Kaliyuga Ekalavya

NCERT Class X Chapter 11: Areas Related To Circles Question: A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding: (i) minor segment (ii) major sector. (Use π = 3.14) Given: Radius of circle, \( r = 10 \) cm Angle subtended by chord at centre, \( \theta = 90^\circ \) \( \pi = 3.14 \) To Find: (i) Area of the minor segment (ii) Area of the major sector Formula: Area of sector of angle \( \theta \) = \( \displaystyle \frac{\theta}{360^\circ} \pi r^2 \) Area of triangle (when two sides and included angle known) = \( \displaystyle \frac{1}{2} ab \sin \theta \) Area of segment = Area of sector − Area of triangle Solution: Step 1: Find the area of the sector with angle \( 90^\circ \). $$ \text{Area of sector} = \frac{90^\circ}{360^\circ} \times \pi r^2 = \frac{1}{4} \times 3.14 \times (10)^2 = \frac{1}{4} \times 3.14 \times 100 = 78.5 \ \text{cm}^2 $$ Step 2: Find the area of the triangle formed by ...

NCERT Class X Chapter 11: Area Related To Circles Question: Find the area of a quadrant of a circle whose circumference is 22 cm. (Use \( \pi = \frac{22}{7} \)) Given: Circumference of the circle = 22 cm \( \pi = \frac{22}{7} \) To Find: Area of a quadrant of the circle Formula: Circumference of a circle: \( 2\pi r \) Area of a circle: \( \pi r^2 \) Area of a quadrant: \( \frac{1}{4} \pi r^2 \) Solution: Step 1: Let the radius of the circle be \( r \) cm. Write the formula for circumference and substitute values. $$ 2\pi r = 22 \\ 2 \times \frac{22}{7} \times r = 22 $$ Step 2: Solve for \( r \). $$ 2 \times \frac{22}{7} \times r = 22 \\ \frac{44}{7} r = 22 \\ r = \frac{22 \times 7}{44} \\ r = \frac{154}{44} \\ r = \frac{7}{2} \\ r = 3.5 \text{ cm} $$ Step 3: Find the area of the circle using the formula \( \pi r^2 \). $$ \text{Area of circle} = \pi r^2 \\ = \frac{22}{7} \times (3.5)^2 \\ = \frac{22...

NCERT Class X Chapter 11: Area Related To Circles Question: The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes. Use \( \pi = \frac{22}{7} \). Given: Length of minute hand (radius), \( r = 14 \) cm Time = 5 minutes \( \pi = \frac{22}{7} \) To Find: Area swept by the minute hand in 5 minutes. Formula: Area of a sector of a circle: $$ \text{Area} = \frac{\theta}{360^\circ} \times \pi r^2 $$ where \( \theta \) is the angle in degrees. Solution: Step 1: Find the angle swept by the minute hand in 5 minutes. In 60 minutes, the minute hand completes a full revolution, i.e., \( 360^\circ \). $$ \text{Angle in 5 minutes} = \frac{5}{60} \times 360^\circ = 30^\circ $$ Step 2: Write the formula for the area swept (sector of circle). $$ \text{Area} = \frac{\theta}{360^\circ} \times \pi r^2 $$ Here, \( \theta = 30^\circ \), \( r = 14 \) cm, \( \pi = \frac{22}{7} \). Step 3: Su...

NCERT Class X Chapter 11: Area Related To Circles Question: Unless stated otherwise, use \( \pi = \frac{22}{7} \). In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: (i) the length of the arc (ii) area of the sector formed by the arc (iii) area of the segment formed by the corresponding chord Given: Radius \( r = 21 \) cm Central angle \( \theta = 60^\circ \) To Find: (i) Length of the arc (ii) Area of the sector (iii) Area of the segment formed by the corresponding chord Formula: (i) Length of arc:   \( L = \frac{\theta}{360^\circ} \times 2\pi r \) (ii) Area of sector:   \( A_{\text{sector}} = \frac{\theta}{360^\circ} \times \pi r^2 \) (iii) Area of segment:   \( A_{\text{segment}} = A_{\text{sector}} - A_{\triangle} \) Solution: Step 1: Find the length of the arc using the formula. $$ L = \frac{\theta}{360^\circ} \times 2\pi r = \frac{60^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 2...

