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1.	If the height and slant height of a cone are 6 cm and 10 cm respecitvely. Then determine the total surface area and the volume of the cone.
Answer» The height of the right circular cone is 6cm and its slant height is 10 cm . Let the radius of the cone be r cm. `therefore 6^2+r^2=10^2rArr r^2=100-36=64rArr r=sqrt(64)=8`. `therefore` the radius of the cone is 8 cm . `therefore` the total surface area of the cone `=pi xx 8(8+10)sq-cm=(22)/(7)xx8xx18sq-cm=144pi sq-cm`. Again , the volume of the cone `=(1)/(3)pi xx (8)^2xx6` cubic - cm `=128pi` cubic - cm. Hence the total surface ara of the cone `=144pi ` sq-cm and volume `=128pi`cubic -cm.

2.	The length of the base diameter of a wooden toy of conical shape uis 10 cm . The expenditure for polishing whole surfaces of the toy at the raye of rupees 2.10 per square - metre is rupees 429. Calculate the height of the toy. Also determine the quantity of wood which is required to make the toy.
Answer» The base diameter of the toy `=10cm`. `therefore` Radius of the base of the toy `=(10)/(2)cm=5cm`. Let the slant height of the toy `=l cm`. `therefore` The curved surface area of the toy `=(220)/(7)xx5xxl sq -cm` rupees 2.10 is expent to polish 1 sq-m of the area . `therefore "rupees " 1 " is " ' " "'" "'(1)/(2.10)" "'" "'" "'" "'` `therefore "rupees " 429 " is " ' " "'" "'" "(1xx429)/(2.10) " sq-m of the area"`. As per condition , `(22)/(7)xx5xxl=(429)/(2.10)rArr l=(429xx7)/(2.10xx22xx5)rArr 1=13` So, the slant height of the toy `=13cm`. `therefore h^2+5^2=(13)^2 or , h^2+25=169 or , h^2=169-25or , h^2=144` `therefore h=sqrt(144)rArr h=12`. `therefore` the toy `=12cm` Again , the volume of the toy `=(1)/(3)xx(22)/(7)xx5^2xx12 c c =(2200)/(7)c c = 314(2)/(7)c c`. Hence , the height of the toy is 12 cm and the quantity of wood required to make the toy is `314(2)/(7)c c`.

3.	The lateral surface area of a right circular cone = (Total surface area)=__________________.
Answer» Answer: (area of base).

4.	The numerical values of the volume and the lateral surface area of a right circular cone are equal. If the height and the radius of the cone are h unit and r unit respectively , then find the value of `(1)/(h^2)+(1)/(r^2)`.
Answer» Let the slant height of the right circular cone is l unit . Also the height and radius of the cone are h and r unit respectively. As per question , `(1)/(3)pi r^2h=pi rl` or `rh=3lrArr r^2h^2=9 l ^2rArr r^2h^2=9 (h^2+r^2)` `rArr (h^2+r^2)/(r^2h^2)=(1)/(9)rArr (h^2)/(r^2h^2)+(r^2)/(r^2h^2)=(1)/(9)rArr (1)/(r^2)+(1)/(h^2)=(1)/(9)rArr (1)/(h^2)+(1)/(r^2)=(1)/(9)` Hence the value of `(1)/(h^2)+(1)/(r^2)=(1)/(9)`.

5.	The difference of the square of the height and the square of the base of a right circular cone of slant height 10 cm is 28cm, then the volume of the cone will beA. `43pi c c`B. `54 pi c c`C. `72pi c c`D. `96pi c c`
Answer» Let the height of the right circular cone h cm and the radius of the base be r cm. The slant height of the cone is 10 cm . `therefore h^2+r^2=(10)^2 or , h^2+r^2=100`................(1) Again , `h^2-r^2=28`...................(2) Adding (1) and (2) we get, `2h^2=128 or , 64 or , h=8` `r^2=100-h^2=100-64=36`. `therefore` the volume of the cone `=(1)/(3) xx pi xx 36 xx 8 c c =96 pi c c`. `therefore` (d) is correct.

6.	Among the total surface area , lateral surface area and the area of the base of right circular cone , the total surface area is the greatest in magnitude.
Answer» Since for a right circular cone , the total surface area = (area of lateral surface)+(area of base), so , the total surface area is the greatest. Hence the given statement is true.

7.	If the numerical values of the area of the base and the volume of a right circular cone of radius 4cm be equal , then the slant height of the cone isA. 3cmB. 4cmC. 5cmD. 6cm
Answer» Let the height of the cone be h cm . As per questions , `(1)/(3)pi.4^2.h=pi.4^2rArr h=3`. `therefore` the slant height of the cone ` =sqrt(4^2+3^2)cm=5cm`. `therefore` (c) is correct.

