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Geometry Problem HELPPPPPPPPPP urgentttt NEWWWWWW

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Geometry Problem HELPPPPPPPPPP urgentttt NEWWWWWW [#permalink] New post 11 Nov 2011, 23:18
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 2 sessions
Q1-In ABC, angle AFB=angle AEC=90. If BF=CE, which of the following must be true?

A)AB=AC
B)AE=CE
C)BF=AF
D)AB=BC
E)none

Q2-In the figure, AD||BC. AB=5 cm and AD=7 cm. If angle BCD=1/2 angle BAD, what is the length of BC in cm?
A)10
B)12
C)17
D)19
E) cannot be determined

Q3-In the figure below, D is the midpoint of AB and E is the midpoint of AC. What is the relation between the area of BFC and the area of ADFE?

A)Area BFC> area ADFE
B)Area BFC= area ADFE
C)Area BFC< area ADFE
D)area ADFE= 1.5 times Area BFC
E) either A or C

please give full explanation and check the figure for reference
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Re: Geometry Problem HELPPPPPPPPPP urgentttt NEWWWWWW [#permalink] New post 21 Nov 2011, 23:37
1) A - since the one of the angles is equivalent and one of the legs are equivalent as well then the entire triangle is equivalent.
2) E (I am less sure on this but think that is correct) - there just because AD and BC are parallel gives you no info on the other angles BCD and BAD can have a relationship but the other two angles could be any number of angles (with the sole constraint that the total figure must sum to 360). So it cannot be determined based on what is given.
3) I am really not sure on this one. I would assume its E - since we do not know where F is in relation to the midpoint. If its way below its greater if its near it it may be less.

Hope this helps.
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Re: Geometry Problem HELPPPPPPPPPP urgentttt NEWWWWWW [#permalink] New post 22 Nov 2011, 02:20
1. 2 angles equal, hence 2 triangles similar. Thus, sides also equal -->
ans A
2. From the angle AH we can drop a bisector to the side BC. one of 2 anlges HAD will be x = angle DCB. As AD||BC and angle x=x, A||CD. Angle DCB = angle AHB = x --> AHB is isosceless and BH = AH = 5. AD =HC=7, hence the whole BC is 7+5 = 12
ans B
3. Lets try this for an equilateral triangle. F would be a centre of inscribed circle with radius a^2 sqrt(3) / 6. Using this we can calculate the area of the bottom triangle CFB and the area of the upper triangles ADE and AFE comprising figure ADFE. the resulting ratio CFB/ADFE = 1/2, therefore
ans C.
Re: Geometry Problem HELPPPPPPPPPP urgentttt NEWWWWWW   [#permalink] 22 Nov 2011, 02:20
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Geometry Problem HELPPPPPPPPPP urgentttt NEWWWWWW

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