Radius of a circle is 10 cm. Measure of an arc of the circle is 54°. Find the area of the sector associated with the arc. ( = 3.14 )
Given radius of circle , r = 10cmMeasure of an arc of the circle , = 54˚= 3.14Area of sector , A = (/360)r2A = (54/360)×3.14×102= (9/60)×3.14×100= 47.1cm2Hence area of the sector is 47.1cm2
The circumferences of circular faces of a frustum are 132 cm and 88 cm and its height is 24 cm. To find the curved surface area of the frustum complete the following activity. ( = 22/7 ).
Circumference1 = 2r1 = 132r1 = 132/2 = 21cm [=22/7]Circumference2 = 2r2 = 88r2 = 88/2 = 14cmSlant height of frustum l = √[h2+(r1-r2)2]= √[242+(21-14)2] [given h =24]= √[576+(7)2]= √[576+49]= √625= 25cmCurved surface area of frustum = (r1+r2)l= ×(21+14)×25= ×(35)×25= 2750sq.cm
The radii of ends of a frustum are 14 cm and 6 cm respectively and its height is 6 cm. Find its
Given r1 = 14cmr2= 6cmHeight , h = 6cmSlant height of frustum l = √[h2+(r1-r2)2]= √[62+(14-6)2]= √[36+(8)2]= √[36+64]= √100 = 10 (i) Curved surface area of frustum = l(r1+r2)= ×10(14+6)= ×10×20= 3.14 ×200= 628cm2Hence curved surface area of frustum is 628cm2. (ii) Total surface area of frustum = l (r1+ r2)+r12+r22= ×10(14+ 6)+×142+×62= ×10×20+×196+×36= 200+196+36= 432= 432×3.14= 1356.48cm2Hence Total surface area of frustum […]
The radii of two circular ends of frustum shape bucket are 14 cm and 7 cm. Height of the bucket is 30 cm. How many liters of water it can hold ? (1 litre = 1000 cm3 )
Given height of bucket , h = 30cmr1 = 14cmr2= 7cmVolume of a frustum = (1/3)h(r12+r22+r1×r2)Volume of bucket = (1/3)×30(142+72+14×7)= 10×(196+49+98)= 3430= 10780cm3= 10.78 litres [∵1 litre = 1000 cm3 ]Hence the bucket can hold 10.78 litres of water.
As shown in the figure, a cylindrical glass contains water. A metal sphere of diameter 2 cm is immersed in it. Find the volume of the water.
Given diameter of cylindrical glass , d = 14cmRadius , R = 14/2 = 7cmHeight of glass in cylindrical glass, H = 30cmGiven diameter of metal sphere = 2cmradius of metal sphere , r = 2/2 = 1cmVolume of sphere = (4/3)r3= (4/3)××13= (4/3)= (4/3)×(22/7)= 4.19cm3Volume of water with sphere in it = R2H= ×72×30= […]
Find the surface area and the volume of a beach ball shown in the figure.
Given diameter of ball , d = 42 cmradius of ball , r = 42/2 = 21cmSurface area of sphere = 4r2= 4××212 = 4×3.14×441 = 5538.96 cm2Volume of sphere = (4/3)r3= (4/3)××213 = (4/3)×3.14×9261 = 38772.72 cm3Hence the surface area and volume of ball are 5538.96 cm2 and 38772.72 cm3 respectively.
Figure 7.13 shows a toy. Its lower part is a hemisphere and the upper part is a cone. Find the volume and the surface area of the toy from the measures shown in the figure.(=3.14)
Given radius of cone ,r = 3 cmHeight of cone ,h = 4 cmRadius of hemisphere, r = 3cmSlant height of cone , l = √(h2+r2)l = √(42+32) = √(16+9) = √25 = 5Curved surface area of cone = rl= ×3×5 = 15cm2Curved surface area of hemisphere = (2/3)r3= (2/3)×33 = 18cm2Surface area of toy = Curved surface area […]
In the figure 7.12, a cylindrical wrapper of flat tablets is shown. The radius of a tablet is 7 mm and its thickness is 5 mm. How many such tablets are wrapped in the wrapper ?
Given radius of tablet, r = 7mmThickness of tablet , h = 5mmVolume of tablet =Radius of cylindrical wrapper , R = 14/2 = 7mmHeight of cylindrical wrapper , H = 10cm = 10×0 = 100mmLet n be the number of tablets that can be wrapped.n = volume of cylindrical wrapper/volume of tablet= R2H/r2h=× 72×100/×72×5= […]
A cylinder and a cone have equal bases. The height of the cylinder is 3 cm and the area of its base is 100 cm2 .The cone is placed upon the cylinder. Volume of the solid figure so formed is 500 cm3 . Find the total height of the figure.
Given height of cylinder , h = 3cmBase area of cylinder , r2 = 100cm2 …………..(i)Volume of solid = 500cm3Given cylinder and cone have equal bases. the radii will be equal.Radius of cone = radius of cylinder = r Let height of cone be HVolume of solid = volume of cylinder + volume of cone500 = r2h+(1/3) […]
Observe the measures of pots in figure 7.8 and 7.9. How many jugs of water can the cylindrical pot hold ?
Given radius of jug , r = 3.5 cmHeight of jug , h = 10 cmVolume of conical jug, V = (1/3)r2h= (1/3)××3.52×10 = (122.5/3)×Given Radius of pot ,R = 7cmHeight of pot, H = 10 cmVolume of pot = R2H= ×72×10 = 490Number of jugs of water the cylindrical pot can hold = Volume […]