Bunuel wrote:
Attachment:
SimilarTriangles2.jpg
In the given figure, ABCD is a parallelogram and E, F, G and H are midpoints of its respective sides. What is the ratio of the shaded area to that of the un-shaded area?
(A) 3:8
(B) 3:5
(C) 5:8
(D) 8:5
(E) 5:3
Kudos for a correct solution.Since all squares are parallelograms, let us assume that the mother figure is also a square for calculation ease.
The figure formed by joining the midpoints of a square is always a square.
The outer square's side (s)=Diagonal of inner square.
Therefore the inner square's side (a) = \(\sqrt{2}\)a=s
a=\(\frac{s}{\sqrt{2}}\)
Area of the inner square = \(a^2\)=\(\frac{s^2}{2}\)...(1)
Now all the triangles formed in the figure are congruent according the the SSS rule.
Each triangle is also an isosceles triangle since 2 of the sides are s/2 and 1 remaining side(the side of the inner square) is \(\frac{s}{\sqrt{2}}\)
Area of an Isosceles Triangle= \(\frac{y}{4} \sqrt{4x^2-y^2}\)
where x=equal side and y=unequal side.
\(\frac{s}{4\sqrt{2}} \sqrt{4*\frac{s^2}{4} - \frac{s^2}{2}}\)
Are of each triangle = \(\frac{s^2}{8}\)...(2)
Shaded Area = (1) + (2)
=\(\frac{s^2}{2}\) + \(\frac{s^2}{8}\)
=\(\frac{5s^2}{8}\)
Unshaded Area = 3* (2)
=\(\frac{3s^2}{8}\)
\(\frac{Shaded Area}{Unshaded Area} = \frac{5}{3}\)
Answer: E