Given perimeter = 32 cm and two sides a = 8 cm, b = 11 cm. Third side c = 32 - (8 + 11) = 32 - 19 = 13 cm. Semi-perimeter s = perimeter / 2 = 32 / 2 = 16 cm. Now apply Heron's formula: Area = √(s x (s - a) x (s - b) x (s - c)) Area = √(16 x (16 - 8) x (16 - 11) x (16 - 13)) Area = √(16 x 8 x 5 x 3)Read more
Given perimeter = 32 cm and two sides a = 8 cm, b = 11 cm.
Third side c = 32 – (8 + 11) = 32 – 19 = 13 cm.
Semi-perimeter s = perimeter / 2 = 32 / 2 = 16 cm.
Now apply Heron’s formula:
Area = √(s x (s – a) x (s – b) x (s – c))
Area = √(16 x (16 – 8) x (16 – 11) x (16 – 13))
Area = √(16 x 8 x 5 x 3)
Area = √(16 x 4 x 2 x 15)
Area = 4 x 2 x √30 = 8√30 cm² (approximately 43.82 cm²).
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
Let the sides be 3x, 5x and 7x meters. Perimeter = 3x + 5x + 7x = 300 m. 15x = 300, which gives x = 20 m. The side lengths are: a = 3 x 20 = 60 m b = 5 x 20 = 100 m c = 7 x 20 = 140 m. Semi-perimeter s = 300 / 2 = 150 m. Using Heron's formula: Area = √(150 x (150 - 60) x (150 - 100) x (150 - 140)) ARead more
Let the sides be 3x, 5x and 7x meters.
Perimeter = 3x + 5x + 7x = 300 m.
15x = 300, which gives x = 20 m.
The side lengths are:
a = 3 x 20 = 60 m
b = 5 x 20 = 100 m
c = 7 x 20 = 140 m.
Semi-perimeter s = 300 / 2 = 150 m.
Using Heron’s formula:
Area = √(150 x (150 – 60) x (150 – 100) x (150 – 140))
Area = √(150 x 90 x 50 x 10)
Area = √(67,500,000) = 1500√3 m² (approx 2598.08 m²).
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
In (i), according to Newton's first law, an object continues moving at constant velocity without external force, so it will remain the same. In (ii), a forward net force causes acceleration in the direction of motion, so the magnitude of velocity will increase. In (iii), an opposing net force createRead more
In (i), according to Newton’s first law, an object continues moving at constant velocity without external force, so it will remain the same. In (ii), a forward net force causes acceleration in the direction of motion, so the magnitude of velocity will increase. In (iii), an opposing net force creates negative acceleration opposing motion, so the magnitude of velocity will decrease.
For more NCERT Solutions of Class 9 Science Exploration Chapter 6 How Forces Affect Motion Question Answer (2026-27)
Since the table moves across the floor at a constant velocity, its acceleration is zero. By Newton's first law of motion, the net external horizontal force acting on the table must be zero. Therefore, the opposing frictional force exerted by the floor on the table must be exactly equal in magnitudeRead more
Since the table moves across the floor at a constant velocity, its acceleration is zero. By Newton’s first law of motion, the net external horizontal force acting on the table must be zero. Therefore, the opposing frictional force exerted by the floor on the table must be exactly equal in magnitude to the applied horizontal force F, acting in the opposite direction.
For more NCERT Solutions of Class 9 Science Exploration Chapter 6 How Forces Affect Motion Question Answer (2026-27)
Using conservation of energy, the velocity before impact equals square root of two times g times h, which is square root of 200, giving 14.14 m s-1. The kinetic energy on impact equals initial potential energy m g h, totaling 150 J. Equating this energy to work done against sand resistance, 150 J diRead more
Using conservation of energy, the velocity before impact equals square root of two times g times h, which is square root of 200, giving 14.14 m s-1. The kinetic energy on impact equals initial potential energy m g h, totaling 150 J. Equating this energy to work done against sand resistance, 150 J divided by 3000 N yields a depression depth of 0.05 m or 5 cm.
For more NCERT Solutions of Class 9 Science Exploration Chapter 7 Work, Energy and Simple Machines Question Answer (2026-27)
Find the area of a triangle, given that its sides are 8 cm and 11 cm long, and its perimeter is 32 cm.
Given perimeter = 32 cm and two sides a = 8 cm, b = 11 cm. Third side c = 32 - (8 + 11) = 32 - 19 = 13 cm. Semi-perimeter s = perimeter / 2 = 32 / 2 = 16 cm. Now apply Heron's formula: Area = √(s x (s - a) x (s - b) x (s - c)) Area = √(16 x (16 - 8) x (16 - 11) x (16 - 13)) Area = √(16 x 8 x 5 x 3)Read more
Given perimeter = 32 cm and two sides a = 8 cm, b = 11 cm.
