To construct a square with area equal to a given polygon 

Step 1 Firstly
we like to construct a triangle with area equals to the area of a
given quadrilateral. The method mentioned here can be extended to any
other polygons. For example, a pentagon can be reduced to a quadrilateral
with equal area. 

Given a quadrilateral
ABCD. 1. Join BD. 2. Draw a line through C parallel to DB. 3. Produce AB to meet the line in 2 at E. 4. Join DE. 5. Then DAED is of equal area to ABCD. 


Reasons: For
DDBC and DDBE, they both have the
same base DB. They also have the same height since they are between two
parallel lines DB and CE. Therefore,
area of DDBC = area of DDBE Area of quadrilateral ABCD
= area of DABD + area of DDBC =
area of DABD + area of DDBE =
area of DAED 

Step 2 We
like to construct a rectangle equal in area to a given triangle.
We use the same triangle produced in Step 1,
that is, DAED, for
demonstration. 

Given DAED, 1. Draw a line through D and parallel to AE. 2. Bisect AE at G. 3. From G construct a line perpendicular
to AE and meet the line in 1 at H. 4. From A construct a line perpendicular
to AE and meet the line in 1 at I. 5. Then the area of rectangle AGHI is equal
to the area of DAED. 


Reasons: Base
of the rectangle is equal to half of the base of the triangle. The height of
the rectangle is equal to the height of the triangle. 

Step 3 Finally
we have to construct a square of area equal to the area of a given rectangle.
We use the same rectangle produced in Step 2, that is, AGHI. 

Given rectangle AGHI, 1. Produce AG to J such that GJ = GH. 2. Find the mid point O, of AJ. 3. Use O as center, OA as radius, draw a semicircle. 4. Produce GH to meet the semicircle at
K. 5. Use GK as side, construct a square
GKLM. 6. Area of AGHI = Area of GKLM. 


Reasons: As in the left diagram, ÐAKG = 90° (Ð in
semicircle) ÐKAG = 90° – ÐAKG =
ÐGKJ \ DAGK ~ DKGJ (AAA) AG/GK = KG/GJ AG ´ GJ = KG^{2} AG ´ HG = KG^{2} \ Area of AGHI = Area of
GKLM. 

