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Triangles

by heshamelaziry » Sun Nov 29, 2009 6:55 pm
Could you expalin why the claimed triangles are similar? the reason does not withstand because the 2 right angles are not shared. In fact the two triangles do not share any angles.

Could provide a simpler way of solving this ?
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by papgust » Sun Nov 29, 2009 9:25 pm
Firstly, in triangle BCE and ABD, angle CEB and angle DAB are 90 degrees.

Secondly, since BC || AD, BD is a traversal across BC and AD. This means that alternate angles EBC and BDA are equal (This is a rule that applies to a traversal crossing 2 or more parallel lines)

In a triangle, if 2 angles are equal then the third angle must also be equal. We have proved that
i. angle CEB = angle DAB
ii. angle EBC = angle BDA

Therefore, angle BCE = angle ABD. This means that triangles BCE and ABD are similar triangles (AAA similarity)

Hope this is what you are expecting.
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by heshamelaziry » Sun Nov 29, 2009 9:35 pm
papgust wrote:Firstly, in triangle BCE and ABD, angle CEB and angle DAB are 90 degrees.

Secondly, since BC || AD, BD is a traversal across BC and AD. This means that alternate angles areEBC and BDA equal (This is a rule that applies to a traversal crossing 2 or more parallel lines)

In a triangle, if 2 angles are equal then the third angle must also be equal. We have proved that
i. angle CEB = angle DAB
ii. angle EBC = angle BDA

Therefore, angle BCE = angle ABD. This means that triangles BCE and ABD are similar triangles (AAA similarity)

Hope this is what you are expecting.
I understood the point that the lines are parallel but don't get why EBC and BDA equal ? Is it more difficult to realize that these two rtiangles are similar, in this problem, than other problems ?
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by papgust » Sun Nov 29, 2009 10:04 pm
This is because "When 2 or more parallel line are cut by a traversal, the alternate interior angles are equal".
Here the 2 alternate interior angles are 1. angle EBC and 2. angle BDA. Hence these 2 angles are equal.

I suggest that you refresh the geometry basics and then start working on these problems. You will then be able to solve with ease.
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by thephoenix » Mon Nov 30, 2009 12:03 am
heshamelaziry wrote:Could you expalin why the claimed triangles are similar? the reason does not withstand because the 2 right angles are not shared. In fact the two triangles do not share any angles.

Could provide a simpler way of solving this ?
BD^2=AB^2+ AD^2(AB=15,AD=20)

SOLVING BD=25

FOR TRI BCD
AREA=1/2(BC*CD)=1/2(CE*BD)

SOLVING CE=12

BE^2=BC^2-CE^2=20^2-12^2=256

BE=16

NOW AREA OF TRI BCE is 1/2(CE*BE)=1/2(12*16)=96
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by Gmatter2.0 » Mon Nov 30, 2009 7:20 pm
Consider Triangles BCD and CDE these two Triangles are Similar.

Because both the Triangles have two Angles and a Side in common.

Now let ED=x
base/base=Hyp/Hyp
x/15=15/25=9
Hence base ED=9.

Now base/base=height/height
9/15=h/20
h=12

Area of BCD=150
Area of CDE =1/2 9*12=54

Hence Area of the Shaded region=150-54=96.

Choice 3 is correct...
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