The possible outcomes are {HHH, TTT, HHT, HTH, THH, TTH, THT, HTT} Number of total possible outcomes = 8 Number of favourable outcomes = 2 {i.e., TTT and HHH}} P (Hanif will win the game) = 2/8 = 1/4 P (Hanif will lose the game) = 1 - 1/4 = 3/4
The possible outcomes are {HHH, TTT, HHT, HTH, THH, TTH, THT, HTT}
Number of total possible outcomes = 8
Number of favourable outcomes = 2 {i.e., TTT and HHH}}
P (Hanif will win the game) = 2/8 = 1/4
P (Hanif will lose the game) = 1 – 1/4 = 3/4
Side of square is 14 cm, therefore, the radius of semicircle is 7 cm. Area of semicircle = 1/2 × πr² = 1/2 × π(7)² = 1/2 × 22/7 × 7 × 7 = 77 cm² Area of square = (side)² = (14)² = 196 cm Area of shaded region = Area of square - Area of two semicircles = 196 - 2 × 77 = 196 - 154 42 cm² See this videoRead more
Side of square is 14 cm, therefore, the radius of semicircle is 7 cm.
Area of semicircle
= 1/2 × πr² = 1/2 × π(7)² = 1/2 × 22/7 × 7 × 7 = 77 cm²
Area of square
= (side)² = (14)² = 196 cm
Area of shaded region = Area of square – Area of two semicircles
= 196 – 2 × 77 = 196 – 154 42 cm²
Radius of smaller circle = 7 cm Radius of larger circle = 14 cm Area of shaded region = Area of sector OAFCO - Area of sector OBEDO = 40°/360° × π(14)² - 40°/360° × π(7)² = 1/9 × 22/7 × 14 × 14 - 1/9 × 22/7 × 7 × 7 = 616/9 - 154/9 = 462/9 = 154/3 cm² Here is the video explanation of the above questiRead more
Radius of smaller circle = 7 cm
Radius of larger circle = 14 cm
Area of shaded region
= Area of sector OAFCO – Area of sector OBEDO
= 40°/360° × π(14)² – 40°/360° × π(7)²
= 1/9 × 22/7 × 14 × 14 – 1/9 × 22/7 × 7 × 7
= 616/9 – 154/9 = 462/9 = 154/3 cm²
Here is the video explanation of the above question😀👇
Radius of each quadrant = 1 cm Area of each quadrant = 90°/360° × πr² = 1/4 × π(1)² = 1/4 × 22/7 × 1 × 1 = 11/14 cm² Area of circle = πr² = π(1)² = 22/7 × 1 × 1 = 22/7 cm² Area of square = (Side)² = (4)² = 16 cm² Area of shaded region = Area of square Area of circle - Area of 4 quadrants 16 - 22/7 -Read more
Radius of each quadrant = 1 cm
Area of each quadrant
= 90°/360° × πr² = 1/4 × π(1)² = 1/4 × 22/7 × 1 × 1 = 11/14 cm²
Area of circle
= πr² = π(1)² = 22/7 × 1 × 1 = 22/7 cm²
Area of square = (Side)² = (4)² = 16 cm²
Area of shaded region = Area of square Area of circle – Area of 4 quadrants
16 – 22/7 – 4 × 11/14 = 16 – 22/7 – 22/7 = (112 – 44)/7 = 68/7 cm²
LHS = sin((B+C)/2) = sin ((180°-A)/2) [∵ A +B +C = 180°] = sin (90°- A/2) = cos A/2 = RHS [∵ sin (90° - 0) = cos 0] Here is the explanation video of the above question😎👇
LHS = sin((B+C)/2)
= sin ((180°-A)/2) [∵ A +B +C = 180°]
= sin (90°- A/2)
= cos A/2 = RHS [∵ sin (90° – 0) = cos 0]
Here is the explanation video of the above question😎👇
A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
The possible outcomes are {HHH, TTT, HHT, HTH, THH, TTH, THT, HTT} Number of total possible outcomes = 8 Number of favourable outcomes = 2 {i.e., TTT and HHH}} P (Hanif will win the game) = 2/8 = 1/4 P (Hanif will lose the game) = 1 - 1/4 = 3/4
The possible outcomes are {HHH, TTT, HHT, HTH, THH, TTH, THT, HTT}
See lessNumber of total possible outcomes = 8
Number of favourable outcomes = 2 {i.e., TTT and HHH}}
P (Hanif will win the game) = 2/8 = 1/4
P (Hanif will lose the game) = 1 – 1/4 = 3/4
Find the area of the shaded region in Figure.
