For a rectangle: if diagonals are equal in a parallelogram, it is a rectangle (True). For a rhombus: if diagonals are perpendicular bisectors, it is a rhombus (True). For a square: if diagonals are equal and perpendicular bisectors, it is ...
Discussion Forum Latest Questions
Piyush365
Asked: In: Class 9 Maths
Rotating forms a cone with height 8 cm, radius 6 cm and slant height 10 cm. Volume is (1/3)π(6²)(8) = 96π ≈ 301.71 cm³. Curved surface area is π(6)(10) = 60π ≈ 188.57 cm². Cbse Class 9 Maths Ganita Manjari Part ...
Piyush365
Asked: In: Class 9 Maths
Slant height is √(8² + 6²) = 10 m. Curved surface area is 3.14 × 6 × 10 = 188.4 m². Dividing by width 3 m gives 62.8 m. Adding 0.2 m wastage gives total length 63 m. Class 9 Ganita ...
a conical tent of height 8 m and base radius 6 massume that the extra length of materialcbse class 9 maths ganita manjari part 2 solutionsclass 9 maths ganita manjari part 2 chapter 14 question answerganita manjari part 2 chapter 14 math of space surface area and volume solutionslength of tarpaulin 3 m wide
Piyush365
Asked: In: Class 9 Maths
Slant height is √(7² + 24²) = 25 cm. One cap needs sheet area πrl = (22/7) × 7 × 25 = 550 cm². For 10 caps, the required sheet area is 10 × 550 = 5500 cm². Class 9 Ganita ...
area of the sheet required to make 10 such capscone of base radius 7 cm and height 24 cmganita manjari part 2 chapter 14 math of space surface area and volume solutionsganita manjari part 2 chapter 14 question answerjoker’s cap is in the form of a right circular conemath of space surface area and volume solutions – ganita manjari part ii chapter 14
Piyush365
Asked: In: Class 9 Maths
Using CSA = πrl = 308, (22/7) × r × 14 = 308 gives base radius r = 7 cm. Total surface area is πrl + πr² = 308 + (22/7) × 7² = 308 + 154 = 462 cm². Ganita ...