10th Maths 12.3
NCERT Class 10th solution of Exercise 12.1
NCERT Class 10th solution of Exercise 12.2
NCERT Class 10th solution of Exercise 13.1
Exercise 12.3
Q1. Find the area of the shaded region in Figure, if `text{PQ = 24 cm, PR = 7 cm}` and `text{O}` is the centre of the circle.
`text{Sol.}`
`text{In right}triangle text{QPR by Pythagoras theorem.}`
`text{RQ = }sqrt((text{PQ})^2 + (text{RP})^2)`
`= sqrt((24)^2+(7)^2)`
`= sqrt(576 + 49)`
`= sqrt(625)`
`= 25`
`text{r} = 25/2`
`text{Area of semicircle QPR = }1/2 pi text{r}^2`
`= 1/2 times 22/7 times(25/2)^2`
`= (11 times 625)/28`
`= 6875/28 text{ cm}^2`
`text{Area of right }triangle text{RPQ} = 1/2 times RP times PQ`
`= 1/2 times 7 times 24`
`= 84 text{ cm}^2`
`text{Area of the shaded region =}``text{ ar of semicircle QPR - ar of }triangle text{RPQ}`
`= 6875/28 - 84`
`= (6875 - 2352)/28`
`= 4523/28 text{ cm}^2`
`text{Answer}`
`text{The area of shaded region is } 4523/28 text{ cm}^2`.
Q2. Find the area of the shaded region in Figure, if radii of the concentric circles with centre `text{O}` are `text{7 cm}` and `text{114 cm}` respectively and `angle text{AOC = 40}^circ.`
`text{Sol.}`
`text{Let radius of concentric circles are}`
`text{ r}_1 = text{OA = OC = 14 cm}`
`text{and r}_2 = text{OB = OD} = 7text{ cm}`
`theta = angle text{AOC} = angle text{BOD} = 40^circ`
`text{area of secter(AOC) = }theta/360^circ times pi r_1^2`
`= 40^circ/360 times 22/7 times(14)^2`
`= (22times28)/9 = 616/9 text{ cm}^2`
`text{area of sector(BOD) = }theta/360^circ times pi r_2^2`
`= 40^circ/360^circ times 22/7 times(7)^2`
`= 154/9 text{ cm}^2`
`text{area of shaded region = ar of AOC - ar of BOD}`
`text{area of ABDC = } 616/9 - 154/9`
`= 154/3 text{ cm}^2`.
`text{Answer}`
`text{The area of the shaded region is }154/3 text{ cm}^2`.
Q3. Find the area of the shaded region in Figure, if `text{ABCD}` is a square of side `text{14 cm}` and `text{APD}` and `text{BPC}` are semicircles.
`text{Sol.}`
`text{Let ABCD is a given square of side (a) = 14 cm}``text{and APD and BPC are two semicircles}`
`text{of diameter (d) = 14 cm}``text{and radius (r) =} 14/2 = 7 text{ cm}`
`text{area of shaded region }``text{= ar of square - ar of the semicircles}`
`= a^2 - pir^2`
`= (14)^2 - 22/7 times (7)^2`
`= 196 - 154`
`= 42 text{ cm}^2`
`text{Answer}`
`text{The area of the shaded region is } 42 text{ cm}^2`.
Q4. Find] the area of the shaded region in Figure, where a circular arc of radius `text{6 cm}` has been drawn with vertex `text{O}` of an equilateral triangle `text{OAB}` of side `text{12 cm}` as centre.
`text{Sol.}`
`text{Let given equilateral } triangle ``text{AOB of side (a) = 12 cm}``text{and a major sector of}`
`text{radius (r) = 6 cm with corresponding angle}``theta = 360^circ - 60^circ = 300^circ`
`text{area of shaded region}``text{= ar AOB + ar of major sector}`
`= sqrt3/4 text{a}^2 + theta/360^circ pi r^2`
`= sqrt3/4 (12)^2 + 300^circ/360^circ times 22/7 times(6)^2`
`= sqrt3/4 times 144 + 30^circ/36^circ times 22/7 times 36`
`= 36sqrt3 + 660/7`
`= (660/7 + 36sqrt3) text{ cm}^2`
`text{Answer}`
`text{The area of the shaded region is }``(660/7 + 36sqrt3) text{ cm}^2`.
