NCERT Class 9 Solutions: Surface Areas and Volumes (Chapter 13) Exercise 13.7 – Part 4

Surface area of cone= πrl

Surface Area of Cone

Surface area of cone= πrl

Q-8 If the triangle BEC in the Question 7 above is revolved about the side 5 cm, and then fined the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.If the triangle is revolved about the side of 5 cm then the 12 cm side becomes the radius of the cylinder and the 13 cm side remains the slant height.

Solution:

Triangle poly1 Triangle poly1: Polygon B_1, B, C_1 Ellipse d Ellipse d: Ellipse with foci B, B_1 passing through D Segment c Segment c: Segment [B_1, B] of Triangle poly1 Segment b Segment b: Segment [B, C_1] of Triangle poly1 Segment a Segment a: Segment [C_1, B_1] of Triangle poly1 Segment f Segment f: Segment [C, E] Point B B = (-1.02, 3.94) Point B B = (-1.02, 3.94) Point B B = (-1.02, 3.94) Point B_1 B_1 = (-1.02, 1.06) Point B_1 B_1 = (-1.02, 1.06) Point B_1 B_1 = (-1.02, 1.06) Point B_1 B_1 = (-1.02, 1.06) Point C Point C: Intersection point of b, a Point C Point C: Intersection point of b, a Point C Point C: Intersection point of b, a Point E Point E: Point on c Point E Point E: Point on c Point E Point E: Point on c 13cm text1 = "13cm" 12cm text2 = "12cm" 5cm text3 = "5cm"

Triangle BEC Rotated to Form a Cone

Triangle BEC, the side 5cm

Given, a right circular cone with

  • Radius 12cm

  • Height =5cm

Therefore, volume =13πr2h

  • =(13×π×12×12×5)cm3

  • =240πcm3

Now, from previous problem, the right circular cone base is 5cm and height =12cm

Therefore, volume =13πr2h

  • =13×π×5×5×12

  • =100πcm3

Ratio of the volumes = 100π240π=512 or 5:12

Q-9 A heap of wheat is in the form of a cone whose diameter is 10.5m and height is 3m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.

Solution:Given,

  • Diameter of the base of the cone =10.5m

  • Radius (r)=10.52m=5.25m

  • Height of the cone =3m

Volume of the heap is 13πr2h

  • =(13×227×5.25×5.25×3)m3

  • =86.625m3

Also to calculate the slant height,

  • l2=h2+r2

  • l2=32+(5.25)2

  • l2=9+27.5625

  • l2=36.5625

  • l=36.5625=6.05m

Area of canvas = Curved surface area of the cone

  • That is, πrl=(227×5.25×6.05)m2

  • 99.825m2 (approx.)

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