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?
(B) To find the number of equilateral triangles of side $1\,cm$ that can fit into the given shapes,we calculate the area of the shapes and divide by the area of one small equilateral triangle.
$1$. For the regular hexagon of side $s = 5\,cm$:
Area $= \frac{3\sqrt{3}}{2} s^2 = \frac{3\sqrt{3}}{2} \times 25 = 37.5\sqrt{3} \approx 64.95\,cm^2$.
Each small triangle of side $1\,cm$ has an area of $\frac{\sqrt{3}}{4} \times 1^2 = 0.25\sqrt{3} \approx 0.433\,cm^2$.
Number of triangles in Fig. $(i) = \frac{37.5\sqrt{3}}{0.25\sqrt{3}} = 150$.
$2$. For the star-shaped figure in Fig. $(ii)$:
This figure consists of a central hexagon of side $5\,cm$ and $6$ equilateral triangles of side $5\,cm$ attached to its sides.
Total area $= \text{Area of hexagon} + 6 \times \text{Area of triangle of side } 5\,cm$.
Total area $= 37.5\sqrt{3} + 6 \times (6.25\sqrt{3}) = 37.5\sqrt{3} + 37.5\sqrt{3} = 75\sqrt{3} \approx 129.9\,cm^2$.
Number of triangles in Fig. $(ii) = \frac{75\sqrt{3}}{0.25\sqrt{3}} = 300$.
Therefore,the star-shaped figure $(ii)$ has more triangles.
$1$. For the regular hexagon of side $s = 5\,cm$:
Area $= \frac{3\sqrt{3}}{2} s^2 = \frac{3\sqrt{3}}{2} \times 25 = 37.5\sqrt{3} \approx 64.95\,cm^2$.
Each small triangle of side $1\,cm$ has an area of $\frac{\sqrt{3}}{4} \times 1^2 = 0.25\sqrt{3} \approx 0.433\,cm^2$.
Number of triangles in Fig. $(i) = \frac{37.5\sqrt{3}}{0.25\sqrt{3}} = 150$.
$2$. For the star-shaped figure in Fig. $(ii)$:
This figure consists of a central hexagon of side $5\,cm$ and $6$ equilateral triangles of side $5\,cm$ attached to its sides.
Total area $= \text{Area of hexagon} + 6 \times \text{Area of triangle of side } 5\,cm$.
Total area $= 37.5\sqrt{3} + 6 \times (6.25\sqrt{3}) = 37.5\sqrt{3} + 37.5\sqrt{3} = 75\sqrt{3} \approx 129.9\,cm^2$.
Number of triangles in Fig. $(ii) = \frac{75\sqrt{3}}{0.25\sqrt{3}} = 300$.
Therefore,the star-shaped figure $(ii)$ has more triangles.
Explore More
Similar Questions
$\Delta ABC$ and $\Delta DBC$ are two isosceles triangles on the same base $BC$ and vertices $A$ and $D$ are on the same side of $BC$ (see Fig). If $AD$ is extended to intersect $BC$ at $P$,show that:
$(i)$ $\Delta ABD \cong \Delta ACD$
$(ii)$ $\Delta ABP \cong \Delta ACP$
$(iii)$ $AP$ bisects $\angle A$ as well as $\angle D$.
$(iv)$ $AP$ is the perpendicular bisector of $BC$.
$(i)$ $\Delta ABD \cong \Delta ACD$
$(ii)$ $\Delta ABP \cong \Delta ACP$
$(iii)$ $AP$ bisects $\angle A$ as well as $\angle D$.
$(iv)$ $AP$ is the perpendicular bisector of $BC$.
Difficult
View SolutionIt is given that $\Delta PQR \cong \Delta EDF$. Is it true to say that $PR = EF$? Give a reason for your answer.
Easy
View Solution$AB$ is a line segment and $P$ is its mid-point. $D$ and $E$ are points on the same side of $AB$ such that $\angle BAD = \angle ABE$ and $\angle EPA = \angle DPB$. Show that:
$(i)$ $\Delta DAP \cong \Delta EBP$
$(ii)$ $AD = BE$
$(i)$ $\Delta DAP \cong \Delta EBP$
$(ii)$ $AD = BE$
Medium
View SolutionIf $\Delta ABC \cong \Delta RPQ,$ then which side of $\Delta PQR$ is equal to side $AB$ of $\Delta ABC$?
Easy
View SolutionIt is given that $\triangle ABC \cong \triangle FDE$. If $AB = 5 \, cm$,$\angle B = 40^{\circ}$,and $\angle A = 80^{\circ}$,then which of the following is true?
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo