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Find the area of trapezium ABCD.

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Find the area of trapezium ABCD.  [#permalink]

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New post Updated on: 12 May 2016, 07:21
2
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A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

68% (02:34) correct 32% (02:56) wrong based on 88 sessions

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Find the area of trapezium ABCD.
Given angle ADB=45,BCN=30 and side AB=4 and BC=10.

A.5/2(9+5√3)
B.5/2*(13+5√3)
C. 13(13+2√3)
D.2/5(13+2√3)
E. None of these

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GEOMETRY1.jpg
GEOMETRY1.jpg [ 13.68 KiB | Viewed 2743 times ]


Originally posted by techiesam on 12 May 2016, 04:57.
Last edited by chetan2u on 12 May 2016, 07:21, edited 3 times in total.
Updated OA and edited the choices
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Re: Find the area of trapezium ABCD.  [#permalink]

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New post 12 May 2016, 06:06
techiesam wrote:
Find the area of trapezium ABCD.
Given angle ADB=45,BCN=30 and side AB=4 and BC=10.

A.5/2(9+5√3)
B.5√3/2(13+5√3)
C. 13(13+2√3)
D.2/5(13+2√3)
E. None of these
Attachments

GEOMETRY1.jpg
GEOMETRY1.jpg [ 13.68 KiB | Viewed 23 times ]


BC = 10
So, we know the hypotenuse of triangle BNC
This is a 30-60-90 degrees triangle
So, L(Side BN) = 1/2 x hypotenuse = 1/2 x 10 = 5 units
L(Side NC) = \sqrt{3}/2 x hypotenuse = \sqrt{3}/2 x 10 = 5\sqrt{3}

Since sides AB and CD are parallel and sides AM and BN are parallel, AB = MN = 4 units
Also, AM = BN = 5
Triangle AMD is a 45-45-90 triangle
which makes AM = MD = 5 units

Area of trapezium ABCD = 1/2 x sum of parallel side AB and CD x height
= 1/2 x (AB + MD + MN + NC) x 5
= 5/2 x (4 + 5 + 4 + 5\sqrt{3} ) = 5/2 x ( 13 + 5\sqrt{3})

E. None of these
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Re: Find the area of trapezium ABCD.  [#permalink]

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New post 12 May 2016, 06:18
I too got E as the answer. Is OA correct?

Experts please help.
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Re: Find the area of trapezium ABCD.  [#permalink]

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New post 12 May 2016, 06:30
2
techiesam wrote:
Find the area of trapezium ABCD.
Given angle ADB=45,BCN=30 and side AB=4 and BC=10.

A.5/2(9+5√3)
B.5√3/2(13+5√3)
C. 13(13+2√3)
D.2/5(13+2√3)
E. None of these



SEE att image
ans B

Always look at the ANGLES.. MORE than often they will be 30-60-90 or 45-45-90

the angle is 30 and opposite to the side will be 5..
so ans = \((4+4+5+5\sqrt{3})*\frac{5}{2}\) = \(\frac{5}{2} * (13+5\sqrt{3})\)
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Re: Find the area of trapezium ABCD.  [#permalink]

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New post 12 May 2016, 06:39
1
chetan2u wrote:
techiesam wrote:
Find the area of trapezium ABCD.
Given angle ADB=45,BCN=30 and side AB=4 and BC=10.

A.5/2(9+5√3)
B.5√3/2(13+5√3)
C. 13(13+2√3)
D.2/5(13+2√3)
E. None of these



SEE att image
ans B

Always look at the ANGLES.. MORE than often they will be 30-60-90 or 45-45-90


chetan2u, isn't the side opposite to 60 degrees be 5 \sqrt{3} and the side opposite to 30 degrees be 5?
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Re: Find the area of trapezium ABCD.  [#permalink]

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New post 12 May 2016, 06:43
Divyadisha wrote:
chetan2u wrote:
techiesam wrote:
Find the area of trapezium ABCD.
Given angle ADB=45,BCN=30 and side AB=4 and BC=10.

