Question

Construct a rectangle ABCD of area 15 sq. cm., construct a square equal in area.

Hint:

This method is a classic geometric solution to the problem \(x^2 = l \times w\), where you find a length \(x\) (the side of the square) that is the geometric mean of length \(l\) and width \(w\). The key is extending the length by the width and using a semicircle.

Answer 1

Step 1: Construct the Rectangle
First, we need a rectangle with an area of 15 sq. cm. We can choose any two factors of 15 for the length and width. Let's use length = 5 cm and width = 3 cm.
Draw a line segment AB of length 5 cm.
At point A, construct a perpendicular line (a 90° angle).
On this perpendicular, measure and mark point D such that AD = 3 cm.
At point B, construct another perpendicular line.
On this perpendicular, measure and mark point C such that BC = 3 cm.
Join points D and C. ABCD is the required rectangle with an area of 5 cm \( \times \) 3 cm = 15 sq. cm.
Step 2: Construct the Square of Equal Area
This construction finds the geometric mean (mean proportional) of the length and width of the rectangle. The result will be the side length of the required square.
Extend the line segment AB to the right.
On this extended line, mark a point E such that the length of BE is equal to the width of the rectangle (BC), which is 3 cm. The total length of AE is now 5 cm + 3 cm = 8 cm.
Find the midpoint of the new line segment AE. To do this, construct its perpendicular bisector. Let the midpoint be M.
With M as the center and MA (or ME) as the radius, draw a semicircle on AE as the diameter.
Extend the side BC of the rectangle upwards until it intersects the semicircle. Label this intersection point F.
The length of the line segment BF is the side length of the required square. (The area of this square will be BF\(^2\), which is equal to AB \( \times \) BE = 5 \( \times \) 3 = 15 sq. cm).
To complete the construction, draw the square. Let's call it BGHF. Draw a line segment BG perpendicular to AB at B, such that BG = BF. Complete the square by drawing sides GH and HF, both equal in length to BF.
The square BGHF has the same area as the rectangle ABCD.