From Fig. 6.42, the trapezium has parallel sides of lengths a and b and perpendicular height h. The segment of length a on the top base partitions the shape into: A parallelogram with base a and height h. Area(parallelogram) = base x height = ah. A triangle with base (b - a) and height h. Area(trianRead more
From Fig. 6.42, the trapezium has parallel sides of lengths a and b and perpendicular height h.
The segment of length a on the top base partitions the shape into:
A parallelogram with base a and height h.
Area(parallelogram) = base x height = ah.
A triangle with base (b – a) and height h.
Area(triangle) = (1/2) x base x height = (1/2)(b – a)h.
Total area = ah + (1/2)(b – a)h
= (1/2)h [2a + b – a]
= (1/2)(a + b)h.
Hence proved.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
Yes, they must be congruent. Let the sides of a rectangle be x and y. Perimeter = 2(x + y) = 2s, so x + y = s. Area = xy = A. Here, x and y are the roots of the quadratic equation: t² - (x + y)t + xy = 0, which is t² - st + A = 0. Since both rectangles have the same perimeter and area, their dimensiRead more
Yes, they must be congruent.
Let the sides of a rectangle be x and y.
Perimeter = 2(x + y) = 2s, so x + y = s.
Area = xy = A.
Here, x and y are the roots of the quadratic equation:
t² – (x + y)t + xy = 0, which is t² – st + A = 0.
Since both rectangles have the same perimeter and area, their dimensions are solutions to the same quadratic equation.
Thus, their side lengths are identical, making the rectangles congruent.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
Radius of the car wheel r = 28 cm. Distance travelled in one complete turn = circumference of wheel: Circumference = 2 x π x r = 2 x (22/7) x 28 = 2 x 22 x 4 = 176 cm = 1.76 m. Total distance of journey = 1 km = 1000 m = 100,000 cm. Number of turns = Total distance / Distance per turn = 100000 / 176Read more
Radius of the car wheel r = 28 cm.
Distance travelled in one complete turn = circumference of wheel:
Circumference = 2 x π x r = 2 x (22/7) x 28 = 2 x 22 x 4 = 176 cm = 1.76 m.
Total distance of journey = 1 km = 1000 m = 100,000 cm.
Number of turns = Total distance / Distance per turn
= 100000 / 176 = 12500 / 22 = 6250 / 11 ≈ 568.18.
Therefore, the wheel turns approximately 568 times during the 1 km journey.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
Circumference of the circle = 2 x π x r = 66 cm. Using π = 22/7: 2 x (22/7) x r = 66 (44/7) x r = 66 r = (66 x 7) / 44 = (3 x 7) / 2 = 21/2 cm = 10.5 cm. A quadrant has a central angle of 90°. Area of a quadrant = (1/4) x π x r² = (1/4) x (22/7) x (21/2) x (21/2) = (1/4) x 11 x 3 x (21/2) = 693 / 8Read more
Circumference of the circle = 2 x π x r = 66 cm.
Using π = 22/7:
2 x (22/7) x r = 66
(44/7) x r = 66
r = (66 x 7) / 44 = (3 x 7) / 2 = 21/2 cm = 10.5 cm.
A quadrant has a central angle of 90°.
Area of a quadrant = (1/4) x π x r²
= (1/4) x (22/7) x (21/2) x (21/2)
= (1/4) x 11 x 3 x (21/2) = 693 / 8 = 86.625 cm².
Hence, the area of the quadrant is 86.625 cm².
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
ECI launched ramps, Braille EVMs and the Saksham App for disabled voters, plus home voting for forty percent benchmark disabilities and senior citizens above eighty-five. Service personnel use the Electronically Transmitted Postal Ballot System from their postings. Convicted prisoners cannot vote byRead more
ECI launched ramps, Braille EVMs and the Saksham App for disabled voters, plus home voting for forty percent benchmark disabilities and senior citizens above eighty-five. Service personnel use the Electronically Transmitted Postal Ballot System from their postings. Convicted prisoners cannot vote by law, but citizens held under preventive detention are legally provided postal ballots to exercise their democratic franchise.
For more NCERT Solutions of Class 9 Social Science Chapter 7 Elections Question Answer (2026-27)
You know that the area of a parallelogram is base x height. Using this and the figure, show that the area of a trapezium is half the sum of the parallel sides x height, i.e., (1/2)(a + b)h.
