Find the area of trapezium ABCD.

Question

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

ronny123 · Answer

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

Divyadisha · Answer

I too got E as the answer. Is OA correct?

Experts please help.

chetan2u · Answer

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})

Divyadisha · Answer

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

chetan2u · Answer

Hi  Divyadisha ,
You are correct. My mistake the answer should be (13+5\sqrt{3}) * 5/2

prabhakar09 · Answer

So E.... not B right.?

prashant212 · Answer

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.

techiesam · Answer

Hi guys ,
The answer is B. I checked.CHECK THE 30-60-90 AGAIN.

chetan2u · Answer

Hi,
your answer was not correct..
B has been edited to be the answer
 \frac{5\sqrt{3}}{2}  has been edited to  \frac{5}{2}

mvictor · Answer

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

Tulkin987 · Answer

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.

Find the area of trapezium ABCD.

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Find the area of trapezium ABCD.

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