Question

Let us draw the squares whose areas equal to the areas of the following triangles:
An equilateral triangle whose side is $6cm$. in length.

Answer 1

Steps of construction:
1) Draw a triangle $PQR$ with sides $PQ=QR=PR=6cm$.
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 $\times$ breadth
$=1/2\times$ base $\times$ 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 $△PQR$