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# What is the area of quadrilateral ABCD?

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22 Jan 2011, 02:41
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Q) In quadrilateral ABCD shown in the figure , BC || AD, AB = 8 and BC = 16. What is the area of quadrilateral ABCD?

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23 Jan 2011, 09:06
Let me first redraw the figure so that it becomes more of a "to scale" figure and we can at least visualize how the actual quadrilateral will look like.

∆ABM is an isosceles right-angled triangle with angle M = 90°
Hence, BM = AM = AB/(√2) = 8/(√2) = 4√2

Hence, CN = AM = 4√2
And, AN = MC = BC - BM = (16 - 4√2)

∆CDN is a 30-60-90 triangle angle N = 90°
Hence, DN = (√3)CN = (√3)*(4√2) = 4√6

Now, area of the quadrilateral ABCD
= (area of ∆ABM) + (area of rectangle AMCN) + (area of ∆CDN)
$$= (\frac{1}{2}*AM*BM) + (AM*MC) + (\frac{1}{2}*CN*DN)$$

$$= (\frac{1}{2}*(4\sqrt{2})*(4\sqrt{2})) + ((4\sqrt{2})*(16 - 4\sqrt{2})) + (\frac{1}{2}*(4\sqrt{2})*(4\sqrt{6}))$$

$$= 16 + 64\sqrt{2} - 32 + 16\sqrt{3}$$

$$= 64\sqrt{2} - 16 + 16\sqrt{3}$$
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Anurag Mairal, Ph.D., MBA
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Last edited by Anurag@Gurome on 24 Jan 2011, 13:10, edited 1 time in total.
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24 Jan 2011, 13:06
The image is not available...can you please post the image again?

Thanks
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25 Sep 2012, 08:13
Anurag@Gurome wrote:
Let me first redraw the figure so that it becomes more of a "to scale" figure and we can at least visualize how the actual quadrilateral will look like.

∆ABM is an isosceles right-angled triangle with angle M = 90°
Hence, BM = AM = AB/(√2) = 8/(√2) = 4√2

Hence, CN = AM = 4√2
And, AN = MC = BC - BM = (16 - 4√2)

∆CDN is a 30-60-90 triangle angle N = 90°
Hence, DN = (√3)CN = (√3)*(4√2) = 4√6

Now, area of the quadrilateral ABCD
= (area of ∆ABM) + (area of rectangle AMCN) + (area of ∆CDN)
$$= (\frac{1}{2}*AM*BM) + (AM*MC) + (\frac{1}{2}*CN*DN)$$

$$= (\frac{1}{2}*(4\sqrt{2})*(4\sqrt{2})) + ((4\sqrt{2})*(16 - 4\sqrt{2})) + (\frac{1}{2}*(4\sqrt{2})*(4\sqrt{6}))$$

$$= 16 + 64\sqrt{2} - 32 + 16\sqrt{3}$$

$$= 64\sqrt{2} - 16 + 16\sqrt{3}$$

HI,

I am unable to understand how you remodeled the figure. According to your figure AB || CD. But in the question AD||BC.

Thanks!
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09 Oct 2012, 08:27
Ah I was using the guys incorrect diagram - thought it was part of the questionI disagree with the answer
In quadrilateral ABCD shown in the figure , BC || AD, AB = 8 and BC = 16. What is the area of quadrilateral ABCD?

if BC is parallel to AD then the shape is a parallelogram, therefore base x height - area. (using the M/N image)

We just need height AM, which comes from a 30.60.90 triangle. The sides are therefore x.sqrt(3)x.2x
as the hypotenuse is 16, 2x = 16 so the height of the parallelogram = 8sqrt(3)
Area is therefore 8x8sqrt(3) or 64sqrt(3)

Is this right?
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01 Feb 2014, 09:07
1
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MichelleSavina wrote:
Q) In quadrilateral ABCD shown in the figure , BC || AD, AB = 8 and BC = 16. What is the area of quadrilateral ABCD?

I agree with the picture posted in the reply with the solution, but the picture posted on the original question is dead wrong, if AB =8 then MB will be 8 sqrt(2) which is larger, this doesn't make any sense. Please refer to the picture below the actual one for a more precise experience

Cheers
J
Re: What is the area of quadrilateral ABCD?   [#permalink] 01 Feb 2014, 09:07
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