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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
If the perimeter of a semicircular protractor is 66 cm, Find the diameter of the protractor. |
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Answer» `Perimeter=2R+piR` `66=R(pi+2)` `R=66/(pi+2)` `R=77/6 cm` D=2R=77/3cm. |
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| 2. |
A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope. Find(i) the area of that part of the field in which the horse can graze.(ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. |
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Answer» Area= area of sector with `(theta=90^0)=theta/360^0*pi*r^2` =90/360*22/7*25 =19.625 `m^2` Increase in area=`theta/360*pi(10m)^2-19.625` =`90/360*3.14*100-19.625` =58.875 `m^2` |
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| 3. |
The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour? |
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Answer» 1 revolution of wheel = distance of `1* `circumference time = `10min = 10/60 hrs` speed `= (66 km)/(1hr) ` distance `= 66*10/60 = 11km = 11000m` revolution = total distance/circumference `= 11000/(2 pi r)` `= 11000/(2*22/7*0.4)` `= 7000/(4*0.4)` `= 70000/(4*4)` `=4375` revolutions answer |
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| 4. |
The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes. |
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Answer» given that `r=14`cm in 60min`-> area= pi r^2` in 5 min `-> area= pi r^2/60 * 5` `= 22/7*14*14*5/60cm^2` `= 22*7/3cm^2` `=154/3cm^2` |
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| 5. |
ABCD is a field in the shape of a trapezium, `AD ||BC`, `angleABC=90^@` and `angleADC=60^@`, Four sectors are formed with centres A, B,C and D, as shown in the figure. The radius of each sector is 14m. Find the following: (i) total area of the four sectors, (ii) area of the remaining portion, given that `AD=55m, BC=45m` and `AB=30m` |
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Answer» 1)Total area of four sector`=pi/2*r^2/2+(pir^2)/4+(2pi)/3*1/2*r^2+pi/3*1/2*r^2` `(pir^2)/2(1/2+1/2+2/3+1/3)` `(pir^2)/22((3+3+4+2)/6)` `((pir^2)/2)*2=pir^2` `r=616m^2` 2)Area of trapezium`=1/2(a+b)*h` `1/2*(45+55)*30=1500m^2` Area of remaining portion`=1500-616=854m^2`. |
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| 6. |
The radii of two circles are 19 cm and 9 cm respectively.Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles. |
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Answer» Circumference of first circle ` = 2pir_1 = 2pi**19 = 38pi` Circumference of second circle ` = 2pir_2 = 2pi**9 = 18pi` Circumference of third circle ` = 18pi+38pi = 56pi` Let radius of third circle is `R` cm. Then,`2piR = 56pi` `=>R = (56pi)/(2pi) = 28cm` |
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| 7. |
In fig. O is the center of a circle such that diameter AB=13cm and AC= 12 cm. BC is joined. Find the area of the shaded region. |
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Answer» Here, `BC = sqrt(13^2-12^2) = 5cm` Radius of circle `(r) = (AB)/2 = 13/2 = 6.5cm` `:.` Area of shaded region `(A_s) = ` Area of semi circle - Area of `Delta ABC` `=>A_s = pi/2(6.5)^2-(1/2**12**5)` `=>A_s = 3.14/2(6.5)^2-30 = 66.3325-30 = 36.333cm^2` |
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| 8. |
Find the cost of fencing a circular field of 560 metres radius at Rs 332 per 10 metres |
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Answer» `r = 560 m` circumference of circulare field = `2 pi r` `= 2 xx 22/7 xx 560 m` `= 44 xx 80 m` 10m cost `= 332 rs` 1 m fencing cost`= 332/10 rs` `44x80` m fencing cost=`332/10 xx 44 xx 80` `=332 xx 352 =116864 rs` Answer |
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| 9. |
PQRS is a diameter of a circle of radius 6 cm. The lengths PQ, QR and RS are equal. Semi-circles are drawn on PQ and QS as diameters as shown in figure. Find the perimeter of the shaded region. |
