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**msbiljanica****Member**- Registered: 2012-01-18
- Posts: 21

Rectangle ABCD, along an AE that is smaller or larger than the page AB, construct an Rectangle AEFG that has the same surface area as the rectangle ABCD, only unmarked ruler and conifer are allowed, preferably in GeoGebra

theme - inverted proportionality solved by geometric process

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 8,914

msbiljanica

(1) Create coordinate axes, with the origin at A, x axis AB, and y axis AD.

(2) Calculate AB x BC = constant, the required area.

(3) Choose a point E, on AB.

(4) Draw EF perpendicular to AE. Let F have coordinates (x,y)

(5) We want AE x EF = constant = x.y, so F lies on the (rectangular) hyperbola xy = constant.

(6) Construct this hyperbola and fix F on it.

Hope that helps,

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**msbiljanica****Member**- Registered: 2012-01-18
- Posts: 21

only ruler and compass , no hyperbola

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 8,914

hi msbiljanica

Sorry, you did say that and I went off on a different route. I have found a way but the construction is complicated. It relies on a property of chords on a circle which is shown in this post:

http://www.mathisfunforum.com/viewtopic … 92#p368592

If AJ is a diameter of a circle, centre I, and a line up from that diameter at B to cut the circle at F then the theorem states

AB x BJ = BF x BF so BF is the side of an equal area square. BFGH.

Then choose your point E ( I have got E inside AB but it will still work is E is outside AB). Construct a copy of AFGH at E so that EK = BF.

Then use the theorem again, but 'in reverse', to find P so that AK x AK = AE x AN. N is the required point for the equal area rectangle with side AE.

Here's the diagram and the steps to take.

(1) Extend the line AB.

(2) Centre B, radius BC draw an arc to cut AB produced at J.

(3) Construct the midpoint of AJ, I.

(4) Centre I, radius IA draw a (part) circle to find F on BC produced.

By the theorem AB x BC = AB x BJ = BF x BF.

(5) So square BFGH is equal in area to ABCD.

(6) Translate BFGH so that BF moves to EK.

As EK = BF, EK x EK = the same area.

(7)Find the midpoint of AK, L

(8) Construct the perpendicular bisector of AK to cut AJ at M

(9) Centre M, radius MA, draw a (part) circle to cut AJ at P

By the theorem, AE x EP = EK x EK

(10) Centre E, radius EP draw a quarter circle to cut EK at N, so that EP = EN

Then AE x EN = require area, so N is the correct position for the second side of the required rectangle.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**msbiljanica****Member**- Registered: 2012-01-18
- Posts: 21

here's the right solution ,

- GeoGebra solution https://onedrive.live.com/?authkey=%21ANn523SIEkl70Bc&cid=48F411219265CF17&id=48F411219265CF17%21194&parId=48F411219265CF17%21105&o=OneUp

for AE greater than AB

- instruction manual, point D changes the length of the side AD, point B changes the length AB, point E changes the length of AE, turn on (design description) and check that I have followed the rules.

q1 q2 ABCD and AEFG rectangle surfaces are accuracy checks

- there is a version when AE is less AB, if you know how to set it

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