ A C E \Delta ACE and Δ A B E \Delta ABE are a a and b b respectively. Find a + b \left\lfloor a+b \right\rfloor where x \left\lfloor x \right\rfloor represents the greatest integer lesser than or equal to x x .


The answer is 25.

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1 solution

Ajit Athle
Oct 11, 2018

There's a typo in the problem. It should read: E is a point on the circumference of the circle with BC as diameter. With the centre of the circle as the origin, the circle is: x² + y² = 9, while BE is: y=√3 x -3. E can now be easily determined as [ 3 3 2 \frac{3√3}{2} , 3 2 \frac{3}{2} ], A:[-6.-3] and C:[0,3]. Sum of the areas of Tr. ACE & Tr. ABE = 25.794

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