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.

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

Find the area of trapezium ABCD.

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards