# NCERT Solutions for Class 10 Maths

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1.    Let R be the radius of the circle which has circumference equal to the sum of circumferences of the two circles, then according to question,

27πR = 2π(19) + 2π(9)      ⇒         R = 19 + 9        ⇒          R = 28 cm

2.   Let R be the radius of the circle which has area equal to the sum of areas of the two circles, then

According to the question,

πR2 = π(8)2 + π(6)2      ⇒     R= (8)+ (6)2     ⇒       R2 = 64 + 36

⇒ R2 = 100              ⇒       R = 10 cm

Area of gold scoring region Red: Area of red scoring region Blue:  Area of blue scoring region Black: Area of black scoring region White: Area of white scoring region 4.    Diameter of wheel = 80 cm    ⇒    Radius of wheel (r) = 40 cm

Distance covered by wheel in one revolution ∵ Distance covered by wheel in 1 hour = 66 km = 66000 m = 6600000 cm

∴ Distance covered by wheel in 10 minutes 5.    (A) Circumference = Area     ⇒     2πr = πr2        ⇒         r = 2 units

6.    Here, r = 6 cm and θ = 600 We know that for quadrant of circle, θ = 900

9.    (i) Here, r = 10 cm and θ = 900  ∴  Area of minor segment = Area of minor sector – Area of Δ OAB

= 78.5 – 50 = 28.5 cm2

(ii) For major sector, radius = 10 cm and θ = 3600 – 900 = 2700

10.   Given, r = 21 cm and θ = 600  ⇒ Area of segment = 231 – Area of Δ OAB           …(i)

In right angled triangle OMA and OMB,

OM = OB             [Radii of the same circle]

OM = OM           [Common]

∴ Δ OMA ≅ Δ OMB                      [ RHS congruency]

∴ AM = BM                      [By CPCT]

∴ M is the mid-point of AB and ∠ AOM = ∠ BOM Therefore, in right angled triangle OMA, Using eq. (i),

Area of segment formed by corresponding chord 11.   Here, r = 15 cm and θ = 600  For, Area of Δ AOB,

Draw OM ⊥ AB.

In right triangles OMA and OMB,

OA = OB                              [Radii of same circle]

OM = OM                            [Common]

∴ Δ OMA ≅ Δ OMB           [RHS congruency]

∴ AM = BM                         [By CPCT] In right angled triangle OMA,  Area of minor segment = Area of minor sector – Area of Δ AOB

= 117.75 – 97.3125 = 20.4375 cm2

And, Area of major segment = πr2Area of minor segment

= 706.5 – 20.4375 = 686.0625 cm2

12.   Here, r = 15 cm and θ =1200  For, Area of Δ AOB,

Draw OM ⊥ AB.

In right triangles OMA and OMB,

OA = OB                             [Radii of same circle]

OM = OM                           [Common]

∴ Δ OMA ≅ Δ OMB          [RHS congruency]

∴ AM = BM                        [By CPCT] In right angled triangle OMA,  ∴ Area of corresponding segment = Area of corresponding sector – Area of Δ AOB

= 150.72 – 62.28 = 88.44 cm2

13.   (i) Area of quadrant with 5 m rope (ii) Area of quadrant with 10 m rope ∴ The increase in grazing area = 78.5 – 19.625 = 58.875 m2

Length of 5 diameters = 35 x 5 = 175 mm    …..(ii)

∴ Total length of the silver wire required = 110 + 175 = 285 mm

The area of each sector of the brooch  Area between two consecutive ribs of the umbrella  16.   Here, r = 25 cm and θ =1150

The total area cleaned at each sweep of the blades =  17.   Here, r = 16.5 km and θ = 800

The area of sea over which the ships are warned =    For, Area of Δ AOB,

Draw OM ⊥ AB.

In right triangles OMA and OMB,

OA = OB                             [Radii of same circle]

OM = OM                           [Common]

∴ Δ OMA ≅ Δ OMB          [RHS congruency]

∴ AM = BM                        [By CPCT] ∴ Area of minor segment = Area of minor sector – Area of Δ AOB = 410.67 – 333.2 = 77.47 cm2

∴ Area of one design = 77.47 cm2

∴ Area of six designs = 77.47 x 6 = 464.82 cm2

Cost of making designs = 464.82 x 0.35 = Rs. 162.68

19.   (D) Given, r = R and θ = p 20.  ∠ RPQ = 900 [Angle in semi circle is 900]

∴ RQ2 = PR2 + PQ2 = (7)2 + (24)2 = 49 + 576 = 625

⇒ RQ = 25 cm         ⇒          Diameter of the circle = 25 cm

∴ Radius of the circle = Area of the semicircle = Area of right triangle RPQ Area of shaded region = Area of semicircle – Area of right triangle RPQ 21.   Area of shaded region = Area of sector OAC – Area of sector OBD 22.   Area of shaded region = Area of square ABCD – (Area of semicircle APD + Area of semicircle BPC) 23.   Area of shaded region = Area of circle + Area of equilateral triangle OAB – Area common to the circle and the triangle 24.   Area of remaining portion of the square = Area of square – (4 x Area of a quadrant + Area of a circle) 25.  Area of design = Area of circular table cover – Area of the equilateral triangle ABC  ∵ G is the centroid of the equilateral triangle.

∴ radius of the circumscribed circle  26.  Area of shaded region = Area of square – 4 x Area of sector 27.   (i) Distance around the track along its inner edge  28.   Area of shaded region = Area of circle + Area of semicircle ACB – Area of Δ ACB 29.  Area of equilateral triangle  Area of shaded region = Area of Δ ABC = 17320.5 – 15700 = 1620.5 cm2

30.  Area of remaining portion of handkerchief = Area of square ABCD – Area of 9 circular designs 31.   (i) Area of quadrant OACB  (ii) Area of shaded region = Area of quadrant OACB – Area of Δ OBD Area of shaded region = Area of quadrant OPBQ – Area of square OABC 33.  Area of shaded region = Area of sector OAB – Area of sector OCD 34.  In right triangle BAC, BC2 = AB2 + AC2    [Pythagoras theorem] ⇒ BC2 = (14)2 + (14)2 = 2(14)2       ⇒      BC = 142 cm

∴ Radius of the semicircle ∴ Required area = Area BPCQB 35.   In right triangle ADC, AC2 = AD2 + CD2         [Pythagoras theorem] ⇒ AC2 = (8)2 + (8)2 = 2(8)2   AC = 128 = 82 cm

Draw BM ⊥ AC. MySchoolPage connects you with exceptional, certified math tutors who help you stay focused, understand concepts better and score well in exams!

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