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In parallelogram ABCD, point E is the midpoint of side AD and point F

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In parallelogram ABCD, point E is the midpoint of side AD and point F  [#permalink]

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New post Updated on: 28 Aug 2018, 04:21
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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.

Originally posted by harshitagarg on 27 Aug 2018, 11:30.
Last edited by Bunuel on 28 Aug 2018, 04:21, edited 1 time in total.
Renamed the topic and edited the question.
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Re: In parallelogram ABCD, point E is the midpoint of side AD and point F  [#permalink]

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New post 27 Aug 2018, 11:34
The explanation provided as per source is:
This is a Value Data Sufficiency question, so use the Pieces of the Puzzle approach to assess the question. Begin by determining “What is known” from the question stem and “What is needed” from the statements to answer the question. Determine “What is known.” The question asks for the area of triangle DGH. Draw and label parallelogram ABCD such that line segment AD represents the base of the parallelogram. Add points E, F, G, and H to the figure, so that line segment GD represents the base of triangle DGH, which occupies the parallelogram’s lower right corner. Whether or not the parallelogram is rectangular, the base of triangle DGH is one-fourth the base of parallelogram ABCD, and the height of triangle DGH is one-fourth the height of parallelogram ABCD. If bh represents the area of parallelogram ABCD, then the area of triangle DGH is , or . Now, determine “What is needed.” In order to answer the question, the statement(s) must provide information sufficient to determine either the area of parallelogram ABCD or the base and height of triangle DGH. Evaluate the statements one at a time.

Evaluate Statement (1). If the area of parallelogram ABCD is 96 square units, then the area of triangle DGH is square units. Statement (1) is sufficient to answer the question, so write down AD.

Now, evaluate Statement (2). If the parallelogram is a rectangle with dimensions 8 by 12, then the base and height of triangle DGH are 2 and 3, and the area of the triangle is square units. However, if parallelogram ABCD is not a rectangle, then its sides are not perpendicular. In this case, the measures of the parallelogram’s interior angles are unknown, so it is impossible to determine the parallelogram’s height, and impossible to determine the triangle’s height. Statement (2) is insufficient to answer the question, so eliminate choice D. The correct answer is choice A.



CAN ANYONE CLARIFY WHY IN OPTION (A) IS THE HEIGHT OF THE PARALLELOGRAM 1/4 EVEN IF IT IS NOT A RECTANGLE?
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Re: In parallelogram ABCD, point E is the midpoint of side AD and point F  [#permalink]

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New post 27 Aug 2018, 21:09
I followed the approach as described here. Construct a //gm and then use the conditions to draw triangles DGH, DEF and DAC. Area of DAC is half that of the //gm since a diagonal divides the //gm in two equal halves. Now consider DAC and DEF which are similar triangles. Area of triangle DEF is half that of DAC (Since all corresponding sides are half that of DAC and all corresponding angles are equal) and similarly area of triangle DGH is half that of DEF. So, Area of //gm is sufficient to answer this question. Hence, A is the answer.
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In parallelogram ABCD, point E is the midpoint of side AD and point F  [#permalink]

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New post 02 Sep 2018, 10:43
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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
Parallelogram ABCD.PNG [ 4.44 KiB | Viewed 240 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
statement 2.PNG [ 8.92 KiB | Viewed 240 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.
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In parallelogram ABCD, point E is the midpoint of side AD and point F &nbs [#permalink] 02 Sep 2018, 10:43
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