In the given figure, ABCD is a parallelogram and E, F, G and H are mid

I approached this problem by proving that each of the smaller triangle is congruent to the other.
1: Tr(AHE) and Tr(CFG)
> AE = CG (since mid points of opp sides of parallelogram, so they are equal)
> AH = CF (same)
> Angle A = Angle C (opp angles of parallelogram are equal)
Thus Tr(AHE) and Tr(CFG) are congruent.

Similarly we can prove that Tr(BFE) and Tr(DHG) are congruent.
Now draw a line from E to G, and consider the triangles EHG and EFG. We can prove that these triangles are also congruent. (Common side and other 2 sides are equal bec of the congruence of other triangles proved above).

In all, we get all the smaller triangles congruent to each other.
Thus the asked ratio is obtained simply by counting the no of triangles that are shaded and that are not shaded.

5 portions are shaded and 3 are not. Thus 5:3.
Ans : E

IMO
The answer is E.
if the area of the shaded triangle is x , we can divide the figure in the same area triangles. So, it becomes 5x/3x which is 5/3.

Refer the triangle GCF. The area is 1/2 * Base * Height. Which is 1/2 * GC * x (let x be the height).

Note that this is the same height for the triangle HDG. Also since DG=GC (mid point), the two triangles have equal area. Lets call it A.

Now consider the triangle HGF (join F & H). The area of that triangle can similarly be written as

1/2*x*2GC. (as CD=FH & CD=2CG).

Which is equal to 2A.

Similarly for the other half.

Adding, we have the shaded portion as 5A & the unshaded portion is 3A, hence the answer is 5:3. That is E.

VERITAS PREP OFFICIAL SOLUTION

There are many ways to do this question but we will look at the method using similar triangles (obviously!).

Assume the area of the parallelogram is 8P. In a parallelogram, the lengths of opposite sides are the same. The two triangles formed by the diagonal and two sides are similar by SSS and the ratio of their sides is 1. So they will have equal areas of 4P each (look at the figures in second row below)

Now look at the original figure.

HE is formed by joining the mid-points of AD and AB. So AH/AD = AE/AB = 1/2 and included angle A is common. Hence by SAS rule, triangle AHE is similar to triangle ADB. If the ratio of sides is 1/2, ratio of areas will be 1/4.

Since area of triangle ADB is 4P, area of AHE is P. We have 3 such triangles, AHE, DHG and CGF which are not shaded so the area of these three triangles together will be 3P.

The total area of parallelogram is 8P and the unshaded region is 3P. So the shaded region must be 5P.

Hence, area of shaded region : Area of unshaded region = 5:3

Answer (E)

not calculating anything, we can eliminate A, B, and C right away.
between D & E, I chose to go with D.
we can aprox. divide everything into 8 small triangles. 5 shaded and 3 un-shaded.
but since the areas of all triangles are not the same, then it is not possible to be 5:3.
therefore D makes more sense.

Hi Sherlocked.

My mistake. The answer is 5:3. My overworked brain has let me do another simple mistake.

Kind Regards,

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

note all the smaller triangles need not to be congruent, it depends on the angles subtended by sides of the ||gram.
however, all the smaller triangles have to have the same area.

Varignon's Theorem:
The midpoints of the sides of an arbitrary quadrangle form a parallelogram.
If the quadrangle is convex or reentrant, i.e. not a crossing quadrangle, then the area of the parallelogram is half the area of the quadrangle.

Let \(x\) be the area of the parallelogram \(ABCD\)
Then
Area of parallelogram \(EFGH\,=\,\frac{x}{2}\)
Area of each triangle \(=\,\frac{1}{4}\,(\frac{x}{2})\)

Area of shaded region:Area of un-shaded region
\(=\) [Area of parallelogram \(EFGH\) + Area of \(\triangle\)EFB]\(:\)[Area of \(\triangle\)s \(AEH\), \(HGD\), \(FGC\)]
\(=\) [\(\frac{x}{2}\,+\,\frac{x}{8}\)]\(:\)[\(3\,*\,\frac{x}{8}\)]
\(=\) \(\frac{5x}{8}\)\(:\)\(\frac{3x}{8}\)
\(=\) \(\,5:3\)

Answer E

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