Circles and Geometric Shapes | EOT

Question 22

"There is no chord of a circle that is longer than its diameter." How do you justify this statement?

Solution

We need to find the perimeter and area of the blue flower shown in the diagram. The square has a side length of 2 units. The diagram shows a common geometric pattern. We will assume the arcs forming the flower are quarter circles centered at the corners of the square, with a radius equal to the side length of the square.

Step 1 — Identify dimensions

Let's identify the side length of the square. The side length of the square is 2 units. The arcs that form the flower are quarter circles. The radius of each quarter circle is equal to the side length of the square.

$\text{Side length of square } (s) = \mathbf{2} \text{ units}$

$\text{Radius of each arc } (r) = s = \mathbf{2} \text{ units}$

Diagram 1

Step 2 — Calculate the perimeter of the flower

The flower has four petals. Each petal is formed by two identical circular arcs. So, the total perimeter of the flower consists of 8 identical arcs. Each arc is a quarter circle of radius 2 units.

$\text{Length of one quarter circle arc} = \frac{1}{4} \times 2\pi r$

$= \frac{1}{2}\pi r$

Substitute $r = \mathbf{2}$ :

$= \frac{1}{2}\pi (\mathbf{2})$

$= \pi \text{ units}$

The total perimeter of the flower is 8 times the length of one arc.

$\text{Perimeter of flower} = 8 \times \pi$

$\boxed{8\pi \text{ units}}$

Step 3 — Calculate the area of the flower

First, let's find the area of the square.

$\text{Area of square} = s^2$

Substitute $s = \mathbf{2}$ :

$= (\mathbf{2})^2$

$= 4 \text{ square units}$

The flower's area is the area of the square minus the area of the four unshaded corner regions. Each unshaded corner region is a circular segment. The area of a circular segment is the area of the sector minus the area of the triangle. Consider one corner of the square. A sector is formed with the corner as the center and radius $r = \mathbf{2}$ . The angle of this sector is $90^\circ$ .

$\text{Area of one sector} = \frac{90}{360} \times \pi r^2$

$= \frac{1}{4}\pi (\mathbf{2})^2$

$= \frac{1}{4}\pi (4)$

$= \pi \text{ square units}$

The corresponding triangle is a right-angled triangle with base and height equal to $r = \mathbf{2}$ .

$\text{Area of this triangle} = \frac{1}{2} \times \text{base} \times \text{height}$

$= \frac{1}{2} \times \mathbf{2} \times \mathbf{2}$

$= 2 \text{ square units}$

Now, let's find the area of one unshaded corner region (segment).

$\text{Area of one segment} = \text{Area of sector} - \text{Area of triangle}$

$= \pi - 2 \text{ square units}$

There are 4 such unshaded corner regions.

$\text{Total unshaded area} = 4 \times (\pi - 2)$

$= 4\pi - 8 \text{ square units}$

Finally, we can find the area of the flower.

$\text{Area of flower} = \text{Area of square} - \text{Total unshaded area}$

$= 4 - (4\pi - 8)$

$= 4 - 4\pi + 8$

$= 12 - 4\pi \text{ square units}$

$\boxed{(12 - 4\pi) \text{ square units}}$

Answer

(i) The perimeter of the flower is $8\pi$ units. (ii) The area of the flower is $(12 - 4\pi)$ square units.

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Q10

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Q11

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Q12

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Q13

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(Hint: Is it a circumcircle of a suitable triangle?)

Q14

Show that rectangle is the only parallelogram that can be inscribed in a circle.

Q15

Show that if a rectangle is inscribed in a circle, then the point of intersection of its diagonals must lie at the centre of the circle.

Q16

Consider all chords of a circle of a fixed length. What is the shape formed by the midpoints of all these chords?

Q17

In a circle with centre O, chords AB and AC are congruent. Explain why this statement is true: "The centre of the circle lies on the angle bisector of ∠BAC".

Q18

Two parallel chords of lengths 10 cm and 24 cm are on the same side of the centre of a circle. The distance between the chords is 7 cm. Find the radius of the circle.

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

A quadrilateral MNOP is inscribed in a circle. If MN is a diameter, what can you say about ∠MOP and ∠MNP? Explain your reasoning.

Q21

Let ABCD be a cyclic quadrilateral. Explain why the exterior angle at any vertex is equal to the interior opposite angle (e.g., ∠CDE = ∠ABC, where E is a point on the extension of side CD).

Q22

"There is no chord of a circle that is longer than its diameter." How do you justify this statement?

Q23

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Q24

How would you use the following figure to justify the statement that the angle in a semicircle is 90°?

Q25

In a circle, two chords CC' and DD' are drawn perpendicular to a diameter AB. Prove that the segment MM' joining the midpoints of the chords CD and C' D' is perpendicular to AB.

Q26

How would you use the following figure to justify the statement that the sum of the opposite angles of a cyclic quadrilateral is 180°?