Класс 10 RD Sharma Solutions — Глава 15 Области, связанные с кругами — Упражнение 15.4 | Набор 3
Вопрос 35. На рисунке AB и CD — два диаметра окружности, перпендикулярные друг другу, а OD — диаметр меньшей окружности. Если ОА = 7 см, найдите площадь заштрихованной области.
Решение:
Given that,
Radius of larger circle(OA) = 7 cm
Diameter of smaller circle(OD) = 7 cm
So, the radius of smaller circle = 7/2 cm
Now we find the area of the shaded region = Area of large circle – Area of small circle
= π(7)2 – π(7/2)2
= π × 49 – π × (49/4)
= 22/7[49 – 49/4]
= 115.5cm2
Hence, the area of the shaded region 115.5 cm2
Вопрос 36. На рисунке PSR, RTQ и PAQ представляют собой три полуокружности диаметром 10 см, 3 см и 7 см соответственно. Найдите периметр заштрихованной области.
Решение:
Given that,
Diameter of semicircle PSR = 10 cm
So, radius of semicircle PSR, (r1) = 10/2 = 5 cm
Diameter of semicircle RTQ = 3 cm
So, radius of semicircle RTQ, (r2) = 3 cm = 3/2 cm
Diameter of semicircle PAQ = 7 cm
So, radius of semicircle PAQ, (r3) = 7/2 cm
Now we find the perimeter of the shaded region = Length of the arc PAQ +
Length of the arc PSR +
Length of the arc RTQ
= πr1 + πr2 + πr3
= π(r1 + r2 + r3)
= π(5 + 3/2 + 7/2)
= π{(10 + 3 + 7)/2}
= π × 20/2
= 10 π
= 10 × 22/7
= 10 × 3.14
= 31.4 cm
Hence, the perimeter of the shaded region is 31.4 cm
Вопрос 37. На рисунке две окружности с центрами А и В касаются друг друга в точке С. Если АС = 8 см и АВ = 3 см, найдите площадь заштрихованной области.
Решение:
Given that,
AC = 8 cm & AB = 3 cm
Here, AC is the radius of the bigger circle and BC is the radius of inner circle.
BC = AC – AB
BC = 8 – 3
BC = 5 cm
Now we find the area of shaded region = Area of bigger circle – Area of inner circle
= πR2 – πr2
= π(R2 – r2)
= 22/7 (82 – 5)
= 22/7 (64 – 25)
= 22/7 × 39
= (22 × 39)/7
= 858/7
= 122.57 cm2
Hence, the area of shaded region is 122.57 cm2
Вопрос 38. На рисунке ABCD — квадрат со стороной 2а. Найдите соотношение между
(i) окружности
(ii) площади вписанной окружности и описанной окружности квадрата.
Решение:
Given that,
Side of a square ABCD = 2a
So, the diameter of incircle = side of a square = 2a
Radius of a incircle(r) = Diameter of incircle/2
r = 2a/2
r = a
Also, the diameter of circumcircle = diagonal of a square = √2 side
So, the radius of circumcircle(R) = √2 side/2
R = (√2 × 2a) /2
R = √2a
R = √2a
(i) Ratio of circumferences of inner circle (C1) and circumcircle (C2)
C1 : C2 = 2πr : 2πR
C1 / C2 = 2πr / 2πR
C1 / C2 = r / R
C1 / C2 = a/√2a
C1 / C2 = 1/√2
C1 : C2 = 1 : √2
(ii) Ratio of Areas of inner circle (A1) and circumcircle (A2) :
A1 : A2 = πr2 : πR2
A1 / A2 = πr2 / πR2
A1 / A2 = r2 /R2
A1 / A2 = a2/(√2a)2
A1 / A2 = a2/2a2
A1 / A2 = 1/2
A1 : A2 = 1 : 2
Hence, the ratio of circumferences of inner circle (A1) and circumcircle (A2) is 1 : √2
and Ratio of Areas of inner circle (A1) and circumcircle (A2) is 1 : 2
Вопрос 39. На рисунке изображены три полукруга А, В и С диаметром по 3 см каждый и еще один полукруг Е с окружностью D диаметром 4,5 см. Рассчитать:
(i) площадь заштрихованной области
(ii) стоимость покраски заштрихованной области из расчета 25 пайсов за см² с точностью до рупии.
Решение:
Given that,
Three semicircles, A, B and C having diameter 3 cm each, and
another semicircle E having a circle D with diameter 4.5 cm.
