The distance travelled by the girl is given by the area under the velocity-time graph. We divide the graph into convenient trapeziums and estimate their areas using the velocities at different times. Adding these areas gives the approximate distance covered during the run. From the values shown in FRead more
The distance travelled by the girl is given by the area under the velocity-time graph. We divide the graph into convenient trapeziums and estimate their areas using the velocities at different times. Adding these areas gives the approximate distance covered during the run. From the values shown in Fig. 4.31, the girl ran approximately 45 km. Therefore, her estimated distance travelled is about 45 km.
For more NCERT Solutions of Class 9 Science Exploration Chapter 4 Describing Motion Around Us Question Answer (2026-27)
From the velocity-time graph: Constant velocity (20–100 s): displacement = area of rectangle = 3 × 80 = 240 m. Decreasing velocity (100–120 s): displacement = area of trapezium = ½(3+2) × 20 = 50 m. Initial part (0–20 s): displacement = area of triangle = ½ × 20 × 3 = 30 m. Therefore, total displaceRead more
From the velocity-time graph:
Constant velocity (20–100 s): displacement = area of rectangle = 3 × 80 = 240 m.
Decreasing velocity (100–120 s): displacement = area of trapezium = ½(3+2) × 20 = 50 m.
Initial part (0–20 s): displacement = area of triangle = ½ × 20 × 3 = 30 m.
Therefore, total displacement = 30 + 240 + 50 = 320 m. Average acceleration = (2−0)/120 = 0.0167 m s⁻².
For more NCERT Solutions of Class 9 Science Exploration Chapter 4 Describing Motion Around Us Question Answer (2026-27)
In the figure, OA, OB, OC and OD are equal radii. Therefore, the triangles formed by these radii are isosceles. Since BOD is a diameter, the angles around the centre give: p + v = 90° and similarly, q + u = 90°. Now, ∠A = p + v = 90° and ∠C = q + u = 90° in the corresponding construction, so the oppRead more
In the figure, OA, OB, OC and OD are equal radii. Therefore, the triangles formed by these radii are isosceles. Since BOD is a diameter, the angles around the centre give:
p + v = 90°
and similarly,
q + u = 90°.
Now, ∠A = p + v = 90° and ∠C = q + u = 90° in the corresponding construction, so the opposite angles together give 180°. Thus, the sum of the opposite angles of a cyclic quadrilateral is 180°.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):
There appears to be a problem in the question as shown in the screenshot. If the two chords CC' and DD' are each drawn perpendicular to the diameter AB, then the perpendicular from the centre to each chord bisects that chord. Since AB passes through the centre, the midpoints M and M' of these chordsRead more
There appears to be a problem in the question as shown in the screenshot. If the two chords CC’ and DD’ are each drawn perpendicular to the diameter AB, then the perpendicular from the centre to each chord bisects that chord. Since AB passes through the centre, the midpoints M and M’ of these chords lie on AB. Therefore, the segment MM’ joining the midpoints lies along AB. Hence, MM’ is parallel to AB, not perpendicular to AB. So the printed statement cannot be proved as written.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):
Let the endpoints of the diameter be B and C, and let A be a point on the semicircle. Join OA. Since OA, OB and OC are radii, OA = OB = OC. Therefore, the two triangles formed with OA are isosceles. If the angles at B and C are a and b, the corresponding central angles are 180° - 2a and 180° - 2b. SRead more
Let the endpoints of the diameter be B and C, and let A be a point on the semicircle. Join OA. Since OA, OB and OC are radii, OA = OB = OC. Therefore, the two triangles formed with OA are isosceles. If the angles at B and C are a and b, the corresponding central angles are 180° – 2a and 180° – 2b. Since BOC is a straight angle, a + b = 90°. Hence, ∠BAC = 90°.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):
A girl is preparing for her first marathon by running on a straight road. She uses a smartwatch to calculate her running speed at different intervals. The graph (Fig. 4.31) depicts her velocity versus time. Estimate the distance she ran based on the graph.
