9 june maths iii

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 www.afterschoool.tk Maths Maths Dr. T.K. Jain. Dr. T.K. Jain. AFTERSCHOOOL AFTERSCHOOOL M: 9414430763 M: 9414430763 [email protected] [email protected] www.afterschool.tk www.afterschool.tk www.afterschoool.tk www.afterschoool.tk

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MathsMathsDr. T.K. Jain.Dr. T.K. Jain.

AFTERSCHOOOLAFTERSCHOOOLM: 9414430763M: 9414430763

[email protected]@yahoo.co.in

www.afterschool.tkwww.afterschool.tk

www.afterschoool.tkwww.afterschoool.tk

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The length of minute hand on a wallclock is 7cms. What is the area swept

by the minute hand in 30 minutes?• This is the question of circle

• Area of circle : pi *r*r =

22/7*7*7=154• But the area in 30 minutes will be

half circle – so it will be 77 sq cm.

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A wheel makes 1000 revolution incovering a distance of 88 km. What is

diameter of the wheel?• The circumference of wheel = 88000

/ 1000 = 88.

• Formulae of circumference of circle= 2pi*r = 2*22/7*R = 88

• 2R = 88*7/22 = 28meters Ans.

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A wire is in the form of an equilateraltriangle with area 5 sq. m. If its shape

is changed to a circle, its radius will be?• Area of Triangle: 3^1/2/4 *side*side• 3^1/2/4 *side*side =5,• side^2 = 20/ 3^1/2 =11.5

Side = 3.4 approx.Length of wire:3*3.4= 10.2Circumference of circle =10.2

2pi *r = 10.2R = 10.2/ 6.28 = 1.63 meters approx.

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at is t e circum erenceof semicircular shape on a

line of length 10 cm?• Circumference = 2pi*r

• =2*22/7*5 =10*3.1428 = 31.42

• Half of circumference = 15.92• Add diameter with this = 25.92 ans.

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T e ratio o t e engt o aside of an equilateral

triangle and its height is?• Let us assume that side of a triangle is X. its areais .433 X^2. (3^1/2/4 *side * side). Let us assumethat we divide it in two triangles of equal sizewith area = .433 X^2/2. This triangle will have

area ½*base*height. (being right angle triangle)• Here base = side /2, so• .433x^2 /2 =1/2*X/2*h• Height = .433x^2*4 /2x = .433 *2 X

• Ratio of side to height = 1: .866 Ans.• Another method = applying pythogorus rule,• X^2 = (x/2)^2 + height ^2 , solving we get ratio of

side to height as 2:3^1/2

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There are 4 circles in a bigger circle.The diameter of biggest circle is 56 but

those of smaller circles is 7. what isarea of bigger circle which has not been

occupied by smaller circles.•  

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Answer:

• Area of bigger circle:• 3.1428*28*28 =2464

• Area of smaller circle =22/7*7/2*7/2 = 38.5 ; there arefour circles so area is 38.5*4 = 154

• Area of uncovered bigger circle• 2464-154 = 2310 ans.

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Inner circumference of a 14 meterwide race track is 440 meters, what

is the radius of outer circle?• Inner radius is

• 2pi*r = 440; r = 440*7/44 = 70

• Radius of outer circle is 70+14 = 84ans.

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A circular wire of radius 42 cm is bentinto the shape of a rectangle. What isthe diagonal of this rectangle, if the

sides are in the ratio of 6:5?• Step 1: find the circumference :• 2*pi*r = 2*22/7*42 =264

• 6x+5x = 264; 11x=264; x =24• Length of rectangle : 144, width is

120.

• Diagonal will be square root of 144^2+120^2= 187.44 approx.

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at is t e area o circ einscribed in an equilateral

triangle of side 24?• We know that radius of incircle is :

• R = Side /( 2*3^1/2)

• Therefore radius is : 24 / (2*1.73)• =6.928

• Area of circle = pie *r *r = 150.8approx. ans.

