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Triangle/Parallelogram

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by ssmiles08 » Sun Sep 13, 2009 6:55 am
I got A for the first one.

The height is the same for both the triangle and the parallelogram.

Area of parallelogram = 12*h
Area of ABE = 1/2*(12 - EC)*h

area of ABE = 3/8(Area of parallelogram)

6h - EC*h = (3/8)(12)*h

6h - EC*h = (9/2)*h

EC*h = 1.5h (h's cancel on both sides and you are left with EC = 1.5 or 3/2)

2) AB is the radius. AD is also the radius.

AB = AD = BC = CD. They all make up a square. the angles are 90 degrees. BD is a bisector which would cut the angle to 45 degrees. IMO C.

-hope I am not missing anything.

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by rohan_vus » Sun Sep 13, 2009 7:03 am
i get D for first one and B for second one.
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by truplayer256 » Sun Sep 13, 2009 7:38 am
I also get D for the first one.

Area of parallelogram ABCD= B*H-->12*H

Area of triangle ABE=3/8(12*H)=36H/8=9H/2

Area of triangle AED= 12*H/2=6*H

Area of triangle ECD=12H-9H/2-6H=7.5H-6H=1.5H

(EC)*(H)/2=3H/2

2*EC*H=6H

EC=3
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by prindaroy » Sun Sep 13, 2009 7:54 am
D for first one. because AD = BC = 12. 2 * 3/8 = 3/4*12 = 9. Therefore 12 - 9 = 3. So 3 is the answer or D. For the second one then answer is 45 degrees, because 2x + 2x + 2x + (180 - 2x) = 360, x =45
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by shanrizvi » Sun Sep 13, 2009 11:14 pm
Ah silly geometry error. For some reason I treated this question like it was a trapezoid and was wondering if AD=BC. Thanks guys.
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by fruti_yum » Tue Sep 15, 2009 11:35 am
truplayer256 wrote:I also get D for the first one.

Area of parallelogram ABCD= B*H-->12*H

Area of triangle ABE=3/8(12*H)=36H/8=9H/2

Area of triangle AED= 12*H/2=6*H

Area of triangle ECD=12H-9H/2-6H=7.5H-6H=1.5H

(EC)*(H)/2=3H/2

2*EC*H=6H

EC=3
Trueplayer256.. how do you get area if AED = 6H?? how do u know that it's half the area of parallelogram?
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by ssmiles08 » Thu Sep 17, 2009 3:17 pm
fruti_yum wrote: Trueplayer256.. how do you get area if AED = 6H?? how do u know that it's half the area of parallelogram?
Yes...you guys are right on the first one. It is 3. My apologies. fruti_yum, You can do it the following way as well. Area of parallelogram = 12*h; Area of triangle ABE = 3/8(12*h) = 9/2*h

Area of a triangle is 1/2*b*h

9/2 = 1/2*b

b = 9

the Base is BE. So EC = 12 - 9 = 3 (D)

2) I guess you can get B for the second one...But I am not sure. AB = BC = AC which would make the triangle ABC equilateral. x = 30.

I was under the assumption that AB = BC = AD = CD which would make it a square (90degrees) and the diagonals of a square would make cut the angle in half which would make x = 45. I am not sure what to make of this question. Suggestions?
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by viju9162 » Thu Sep 17, 2009 8:35 pm
Isn't answer "B" for the second one?

AB = BC ( AB is radius from the figure) ...Join B and C ..

So in triangle ABC, AB = BC = AC ..because AC is also the radius .. therefore each angle is 60 degrees..

Similarly, in triangle ACD, each angle constitutes 60 degrees..

therefore, consider the triangle ABD. where Angle A = 60+60 = 120 degrees..

hence, X + X + 120 = 180
2x = 60
X = 30 degrees...

Please let me know if I am wrong here...

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