Question

Draw an equilateral triangle of sides 5 cm. Construct a square having the same area as that of this triangle.

Answer 1

Steps of construction:
1. Draw equilateral $\Delta$ABC according to the given measures.
2. Draw the perpendicular CD from C to AB.
Now CD is the height $h=\sqrt{5^2-{\left(\frac{5}{2}\right)}^2}=\frac{5\sqrt{3}}{2}=4.3cm$ of the triangle. Draw perpendicular bisector of CD to get DE = h/2.
Area of $\Delta ABC=\frac{1}{2}\times AB\times h=AB\times \frac{h}{2}$
To draw a square of area $AB\times \left(h/2\right)$, the side of the square should be $\sqrt{AB\times \frac{h}{2}}=\frac{5(3{)}^{\frac{1}{4}}}{2}=3.3cm$.

Steps of construction:
1. Draw $PQ=AB+\frac{h}{2}=3.3cm$.

2. At the point $Q$ and $P$, draw angles of $90˚$ on the same side of $PQ$.

3. Cut those perpendicular lines at $3.3cm$ at $R$ and $S$.

4. Join $RS$.

5. Area of square $PQ\times QR=3.3\times 3.3{cm}^2\approx 10.89{cm}^2$