Third side c = 32 – (8 + 11) = 32 – 19 = 13 cm.
Semi-perimeter s = perimeter / 2 = 32 / 2 = 16 cm.
Now apply Heron’s formula:
Area = √(s x (s – a) x (s – b) x (s – c))
Area = √(16 x (16 – 8) x (16 – 11) x (16 – 13))
Area = √(16 x 8 x 5 x 3)
Area = √(16 x 4 x 2 x 15)
Area = 4 x 2 x √30 = 8√30 cm² (approximately 43.82 cm²).
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-6/
See lessThe sides of a triangular plot are in the ratio 3: 5: 7; its perimeter is 300 m. Find its area.
Let the sides be 3x, 5x and 7x meters. Perimeter = 3x + 5x + 7x = 300 m. 15x = 300, which gives x = 20 m. The side lengths are: a = 3 x 20 = 60 m b = 5 x 20 = 100 m c = 7 x 20 = 140 m. Semi-perimeter s = 300 / 2 = 150 m. Using Heron's formula: Area = √(150 x (150 - 60) x (150 - 100) x (150 - 140)) ARead more
Let the sides be 3x, 5x and 7x meters.
Perimeter = 3x + 5x + 7x = 300 m.
15x = 300, which gives x = 20 m.
The side lengths are:
a = 3 x 20 = 60 m
b = 5 x 20 = 100 m
c = 7 x 20 = 140 m.
Semi-perimeter s = 300 / 2 = 150 m.
Using Heron’s formula:
Area = √(150 x (150 – 60) x (150 – 100) x (150 – 140))
Area = √(150 x 90 x 50 x 10)
Area = √(67,500,000) = 1500√3 m² (approx 2598.08 m²).
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-6/
See lessFor a ball moving on a smooth frictionless surface, choose the appropriate option that will make the following statements physically correct. (i) If no net force is applied on the ball, the velocity of the ball will remain the same/increase/decrease. (ii) If a net force is applied on the ball in the direction of its motion, the magnitude of the velocity of the ball will remain the same/increase/decrease. (iii) If a net force is applied on the ball in a direction opposite to the direction of its motion, the magnitude of the velocity of the ball will remain the same/increase/decrease.
In (i), according to Newton's first law, an object continues moving at constant velocity without external force, so it will remain the same. In (ii), a forward net force causes acceleration in the direction of motion, so the magnitude of velocity will increase. In (iii), an opposing net force createRead more
In (i), according to Newton’s first law, an object continues moving at constant velocity without external force, so it will remain the same. In (ii), a forward net force causes acceleration in the direction of motion, so the magnitude of velocity will increase. In (iii), an opposing net force creates negative acceleration opposing motion, so the magnitude of velocity will decrease.
For more NCERT Solutions of Class 9 Science Exploration Chapter 6 How Forces Affect Motion Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-6/
See lessUsing a horizontal force F, a table is moved across the floor at a constant velocity. How much is the frictional force exerted by the floor on the table?
Since the table moves across the floor at a constant velocity, its acceleration is zero. By Newton's first law of motion, the net external horizontal force acting on the table must be zero. Therefore, the opposing frictional force exerted by the floor on the table must be exactly equal in magnitudeRead more
Since the table moves across the floor at a constant velocity, its acceleration is zero. By Newton’s first law of motion, the net external horizontal force acting on the table must be zero. Therefore, the opposing frictional force exerted by the floor on the table must be exactly equal in magnitude to the applied horizontal force F, acting in the opposite direction.
For more NCERT Solutions of Class 9 Science Exploration Chapter 6 How Forces Affect Motion Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-6/
See lessA coconut of mass 1.5 kg falls from the top of a coconut tree onto the wet sand on a beach. The height of the tree is 10 m. On impact, the coconut comes to rest by making a depression in the sand. (i) Calculate the velocity of the coconut just before it hits the sand. (ii) Assume that the average resistive force of sand is 3000 N and all of the coconut’s energy is used to create the depression in the sand. Calculate the depth of the depression the coconut makes in the sand. Assume g = 10 m s-2.
Using conservation of energy, the velocity before impact equals square root of two times g times h, which is square root of 200, giving 14.14 m s-1. The kinetic energy on impact equals initial potential energy m g h, totaling 150 J. Equating this energy to work done against sand resistance, 150 J diRead more
Using conservation of energy, the velocity before impact equals square root of two times g times h, which is square root of 200, giving 14.14 m s-1. The kinetic energy on impact equals initial potential energy m g h, totaling 150 J. Equating this energy to work done against sand resistance, 150 J divided by 3000 N yields a depression depth of 0.05 m or 5 cm.
For more NCERT Solutions of Class 9 Science Exploration Chapter 7 Work, Energy and Simple Machines Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-6/
See less