Q5. From each corner, a square of side `text{4 cm}` a quadrant of a circle of radius `text{1 cm}` is cut and also a circle of diameter `text{2 cm}` is cut as shown in Figure. Find the area of the remaining portion of the square.
`text{Sol.}`
`text{Let given square of side (a) = 4 cm,}``text{quadrant of radius (r) = 1 cm,}``text{and a circle of diameter 2 cm and r = 1 cm.}`
`text{area of shaded region}`
`= text{ar of square}- 2 times``text{ar of circle}`
`= a^2 - 2 times pi r^2`
`= (4)^2 - 2 times 22/7 times (1)^2`
`= 16 - 44/7`
`= (112 - 44)/7`
`= 68/7 text{ cm}^2`
`text{Answer}`
`text{The area of remaining portion of the square is }``68/7 text{ cm}^2`.
Q6. In a circular table cover of radius `text{32 cm},` a design is formed leaving an equilateral triangle `text{ABC}` in the middle as shown in Figure. Find the area of the design.
`text{Sol.}`
`text{radius of the circle (r) = 32 cm}`
`text{side of equilateral}triangle = (a) = 32sqrt3 text{ cm}`
`text{Area of design = ar of circle - ar of equilateral}triangle`
`= pir^2 - sqrt3/4a^2`
`= 22/7times(32)^2 - sqrt3/4(32sqrt3)^2`
`= 1024times22/7 - 768sqrt3`
`= [22528/7 - 768sqrt3]text{ cm}^2`
`text{Answer}`
`text{The area of the shaded region is }``[22528/7 - 768sqrt3]``text{ cm}^2`.
Q7. In Figure `text{ABCD}` is a square of side `text{14 cm}.` With centres `text{A,B,C}` and `text{D}` four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.
`text{Sol.}`
`text{Given square ABCD of sides (a) = 14 cm}``text{and four quadrant of radius (r) = 7 cm}`
`text{area of shaded region = ar of square - ar of circle}`
`= a^2 - pi r^2`
`= (14)^2 - 22/7 times (7)^2`
`= 196 - 154`
`= 42 text{ cm}^2`
`text{Answer}`
`text{The area of the shaded region is }42 text{ cm}^2`.
Q8. Figure depicts a racing track whose left and right ends are semicircular.
The distance between the two inner parallel line segments is `text{60 m}` and they are each `text{106 m}` long. If the track is `text{10 m}` wide, find :
i) the distance around the track along its inner edge
ii) the area of the track.
`text{Sol.}`
`text{The track is 10 m wide, two semicircular paths}``text{radius r}_1 = 30 + 10 = 40 text{ cm}`
`text{and r}_2 = 30 text{ cm}``text{each length 106 m}`
`text{The distance around the track along its }``text{inner edge}`
`= text{ inner circumference of the circle + }``2times text{length of parallel sids}`
`= 2pitext{r + 2l}`
`= 2times22/7times30 + 2times106`
`= 1320/7 + 212`
`= (1320 + 1448)/7`
`= 2804/7 text{ m}`
`text{Answer}`
`text{The distance around the track along its inner}``text{edges is}= 2804/7 text{ m}`.
ii)
`text{The area of the track}``text{= area of circular ring + 2 area of the rectangle}`
`= pi(r_1^2 - r_2^2)+2times l timesb`
`=22/7[(40)^2 - (30)^2]+2times106times10`
`= 22/7[1600 - 900]+2120`
`= 22/7times700+2120`
`= 2200 + 2120`
`= 4320 text{ m}`
`text{Answer}`
`text{The area of the track is 4320 m}^2`.