From Fig. 6.42, the trapezium has parallel sides of lengths a and b and perpendicular height h. The segment of length a on the top base partitions the shape into: A parallelogram with base a and height h. Area(parallelogram) = base x height = ah. A triangle with base (b - a) and height h. Area(trianRead more
From Fig. 6.42, the trapezium has parallel sides of lengths a and b and perpendicular height h.
The segment of length a on the top base partitions the shape into:
A parallelogram with base a and height h.
Area(parallelogram) = base x height = ah.
A triangle with base (b – a) and height h.
Area(triangle) = (1/2) x base x height = (1/2)(b – a)h.
Total area = ah + (1/2)(b – a)h
= (1/2)h [2a + b – a]
= (1/2)(a + b)h.
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 lessTwo rectangles have the same area and the same perimeter. Does this mean that they are congruent to each other?
Yes, they must be congruent. Let the sides of a rectangle be x and y. Perimeter = 2(x + y) = 2s, so x + y = s. Area = xy = A. Here, x and y are the roots of the quadratic equation: t² - (x + y)t + xy = 0, which is t² - st + A = 0. Since both rectangles have the same perimeter and area, their dimensiRead more
Yes, they must be congruent.
Let the sides of a rectangle be x and y.
Perimeter = 2(x + y) = 2s, so x + y = s.
Area = xy = A.
Here, x and y are the roots of the quadratic equation:
t² – (x + y)t + xy = 0, which is t² – st + A = 0.
Since both rectangles have the same perimeter and area, their dimensions are solutions to the same quadratic equation.
Thus, their side lengths are identical, making the rectangles congruent.
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 wheel of a car has an outer radius of 28 cm. Calculate how far the car travels after one complete turn of the wheel and how many times the wheel turns during a journey of 1 km.
Radius of the car wheel r = 28 cm. Distance travelled in one complete turn = circumference of wheel: Circumference = 2 x π x r = 2 x (22/7) x 28 = 2 x 22 x 4 = 176 cm = 1.76 m. Total distance of journey = 1 km = 1000 m = 100,000 cm. Number of turns = Total distance / Distance per turn = 100000 / 176Read more
Radius of the car wheel r = 28 cm.
Distance travelled in one complete turn = circumference of wheel:
Circumference = 2 x π x r = 2 x (22/7) x 28 = 2 x 22 x 4 = 176 cm = 1.76 m.
Total distance of journey = 1 km = 1000 m = 100,000 cm.
Number of turns = Total distance / Distance per turn
= 100000 / 176 = 12500 / 22 = 6250 / 11 ≈ 568.18.
Therefore, the wheel turns approximately 568 times during the 1 km journey.
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 lessFind the area of a quadrant of a circle whose circumference is 66 cm.
Circumference of the circle = 2 x π x r = 66 cm. Using π = 22/7: 2 x (22/7) x r = 66 (44/7) x r = 66 r = (66 x 7) / 44 = (3 x 7) / 2 = 21/2 cm = 10.5 cm. A quadrant has a central angle of 90°. Area of a quadrant = (1/4) x π x r² = (1/4) x (22/7) x (21/2) x (21/2) = (1/4) x 11 x 3 x (21/2) = 693 / 8Read more
Circumference of the circle = 2 x π x r = 66 cm.
Using π = 22/7:
2 x (22/7) x r = 66
(44/7) x r = 66
r = (66 x 7) / 44 = (3 x 7) / 2 = 21/2 cm = 10.5 cm.
A quadrant has a central angle of 90°.
Area of a quadrant = (1/4) x π x r²
= (1/4) x (22/7) x (21/2) x (21/2)
= (1/4) x 11 x 3 x (21/2) = 693 / 8 = 86.625 cm².
Hence, the area of the quadrant is 86.625 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 lessWhat reforms have been introduced by the ECI to make voting more inclusive for the following groups a. People with Disabilities b. Service Voters c. Senior Citizens 60 years and above and 80 years and above d. Prisoners e. Persons in preventive detention
ECI launched ramps, Braille EVMs and the Saksham App for disabled voters, plus home voting for forty percent benchmark disabilities and senior citizens above eighty-five. Service personnel use the Electronically Transmitted Postal Ballot System from their postings. Convicted prisoners cannot vote byRead more
ECI launched ramps, Braille EVMs and the Saksham App for disabled voters, plus home voting for forty percent benchmark disabilities and senior citizens above eighty-five. Service personnel use the Electronically Transmitted Postal Ballot System from their postings. Convicted prisoners cannot vote by law, but citizens held under preventive detention are legally provided postal ballots to exercise their democratic franchise.
For more NCERT Solutions of Class 9 Social Science Chapter 7 Elections Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/social-science/
See less