Question

Complete the hexagonal and star shaped Rangolies [see Fig. (i) and (ii)] by filling them with as many equilateral triangles of side $1$ cm as you can. Count the number of triangles in each case. Which has more triangles?

Answer 1

$\left(i\right)$ From the figure, we can say that the rangoli is in the shape of a regular hexagon.

Let the area of hexagon be $P$

$P=\frac{3\sqrt{3}}{2}(side{)}^2=\frac{3\sqrt{3}}{2}\times 5^2$

$P=\frac{75\sqrt{3}}{2}c{m}^2$

$\therefore A\left(Rangoli\right)=\frac{75\sqrt{3}}{2}c{m}^2$

Let area of equilateral triangle of side $1cm$ be $A^{\prime }$

$A^{\prime }=\frac{\sqrt{3}}{4}(1{)}^2=\frac{\sqrt{3}}{4}c{m}^2$

Let no. of equilateral triangles in rangoli be $n$

$n=\frac{A\left(Rangoli\right)}{A(eq.\Delta of1cm)}=\frac{\frac{150\sqrt{3}}{4}}{\frac{\sqrt{3}}{4}}=150$

There can be $150$ equilateral triangles each of side $1$cm in the hexagonal rangoli.

$\left(ii\right)$ From the figure, we can say that the rangoli is in the shape of a star.

Hence, the figure consist of $12$ equilateral triangles each of side $5$cm.

$\therefore A\left(Rangoli\right)=12\times \frac{\sqrt{3}}{4}(5{)}^2=75\sqrt{3}c{m}^2$

Let area of equilateral triangle of side $1cm$ be $A^{\prime }$

$A^{\prime }=\frac{\sqrt{3}}{4}(1{)}^2=\frac{\sqrt{3}}{4}c{m}^2$

$\text{No. of equilateral triangles in rangoli}=\frac{A\left(Rangoli\right)}{A(eq.\Delta of1cm)}=\frac{75\sqrt{3}}{\frac{\sqrt{3}}{4}}=300$

There can be $300$ equilateral triangles each of side $1$cm in the hexagonal rangoli.

Hence, star shaped rangoli has more equilateral triangles in it.