The perpendicular OV drown form O to chord ST bisects ST. Therefore, SV = VT In △OVS, OV/OS = cos 60° ⇒ OV/12 = 1/2 ⇒ OV = 6 cm SV/OS = sin 60° ⇒ SV/12 = √3/2 ⇒ OV = 6√3 cm ST = 2 x SV = 2 x 6√3 = 12√3 Area of △OST = 1/2 × ST × OV = 1/2 × 12√3 × 6 = 36√3 = 36 x 1.73 = 62.28 cm² Area of sector OSUT =Read more
The perpendicular OV drown form O to chord ST bisects ST.
Therefore, SV = VT
In △OVS,
OV/OS = cos 60° ⇒ OV/12 = 1/2 ⇒ OV = 6 cm
SV/OS = sin 60° ⇒ SV/12 = √3/2 ⇒ OV = 6√3 cm
ST = 2 x SV = 2 x 6√3 = 12√3
Area of △OST
= 1/2 × ST × OV = 1/2 × 12√3 × 6 = 36√3 = 36 x 1.73 = 62.28 cm²
Area of sector OSUT
= 120°/360° x πr² = 1/3 x π(12)²
= 1/3 x 3.14 x 12 x 12 = 150.72 cm²
Area of minor segment
= Area of sector OSUT – Area of △OST
= (150.72 – 62.28) cm²
= 88.44 cm²
The shape of grass field, where the horse can graze is a sector with central angle 90°. (i) The area of the field, where the horse can graze = Area of sector OABO with radius 5 = 90°/360° × πr² = 1/4 × (5)² = 1/4 × 3.14 × 25 = 19.625 m² (ii) If the rope were 10 m long instead of 5 m, the area of fieRead more
The shape of grass field, where the horse can graze is a sector with central angle 90°.
(i) The area of the field, where the horse can graze
= Area of sector OABO with radius 5
= 90°/360° × πr² = 1/4 × (5)²
= 1/4 × 3.14 × 25 = 19.625 m²
(ii) If the rope were 10 m long instead of 5 m, the area of field where the horse can graze = Area of sector 0ABO with radius 10
= 90°/360° × πr² = 1/4 × (10)²
= 1/4 × 3.14 × 100 = 78.50 m²
The increase in grazing area = (78.50 – 19.625) m² = 58.875 m²
Diameter = 35 mm Therefore, radius = 35/2 mm (i) Total length of wire = 5 × Diameter + Circumference = 5 × 35 + 2πr = 175 + 2 × 22/7 × 35/2 = 175 + 110 = 285 mm (ii) There are total 10 sectors. Therefore, the angle of each sector = 360°/10° = 36° Area of each sector = 36°/360° × πr² = 1/10 × π(35/2)Read more
Diameter = 35 mm
Therefore, radius = 35/2 mm
(i) Total length of wire
= 5 × Diameter + Circumference = 5 × 35 + 2πr
= 175 + 2 × 22/7 × 35/2
= 175 + 110 = 285 mm
(ii) There are total 10 sectors.
Therefore, the angle of each sector = 360°/10° = 36°
Area of each sector
= 36°/360° × πr²
= 1/10 × π(35/2)²
= 1/10 × 22/7 × 35/2 × 35/2
= 385/4 mm²
Diameter of wheel = 80 cm Radius of wheel (r) = 40 cm Circumference of wheel = 2πr = 2π (40) = 80m cm Speed of car = 66 km/hour = (66 × 10000)/60 cm/min = 110000 cm/min Distance travelled in 10 minutes = 110000 x 10 1100000 cm Let the wheel makes n revolutions in 10 minutes. Therefore, n × DistanceRead more
Diameter of wheel = 80 cm
Radius of wheel (r) = 40 cm
Circumference of wheel = 2πr = 2π (40) = 80m cm
Speed of car = 66 km/hour = (66 × 10000)/60 cm/min = 110000 cm/min
Distance travelled in 10 minutes = 110000 x 10 1100000 cm
Let the wheel makes n revolutions in 10 minutes. Therefore,
n × Distance travelled in one revolution (circumference) = Distance travelled in 10 minutes
⇒ n × 80π = 1100000
⇒ n = (1100000)/80π = (1100000 x 7)/(80 x 22) = (35000)/8 = 4375
Hence, the wheel takes 4375 revolutions in 10 minutes.
Let the radius of circle = r, Circumference of circle (perimeter) = 2πr, Area of circle = πr² Given that the perimeter and area are equal in magnitude. Therefore 2πr = πr² ⇒ 2 = r Therefore, the radius of circle is 2 units. Hence, the option (A) is correct. See this 😀👇
Let the radius of circle = r, Circumference of circle (perimeter) = 2πr, Area of circle = πr²
Given that the perimeter and area are equal in magnitude. Therefore
2πr = πr²
⇒ 2 = r
Therefore, the radius of circle is 2 units.
Hence, the option (A) is correct.
