Question

Suppose, ABCD is a rectangle and that E is a point on CD. Let x be the area of AED, y be the area of BCE and z be the area of ABE and suppose y^2 = xz. What is the value of DE/EC ?

Answer 1

question looks incomplete ?

Answer 2

yes diagram

Answer 3

it takes so much time to load openstudy page.so coudnt able to post properly

Answer 4

e is point on cd.what is the value of de/bc?

Answer 5

I have modified the question, see if it looks correct now ?

Answer 6

\[\large y = \dfrac{EC\times BC}{2}\]
\[\large x = \dfrac{DE\times BC}{2}\]
\[\large z = \dfrac{(DE+EC)\times BC}{2}\]

Answer 7

plug them in the given relation : \(\large y^2 = xz\)

Answer 8

\[\large \left(\dfrac{EC\times BC}{2}\right)^2 = \left(\dfrac{DE\times BC}{2}\right)\left(\dfrac{(DE+EC)\times BC}{2}\right)\]

Answer 9

canceling BC^2/4 both sides gives you :


\[\large EC^2 = DE(DE+EC)\]

Answer 10

\[\large EC^2 = DE^2+DE\times EC\]

\[\large 1 = \dfrac{DE^2}{EC^2}+\dfrac{DE}{ EC}\]

\[\large 1 = \left(\dfrac{DE}{EC}\right)^2+\dfrac{DE}{ EC}\]

Answer 11

its a quadratic, you can solve it using quadratic formula

Answer 12

\[\large 1 = x^2 + x\]

\[\large x^2 + x-1=0\]

\[\large x = \dfrac{-1\pm \sqrt{1^2 - 4(1)(-1)}}{2(1)}\]

\[\large = \dfrac{-1\pm \sqrt{5}}{2}\]

Answer 13

hey how u modified my question ?

Answer 14

moderators can modify/delete questions :)

Answer 15

hmmmmmmmmmmm.........great

Answer 16

since \(\large \dfrac{DE}{EC}\) is a ratio of distances, it cannot be negative,
so the ratio has to be : \(\large \dfrac{-1+\sqrt{5}}{2}\)

Answer 17

see if that makes more or less sense..

Answer 18

yes.how u are saying ec*bc/2

Answer 19

\(y\) is the area of right most triangle :
base = \(EC\)
height = \(BC\)

Answer 20

u rocked!

Answer 21

r u a maths professor.

Answer 22

lol no, that title was just a joke :)

Answer 23

why u are giving medal to me.i am the questionnaire.:-)

Answer 24

cos i notice that you always participate actively in discussions about ur question and ask back good good questions :) thats the reason i gave u full solution

Answer 25

no,no really i asked ru.or u should loved the maths most than any other.please tell me some tips to get threw.or suggest me some books

Answer 26

ughh..my page is not loading.thats why it takes time for my reply

Answer 27

I think, taking time to understand fully below things help you in approaching any problem in geometry confidently :

1) how/why similar triangle ratios work (AA/SAS/SSS/AAS congruence and similarity applications)

2) making sense of all the area/volume formulas (triangles, six types of quadrilaterals : [parallelogram, rhombus, rectangle, square, trapezoid, kite], prisms, pyramids etc)

3) full understanding of four centers of triangle : centroid, incenter, circumcenter, orthocenter

Answer 28

50% of medium-to-hard problems in geometry involve dealing with similar/congruent triangles !

Answer 29

circles and inscribed polygons is also an interesting topic worth spending time

Answer 30

ya came through these topic formulae,definition n all.but still

Answer 31

that `but still...` will disappear if we try to derive everything on our own and stop trusting all the formulas at their face value :P

Answer 32

when u have time, try to derive below well known formula for area of triangle :



\(\large \text{Area} = \dfrac{b*h}{2}\)