NCERT Class X Chapter 11: Area Related To Circles Question: A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use \( \pi = 3.14 \) and \( \sqrt{3} = 1.73 \)) Given: Radius of the circle, \( r = 15\,\text{cm} \) Angle subtended by the chord at the centre, \( \theta = 60^\circ \) \( \pi = 3.14 \) \( \sqrt{3} = 1.73 \) To Find: Areas of the minor and major segments of the circle. Formula: Area of sector:   \( \displaystyle \text{Area}_{\text{sector}} = \frac{\theta}{360^\circ}\pi r^2 \) Area of triangle (using sine rule):   \( \displaystyle \text{Area}_{\triangle} = \frac{1}{2} r^2 \sin\theta \) Area of segment:   \( \displaystyle \text{Area}_{\text{segment}} = \text{Area}_{\text{sector}} - \text{Area}_{\triangle} \) Solution: Step 1: Find the area of the sector with angle \( 60^\circ \). $$ \text{Area}_{\text{sector}} = \frac{60^...

NCERT Class X Chapter 12: Areas Related to Circles Question: A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14 and √3 = 1.73) Given: Radius of the circle, \( r = 12 \) cm Angle subtended at the centre, \( \theta = 120^\circ \) To Find: Area of the corresponding segment of the circle Formula: Area of segment = Area of sector − Area of triangle $$ \text{Area of sector} = \frac{\theta}{360} \times \pi r^2 $$ $$ \text{Area of triangle} = \frac{1}{2} r^2 \sin \theta $$ Solution: Step 1: Find the area of the sector of the circle. $$ \text{Area of sector} = \frac{120}{360} \times 3.14 \times 12^2 = \frac{1}{3} \times 3.14 \times 144 = 150.72 \text{ cm}^2 $$ Step 2: Find the area of the triangle formed by the radii and the chord. $$ \sin 1...

NCERT Class X Chapter 11: Area Related To Circles Question: A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope. (i) Find the area of that part of the field in which the horse can graze. (ii) Find the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use π = 3.14) Given: Side of square field = 15 m Length of rope = 5 m (initially), then 10 m π = 3.14 To Find: (i) Area of field in which horse can graze with 5 m rope (ii) Increase in grazing area if rope length is increased to 10 m Formula: Area of a circle:   \( A = \pi r^2 \) Area of a quarter circle:   \( A = \dfrac{\pi r^2}{4} \) Solution: Step 1: Area grazed when rope is 5 m The horse grazes a quarter circle of radius 5 m. $$ \text{Area}_1 = \frac{\pi r^2}{4} = \frac{3.14 \times (5)^2}{4} $$ Step 2: Calculate the value for 5 m rope $$ \text{Area}_1 = \frac{3....

NCERT Class X Chapter 11: Area Related To Circles Question: A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors. Find: (i) the total length of the silver wire required. (ii) the area of each sector of the brooch. (Use \( \pi = 3.14 \)) Given: Diameter of the circle, \( d = 35 \) mm Number of diameters used = 5 Number of sectors = 10 \( \pi = 3.14 \) To Find: (i) Total length of silver wire required (ii) Area of each sector of the brooch Formula: Circumference of a circle:   \( C = \pi d \) Area of a circle:   \( A = \pi r^2 \) Area of each sector:   \( \text{Area of sector} = \dfrac{\text{Area of circle}}{\text{Number of sectors}} \) Solution: Step 1: Find the circumference of the circle. $$ \text{Circumference} = \pi d = 3.14 \times 35 = 109.9 \ \text{mm} $$ Step 2: Calculate th...

NCERT Class X Chapter 11: Area Related To Circles Question: An umbrella has 8 ribs which are equally spaced. Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella. Given: Number of ribs \( = 8 \) Radius of the umbrella \( r = 45\, \text{cm} \) To Find: Area between two consecutive ribs of the umbrella. Formula: Area of a circle:   \( A = \pi r^2 \) Area between two consecutive ribs:   \( \dfrac{\text{Area of circle}}{\text{Number of ribs}} \) Solution: Step 1: Write the formula for the area of the umbrella (circle). $$ \text{Area of umbrella} = \pi r^2 $$ Step 2: Substitute the value of \( r = 45\, \text{cm} \). $$ \text{Area} = \pi \times (45)^2 = \pi \times 2025 = 2025\pi\ \text{cm}^2 $$ Step 3: Find the area between two consecutive ribs. $$ \text{Area between two ribs} = \frac{2025\pi}{8}\ \text{cm}^2 $$ Step 4: Calculate the value using \( \pi = \frac{22}{7} \)....

