Name: Areas Related to a Circle
Uploaded: 2019-07-19
Duration: 18 min 34 s
Description: Have you ever wondered why the area of the circle is πr^2? Let's learn more about the formula related to the area, circumferences, arc, and sector of a circle.

Question

The perimeter of a regular hexagon with vertices on a circle is

24

centimetres. What is the perimeter of an equilateral triangle with vertices on a circle of diameter half that of the first circle? Write the answer in centimetre.

Accepted Answer

Given that the perimeter of the regular hexagon with vertices on a circle is $24 cm$ .

Here, $A B C D E F$ is a regular hexagon, with its vertices on the circle with center $O$ .

Let, $a cm$ be the length of each side of the regular hexagon. So, perimeter of the hexagon $= 6 a$ .

Hence, $6 a = 24 \Rightarrow a = \frac{24}{6} = 4 cm$ .

So, we get, $A B = B C = C D = D E = E F = F A = 4 cm$ .

We join $A D, B E$ and $C F$ .

From the symmetry of the figure, we can say that the diagonals $A D, B E$ and $C F$ are the diameters of the circle, passing through the center of the circle.

Again, because of the symmetry of the figure, we can prove that the hexagon is divided into $6$ equilateral triangles, each of which is congruent to the other.

So, in equilateral triangle $O C D$ , we have, $C D = O C = 4 cm$ .

Now, the diameter of the circle $= C F = 2 \times C O = 2 \times 4 = 8 cm$ .

Half the diameter of the circle $= \frac{8}{2} = 4 cm$ .

Now, we have to find the perimeter of an equilateral triangle with vertices on this circle with diameter $4 cm$ .

Here, $△ P Q R$ is the equilateral triangle. We draw medians $P S, R U$ and $Q T$ of the triangle. They intersect at the point $W$ . This point is called the centroid of the triangle. [The point of intersection of the medians of a triangle is called the centroid]

This point $W$ is also called the circumcentre of the equilateral triangle $△ P Q R$ . [In an equilateral triangle, the centroid and the circumference are the same point]

Now, we extend $P S$ , such that it intersects the circumference of the circle at the point $V$ . So, $P V$ is the diameter of the circle. So, $P V = 4 cm$ .

So, $P W =$ radius of the circle $= \frac{4}{2} = 2 cm$ .

In this circle, the point $W$ is the centre as well as the centroid of the circle.

So, $P W : S W = 2 : 1$

$\Rightarrow \frac{P W}{S W} = \frac{2}{1}$

$\Rightarrow \frac{2}{S W} = \frac{2}{1}$

$\Rightarrow 2 \times S W = 1 \times 2$

$\Rightarrow S W = \frac{2}{2} = 1 cm$ .

So, $P S = P W + S W = 2 + 1 = 3 cm$ .

Let, the side of the equilateral triangle be $a cm$ .

So, $P Q = a cm$ and $Q S = \frac{Q R}{2} = \frac{P Q}{2} = \frac{a}{2} cm$ .

Now, applying Pythagoras theorem in $△ P Q S$ , we get,

$P Q^{2} = Q S^{2} + P S^{2}$

$\Rightarrow a^{2} = {(\frac{a}{2})}^{2} + 3^{2}$

$\Rightarrow a^{2} = \frac{a^{2}}{4} + 9$

$\Rightarrow a^{2} - \frac{a^{2}}{4} = 9$

$\Rightarrow \frac{4 a^{2} - a^{2}}{4} = 9$

$\Rightarrow \frac{3 a^{2}}{4} = 9$

$\Rightarrow a^{2} = 9 \times \frac{4}{3} = 12$

$\Rightarrow a = \sqrt{12} = 2 \sqrt{3} cm$ .

Hence, the length of the side of the triangle is $2 \sqrt{3} cm$ .

Now, the perimeter of the side of the equilateral triangle $= 3 \times side = 3 \times 2 \sqrt{3} = 3 \times 2 \times 1.732 = 10.39 cm$ .

Therefore, the required perimeter of the triangle is $10.39 cm$ .

The perimeter of a regular hexagon with vertices on a circle is $24$ centimetres. What is the perimeter of an equilateral triangle with vertices on a circle of diameter half that of the first circle? Write the answer in centimetre.

Important Questions on Circle Measures

Learn and Prepare for any exam you want !