Question

Find the ratio of the area of triangle XBY to the area of triangle ABC for the given measurements, if XY l l AC

XY = 2, AC = 3

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Answer 1

where is the pic?

Answer 2

Interesting question.
The ratio is \[(AC/XY)^{2}\] = \[(3/2)^{2}\] = 9/4
Why?
Area is (base x height)/2.
Finding height of isosceles triangle...
cut the triangle in two parts. You will find two pitagoric triangles. Formula for this case is h\[h^{2} = side^{2} - (base/2)^{2}\]
base/2 because new triangles has this base. Solving, \[h= \sqrt{lado^{2}-(base/2)^{2}}\]
Area is \[1/2*base*\sqrt{lado^{2}-(base/2)^{2}}\]
When you increase sidesproportionaly the new area is \[1/2*3/2*base*\sqrt{(3/2lado)^{2}-(3/2*base/2)^{2}} =\]
\[1/2*3/2*base*3/2*\sqrt{(lado)^{2}-(base/2)^{2}}\], jumping out 3/2 of square root.
The diference betwenn cases is 3/2 and another 3/2. Finally 9/4.