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What is the area of circlecircumscribed in an equilateral triangle

of side 24?• We know that radius of circumcircle

is :

• R = Side /3^1/2

• Therefore radius is : 24 / 1.73• =14aprox

• Area of circle = pie *r *r• = 196 pi. Ans.

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A man builds a circular pool of radius 5 inside acircular garden of radius 12. In order tocompensate for area lost, he extends the

garden, What is radius of garden, if its area is

same as earlier?• Area of garden: 144 pi. Area of pool=25 pi• Let us assume that he increases radius by x• Pi * (12+x)^2 -144pi = 25 pi.• 144+24x+x^2-144 = 25• X^2+24x-25 =0• X^2 +25x-x-25; x(x+25)-1(x+25)• X=1• New radius of the garden is 13 Ans.

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What is the area of largest trianglethat can be inscribed in a semi-circle of

radius 7?• Here we will keep the base as

diameter of the circle = 14. Height

will be 7 (as radius).• Thus the area will be ½*7*14

• =49 ans.

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What is the area of largesttriangle that can be inscribed in a

circle of radius 7?• Radius = side / 3^1/2

• Side= Radius * 1.73 = 7*1.73 =12.12

• Area of triangle:• 3^1/2 /4 * 12.12*12.12

• =63 approx. ans.

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What is the area of largest squarethat can be inscribed in a circle of

radius 7?• Diagonal of square is 14.

• Area of square is ½ *Diagonal *

diagonal.• ½*14*14 = 98 ans.

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W d ’t t t th

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We don’t want a cow to graze more than9856 sq m area in circular form. Whatshould be the length of rope by which

we should tether the cow?• We have to calculate the radius in

this question.

• Pi*r *r = 9856• R^2 =9856 / pi = 3136

• R = 56 ans.

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The ratio of radii of two circles is inthe ratio of 1:2, what is the ratio of

their area?• 1:4 ans.

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The ratio of radii of two circles isin the ratio of 1:2, what is the

ratio of their circumference?• 1:2 ans.

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The ratio of radii of two spheresin the ratio of 1:2, what is the

ratio of their volume?• 1:8 ans.

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ThanksAfterschoool conducts three year integrated PGPSE(after class 12th  along with IAS / CA / CS) and 18month PGPSE (Post Graduate Programme in Social

Entrepreneurship) along with preparation for CS /CFP / CFA /CMA / FRM. This course is alsoavailable online also. It also conducts workshops onsocial entrepreneurship in schools and colleges allover India – start social entrepreneurship club in

  your institution today with the help fromafterschoool and help us in developing society.

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About AFTERSCHOOOL PGPSE –the best programme for developing

great entrepreneurs• Most flexible, adaptive but rigorous programme• Available in distance learning mode• Case study focused- latest cases

• Industry oriented practical curriculum• Designed to make you entrepreneurs – not just an

employee• Option to take up part time job – so earn while

 you learn• The only absolutely free course on internet

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Workshops from

AFTERSCHOOOL• IIF, Delhi• CIPS, Jaipur• ICSI Hyderabad Branch

• Gyan Vihar, Jaipur• Apex Institute of Management, Jaipur• Aravali Institute of Management, Jodhpur• Xavier Institute of Management, Bhubaneshwar

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Flexible Specialisations:• Spiritualising business and society• Rural development and transformation• HRD and Education, Social Development

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Pedagogy• Case analysis• Quiz, seminars, workshops, games,• Visits to entrepreneurs and industrial

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Branches• AFTERSCHOOOL will shortly openits branches in important cities inIndia including Delhi, Kota, Mumbai,Gurgaon and other important cities.Afterschooolians will be responsiblefor managing and developing these

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Case Studies• We want to write case studies onsocial entrepreneurs, first

generation entrepreneurs, ethicalentrepreneurs. Please help us in thisprocess. Help us to be in touch withentrepreneurs, so that we maydevelop entrepreneurs.

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www.a ersc ooo .  social entrepreneurship for

better society