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 a square with vertices on a circle of double the diameter?

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 $= C F = 2 \times C O = 2 \times 4 = 8 cm$ .

Double the diameter of the circle $= 2 \times 8 = 16 cm$ .

Now, we have to find the perimeter of a square with vertices on this circle with diameter $16 cm$ .

Let the side of the square be $a cm$ .

Here, $△ P Q S$ is a right-angled triangle. $Q S$ is the hypotenuse of this triangle and also, $Q S$ is the diameter of the circle.

Hence, $Q S = 16 cm$ .

Applying Pythagoras theorem on $△ P Q S$ , we get,

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

$\Rightarrow a^{2} + a^{2} = 16^{2}$

$\Rightarrow 2 a^{2} = 256$

$\Rightarrow a^{2} = \frac{256}{2} = 128$

$\Rightarrow a = \sqrt{128} = 8 \sqrt{2} cm$ .

So, the side of the square is $8 \sqrt{2} cm$ .

We know that the perimeter of a square $= 4 \times side$ .

Hence, the perimeter of the square is $= 4 \times 8 \sqrt{2} = 32 \sqrt{2} = 32 \times 1.414 = 45.248 cm$ .

Therefore, the required perimeter of the square is $45.248 cm$ .

The perimeter of a regular hexagon with vertices on a circle is $24$ centimetres. What is the perimeter of a square with vertices on a circle of double the diameter?

Important Questions on Circle Measures

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