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.

Arnab Kumar Debnath by Arnab Kumar Debnath , 21, Bangladesh

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

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

3 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...