Measuring Space: Perimeter and Area · Lesson 7 of 10
Squaring a Rectangle
“Turning a rectangle into a square sounds like magic, but area keeps the receipts.”
• Understand what it means to square a rectangle. • Follow Baudhāyana's equal-area construction conceptually. • Connect geometric construction with algebraic identities. • Use the Baudhāyana–Pythagoras theorem in the construction. • Recognise equivalence of area under rearrangement and construction.
A classic problem in geometry asks whether a rectangle can be replaced by a square having exactly the same area. Suppose the rectangle has side lengths a and b. Its area is therefore ab. If a square has the same area, then the square of its side length must also be equal to ab.
So, if the side of the required square is x, then x² = ab. This means x = √(ab). The interesting part of the problem is not just finding this value algebraically, but constructing the length √(ab) using only geometric ideas. This connects algebra, square roots, and geometric construction in a very direct way.
Baudhāyana's Construction Idea
The chapter presents a construction based on extending and rearranging lengths from a rectangle and then using a right triangle. The construction converts the product ab into a difference of two squares.
Why does this construction work?
To square a rectangle means to construct a square having exactly the same area as the rectangle. Rectangle ABCD has sides a and b, so its area is ab.
Point E is chosen on AD so that AE = AB = b. Since AD = a, the remaining segment ED has length a − b.
Point F is the midpoint of ED. Therefore, EF and FD are equal, and each has length (a − b)/2.
The side AF consists of AE followed by EF. Substituting their lengths gives AF = (a + b)/2.
Square AFGH is constructed on AF. All its sides are equal, so AH and HG also have length (a + b)/2. The arc AG is drawn with H as its centre. Since K lies on this arc, HK is a radius and is equal to HG.
The line through K is drawn parallel to AH and meets GH at P. This makes BHPK a rectangle, so PK = BH. Since BH is the part of AH remaining after removing AB, its length is (a − b)/2.
Triangle HKP is right-angled at P. Applying the Baudhāyana–Pythagoras theorem gives HP² = HK² − PK².
Square HPQS is constructed using HP as its side. Its area is therefore HP², which we have shown to be ab.
The construction produces the length √(ab) geometrically. A square drawn on this length has area ab, exactly equal to the area of the original rectangle.
The Algebra Hidden in the Geometry
The construction creates two key lengths: (a+b)/2 and (a−b)/2. A right triangle then gives the side x of the new square through the Baudhāyana–Pythagoras theorem.
Expanding both squares gives
Therefore the constructed square has area x²=ab, exactly equal to the original rectangle.
Geometry and Algebra Say the Same Thing
This construction is a good example of two mathematical languages describing the same idea. Geometry constructs the required length; algebra proves why that length has the right square.
Practice Problems
- A rectangle has sides 9 cm and 4 cm. What should be the side of an equal-area square?
- A rectangle has sides 18 cm and 8 cm. Find the side of the equal-area square.
- Verify algebraically that [(a+b)/2]²−[(a−b)/2]²=ab.
- Explain where the Baudhāyana–Pythagoras theorem enters the construction.
- Why is an equal-area square not required to have the same perimeter as the rectangle?
Key Takeaways
• Squaring a rectangle means constructing an equal-area square. • The required side length is √ab. • Baudhāyana's construction turns the problem into a right-triangle calculation. • A difference-of-squares identity explains the construction algebraically. • Geometry and algebra can encode the same mathematical relationship.
Next, we return to circles and derive the formula A=πr² from historical and visual ideas.