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O is any point on the diagonal PR of a parallelogram PQRS. Prove that the areas of triangles PSO and PQO are equal.
Diagonal PR divides parallelogram PQRS into two triangles of equal area: Area(ΔPSR) = Area(ΔPQR) ... (Equation 1). Draw perpendiculars from vertices S and Q to diagonal PR. Since PQRS is a parallelogram, these two altitudes are equal. Triangles OSR and OQR share base OR and have equal altitudes fromRead more
Diagonal PR divides parallelogram PQRS into two triangles of equal area:
Area(ΔPSR) = Area(ΔPQR) … (Equation 1).
Draw perpendiculars from vertices S and Q to diagonal PR. Since PQRS is a parallelogram, these two altitudes are equal.
Triangles OSR and OQR share base OR and have equal altitudes from S and Q:
Area(ΔOSR) = Area(ΔOQR) … (Equation 2).
Subtracting Equation 2 from Equation 1:
Area(ΔPSR) – Area(ΔOSR) = Area(ΔPQR) – Area(ΔOQR)
Area(ΔPSO) = Area(ΔPQO).
Hence proved.
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 lessABCD is a parallelogram. P and Q are any two points on side AB. What can you say about the ratio area (ΔPCD): area (ΔQCD)?
Triangles PCD and QCD stand on the exact same base CD. Because ABCD is a parallelogram, the line AB containing points P and Q is parallel to side CD. Triangles on the same base and between the same parallel lines have identical altitudes (perpendicular heights) to that base. Area(ΔPCD) = (1/2) x CDRead more
Triangles PCD and QCD stand on the exact same base CD.
Because ABCD is a parallelogram, the line AB containing points P and Q is parallel to side CD.
Triangles on the same base and between the same parallel lines have identical altitudes (perpendicular heights) to that base.
Area(ΔPCD) = (1/2) x CD x height
Area(ΔQCD) = (1/2) x CD x height
Since bases and heights are identical, Area(ΔPCD) = Area(ΔQCD).
Thus, the ratio area(ΔPCD) : area(ΔQCD) is 1:1.
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 lessA bullet of mass 50 g moving with a speed of 100 m s-1 enters a heavy stationary wooden block and stops after penetrating a distance of 50 cm. Estimate the stopping force acting on the bullet (assume that the bullet undergoes constant acceleration within the block).
Given mass is 0.05 kilograms, initial velocity is 100 meters per second, final velocity is zero and displacement is 0.5 meters. Using the third kinematic formula, acceleration equals zero minus 10000 divided by twice 0.5, which is minus 10000 meters per second squared. From Newton second law, forceRead more
Given mass is 0.05 kilograms, initial velocity is 100 meters per second, final velocity is zero and displacement is 0.5 meters. Using the third kinematic formula, acceleration equals zero minus 10000 divided by twice 0.5, which is minus 10000 meters per second squared. From Newton second law, force equals mass times acceleration, giving 0.05 multiplied by 10000, yielding an opposing stopping force of 500 newtons.
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 lessThe velocity-time graph of an object of mass 10 kg moving along a straight line is shown in Fig. 6.41. Calculate the force acting on the object by using the graph.
Using the velocity-time graph in Fig. 6.41, the initial velocity at 0 s is 10 m s-1 and final velocity at 8 s is 30 m s-1. The acceleration equals the slope, calculated as 30 minus 10 divided by 8, giving 2.5 m s-2. According to Newton's second law, force equals mass times acceleration, giving 10 kgRead more
Using the velocity-time graph in Fig. 6.41, the initial velocity at 0 s is 10 m s-1 and final velocity at 8 s is 30 m s-1. The acceleration equals the slope, calculated as 30 minus 10 divided by 8, giving 2.5 m s-2. According to Newton’s second law, force equals mass times acceleration, giving 10 kg multiplied by 2.5 m s-2, which equals 25 N.
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 lessThe acceleration-mass graph for the acceleration produced by a force on objects of different masses is plotted in Fig. 6.40. Plot the force-mass graph for this case.
From Fig. 6.40, multiplying mass by acceleration gives 1 kg times 10 m s-2 equaling 10 N, 2 kg times 5 m s-2 equaling 10 N, 4 kg times 2.5 m s-2 equaling 10 N and 5 kg times 2 m s-2 equaling 10 N. Because the force is constant at 10 N for all masses, the force-mass graph is a horizontal straight linRead more
From Fig. 6.40, multiplying mass by acceleration gives 1 kg times 10 m s-2 equaling 10 N, 2 kg times 5 m s-2 equaling 10 N, 4 kg times 2.5 m s-2 equaling 10 N and 5 kg times 2 m s-2 equaling 10 N. Because the force is constant at 10 N for all masses, the force-mass graph is a horizontal straight line at F equals 10 N.
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 less