### geometric seq: area of hexagons

Question Sample Titled 'geometric seq: area of hexagons'

Destiny was making a paper figure. At first, she cut regular hexagon from paper. The length of each side is ${a}$ $\text{cm}$.

 (a) Find the area of the hexagon in terms of ${a}$. (2 marks) (b) After she cut the hexagon, she cut another smaller regular hexagon from paper and then placed it onto the first paper hexagon such that the vertices of the second hexagon lied on the mid-points of the sides of the first hexagon. She repeated this process for a number of times as shown in the figure.

$\ldots$

 (i) Find the area of the second and the third regular hexagons in terms of ${a}$. (ii) If the total sum of area of all regular hexagons that Destiny cut is $\dfrac{{{42591}\sqrt{{3}}}}{{8192}}{a}^{{2}}$ $\text{cm}$ , find the number of hexagons that she cut.  (5 marks)

 (a) The hexagon consists of six equilateral triangles with side ${a}$ $\text{cm}$ . Required area $={6}\times{\left(\dfrac{{1}}{{2}}\times{a}^{{2}}{\sin{{60}}}^{\circ}\right)}$ 1M  $=\dfrac{{{3}\sqrt{{3}}}}{{2}}{a}^{{2}}$ $\text{cm}^{{2}}$ 1A  (bi) By the cosine formula, the side length of the second hexagon $=\sqrt{{{\left(\dfrac{{a}}{{2}}\right)}^{{2}}+{\left(\dfrac{{a}}{{2}}\right)}^{{2}}-{2}{\left(\dfrac{{a}}{{2}}\right)}{\left(\dfrac{{a}}{{2}}\right)}{\cos{{120}}}^{\circ}}}$ 1M $=\dfrac{{\sqrt{{3}}{a}}}{{2}}$ $\text{cm}$  Thus, Side length of ${\left({n}+{1}\right)}^{{{t}{h}}}$ hexagon$:$Side length of ${n}^{{{t}{h}}}$ hexagon $=\dfrac{{\sqrt{{3}}{a}}}{{2}}:{a}$ i.e. Common ratio of sides  $=\dfrac{{\sqrt{{3}}}}{{2}}$ So, common ratio of areas  $={\left(\dfrac{{\sqrt{{3}}}}{{2}}\right)}^{{2}}=\dfrac{{3}}{{4}}$  ∴  The area of the second hexagon $=\dfrac{{{3}\sqrt{{3}}}}{{2}}{a}^{{2}}\times{\left(\dfrac{{3}}{{4}}\right)}$  $=\dfrac{{{9}\sqrt{{3}}}}{{8}}{a}^{{2}}$ $\text{cm}^{{2}}$ 1A The area of the third hexagon $=\dfrac{{{3}\sqrt{{3}}}}{{2}}{a}^{{2}}\times{\left(\dfrac{{3}}{{4}}\right)}^{{2}}$  $=\dfrac{{{27}\sqrt{{3}}}}{{32}}{a}^{{2}}$ $\text{cm}^{{2}}$ 1A  (bii) Let ${n}$ be the number of hexagons.  $\dfrac{{{\left(\dfrac{{{3}\sqrt{{3}}}}{{2}}{a}^{{2}}\right)}{\left({1}-{\left(\dfrac{{3}}{{4}}\right)}^{{n}}\right)}}}{{{1}-\dfrac{{3}}{{4}}}}$ $=\dfrac{{{42591}\sqrt{{3}}}}{{8192}}{a}^{{2}}$ 1M ${1}-{\left(\dfrac{{3}}{{4}}\right)}^{{n}}$ $=\dfrac{{14197}}{{16384}}$ ${\log{{\left(\dfrac{{2187}}{{16384}}\right)}}}$ $={\log{{\left({\left(\dfrac{{3}}{{4}}\right)}^{{n}}\right)}}}$ ${\log{{\left(\dfrac{{2187}}{{16384}}\right)}}}$ $={n}{\log{{\left(\dfrac{{3}}{{4}}\right)}}}$ ${n}$ $={7}$ ∴  Destiny cut ${7}$ hexagons. 1A

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