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1. We know Angle B=Angle C, BE=5, CE=4  If the area of AEB=50 what is the area of CED?

2. The area of square ABCD is 36. E is a point on line CD such that CE=2ED. The extensions of line AE and line BE intersect at F.  Find the area of DEF.

 

 

3. In the diagram below, we have angle ABC=angle CAB=angle DEB=angle BDE. Given that AE=21 and ED=27, find BD.
 

 

4. In the diagram below, we have angle ABC=angle CAB=angle DEB=angle BDE. Given that AE=21 and ED=27, find CA.

 

 Oct 22, 2018
 #1
avatar+624 
+9

I'd love to answer these except for the fact that they're under moderation right now...It says that moderation is at 74%, then it goes back down to 8%, and so on. I've been seeing this on almost every graph I see here for the past two days...weird.

 Oct 22, 2018
 #4
avatar+98130 
0

Note, Knockout....if you right click on the pic  and select  "Copy Image Address"....you can paste this address in a new browser window  and view the image

 

 

cool cool cool

CPhill  Oct 22, 2018
 #6
avatar+17331 
+2

Thanx....that is cool!

('butt' a PIA)   cheeky

ElectricPavlov  Oct 22, 2018
 #7
avatar+98130 
+1

HAHAHA, EP!!!!!......hectictar gave me that tip...!!!

 

 

cool  cool cool

CPhill  Oct 22, 2018
 #2
avatar+17331 
+1

....but apparently the moderators can see the submissions....cphil has answered several that still say 'waiting for moderator'

 Oct 22, 2018
 #3
avatar+98130 
+2

First one

 

Triangle AEB  is similar to  triangle DEC

 

Note that   EC /EB  =    4/5   so...this is the scale factor of triangle DEC to triangle AEB

 

So......the area  of triangle DEC  =

 

[ scale factor  of  triangle DEC to triangle AEB ]^ 2 *  area of triangle AEB  =

 

[ 4/5] ^2  *  50  =

 

[ 16 / 25 ] * 50  =

 

16 *  2  =

 

32

 

cool cool cool

 Oct 22, 2018
 #5
avatar+98130 
+2

Second one

 

Because DC is parallel o AB, then triangle BAF  is similar to triangle EDF

 

And   since   the side of the square =  sqrt (36)  = 6

Then DE  + EC  = 6

DE + 2DE  = 6

3DE = 6

DE  = 2

 

So 

 

[AB ] / [AD + DF ]  =  DE / DF

 

6 / [ 6 + DF ]  =  2 / [  DF ]     cross-multiply

 

6 [  DF ] = 2 [ 6 + DF ]

 

 6DF  =  12 + 2DF    subtract 2DF from  both sides

 

4DF  = 12    divide both sides by 4

 

DF  = 3  =  height of triangle    DEF

 

So  area of DEF  = (1/2)DE * DF  =  (1/2) * 2 * 3   =  3 units^2

 

 

cool cool cool

 Oct 22, 2018
 #8
avatar+98130 
+2

I believe I answered the last two, here :

 

https://web2.0calc.com/questions/more-questions

 

 

cool cool cool

 Oct 22, 2018

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