Side of square is 14 cm, therefore, the radius of semicircle is 7 cm. Area of semicircle = 1/2 × πr² = 1/2 × π(7)² = 1/2 × 22/7 × 7 × 7 = 77 cm² Area of square = (side)² = (14)² = 196 cm Area of shaded region = Area of square - Area of two semicircles = 196 - 2 × 77 = 196 - 154 42 cm² See this videoRead more
Side of square is 14 cm, therefore, the radius of semicircle is 7 cm.
Area of semicircle
= 1/2 × πr² = 1/2 × π(7)² = 1/2 × 22/7 × 7 × 7 = 77 cm²
Area of square
= (side)² = (14)² = 196 cm
Area of shaded region = Area of square – Area of two semicircles
= 196 – 2 × 77 = 196 – 154 42 cm²
See this video explanation 😎👇
See lessFind the area of the shaded region in Figure.
Radius of smaller circle = 7 cm Radius of larger circle = 14 cm Area of shaded region = Area of sector OAFCO - Area of sector OBEDO = 40°/360° × π(14)² - 40°/360° × π(7)² = 1/9 × 22/7 × 14 × 14 - 1/9 × 22/7 × 7 × 7 = 616/9 - 154/9 = 462/9 = 154/3 cm² Here is the video explanation of the above questiRead more
Radius of smaller circle = 7 cm
Radius of larger circle = 14 cm
Area of shaded region
= Area of sector OAFCO – Area of sector OBEDO
= 40°/360° × π(14)² – 40°/360° × π(7)²
= 1/9 × 22/7 × 14 × 14 – 1/9 × 22/7 × 7 × 7
= 616/9 – 154/9 = 462/9 = 154/3 cm²
Here is the video explanation of the above question😀👇
See lessFrom each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in Figure. Find the area of the remaining portion of the square.
Radius of each quadrant = 1 cm Area of each quadrant = 90°/360° × πr² = 1/4 × π(1)² = 1/4 × 22/7 × 1 × 1 = 11/14 cm² Area of circle = πr² = π(1)² = 22/7 × 1 × 1 = 22/7 cm² Area of square = (Side)² = (4)² = 16 cm² Area of shaded region = Area of square Area of circle - Area of 4 quadrants 16 - 22/7 -Read more
Radius of each quadrant = 1 cm
See lessArea of each quadrant
= 90°/360° × πr² = 1/4 × π(1)² = 1/4 × 22/7 × 1 × 1 = 11/14 cm²
Area of circle
= πr² = π(1)² = 22/7 × 1 × 1 = 22/7 cm²
Area of square = (Side)² = (4)² = 16 cm²
Area of shaded region = Area of square Area of circle – Area of 4 quadrants
16 – 22/7 – 4 × 11/14 = 16 – 22/7 – 22/7 = (112 – 44)/7 = 68/7 cm²
If A, B and C are interior angles of a triangle ABC, then show that Sin(B+C/2) = cos A/2.
LHS = sin((B+C)/2) = sin ((180°-A)/2) [∵ A +B +C = 180°] = sin (90°- A/2) = cos A/2 = RHS [∵ sin (90° - 0) = cos 0] Here is the explanation video of the above question😎👇
LHS = sin((B+C)/2)
= sin ((180°-A)/2) [∵ A +B +C = 180°]
= sin (90°- A/2)
= cos A/2 = RHS [∵ sin (90° – 0) = cos 0]
Here is the explanation video of the above question😎👇
See less