harshitagarg wrote:

In parallelogram ABCD, point E is the midpoint of side AD and point F is the midpoint of side CD. If point G is the midpoint of line segment ED and point H is the midpoint of line segment DF, what is the area of triangle DGH ?

(1) Parallelogram ABCD has an area of 96.

(2) The parallelogram has sides of lengths 8 and 12.

OA: A

Given : ABCD is parallelogram, \(DG = \frac{1}{4} AD\) and \(DH = \frac{1}{4} DC\)

Attachment:

Parallelogram ABCD.PNG [ 4.44 KiB | Viewed 485 times ]
Joining AC ,we get \(\triangle\) ADC.

Area of \(\triangle\) ADC = \(\frac{1}{2}\) Area of parallelogram ABCD (Diagonal of parallelogram divides it into two triangles of equal area.)

As GH divides the 2 sides of the triangle ADC in same ratio (\(DG = \frac{1}{4} AD\) and \(DH = \frac{1}{4} DC\)),GH will be parallel to AC.

(if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.)

\(\triangle\) GDH ~ \(\triangle\) ADC (AAA triangle similarity)

\(\frac{Area \quad \triangle GDH}{Area \quad \triangle ADC}=\frac{DG^2}{AD^2}\)

( If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.)

\(\frac{Area \quad \triangle GDH}{Area \quad \triangle ADC}=\frac{1}{16}\) (As \(DG = \frac{1}{4} AD\))

Area\(\quad \triangle\)GDH \(=\frac{1}{2}*\frac{1}{16}\) Area parallelogram ABCD (As Area of \(\triangle\) ADC = \(\frac{1}{2}\) Area of parallelogram ABCD)

Area\(\quad \triangle\)GDH \(=\frac{1}{32}\) Area parallelogram ABCD

Question Stem reduces to

What is area of parallelogram ABCD?(1) Parallelogram ABCD has an area of 96.

Statement 1 gives the area of Parallelogram ABCD, So Statement 1 alone is sufficient.

(2) The parallelogram has sides of lengths 8 and 12.

Let us consider ABCD as rectangle, then Area of Parallelogram ABCD \(= 8*12 = 96\)

Attachment:

statement 2.PNG [ 8.92 KiB | Viewed 485 times ]
Now consider \(\angle ADC\) as \(60^{\circ}\), and \(AD =8\) and \(CD = 12\), then Height(h) \(= \frac{\sqrt{3}AD}{2} = 4\sqrt{3}\) (Using ratio of \(30^{\circ}-60^{\circ}-90^{\circ} \triangle\))

Area of Parallelogram ABCD ( \(\angle ADC = 60^{\circ}\)) \(= CD * Height(h) = 12 * 4\sqrt{3} = 48\sqrt{3}\)

As there is no unique value of area of parallelogram ABCD, Statement 2 alone is not sufficient.