To find the class mark (xᵢ) for each interval, the folowing relation is used. xᵢ = (Upper limit + Lower limit)/2 Given that the mean nocket allawance is ₹18 Taking 18 as.assured mean (a), dᵢ and fᵢdᵢ are calculated as follows: From the table, we obtain ∑fᵢ = 44 + f, ∑fᵢdᵢ = 2f - 40 and a = 18 mean(XRead more
To find the class mark (xᵢ) for each interval, the folowing relation is used.
xᵢ = (Upper limit + Lower limit)/2
Given that the mean nocket allawance is ₹18 Taking 18 as.assured mean (a), dᵢ and fᵢdᵢ are calculated as follows:
From the table, we obtain
∑fᵢ = 44 + f, ∑fᵢdᵢ = 2f – 40 and a = 18
mean(X̄) = a + (∑fᵢdᵢ)/∑fᵢ ⇒ 18 = 18 + ((2F – 40)/(44 + F)) ⇒ 0 = (2F – 40)/(44 + f) ⇒ 2f – 40 = 0 ⇒ f = 20
Hence, the value of missing frequency f is 20.
We know that, the minute hand makes an angle of 360° in one hour. Hence, angle formed in 5 minutes = 360°/60° x 5 = 30° Therefore, area swept be the minute hand in 5 minutes. = Area of sector of angle 30° = 30° x πr² = 1/12 x π(14) ² = 1/12 x 22/7 x 14 x 14 = 154/3 cm²
We know that, the minute hand makes an angle of 360° in one hour.
Hence, angle formed in 5 minutes
= 360°/60° x 5 = 30°
Therefore, area swept be the minute hand in 5 minutes.
= Area of sector of angle 30°
= 30° x πr²
= 1/12 x π(14) ²
= 1/12 x 22/7 x 14 x 14
= 154/3 cm²
(i) Let the chord AB subtends right angle at the centre O. Area of segment with an angle 90° = 90°/360° × πr² = 1/4 × π(10)² = 1/4 × 3.14 × 10 × 10 = 78.5 cm² Area of triangle OAB = 1/2 × OA × OB = 1/2 × 10 × 10 = 50 cm² Area of minor segment = Area of minor sector Area of triangle OAB = 78.5 cm² -Read more
(i) Let the chord AB subtends right angle at the centre O.
Area of segment with an angle 90°
= 90°/360° × πr² = 1/4 × π(10)² = 1/4 × 3.14 × 10 × 10 = 78.5 cm²
Area of triangle OAB
= 1/2 × OA × OB = 1/2 × 10 × 10 = 50 cm²
Area of minor segment = Area of minor sector Area of triangle OAB
= 78.5 cm² – 50 cm² = 28.5 cm²
(ii) Area of major segment = Area of cirlce Area of minor segment
= πr² – 1/4πr² = (1 – 1/4)πr² = 3/4πr²
= 3/4 × π(10)² = 3/4 × 3.14 × 10 × 10 = 235.5 cm²
Radius of circle = 21 cm (i) The length of arc = θ/360° × 2πr = 60°/360° × 2πr = 1/6 × 2 × π × 21 = 1/6 × 2 × 22/7 × 21 = 22cm (ii) Area of the sector formed by the arc = 60°/360° × πr² = 1/6 × π(21)² = 1/6 × 22/7 × 21 × 21 = 231 cm² (iii) Area of the segment formed by the corresponding chord Area oRead more
Radius of circle = 21 cm
(i) The length of arc
= θ/360° × 2πr = 60°/360° × 2πr
= 1/6 × 2 × π × 21 = 1/6 × 2 × 22/7 × 21 = 22cm
(ii) Area of the sector formed by the arc
= 60°/360° × πr²
= 1/6 × π(21)²
= 1/6 × 22/7 × 21 × 21 = 231 cm²
(iii) Area of the segment formed by the corresponding chord
Area of sector – Area of △0AB
= 231 cm² – √3/4(21)²cm² [As triangle OAB is an equilateral triangle]
= 231 cm² – 441 √3/4cm²
= [231 – (441√3)/4]cm²
Radius of circle 15 cm Area of sector OPRQ = 60°/360° × πr² = 1/6 × π(15)² = 1/6 × 3.14 × 15 × 15 = 117.75 cm² Area of minor segment = Area of minor sector OPRQ - Area of △0PQ = 117.75 cm² - √3/4(15) cm² [As triangle OPQ is an equilateral triangle] = 117.75 cm² - 225√3/4 cm² = 117.75 cm² - 56.25 x 1Read more
Radius of circle 15 cm
Area of sector OPRQ
= 60°/360° × πr² = 1/6 × π(15)² = 1/6 × 3.14 × 15 × 15 = 117.75 cm²
Area of minor segment
= Area of minor sector OPRQ – Area of △0PQ
= 117.75 cm² – √3/4(15) cm² [As triangle OPQ is an equilateral triangle]
= 117.75 cm² – 225√3/4 cm²
= 117.75 cm² – 56.25 x 1.73 cm²
= 231 cm² – 97.3125 cm²
= 20.4375 cm²
Area of major segment = Area of circle – Area of minor segment
= πr² – 20.4375 cm²
= π(15)² – 20.4375] cm²
= [3.14 x 15 x 15 – 20.4375] cm²
= [706.5 – 20.4375] cm²
= 686.0625 cm²
For better understanding to the above question see here😎👇
The following distribution shows the daily pocket allowance of children of a locality.