Q9. In figure, `text{AB}` and `text{CD}` are two diameters of a circle (with centre `text{O}`) perpendicular to each other and `text{OD}` is the diameter of the smaller circle. If `text{OA = 7 cm},` find the area of the shaded region.
`text{Sol.}`
`text{Large circle of radius r}_1 =`` text{OA = OB = OC = OD = 7 cm}`
`text{and small circle of diameter d} =`` text{OD = OA = 7 cm}`
`text{small circle of radius r}_2 =``d/2 = 7/2 text{ cm}`
`text{base of }triangle text{ABC}``text{ = AB = OA + OB = 7 + 7 = 14 cm}`
`text{Height of } triangle text{ABC = OC = OB = 7 cm}`
`text{ar of shaded region}``text{ = ar of small circle}``text{ + ar of large semicircle - ar ABC}`
`= pir_2^2 + 1/2pir_1^2 - 1/2ABtimesOC`
`= 22/7times(7/2)^2+1/2times22/7(7)^2-1/2times14times7`
`= 77/2 + 77 - 49`
`= 38.5 + 77 - 49`
`= 115.5 - 49`
`= 66.5 text{ cm}^2`
`text{Answer}`
`text{The area of the shaded region is }66.5 text{ cm}^2`.
Q10. The area of an equilateral triangle `text{ABC}` is `17320.5 text{cm}^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 figure). Find the area of the shaded region. (Use `pi = 3.14` and `sqrt 3 = 1.732`)
`text{Sol.}`
`text{Let the side of equilateral }triangle text{ a cm}`
`text{area of }triangle text{ABC} = sqrt3/4a^2`
`17320.5 = 1.73205/4a^2`
`text{a}^2 = (17320.5times4)/1.73205`
`text{a}^2 = 40000`
`text{a} = sqrt(40000)`
`text{a = 200 cm}`
`text{radius of circle r =} (text{a})/2 = 200/2 = 100 text{ cm}`
`triangle text{ABC is equilateral hence}`
`text{and angle of centre } theta = 60^circ`
`text{Area of 3 sectors = }3timestheta/360^circ pi r^2`
`= 3times60^circ/360^circtimes3.14times(100)^2`
`= 1.57 times 10000`
`= 15700 text{ cm}^2`
`text{area of shaded region }``text{= ar of }triangle text{ABC - 3 area of sector}`
`= 17320.5 - 15700`
`= 1620.5 text{ cm}^2`
`text{Answer}`
`text{The area of the shaded region is }1620.5 text{ cm}^2`.
Q11. On a square handkerchief, nine circular designs each of radius `text{7 cm}` are made (see figure). Find the area of the remaining portion of the handkerchief.
`text{Sol.}`
`text{circular designs of radius (r) = 7 cm}`
`text{Square handkerchief of side (a)}``= 7times6 = 42 text{cm}`
`text{Area of remaining portion }``text{= ar of square - 9 ar of circle}`
`= (a)^2 - 9 pi r^2`
`= (42)^2 - 9times22/7times (7)^2`
`= 1764 - 1386`
`= 378 text{ cm}^2`
`text{Answer}`
`text{The area of the remain portion is }378 text{ cm}^2`.
Q12. In Figure , `text{OACB}` is a circle with centre `text{O}` and radius `text{3.5 cm}`. If `text{OD = 2 cm},`
find the area of the
i) quadrant `text{OACB}`,
ii) shaded region.
`text{Sol.}`
`text{Radius of quadrant (r) = 3.5 cm and OD = 2 cm}`
`text{area of quadrant =}1/4 pi r^2`
`= 1/4 times 22/7times(3.5)^2`
`= 77/8 text{ cm}^2`
`text{Answer}`
`text{The area of the quadrant} 77/8 text{ cm}^2`.
ii)
`text{area of shaded region }``text{= ar of quadrant OACB - ar of right }triangle text{DOB}`
`= 77/8 - 1/2timesOBtimesOD`
`= 77/8 - 1/2times3.5times2`
`= 77/8 - 7/2`
`= (77-28)/8`
`= 49/8 text{ cm}^2`
`text{Answer}`
`text{The area of the shaded region } 49/8 text{ cm}^2`.
