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10 Nov 2011, 08:54
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Q1 .In this diagram, AB=AC. and BD=CD. Which of the following statements is true?
A. BE=EC
B. AD is perpendicular to BC
C. triangle BDE and CDE are congruent
D. angle ABD equal angle ACD
E. All of these
Figure: SEE ATTACHMENT
Please give explanation for each of these why something is not true
Q2. ABCD is a rectangle; the diagonals AC and BD intersect at E. Which of the following statements is not necessarily true?
A. AE=BE
B. Angle AEB equals angle CED
C. AE is perpendicular to BD
D. Triangles AED and AEB are equal in area
E. Angle BAC equals angle BDC
Pls give full explanation
A. BE=EC
B. AD is perpendicular to BC
C. triangle BDE and CDE are congruent
D. angle ABD equal angle ACD
E. All of these
Figure: SEE ATTACHMENT
Please give explanation for each of these why something is not true
Q2. ABCD is a rectangle; the diagonals AC and BD intersect at E. Which of the following statements is not necessarily true?
A. AE=BE
B. Angle AEB equals angle CED
C. AE is perpendicular to BD
D. Triangles AED and AEB are equal in area
E. Angle BAC equals angle BDC
Pls give full explanation
Attachments
File comment: ATTACHMENT FOR QUESTION 1
ar.JPG [ 14.25 KiB | Viewed 4947 times ]
11 Nov 2011, 17:02
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E / C
Q1.
AB=AC, BD=CD
-> ABC, DBC are isosceles triangles respectively.
-> Angle ABD = Angle ACD
-> Triangle ABD = Triangle ACD
-> Angle BAD = Angle CAD
-> BC is perpendicular to AD (because Triangle ABC is isosceles.)
-> Triangel ABE=ACE, DBE=BCE,
and Angle AEB=AEC=DEB=DEC=90
A. BE=EC <- true, Triangel AEB=AEC
B. AD is perpendicular to BC <- true
C. Triangel BDE=CDE <- true
D. Angle ABD=ACD <- true
E. all true <- true
answer : E
Q2.
Rect.ABCD -> AB=CD, AD=BC,
and, according to the symmetricalness of rect., AE=BE=CE=DE
-> tri. AEB=CED : isosceles,
tri. BEC=AED : isosceles.
A. AE=BE <- true
B. ang. AEB=CED <- true; vertical angles.
C. AE perpend. BD <- not necessarilly; true only when the rect. is a square.
D. tri. AED and AEB are equal in area <- true; these 2 triangles are 2 parts of tri. ABD. When you assume BE and DE are the bases of each tri., they are sharing the height. So, they have the same height and the same length of the base, and thus the same area.
E. ang. BAC=BDC <- true; tri. EAB=EDC.
answer : C
Q1.
AB=AC, BD=CD
-> ABC, DBC are isosceles triangles respectively.
-> Angle ABD = Angle ACD
-> Triangle ABD = Triangle ACD
-> Angle BAD = Angle CAD
-> BC is perpendicular to AD (because Triangle ABC is isosceles.)
-> Triangel ABE=ACE, DBE=BCE,
and Angle AEB=AEC=DEB=DEC=90
A. BE=EC <- true, Triangel AEB=AEC
B. AD is perpendicular to BC <- true
C. Triangel BDE=CDE <- true
D. Angle ABD=ACD <- true
E. all true <- true
answer : E
Q2.
Rect.ABCD -> AB=CD, AD=BC,
and, according to the symmetricalness of rect., AE=BE=CE=DE
-> tri. AEB=CED : isosceles,
tri. BEC=AED : isosceles.
A. AE=BE <- true
B. ang. AEB=CED <- true; vertical angles.
C. AE perpend. BD <- not necessarilly; true only when the rect. is a square.
D. tri. AED and AEB are equal in area <- true; these 2 triangles are 2 parts of tri. ABD. When you assume BE and DE are the bases of each tri., they are sharing the height. So, they have the same height and the same length of the base, and thus the same area.
E. ang. BAC=BDC <- true; tri. EAB=EDC.
answer : C
