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by Rahul@gurome » Tue Nov 23, 2010 12:28 am
Let BM = x. So MC = 2x.
Or BC = x + 2x = 3x = AD.
Or BM/AD is x/3x = 1/3.
Now triangle BGM and AGD are similar.
So (Area of BGM)/(Area of AGD) = (BM)^2/(AD)^2 = 1/9.
Or Area of AGD is (9* area of BGM) = 9*1 = 9.
Next consider triangles ABM and BDC.
Now their heights will be same because they are between same two parallel lines.
So their areas will correspond to their bases which are in the ratio 1 : 3.
Hence the areas of ABM and BDC will be in the ratio 1 : 3.
Let the area of triangle ABG be m and the area of quadrilateral GMCD be n.
So m+1 = area of ABM and n+1 = area of BDC.
Or m+1 = 1/3*(n+1). (equation 1)
Also since area of ABD = area of BDC, m+9 = n+1. (equation 2).
From the two equations we get m = 3 and n = 11.
So area of parallelogram is 3+9+11+1 = 24.
The correct answer is D.
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by goyalsau » Tue Nov 23, 2010 1:34 am
Image


Rahul Thanks for the Above explanation, Can we do the problem Like this,

We know the Area of Traingle BGM is 1

= 1/2 ( BM * Height of traingle )

Now BM * Height of Traingle Must be equal to 2

Only then the Area of Triangle will be 1 .

IF we assume that Length of BM is 2 and Height Lets say GX is 1

We know Triangles BGM and ADG is similar

BM / GX = AD / GY { Y is the point of line AD }

2/1 = 6 / GY

GY = 3

We know formula of parallelogram is Base and Height

Total Height is 1 + 3 = 4
Base is 6

answer is 24

I am not sure can we do it this way for Not Because i assumed everything from like value of BM , Value of GX But Got the same answer ...
Please tell me Can we do it this way............
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by goyalsau » Tue Nov 23, 2010 1:41 am
Rahul@gurome wrote:Let BM = x. So MC = 2x.
Or BC = x + 2x = 3x = AD.
Or BM/AD is x/3x = 1/3.
Now triangle BGM and AGD are similar.
So (Area of BGM)/(Area of AGD) = (BM)^2/(AD)^2 = 1/9. { why squaring }

Or Area of AGD is (9* area of BGM) = 9*1 = 9.

Next consider triangles ABM and BDC.

Now their heights will be same because they are between same two parallel lines.

So their areas will correspond to their bases which are in the ratio 1 : 3.

Hence the areas of ABM and BDC will be in the ratio 1 : 3.

Let the area of triangle ABG be m and the area of quadrilateral GMCD be n.

So m+1 = area of ABM and n+1 = area of BDC.
Or m+1 = 1/3*(n+1). (equation 1)
Also since area of ABD = area of BDC, m+9 = n+1. (equation 2).
From the two equations we get m = 3 and n = 11.
So area of parallelogram is 3+9+11+1 = 24.
The correct answer is D.
Saurabh Goyal
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by frank1 » Tue Nov 23, 2010 2:28 am
goyalsau wrote:
BM / GX = AD / GY { Y is the point of line AD }

2/1 = 6 / GY
i understood every thing except this
we have considered BM=1 and MC=2
thats mean BC=3
as it is parallelogram AD should be 3 how is it 6? (i dont see any other relations)
Bit confused...

thanks
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by Rahul@gurome » Tue Nov 23, 2010 2:54 am
In similar triangles, the ratio of areas of two triangles is the square of the ratio of corresponiding sides.
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by goyalsau » Tue Nov 23, 2010 4:29 am
frank1 wrote:
goyalsau wrote:
BM / GX = AD / GY { Y is the point of line AD }

2/1 = 6 / GY
i understood every thing except this
we have considered BM=1 and MC=2
thats mean BC=3
as it is parallelogram AD should be 3 how is it 6? (i dont see any other relations)
Bit confused...

thanks
I am considering BM is equal to 2 and MC is equal to 4

Then AD is equal to 6
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by goyalsau » Tue Nov 23, 2010 4:32 am
Rahul@gurome wrote:In similar triangles, the ratio of areas of two triangles is the square of the ratio of corresponiding sides.
This means If we have two corresponding sides of similar triangles we can determine the ratio of the Areas of two triangles

and this will not be the actual Area it will only be the ratio of AREAS.
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by GMATGuruNY » Tue Nov 23, 2010 5:12 am
goyalsau wrote:Image
I solved this very quickly by guessing and checking.

Let's try BM=1. Then MC = 2 and BC=3. This means that AD=3.
Triangles BMG and AGD are similar because all their corresponding angles are equal.
Since BM=1 and AD=3, the ratio of the heights of triangles BMG and AGD is 1:3.
Since the area of triangle BMG=1 and BM=1, h = area/(1/2*b) = 1(1/2*1) = 2.
Since the ratio of the heights of the two triangles is 1:3, the height of triangle AGD = 3*2 = 6.
Thus, the height of parallelogram ABCD = height of BMG + height of AGD = 2+6 = 8.
Area of parallelogram ABCD = bh = 3*8 = 24.

The correct answer is D.
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