In a circle, two chords and are drawn perpendicular to a diameter . Prove that the segment joining the midpoints of the chords and is perpendicular to .

Understand the Question Two identical circles with radius pass through each other's centres, and . Thus, the distance between their centres is . The region enclosed between the two circles is symmetrical and consists of two identical circular segments on the common chord . Triangles and are equilateral triangles with all side lengths equal to , each having an interior angle of . The central angle subtended by chord at centre is . The area of the overlapping region is found by calculating the area of one circular segment and multiplying it by . Step 1 · Find Central Angle Let the centres of the two circles be and , each with radius . Since each circle passes through the other's centre, . Let and be the points of intersection of the two circles. In : Therefore, is equilateral, which gives . Similarly, is equilateral, which gives . The central angle is: Step 2 · Calculate Area of One Circular Segment The enclosed region consists of two identical circular segments bounded by chord . First, find the area of sector : Next, find the area of : Since : Now, find the area of one segment: Step 3 · Calculate Total Enclosed Area The total area enclosed by both circles is twice the area of one segment: Answer > Common Mistakes Computing Only One Segment: Finding the area of a single circular segment and forgetting to multiply by to account for both overlapping segments. Incorrect Central Angle: Taking the central angle as from a single triangle instead of summing the two equilateral triangle angles (). Trigonometric Value Error: Miscalculating ; remember that .

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Question 25

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 $AB$ .

Solution

Understand the Question

Two identical circles with radius $r$ pass through each other's centres, $O_1$ and $O_2$ . Thus, the distance between their centres is $O_1O_2 = r$ .
The region enclosed between the two circles is symmetrical and consists of two identical circular segments on the common chord $CD$ .
Triangles $\Delta O_1CO_2$ and $\Delta O_1DO_2$ are equilateral triangles with all side lengths equal to $r$ , each having an interior angle of $60^\circ$ .
The central angle subtended by chord $CD$ at centre $O_1$ is $\angle CO_1D = 60^\circ + 60^\circ = 120^\circ$ .
The area of the overlapping region is found by calculating the area of one circular segment and multiplying it by $2$ .

Step 1 · Find Central Angle $\angle CO_1D$

Let the centres of the two circles be $O_1$ and $O_2$ , each with radius $r$ .

Since each circle passes through the other's centre, $O_1O_2 = r$ .

Let $C$ and $D$ be the points of intersection of the two circles. Diagram 1

In $\Delta O_1CO_2$ : $O_1C = O_2C = O_1O_2 = r$

Therefore, $\Delta O_1CO_2$ is equilateral, which gives $\angle CO_1O_2 = 60^\circ$ .

Similarly, $\Delta O_1DO_2$ is equilateral, which gives $\angle DO_1O_2 = 60^\circ$ .

The central angle $\angle CO_1D$ is:

\begin{aligned} \angle CO_1D &= 60^\circ + 60^\circ \\ &= 120^\circ \end{aligned}

Step 2 · Calculate Area of One Circular Segment

The enclosed region consists of two identical circular segments bounded by chord $CD$ .

First, find the area of sector $O_1CD$ :

\begin{aligned} \text{Area of sector} &= \dfrac{\text{angle}}{360^\circ} \times \pi r^2 \\[0.6em] &= \dfrac{120^\circ}{360^\circ} \times \pi r^2 \\[0.6em] &= \dfrac{1}{3} \pi r^2 \end{aligned}

Next, find the area of $\Delta O_1CD$ :

\begin{aligned} \text{Area of triangle} &= \dfrac{1}{2} \times O_1C \times O_1D \times \sin(\angle CO_1D) \\[0.6em] &= \dfrac{1}{2} \times r \times r \times \sin(120^\circ) \end{aligned}

Since $\sin(120^\circ) = \sin(60^\circ) = \dfrac{\sqrt{3}}{2}$ :

\begin{aligned} \text{Area of triangle} &= \dfrac{1}{2} r^2 \times \dfrac{\sqrt{3}}{2} \\[0.6em] &= \dfrac{\sqrt{3}}{4} r^2 \end{aligned}

Now, find the area of one segment:

\begin{aligned} \text{Area of segment} &= \text{Area of sector} - \text{Area of triangle} \\[0.6em] &= \dfrac{1}{3} \pi r^2 - \dfrac{\sqrt{3}}{4} r^2 \\[0.6em] &= r^2 \left(\dfrac{\pi}{3} - \dfrac{\sqrt{3}}{4}\right) \end{aligned}

Step 3 · Calculate Total Enclosed Area

The total area enclosed by both circles is twice the area of one segment:

\begin{aligned} \text{Total Area} &= 2 \times r^2 \left(\dfrac{\pi}{3} - \dfrac{\sqrt{3}}{4}\right) \\[0.6em] &= r^2 \left(\dfrac{2\pi}{3} - \dfrac{2\sqrt{3}}{4}\right) \\[0.6em] &= r^2 \left(\dfrac{2\pi}{3} - \dfrac{\sqrt{3}}{2}\right) \end{aligned}

Answer

$r^2 \left(\dfrac{2\pi}{3} - \dfrac{\sqrt{3}}{2}\right)$

Common Mistakes

Computing Only One Segment: Finding the area of a single circular segment and forgetting to multiply by $2$ to account for both overlapping segments.
Incorrect Central Angle: Taking the central angle as $60^\circ$ from a single triangle instead of summing the two equilateral triangle angles ( $60^\circ + 60^\circ = 120^\circ$ ).
Trigonometric Value Error: Miscalculating $\sin(120^\circ)$ ; remember that $\sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ) = \dfrac{\sqrt{3}}{2}$ .

Question 25

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