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