Circles and Geometric Shapes | EOT

Question 19

A regular hexagon is inscribed in a circle of radius $r$ . Find the length of the sides of the hexagon and the distance of each side from the centre of the circle.

Solution

Understand the Question

Nine identical rectangles of length $l$ and width $w$ fit together to form a large rectangle of area $72\text{ cm}^2$ .
The top row consists of $4$ rectangles laid horizontally along their lengths, giving a total width of $4l$ and height $w$ .
The bottom row consists of $5$ rectangles placed vertically along their widths, giving a total width of $5w$ and height $l$ .
Since the top and bottom rows span the same large rectangle, their total widths are equal: $4l = 5w$ .
We use this relation along with the total area $(4l)(w+l) = 72\text{ cm}^2$ to find the dimensions $l$ and $w$ , and subsequently the perimeter $P = 2(l+w)$ of each small rectangle.

Step 1 · Set Up Equations from Rectangle Dimensions

Let the length and width of each small rectangle be $l$ and $w$ respectively. Diagram 1

Equating the total width of the top and bottom rows: $4l = 5w \implies w = \dfrac{4}{5}l$

The total height of the large rectangle is: $\text{Total height} = w + l$

Given that the total area is $72\text{ cm}^2$ : $\text{Area} = \text{Total width} \times \text{Total height}$ $72 = (4l)(w+l)$

Step 2 · Solve for Dimensions $l$ and $w$

Substitute $w = \dfrac{4}{5}l$ into the area equation:

\begin{aligned} 72 &= 4l\left(\dfrac{4}{5}l + l\right) \\[0.6em] &= 4l\left(\dfrac{4l+5l}{5}\right) \\[0.6em] &= 4l\left(\dfrac{9l}{5}\right) \\[0.6em] 72 &= \dfrac{36l^2}{5} \end{aligned}

Solve for $l$ :

\begin{aligned} 36l^2 &= 72 \times 5 \\[0.6em] 36l^2 &= 360 \\[0.6em] l^2 &= \dfrac{360}{36} \\[0.6em] l^2 &= 10 \\[0.6em] l &= \sqrt{10}\text{ cm} \end{aligned}

Substitute $l = \sqrt{10}$ to find $w$ : $w = \dfrac{4}{5}l = \dfrac{4}{5}\sqrt{10}\text{ cm}$

Step 3 · Calculate the Perimeter of a Small Rectangle

The perimeter $P$ of each small rectangle is given by $2(l+w)$ :

\begin{aligned} P &= 2\left(\sqrt{10} + \dfrac{4}{5}\sqrt{10}\right) \\[0.6em] &= 2\left(\dfrac{5\sqrt{10} + 4\sqrt{10}}{5}\right) \\[0.6em] &= 2\left(\dfrac{9\sqrt{10}}{5}\right) \\[0.6em] &= \dfrac{18\sqrt{10}}{5}\text{ cm} \end{aligned}

Answer

$\dfrac{18\sqrt{10}}{5} \text{ cm}$

Common Mistakes

Overlooking the Width Equivalence: Forgetting to equate the total width of both rows ( $4l = 5w$ ) to establish the relation between $l$ and $w$ .
Area of Single Rectangle Method: Finding the area of one rectangle as $\dfrac{72}{9} = 8\text{ cm}^2$ is valid ( $lw = 8$ ), but failing to substitute $w = \dfrac{4}{5}l$ into $l \left(\dfrac{4}{5}l\right) = 8$ leads to errors.
Perimeter Formula Error: Forgetting to multiply $(l+w)$ by $2$ when calculating perimeter.

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Q10

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Q11

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Q12

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Q13

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Q14

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Q15

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Q16

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Q17

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Q18

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Q19

A regular hexagon is inscribed in a circle of radius $r$ . Find the length of the sides of the hexagon and the distance of each side from the centre of the circle.

Q20

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Q21

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Q22

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Q23

Let $A$ be any point within a given circle with centre $O$ . Show that the shortest chord of the circle that passes through point $A$ is the one that is perpendicular to $OA$ .

Q24

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Q25

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Q26

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