## Some plastic balls of radius 1 cm were melted and cast into a tube. The thickness, length and outer radius of the tube were 2 cm, 90 cm and 30 cm respectively. How many balls were melted to make the tube ?

Given radius of plastic ball, r = 1cmThickness of tube = 2cmHeight of tube, h = 90cmOuter radius, R = 30cmInner radius of tube ,r1 = outer radius – thickness= 30-2 = 28cmVolume of plastic needed for tube = volume of outer tube – volume of inner tube= R2h -r12h= h(R2-r12)= ×90(302-282)= ×90(900-784)= ×90×116= 10440cm3Volume of […]

## A washing tub in the shape of a frustum of a cone has height 21 cm. The radii of the circular top and bottom are 20 cm and 15 cm respectively. What is the capacity of the tub ?( = 22/7)

Given r1 = 20cmr2 = 15cmHeight , h = 21cmVolume of frustum ,V = (1/3)h(r12+r22+r1×r2)V = (1/3)×(22/7)×21×(202+152+20×15)V = 22×(400+225+300)V = 22×925V = 22×925= 20350cm3= 20.34litres [∵ 1 litre = 1000cm3]Hence the capacity of the tub is 20.34 litres.

## Choose the correct alternative answer for each of the following questions.

(1) The ratio of circumference and area of a circle is 2:7. Find its circumference.(A) 14 (B) 7/ (C) 7 (D) 14/ Solution:Circumference of circle = 2rArea of circle = r2Given ratio of circumference and area = 2:72r/r2 = 2/72/r = 2/7r = 7cmCircumference = 2r = 2×7 = 14.Hence option A is the answer. (2) […]

## A chord PQ of a circle with radius 15 cm subtends an angle of 60° with the centre of the circle. Find the area of the minor as well as the major segment. ( = 3.14, √3 = 1.73)

Given central angle = 60˚Radius ,r = 15cmLet chord PQ subtend ∠POQ = 60° at centre.∴ θ = 60°Area of minor segment = r2[(/360) – (sin/2)]= 152[3.14×(60/360) – sin60˚/2]= 225[3.14×(1/6) – √3/4]= 225[(3.14/6) – 1.73/4]= 225[(6.28- 5.19)/12]= 20.44Hence area of minor segment is 20.44cm2.Area of circle = r2= 3.14×152= 3.14×225= 706.5cm2Area of major segment = […]

## In the figure 7.46, if O is the centre of the circle, PQ is a chord. POQ = 90°, area of shaded region is 114 cm2 , find the radius of the circle. ( = 3.14)

Given POQ = = 90°,area of shaded region PRQ = 114cm2Area of segment PRQ = r2[(/360) – (sin/2)]114 = r2[3.14×90/360 – sin90˚/2]114 = r2[3.14×(1/4) – 1/2]114 = r2[0.785-0.5]114 = r2×0.285r2 = 114/0.285 = 400Taking square root on both sidesr = 20cmHence the radius of the circle is 20cm.

## In the figure 7.45, if A is the centre of the circle. PAR = 30°, AP = 7.5, find the area of the segment PQR ( = 3.14)

Given radius AP , r = 7.5Central angle PAR = = 30Area of segment PQR = r2[(/360) – (sin/2)]= 7.52[3.14×(30/360)- sin30˚/2]= 56.25[3.14×(1/12)- 1/4]= 56.25[(3.14/12)- (3/12)]= 56.25×0.14/12= 0.65625cm2Hence area of segment PQR is 0.65625cm2.

## In the figure 7.44, O is the centre of the circle. m(arc PQR) = 60° OP = 10 cm. Find the area of the shaded region. ( = 3.14, √3 = 1.73)

Given radius OP , r = 10cmm(arc PQR) , = 60°Area of segment PQR = r2[(/360) – (sin/2)]= 102[3.14×(60/360)- sin60˚/2]= 100[3.14×(1/6)- √3/4]= 100[(3.14/6)- 1.73/4]= (314/6)-(173/4)= 52.33- 43.25= 9.08cm2Hence area of shaded region is 9.08cm2.

## In figure 7.43, A is the centre of the circle. ABC = 45° and AC = 7√2 cm. Find the area of segment BXC.

Given radius of circle, r = 7√2cmAB = AC [radii of same circle]ABC = ACB = 45° [isosceles triangle theorem]InABC,A = = 90˚ [Angle sum property of triangle]Area of segment BXC = r2[(/360) – (sin/2)]= (7√2)2[3.14×90/360 – (sin90˚)/2]= 98×[(3.14/4)-(1/2)]= 98×[0.785-0.5]= 27.93cm2Hence area of segment BXC is 27.93cm2.

## LMN is an equilateral triangle. LM = 14 cm. As shown in figure, three sectors are drawn with vertices as centres and radius7 cm. Find,

(1) Given LMN is an equilateral triangle .LM = 14 cmArea of LMN = (√3/4)a2Here a represents the side of equilateral triangle.a = 14Area of LMN = (√3/4)×142= 49√3= 84.87cm2Hence area of LMN is 84.87cm2 (2) Since LMN is equilateral, L = M = N = 60˚= 60˚Given r = 7Area of sector = (/360)r2Area […]

## In figure 7.35, PQRS is a rectangle. If PQ = 14 cm, QR = 21 cm, find the areas of the parts x, y and z.

Given PQ = 14cmQR = 21cmQ = = 90˚Area of part x = (/360)r2Area of part x = (90/360)×(22/7)×142= 11×14= 154cm2Consider sector (R-BYA)QR = QB+BRBR = 21-14 = 7cmAR = 7cm [radius of same circle]Area of part y = (/360)r2Area of part y = (90/360)×(22/7)×72= 11×7/2= 38.5cm2Area of rectangle PQRS = length ×breadth= QR×PQ= 21×14= […]