+0  
 
0
56
8
avatar

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.

 

Guest Oct 22, 2018
 #1
avatar+406 
+8

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.

KnockOut  Oct 22, 2018
 #4
avatar+90968 
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+13567 
+2

Thanx....that is cool!

('butt' a PIA)   cheeky

ElectricPavlov  Oct 22, 2018
 #7
avatar+90968 
+1

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

 

 

cool  cool cool

CPhill  Oct 22, 2018
 #2
avatar+13567 
+1

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

ElectricPavlov  Oct 22, 2018
 #3
avatar+90968 
+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

CPhill  Oct 22, 2018
 #5
avatar+90968 
+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

CPhill  Oct 22, 2018
 #8
avatar+90968 
+2

I believe I answered the last two, here :

 

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

 

 

cool cool cool

CPhill  Oct 22, 2018

36 Online Users

avatar
avatar
avatar
avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.