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**+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.

KnockOut Oct 22, 2018

#2**+1 **

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

ElectricPavlov Oct 22, 2018

#3**+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

CPhill Oct 22, 2018

#5**+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

CPhill Oct 22, 2018