The distance travelled by the girl is given by the area under the velocity-time graph. We divide the graph into convenient trapeziums and estimate their areas using the velocities at different times. Adding these areas gives the approximate distance covered during the run. From the values shown in FRead more
The distance travelled by the girl is given by the area under the velocity-time graph. We divide the graph into convenient trapeziums and estimate their areas using the velocities at different times. Adding these areas gives the approximate distance covered during the run. From the values shown in Fig. 4.31, the girl ran approximately 45 km. Therefore, her estimated distance travelled is about 45 km.
For more NCERT Solutions of Class 9 Science Exploration Chapter 4 Describing Motion Around Us Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-4/
See lessThe velocity-time graph from 0 s to 120 s for a cyclist is shown in Fig. 4.30. Shade the areas (in different colours) representing the displacement of the cyclist (i) while cyclist is moving with constant velocity. (ii) when the velocity of cyclist is decreasing. Also, calculate the displacement and average acceleration in the 120 s time interval.
From the velocity-time graph: Constant velocity (20–100 s): displacement = area of rectangle = 3 × 80 = 240 m. Decreasing velocity (100–120 s): displacement = area of trapezium = ½(3+2) × 20 = 50 m. Initial part (0–20 s): displacement = area of triangle = ½ × 20 × 3 = 30 m. Therefore, total displaceRead more
From the velocity-time graph:
Therefore, total displacement = 30 + 240 + 50 = 320 m. Average acceleration = (2−0)/120 = 0.0167 m s⁻².
For more NCERT Solutions of Class 9 Science Exploration Chapter 4 Describing Motion Around Us Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-4/
See lessHow would you use the following figure to justify the statement that the sum of the opposite angles of a cyclic quadrilateral is 180°?
In the figure, OA, OB, OC and OD are equal radii. Therefore, the triangles formed by these radii are isosceles. Since BOD is a diameter, the angles around the centre give: p + v = 90° and similarly, q + u = 90°. Now, ∠A = p + v = 90° and ∠C = q + u = 90° in the corresponding construction, so the oppRead more
In the figure, OA, OB, OC and OD are equal radii. Therefore, the triangles formed by these radii are isosceles. Since BOD is a diameter, the angles around the centre give:
p + v = 90°
and similarly,
q + u = 90°.
Now, ∠A = p + v = 90° and ∠C = q + u = 90° in the corresponding construction, so the opposite angles together give 180°. Thus, the sum of the opposite angles of a cyclic quadrilateral is 180°.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-5/
See lessIn a circle, two chords CC’ and DD’ are drawn perpendicular to a diameter AB. Prove that the segment MM’ joining the midpoints of the chords CD and C’D’ is perpendicular to AB.
There appears to be a problem in the question as shown in the screenshot. If the two chords CC' and DD' are each drawn perpendicular to the diameter AB, then the perpendicular from the centre to each chord bisects that chord. Since AB passes through the centre, the midpoints M and M' of these chordsRead more
There appears to be a problem in the question as shown in the screenshot. If the two chords CC’ and DD’ are each drawn perpendicular to the diameter AB, then the perpendicular from the centre to each chord bisects that chord. Since AB passes through the centre, the midpoints M and M’ of these chords lie on AB. Therefore, the segment MM’ joining the midpoints lies along AB. Hence, MM’ is parallel to AB, not perpendicular to AB. So the printed statement cannot be proved as written.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-5/
See lessHow would you use the following figure to justify the statement that the angle in a semicircle is 90°?
Let the endpoints of the diameter be B and C, and let A be a point on the semicircle. Join OA. Since OA, OB and OC are radii, OA = OB = OC. Therefore, the two triangles formed with OA are isosceles. If the angles at B and C are a and b, the corresponding central angles are 180° - 2a and 180° - 2b. SRead more
Let the endpoints of the diameter be B and C, and let A be a point on the semicircle. Join OA. Since OA, OB and OC are radii, OA = OB = OC. Therefore, the two triangles formed with OA are isosceles. If the angles at B and C are a and b, the corresponding central angles are 180° – 2a and 180° – 2b. Since BOC is a straight angle, a + b = 90°. Hence, ∠BAC = 90°.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-5/
See less