Let us draw the squares whose areas equal to the areas of the following triangles:

An equilateral triangle whose side is 6cm. in length.

Steps of construction:

1) Draw a triangle PQR with sides PQ = QR = PR = 6 cm.

2) Draw a rectangle ARCP with area equal to that of triangle PQR, by choosing length of rectangle as half of base and height same as height of triangle.

[Here, area of rectangle = length × breadth

= 1/2 × base × height

= area of triangle]

Now, we can make a square with area equal to that of rectangle and hence equal to area of triangle.

3) Extend CP to CE, such that CE = BC

4) Draw the perpendicular bisector of PE which bisects PE at O.

5) Taking O as center and OD = OE as radius, draw a semicircle.

6) Extend BC which intersects semicircle at F.

7) Draw a square CFGH taking CF as side.

Here,

Area of square CFGH = area of rectangle ABCD

= area of ΔPQR