To find the class mark (xᵢ) for each interval, the folowing relation is used. xᵢ = (Upper limit + Lower limit)/2 Given that the mean nocket allawance is ₹18 Taking 18 as.assured mean (a), dᵢ and fᵢdᵢ are calculated as follows: From the table, we obtain ∑fᵢ = 44 + f, ∑fᵢdᵢ = 2f - 40 and a = 18 mean(XRead more
To find the class mark (xᵢ) for each interval, the folowing relation is used.
xᵢ = (Upper limit + Lower limit)/2
Given that the mean nocket allawance is ₹18 Taking 18 as.assured mean (a), dᵢ and fᵢdᵢ are calculated as follows:
From the table, we obtain
∑fᵢ = 44 + f, ∑fᵢdᵢ = 2f – 40 and a = 18
mean(X̄) = a + (∑fᵢdᵢ)/∑fᵢ ⇒ 18 = 18 + ((2F – 40)/(44 + F)) ⇒ 0 = (2F – 40)/(44 + f) ⇒ 2f – 40 = 0 ⇒ f = 20
Hence, the value of missing frequency f is 20.
See this video explanation 😮✌👇
See lessThe length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
We know that, the minute hand makes an angle of 360° in one hour. Hence, angle formed in 5 minutes = 360°/60° x 5 = 30° Therefore, area swept be the minute hand in 5 minutes. = Area of sector of angle 30° = 30° x πr² = 1/12 x π(14) ² = 1/12 x 22/7 x 14 x 14 = 154/3 cm²
We know that, the minute hand makes an angle of 360° in one hour.
See lessHence, angle formed in 5 minutes
= 360°/60° x 5 = 30°
Therefore, area swept be the minute hand in 5 minutes.
= Area of sector of angle 30°
= 30° x πr²
= 1/12 x π(14) ²
= 1/12 x 22/7 x 14 x 14
= 154/3 cm²
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)
(i) Let the chord AB subtends right angle at the centre O. Area of segment with an angle 90° = 90°/360° × πr² = 1/4 × π(10)² = 1/4 × 3.14 × 10 × 10 = 78.5 cm² Area of triangle OAB = 1/2 × OA × OB = 1/2 × 10 × 10 = 50 cm² Area of minor segment = Area of minor sector Area of triangle OAB = 78.5 cm² -Read more
(i) Let the chord AB subtends right angle at the centre O.
See lessArea of segment with an angle 90°
= 90°/360° × πr² = 1/4 × π(10)² = 1/4 × 3.14 × 10 × 10 = 78.5 cm²
Area of triangle OAB
= 1/2 × OA × OB = 1/2 × 10 × 10 = 50 cm²
Area of minor segment = Area of minor sector Area of triangle OAB
= 78.5 cm² – 50 cm² = 28.5 cm²
(ii) Area of major segment = Area of cirlce Area of minor segment
= πr² – 1/4πr² = (1 – 1/4)πr² = 3/4πr²
= 3/4 × π(10)² = 3/4 × 3.14 × 10 × 10 = 235.5 cm²
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
Radius of circle = 21 cm (i) The length of arc = θ/360° × 2πr = 60°/360° × 2πr = 1/6 × 2 × π × 21 = 1/6 × 2 × 22/7 × 21 = 22cm (ii) Area of the sector formed by the arc = 60°/360° × πr² = 1/6 × π(21)² = 1/6 × 22/7 × 21 × 21 = 231 cm² (iii) Area of the segment formed by the corresponding chord Area oRead more
Radius of circle = 21 cm
(i) The length of arc
= θ/360° × 2πr = 60°/360° × 2πr
= 1/6 × 2 × π × 21 = 1/6 × 2 × 22/7 × 21 = 22cm
(ii) Area of the sector formed by the arc
= 60°/360° × πr²
= 1/6 × π(21)²
= 1/6 × 22/7 × 21 × 21 = 231 cm²
(iii) Area of the segment formed by the corresponding chord
Area of sector – Area of △0AB
= 231 cm² – √3/4(21)²cm² [As triangle OAB is an equilateral triangle]
= 231 cm² – 441 √3/4cm²
= [231 – (441√3)/4]cm²
See here for video explanation😎👇
See lessA chord of a circle of radius 15 cm subtends an angle of 60° at the Centre. Find the areas of the corresponding 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 π= 3.14 and √3 = 1.73)minor and major segments of the circle. (Use π= 3.14 and √3 = 1.73)
Radius of circle 15 cm Area of sector OPRQ = 60°/360° × πr² = 1/6 × π(15)² = 1/6 × 3.14 × 15 × 15 = 117.75 cm² Area of minor segment = Area of minor sector OPRQ - Area of △0PQ = 117.75 cm² - √3/4(15) cm² [As triangle OPQ is an equilateral triangle] = 117.75 cm² - 225√3/4 cm² = 117.75 cm² - 56.25 x 1Read more
Radius of circle 15 cm
Area of sector OPRQ
= 60°/360° × πr² = 1/6 × π(15)² = 1/6 × 3.14 × 15 × 15 = 117.75 cm²
Area of minor segment
= Area of minor sector OPRQ – Area of △0PQ
= 117.75 cm² – √3/4(15) cm² [As triangle OPQ is an equilateral triangle]
= 117.75 cm² – 225√3/4 cm²
= 117.75 cm² – 56.25 x 1.73 cm²
= 231 cm² – 97.3125 cm²
= 20.4375 cm²
Area of major segment = Area of circle – Area of minor segment
= πr² – 20.4375 cm²
= π(15)² – 20.4375] cm²
= [3.14 x 15 x 15 – 20.4375] cm²
= [706.5 – 20.4375] cm²
= 686.0625 cm²
For better understanding to the above question see here😎👇
See less