A geometry problem by Arnab Kumar Debnath

Geometry Level pending

The above shows a triangle A B C ABC with area 15, and given side length B C = 10 BC = 10 .

A square D E F G DEFG is inscribed inside this triangle such that E E and F F lies on the line E F EF .

What is the sum of areas of the triangles D E B DEB and G F C GFC ?

Give your answer to 2 decimal places.


The answer is 8.88.

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1 solution

A T = 1 2 b h A_T=\frac{1}{2}bh

15 = 1 2 ( 10 ) h 15=\frac{1}{2}(10)h

h = 3 h=3

x = b h b + h = 30 13 x=\frac{bh}{b+h}=\frac{30}{13}

A s h a d e d = A_{shaded}= A T A A D G A s q u a r e = 15 [ 1 2 ( 30 13 ) ( 3 30 13 ) ] + ( 30 13 ) 2 = 8.88 A_T - A_{ADG} - A{square}=15 - [\frac{1}{2}(\frac{30}{13})(3 - \frac{30}{13})] + (\frac{30}{13})^2=8.88 s q u a r e square u n i t s units

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