NCERT Class X Chapter 11: Area Related To Circles Question: A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades. Given: Length of each wiper blade, \( r = 25 \) cm Angle swept by each wiper, \( \theta = 115^\circ \) To Find: Total area cleaned at each sweep of the blades. Formula: Area of a sector of a circle: $$ \text{Area} = \frac{\theta}{360^\circ} \times \pi r^2 $$ where \( \theta \) is in degrees. Solution: Step 1: Find the area cleaned by one wiper using the sector formula. $$ \text{Area by one wiper} = \frac{115^\circ}{360^\circ} \times \pi \times (25)^2 $$ Step 2: Substitute the values and simplify. $$ = \frac{115}{360} \times \pi \times 625 $$ Step 3: Calculate the numerical value. $$ = 0.3194 \times 3.1416 \times 625 $$ Step 4: Multiply to get the area cleaned by one wiper. $$ ...

NCERT Class X Chapter 11: Area Related To Circles Question: To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use $\pi = 3.14$) Given: Angle of the sector, $\theta = 80^\circ$ Radius, $r = 16.5$ km $\pi = 3.14$ To Find: Area of the sector (area of the sea over which ships are warned) Formula: Area of a sector of a circle: $$ \text{Area} = \frac{\theta}{360^\circ} \times \pi r^2 $$ where $\theta$ is the angle of the sector (in degrees), $r$ is the radius. Solution: Step 1: Write the formula for the area of a sector and substitute the given values. $$ \text{Area} = \frac{80^\circ}{360^\circ} \times 3.14 \times (16.5)^2 $$ Step 2: Simplify the fraction $\frac{80}{360}$. $$ \frac{80}{360} = \frac{8}{36} = \frac{2}{9} $$ Step 3: Calculate $(16.5)^2$. $$ (16.5)^2 = 272.25 $$ Step 4: Substitute the values i...

NCERT Class X Chapter 12: Areas Related to Circles Question: The area of a sector of angle p (in degrees) of a circle with radius R is: (A) \( \dfrac{p}{180} \times 2\pi R \) (B) \( \dfrac{p}{180} \times \pi R^2 \) (C) \( \dfrac{p}{360} \times 2\pi R \) (D) \( \dfrac{p}{720} \times 2\pi R^2 \) Given: Radius of the circle \( = R \) Angle of the sector \( = p^\circ \) To Find: Area of the sector of the circle. Formula: Area of a sector of angle \( \theta^\circ \) and radius \( r \) is: $$ \text{Area} = \dfrac{\theta}{360} \times \pi r^2 $$ Solution: Step 1: Area of a complete circle of radius \( R \) is: $$ \pi R^2 $$ Step 2: A complete circle corresponds to an angle of: $$ 360^\circ $$ Step 3: Area of a sector of angle \( p^\circ \) is: $$ \dfrac{p}{360} \times \pi R^2 $$ Step 4: Rewriting the expression: $$ \dfrac{p...

NCERT Class X Chapter 12: Areas Related to Circles Question: A round table cover has six equal designs as shown in Fig. 11.11. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹ 0.35 per cm 2 . (Use \( \sqrt{3} = 1.7 \)) Given: Radius of the circular table cover, \( r = 28 \) cm Number of designs = 6 Rate of making designs = ₹ 0.35 per cm 2 \( \sqrt{3} = 1.7 \) To Find: Cost of making the six designs on the table cover Formula: Area of a circle: $$ A = \pi r^2 $$ Area of a regular hexagon of side \( a \): $$ A = \frac{3\sqrt{3}}{2} a^2 $$ Solution: Step 1: The six designs together form the region between the circular table cover and the regular hexagon inscribed in it. Step 2: Find the area of the circular table cover. $$ A_{\text{circle}} = \pi r^2 = \frac{22}{7} \times 28 \times 28 = 2464 \text{ cm}^2 $$ ...

NCERT Class X Chapter 11: Area Related To Circles Question: Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°. Given: Radius, \( r = 6 \) cm Angle of sector, \( \theta = 60^\circ \) To Find: Area of the sector Formula: Area of a sector = \( \frac{\theta}{360^\circ} \times \pi r^2 \) Solution: Step 1: Write down the given values. $$ r = 6 \text{ cm}, \quad \theta = 60^\circ $$ Step 2: Write the formula for the area of a sector. $$ \text{Area of sector} = \frac{\theta}{360^\circ} \times \pi r^2 $$ Step 3: Substitute the given values into the formula. Use \( \pi = \frac{22}{7} \). $$ \text{Area} = \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times (6)^2 $$ Step 4: Simplify the fraction \( \frac{60}{360} = \frac{1}{6} \) and calculate \( (6)^2 = 36 \). $$ \text{Area} = \frac{1}{6} \times \frac{22}{7} \times 36 $$ Step 5: Multiply the numerators and denominators. $...