A.5/2(9+5√3)
B.5√3/2(13+5√3)
C. 13(13+2√3)
D.2/5(13+2√3)
E. None of these



SEE att image
ans B

Always look at the ANGLES.. MORE than often they will be 30-60-90 or 45-45-90




chetan2u, isn't the side opposite to 60 degrees be 5 \sqrt{3} and the side opposite to 30 degrees be 5?


Hi Divyadisha,
You are correct. My mistake the answer should be (13+5\sqrt{3}) * 5/2
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Re: Find the area of trapezium ABCD.  [#permalink]

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New post 12 May 2016, 07:01
Quote:
,
You are correct. My mistake the answer should be (13+5\sqrt{3}) * 5/2


So E.... not B right.?
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Find the area of trapezium ABCD.  [#permalink]

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New post 12 May 2016, 07:46
Divyadisha wrote:
chetan2u wrote:
techiesam wrote:
Find the area of trapezium ABCD.
Given angle ADB=45,BCN=30 and side AB=4 and BC=10.

A.5/2(9+5√3)
B.5√3/2(13+5√3)
C. 13(13+2√3)
D.2/5(13+2√3)
E. None of these



SEE att image
ans B

Always look at the ANGLES.. MORE than often they will be 30-60-90 or 45-45-90


chetan2u, isn't the side opposite to 60 degrees be 5 \sqrt{3} and the side opposite to 30 degrees be 5?


If a right triangle is 1,(sqrt)3 and 2, then triangle is 30-60-90 and side opposite to 30 is 1 , side opposite to 60 is (sqrt)3 and hypotenuse is 2.
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Re: Find the area of trapezium ABCD.  [#permalink]

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New post 12 May 2016, 09:29
Hi guys ,
The answer is B. I checked.CHECK THE 30-60-90 AGAIN.
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Re: Find the area of trapezium ABCD.  [#permalink]

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New post 12 May 2016, 09:35
techiesam wrote:
Hi guys ,
The answer is B. I checked.CHECK THE 30-60-90 AGAIN.


Hi,
your answer was not correct..
B has been edited to be the answer
\(\frac{5[square_root]3[}{square_root]/2}\) has been edited to \(\frac{5}{2}\)
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Re: Find the area of trapezium ABCD.  [#permalink]

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New post 05 Oct 2016, 07:58
techiesam wrote:
Find the area of trapezium ABCD.
Given angle ADB=45,BCN=30 and side AB=4 and BC=10.

A.5/2(9+5√3)
B.5/2*(13+5√3)
C. 13(13+2√3)
D.2/5(13+2√3)
E. None of these


good question...had to refresh my memory on geometry...
knowing the angles, we can find the BN and NC (applying 30-60-90 right triangle property). BN = 5, NC=5*sqrt(3)
since AM=BN, and since ADM is a 45-45-90 right triangle, we can deduce that DM is 5.
now.
Area of ADM = 5*5/2 = 25/2
Area of AMNB = 4*5 = 20
Area of BNC = 5*5*sqrt(3)/2 = 25*sqrt(3)/2
let's write 20 in the form of 40/2
25/2 +40/2 = 65/2
65/2 + 25*sqrt(3)/2
factor out a 5.
5*13/2 + 5*5*sqrt(3)/2
put 5/2 in the front
5/2 * (13+5*sqrt(3))

B
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Re: Find the area of trapezium ABCD.  [#permalink]

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New post 20 Apr 2018, 00:34
techiesam wrote:
Find the area of trapezium ABCD.
Given angle ADB=45,BCN=30 and side AB=4 and BC=10.

A.5/2(9+5√3)
B.5/2*(13+5√3)
C. 13(13+2√3)
D.2/5(13+2√3)
E. None of these


Answer B is correct choice, however 5/2*(13+5√3) implies that \(\frac{5}{2*(13+5√3)}\).
Answer should be \(\frac{5}{2}\)*(13+5√3)
Please correct that minor mistake.
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Re: Find the area of trapezium ABCD. &nbs [#permalink] 20 Apr 2018, 00:34
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