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One diagonal of a rhombus is twice as long as the other diagonal. If the rhombus has area 128 cm², find the length of the shorter diagonal.
Let the length of the shorter diagonal be d cm. Then the longer diagonal is 2d cm. The formula for the area of a rhombus is: Area = (1/2) x diagonal 1 x diagonal 2 Given Area = 128 cm²: (1/2) x d x 2d = 128 d² = 128 Taking square root on both sides: d = √128 = √(64 x 2) = 8√2 cm. In decimal form, 8√Read more
Let the length of the shorter diagonal be d cm.
Then the longer diagonal is 2d cm.
The formula for the area of a rhombus is:
Area = (1/2) x diagonal 1 x diagonal 2
Given Area = 128 cm²:
(1/2) x d x 2d = 128
d² = 128
Taking square root on both sides:
d = √128 = √(64 x 2) = 8√2 cm.
In decimal form, 8√2 is approximately 8 x 1.414 = 11.31 cm.
Therefore, the shorter diagonal is 8√2 cm.
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 lessIf the diameter of a car tyre is 56 cm, then: (i) How far does the car need to travel for the tyre to complete one revolution? (ii) How many revolutions does the tyre make if the car travels 10 km?
Diameter of tyre d = 56 cm, so radius r = 28 cm. (i) Distance covered in one revolution equals the tyre's circumference: Distance = π x d = (22/7) x 56 = 22 x 8 = 176 cm = 1.76 m. (ii) Total distance = 10 km = 10 x 1000 x 100 cm = 1,000,000 cm. Number of revolutions = Total distance / CircumferenceRead more
Diameter of tyre d = 56 cm, so radius r = 28 cm.
(i) Distance covered in one revolution equals the tyre’s circumference:
Distance = π x d = (22/7) x 56 = 22 x 8 = 176 cm = 1.76 m.
(ii) Total distance = 10 km = 10 x 1000 x 100 cm = 1,000,000 cm.
Number of revolutions = Total distance / Circumference
= 1000000 / 176 = 62500 / 11 = 5681.82.
Hence, the tyre completes approximately 5682 revolutions.
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 total perimeter of all the petals in each of the given flowers. (i) Fig. 6.15A: The centres of the arcs are the midpoints of the sides of the square. (ii) Fig. 6.15B: The centres of the arcs are the vertices of the hexagon.
(i) In Fig. 6.15A, the 4 petals are formed by 8 quarter-circle arcs of radius r = 14 / 2 = 7 cm. Length of one arc = (1/4) x 2 x π x r = (1/4) x 2 x (22/7) x 7 = 11 cm. Total perimeter of 4 petals (8 arcs) = 8 x 11 = 88 cm. (ii) In Fig. 6.15B, the 6 petals are formed by 12 arcs with centres at the vRead more
(i) In Fig. 6.15A, the 4 petals are formed by 8 quarter-circle arcs of radius r = 14 / 2 = 7 cm.
Length of one arc = (1/4) x 2 x π x r = (1/4) x 2 x (22/7) x 7 = 11 cm.
Total perimeter of 4 petals (8 arcs) = 8 x 11 = 88 cm.
(ii) In Fig. 6.15B, the 6 petals are formed by 12 arcs with centres at the vertices of a regular hexagon of side 42 cm (central angle = 60°, radius r = 42 cm).
Length of one arc = 2 x (22/7) x 42 x (60/360) = 264 x (1/6) = 44 cm.
Total perimeter of 6 petals (12 arcs) = 12 x 44 = 528 cm.
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 ratio of the perimeters of two circles is 5:4. What is the ratio of their radii?
Let the radii of the two circles be r1 and r2, and their respective perimeters (circumferences) be C1 and C2. Formula for circumference is: C = 2 x π x r Given ratio of perimeters: C1 / C2 = 5 / 4 Substituting circumference formulas: (2 x π x r1) / (2 x π x r2) = 5 / 4 Cancelling common factors 2 anRead more
Let the radii of the two circles be r1 and r2, and their respective perimeters (circumferences) be C1 and C2.
Formula for circumference is:
C = 2 x π x r
Given ratio of perimeters:
C1 / C2 = 5 / 4
Substituting circumference formulas:
(2 x π x r1) / (2 x π x r2) = 5 / 4
Cancelling common factors 2 and π:
r1 / r2 = 5 / 4.
Thus, the ratio of their radii is also 5:4.
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 triangle ADE in Fig. 6.31.
From rectangle ABCD shown in Fig. 6.31, side AD = BC = 8 cm and side AB = CD = 10 cm. For triangle ADE, take side AD as the base: Base = AD = 8 cm. Vertex E lies on the opposite side BC. The perpendicular distance from E to AD equals the width of the rectangle, which is CD = 10 cm. Height = 10 cm. ARead more
From rectangle ABCD shown in Fig. 6.31, side AD = BC = 8 cm and side AB = CD = 10 cm.
For triangle ADE, take side AD as the base:
Base = AD = 8 cm.
Vertex E lies on the opposite side BC. The perpendicular distance from E to AD equals the width of the rectangle, which is CD = 10 cm.
Height = 10 cm.
Area of triangle ADE = (1/2) x base x height = (1/2) x 8 x 10 = 40 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 less