8.	The ratio of the volumes of two right circular cones is `1:4` and the ratio of their lengths of radii of the bases is `4:5` then the ratio of their heights isA. `1:5`B. `5:4`C. `25:16`D. `25:64`
Answer» Let the lengths of the radii of the bases of the right circular cone area `4x` unit and `5x` unit. Also , let heights of the two right circular cones are `h_1` unit and `h_2` unit. So, the volume of the cones are `(1)/(3)pi(4x)^2h_1` cubic - units and `(1)/(3)pi(5x)^2h_2` cubic-unit. As per question, `(1)/(3)pi(4x)^2h_1:(1)/(3)(5x)^2h_2` cubic - unit. `rArr 16x^2h_1:25x^2h_2=1:4rArr (16x^2h_1)/(25x^2h_2)=(1)/(4)rArr (h_1)/(h_2)(1)/(4)xx(25)/(64)rArr h_1:h_2=25:64` Hence the ratio of the heights of the two right circularf cones is `25:64` . `therefore` (d) is correct.

9.	If the height , curved surface area and volume of a right circular cone be h, c and v respectively , then that `3pi vh^3-c^2h^2+9v^2=0`.
Answer» Let the radius of base and slant height of the cone be r unit and l unit respecitvely. `therefore l^2=h^2+r^2`....................(1) Again, `v=(1)/(3)pi r^2h` .................(2) and `c=pi r l`................(3) `therefore 3 pi vh^3-c^2h^2+9v^2` `=3pi xx(1)/(3)pi r^2h^3-(pi r l)^2xxh^2+9xx((1)/(3)pi r^2h)^2`[by (2) and (3)]. `=pi^2r^2h^4-pi^2r^2l^2h^2+pi^2r^4h^2` `=pi^2r^2h^4-pi^2r^2(r^2+h^2).h^2+pi^2r^4h^2`[by (1)]. `= pi ^2r^2h^4-pi^2r^4h^2-pi^2r^2h^4+pi^2r^4h^2=0` Hence `3pi vh^3-c^2h^2+9v^2=0` .

10.	The quantity of iron - sheet to make a boy a of right circular conical shape is `75(3)/(7)` sq-m. if the slant height of it 5 m, then calculate the volume of air in the boya and its height. Determine of the expenditure to colour the whole surface of the boya at the rate of rupees 2.80 per square - metres. [The width of the iron - sheet not to be considered while calculating.]
Answer» The slant height of the boya is 5 metres. Let the radius of the boya be r metres. `therefore` the total surface area of the boya ` =(22)/(7) r(5+r)sq-m` `therefore (22)/(7)r(5+r)=75(3)/(7) or , (22)/(7)r(5+r)=(528)/(7)or , (r+5)=(528xx7)/(7xx22) or , r(r+5)=24` `or , r^2+5r-24=0 or , r^2+8r-3r-24=0 or , r(r+8)-3(r+8)=0 or , (r+8)(r-3)=0` `therefore` either `r+8=0, or r-3=0` `rArr r=-8 rArr r =3`. Since the radius can never be negative , `therefore r=3`. Now , if the height of the boya be h m , then `h^2+3^2=5^2or , h^2=25-9 or , h^2=16 or , h=4`. `therefore ` the height of the boya is 4 metres. `therefore` the volume of boya `=(10)/(3)xx(22)/(7)xx3^2xx4` cubic - metres `=(264)/(7)`cubic - metres `=37(5)/(7)` cubic- metres. Hence the height of the boya =4 metres and the volume of air in the boya `=37(5)/(7)` cubic - metres. Also , at the rate of rupees 2.80 per metre, the cost of colouring the whole surface of the boya ` "rupees" 2.80xx(528)/(7)= "rupees"211.20`.

11.	77 sq-cm tripal is required to make a right circular conical tent. If the slant height of the tent is 7m, then calculate the base area of the tent.
Answer» The slant height of the tent is 7 metres. Let the radius of the base of the tent be r metres. `therefore` the curved surface area of the tent `=(22)/(7)xx r xx 7 sq-m =22r sq-cm`. As per question , `22r =77rArr r=(77)/(22)=(7)/(2)`. `therefore` The area of the base of the tent `=(22)/(7)xx((7)/(2))^2sq-m=(22)/(7)xx(7)/(2)xx(7)/(2)sq-m` `=(77)/(2)sq-m=38.5 sq-cm`. `therefore` the required base area of the tent `=38.5sq-m`.