Q13. In Figure, a square `text{OABC}` is inscribed in a quadrant `text{OPBQ}.` If `text{OA = 20 cm}`, find the area of the shaded region. (Use `pi = 3.14`)
`text{Sol.}`
`text{square OABC of side OA (a) = 20 cm}`
`text{The radius of quadrant is r = diagnal of square}`
`r = asqrt2`
`r = 20sqrt2 text{ cm}`
`text{area of shaded region}``text{ = ar of quadrant - ar of square}`
`= 1/4 pi r^2 - (a)^2`
`= 1/4times3.14times(20sqrt2)^2 - (20)^2`
`= 1/4times3.14times800 - 400`
`= 3.14times200 - 400`
`= 628 - 400`
`=228 text{ cm}^2`
`text{Answer}`
`text{The area of the shaded region is }228 text{ cm}^2`.
Q14. `text{AB}` and `text{CD}` are respectively arcs of two concentric circles of radii `text{21 cm}` and `text{7 cm}` and centre `text{O}`(see Figure). If `angle text{AOB = 30}^circ`, find the area of the shaded region.
`text{Sol.}`
`text{The radius of sector AOB = r}_1 = 21 text{cm}`
`text{and radius of sector COD = r}_2 = 7 text{cm}`
`text{and angle of AOB and COD }theta = 30^circ`
`text{area of shaded region = ar of AOB - ar of COD}`
`= theta/360^circ times pir_1^2 - theta/360^circ times pir_2^2`
`= theta/360^circ pi (r_1^2 - r_2^2)`
`= 30^circ/360^circ times22/7[(21)^2 - (7)^2]`
`= 1/12times22/7[441 - 49]`
`= (11times392)/(6times7)`
`= 308/3 text{ cm}`
`text{Answer}`
`text{The area of the shaded region is }308/3 text{ cm}^2`.
Q15. In Figure, `text{ABC}` is a quadrant of a circle of radius `text{14 cm}` and a semicircle is drawn with `text{BC}` as diameter. Find the area of the shaded region.
`text{Sol.}`
`text{The radius of quadrant r = 14 cm}`
`text{and angle of quadrant }theta = 90^circ`
`text{In right}triangle text{ABC by Pythagoras theorem}`
`BC = sqrt(AB^2+AC^2)`
`BC = sqrt((14)^2 + (14)^2)`
`BC = 14sqrt2 text{ cm = diameter}`
`r_2 = (14sqrt2)/2`
`r_2 = 7sqrt2 text{ cm}`
`text{area of shaded region}``text{ = ar}triangle text{ABC + ar semicircle - ar quadrant}`
`= 1/2timesACtimesAB + 1/2pir_2^2 - 1/4pir_1^2`
`= 1/2timesACtimesAB + 1/2pir_2^2 - 1/4pir_1^2`
`= 1/2times14times14 + 1/2(22/7)(14)^2`` - 1/4(22/7)(7sqrt2)^2`
`= 98 +154 - 154`
`= 98 text{ cm}^2`
`text{Answer}`
`text{The area of the shaded region is }98 text{ cm}^2`.
Q16. Calculate the area of the designed region in Figure common between the two quadrants of circles of radius `text{8 cm}` each.
`text{Sol.}`
`text{side of square (a) = 8 cm}`
`text{radius of quadrant (r) = 8 cm}`
`text{area of design = 2 ar of quadrant - ar of square}`
`= 2times1/4pir^2 - (a)^2`
`= 2times1/4times22/7times(8)^2 - (8)^2`
`= 704/7 - 64`
`= (704 - 448)/7`
`= 256/7 text{ cm}^2`
`text{Answer}`
`text{The area of the shaded region is }256/7 text{ cm}^2`.
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