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)
The perpendicular OV drown form O to chord ST bisects ST. Therefore, SV = VT In △OVS, OV/OS = cos 60° ⇒ OV/12 = 1/2 ⇒ OV = 6 cm SV/OS = sin 60° ⇒ SV/12 = √3/2 ⇒ OV = 6√3 cm ST = 2 x SV = 2 x 6√3 = 12√3 Area of △OST = 1/2 × ST × OV = 1/2 × 12√3 × 6 = 36√3 = 36 x 1.73 = 62.28 cm² Area of sector OSUT =Read more
The perpendicular OV drown form O to chord ST bisects ST.
Therefore, SV = VT
In △OVS,
OV/OS = cos 60° ⇒ OV/12 = 1/2 ⇒ OV = 6 cm
SV/OS = sin 60° ⇒ SV/12 = √3/2 ⇒ OV = 6√3 cm
ST = 2 x SV = 2 x 6√3 = 12√3
Area of △OST
= 1/2 × ST × OV = 1/2 × 12√3 × 6 = 36√3 = 36 x 1.73 = 62.28 cm²
Area of sector OSUT
= 120°/360° x πr² = 1/3 x π(12)²
= 1/3 x 3.14 x 12 x 12 = 150.72 cm²
Area of minor segment
= Area of sector OSUT – Area of △OST
= (150.72 – 62.28) cm²
= 88.44 cm²
See here for video explanation👇😇
See lessA 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 Find.
The shape of grass field, where the horse can graze is a sector with central angle 90°. (i) The area of the field, where the horse can graze = Area of sector OABO with radius 5 = 90°/360° × πr² = 1/4 × (5)² = 1/4 × 3.14 × 25 = 19.625 m² (ii) If the rope were 10 m long instead of 5 m, the area of fieRead more
The shape of grass field, where the horse can graze is a sector with central angle 90°.
See less(i) The area of the field, where the horse can graze
= Area of sector OABO with radius 5
= 90°/360° × πr² = 1/4 × (5)²
= 1/4 × 3.14 × 25 = 19.625 m²
(ii) If the rope were 10 m long instead of 5 m, the area of field where the horse can graze = Area of sector 0ABO with radius 10
= 90°/360° × πr² = 1/4 × (10)²
= 1/4 × 3.14 × 100 = 78.50 m²
The increase in grazing area = (78.50 – 19.625) m² = 58.875 m²
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 as shown in Find.
Diameter = 35 mm Therefore, radius = 35/2 mm (i) Total length of wire = 5 × Diameter + Circumference = 5 × 35 + 2πr = 175 + 2 × 22/7 × 35/2 = 175 + 110 = 285 mm (ii) There are total 10 sectors. Therefore, the angle of each sector = 360°/10° = 36° Area of each sector = 36°/360° × πr² = 1/10 × π(35/2)Read more
Diameter = 35 mm
See lessTherefore, radius = 35/2 mm
(i) Total length of wire
= 5 × Diameter + Circumference = 5 × 35 + 2πr
= 175 + 2 × 22/7 × 35/2
= 175 + 110 = 285 mm
(ii) There are total 10 sectors.
Therefore, the angle of each sector = 360°/10° = 36°
Area of each sector
= 36°/360° × πr²
= 1/10 × π(35/2)²
= 1/10 × 22/7 × 35/2 × 35/2
= 385/4 mm²
The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?
Diameter of wheel = 80 cm Radius of wheel (r) = 40 cm Circumference of wheel = 2πr = 2π (40) = 80m cm Speed of car = 66 km/hour = (66 × 10000)/60 cm/min = 110000 cm/min Distance travelled in 10 minutes = 110000 x 10 1100000 cm Let the wheel makes n revolutions in 10 minutes. Therefore, n × DistanceRead more
Diameter of wheel = 80 cm
Radius of wheel (r) = 40 cm
Circumference of wheel = 2πr = 2π (40) = 80m cm
Speed of car = 66 km/hour = (66 × 10000)/60 cm/min = 110000 cm/min
Distance travelled in 10 minutes = 110000 x 10 1100000 cm
Let the wheel makes n revolutions in 10 minutes. Therefore,
n × Distance travelled in one revolution (circumference) = Distance travelled in 10 minutes
⇒ n × 80π = 1100000
⇒ n = (1100000)/80π = (1100000 x 7)/(80 x 22) = (35000)/8 = 4375
Hence, the wheel takes 4375 revolutions in 10 minutes.
Video explanation of the above question✌😀
See lessTick the correct answer in the following and justify your choice : If the perimeter and the area of a circle are numerically equal, then the radius of the circle is
Let the radius of circle = r, Circumference of circle (perimeter) = 2πr, Area of circle = πr² Given that the perimeter and area are equal in magnitude. Therefore 2πr = πr² ⇒ 2 = r Therefore, the radius of circle is 2 units. Hence, the option (A) is correct. See this 😀👇
Let the radius of circle = r, Circumference of circle (perimeter) = 2πr, Area of circle = πr²
Given that the perimeter and area are equal in magnitude. Therefore
2πr = πr²
⇒ 2 = r
Therefore, the radius of circle is 2 units.
Hence, the option (A) is correct.
See this 😀👇
See less