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Answer» perimeter of shaded region is `(2pi ((PS)/2))/2 + (2pi((PQ)/2))/2 + (2pi((QS)/2))/2` `= pi(PS)/2 + pi(PQ)/2 + pi(QS)/2` `=3PQ = PS = 2*6= 12 `cm `=PQ =4`cm `QS = QR + RS = 2*4 = 8` putting in eqn `pi*12/2 + pi*4/2 + pi*8/2 ` `= 6pi + 2pi+ 4 pi = 12 pi`cm |
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| 10. |
Figure shows a sector of a circle, centre O, containing an angle `theta`. Prove that (i) Perimeter of the shaded region is `r ( tan theta + sec theta + (pi theta)/180 -1)` (ii) Area of shaded region is `r^2/2 ( tan theta - pi theta/180)` |
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Answer» `tantheta=(AB)/r` `AB=rtantheta` `sectheta=(OB)/(OA)=(OB)/r` `OB=rsectheta` `PB=OB-OP=rsectheta-r` `arcAB=pirtheta/180^o` Perimeter=`rtantheta+rsectheta-r+(pirtheta)/180^0` `=r(tantheta+sectheta+pitheta/180^o-1)` Area of `/_OAB=1/2*r^2tantheta` Area of shaded region=`1/2r^2tantheta-pir^2theta/360` `=r^2/2(tantheta-pitheta/180^0)` |
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| 11. |
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (i) minor segment (ii) major sector. |
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Answer» 1) area(minor segment)=area(sectorOAB)-area(triangleOAB) =`theta/360pir^2-1/2OA*ob` =`90/360*22/7*100^2-1/2100^2` =28.25 `cm^2` 2) area(major segment)=`theta/360*3.14*100^2`=`270/360*3.14*100^2`=235.5 `cm^2` |
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| 12. |
A chord of a circle of radius 15 cm subtends an angle of `60o`at the centre. Find the areas of the corresponding minor and major segments of the circle. |
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Answer» area(minor segment)=area(sector OAB)-area(triangleOAB) =`theta/360^0pir^2-sqrt3/4r^2` =`60/360*pi 15^2-sqrt3/4 15^2` =20.4375 `cm^2` area(major segment)=`pir^2-20.4375` =`3.14*225-20.4375` =686.0625 `cm^2` |
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| 13. |
The area of an equilateral triangle ABC is `17320. 5 c m^2`. With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (see Fig. 12.28). Find the area of the shaded region. (`U |
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Answer» side of triangle=2r area of triangle=`sqrt3/4r^2=17320.5` `r^2=100^2` r=100 area(shaded)=area triangle -3 area(sector)=`17320.5-3*(60/360)*3.14*100^2`=`1620.5 cm^2` |
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| 14. |
Find the area of the shaded design in Fig. 12.17, where ABCD is a square of side 10 cm and semicircles are drawn with each side of the square asdiameter. |
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Answer» If we look closely at the given figure, we can see, Area of 4 semi circles = Area of shaded region + Area of square `ABCD` So, Area of shaded region,`A_S` = Area of 4 semi circles - Area of square `ABCD` Here, Radius of circle, `r = (CD)/2 =10/2 = 5cm` Area of semi-circle, `= 1/2pir^2 ` So, area of 4 semi-circles, `A_(SC) = 4*1/2pir^2 = 2pir^2 = 2**3.14**5**5` `A_(SC) = 157cm^2` Now, Area of square, `A_(SQ) = CD^2 = 100cm^2` So, Area of shaded region, `A_(S) = 157-100 = 57cm^2` |
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| 15. |
Calculate the area of the designed region in Fig. 12.34 common between the two quadrants of circles of radius 8 cm each. |
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Answer» as we can see that ar(2 quad) = area (square) + ar(design) ar(design) = 2* ar(quadrant) - ar(square) `= 2*1/4* pi r^2 - r^2` `= (pi r^2)/2 - r^2` `= r^2(pi/2 - 1)` `=8^2(1/2*22/7 - 1)` `= 64(11-7)/7` `= 256/7 cm^2` answer |
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| 16. |
An umbrella has 8 ribs which are equally spaced. Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella. |
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Answer» given that `r=45cm` `theta= 360/8` area of sector`= theta/360 pi r^2` `= 360/8*1/360*22/7*45*45 cm^2` `= 2025*11/28` `= 22275/28 = 795.54 cm^2` answer |
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