(i) Now area of shaded region = Area of the semicircle with diameter 9 cm –
Area of two semicircles with radius 3 cm –
Area of the circle with centre D +
Area of semicircle with radius 3 cm
= 1/2 π(9/2)2 – 2 × 1/2 π(3/2)2 – π(4.5/2)2 + 1/2 π(3/2)2
= 1/2 π(4.5)2 – π(1.5)2 – π(2.25)2 + 1/2 π(1.5)2
= 1/2 π(4.5)2 – π(1.5)2 + 1/2 π(1.5)2 – π(2.25)2
= 1/2 π(4.5)2 – 1/2 π(1.5)2 – π(2.25)2
= 1/2π(4.52 – 1.52) – π 2.252
= 1/2 π (20.25 – 2.25) – π × 5.0625
= 1/2 π(18) – π × 5.0625
= 9π – π 5.0625
= π(9 – 5.0625)
= π × 3.9375
= 22/7 × 3.9375
= 0.5625 × 22
= 12.375 cm2
(ii) Cost of painting 1 cm² Shaded Region = 25 p
Cost of painting 13.275 cm² Shaded Region = 25 p × 13.275
= 309.375 paise
= ₹ 309.375 /100
= ₹ 3 (nearest rupee)
Cost of painting the Shaded Region = ₹ 3.
Hence, the area of the shaded region is12.375 cm2 and the cost of painting the Shaded Region is ₹ 3.
Вопрос 40. На рисунке треугольник ABC прямоугольный, ∠B = 90°, AB = 28 см и BC = 21 см. При диаметре AC начерчена полуокружность, а при радиусе BC начерчена четверть окружности. Найдите площадь заштрихованной области с точностью до двух знаков после запятой.
Решение:
Given that,
ABC is a right-angled triangle, ∠B = 90°, AB = 28 cm and BC = 21 cm.
AC as diameter a semicircle is drawn and with BC as radius a quarter circle is drawn
So, the area of triangle = 1/2 x 21 x 28 = 294 cm2
Now in ΔABC,
By using Pythagoras Theorem
AC2 = 282 + 212
AC = √1225
AC = 35
So, the radius of semi-circle = 35/2 = 17.5
Now, the area of semi-circle
= 1/2 x 3.14 x (17.5)2 = 480.8 cm2
Area of quarter circle = 1/4 x 3.14 x 212 = 346.2 cm2
Now we find the area of the shaded region = Area of Semi-circle + Area of triangle ABC –
Area of quarter circle
= 294 + 480.8 – 346.2 = 428.75cm2
Hence, the area of shaded region is 428.75cm2
Вопрос 41. На рисунке О — центр дуги окружности, АОВ — прямая. Найдите периметр и площадь заштрихованной области с точностью до одного десятичного знака. (Возьмите π = 3,142)
Решение:
Given that,
∆ACB is a right-angled triangle, in which AC = 12 cm, BC = 16 cm, ∠C = 90°
Now In ∆ACB,
By using Pythagoras theorem, we get
AB2 = AC2 + BC2
AB2 = 122 + 162
AB2 = 144 + 256
AB2 = 400
AB = √400
AB = 20 cm
So, the diameter of a semicircle = 20 cm
So, the radius of semicircle, r = 10 cm
Now we find the perimeter of a Shaded region = circumference of semicircle + AC + AB
= πr + 12 + 16
= 3.142 × 10 + 28
= 31.42 + 28
= 59.42 cm
Now we find the area of the shaded region = Area of a semicircle – Area of a right angle triangle
= 1/2 πr2 – 1/2 × base × height
= 1/2 × 3.142 × 102 – 1/2 × AC × BC
= 1/2 × 3.142 × 100 – 1/2 × 12 × 16
= 3.142 × 50 – 6 × 16
= 157.1 – 96
= 61.1 cm2
Hence, the required Perimeter of a Shaded region is 59.4 cm and area of the shaded region is 61.1 cm2.
Вопрос 42. На рисунке граница заштрихованной области состоит из четырех полуокружностей, причем две наименьшие равны. Найдите диаметр наибольшего из них 14 см, а наименьшего 3,5 см.
(i) длина границы,
(ii) площадь заштрихованной области.
Решение:
Given that,
The shaded region consists of four semi-circular arcs,
the smallest two being equal.
If the diameter of the largest is 14 cm and of the smallest is 3.5 cm.
(i) Length of the boundary = (boundary of bigger semicircle +
boundary of smaller semicircle +
2 × boundary of smallest semicircle)
= π(14/2) + π(7/2) + 2× π(3.5/2)
= 7π + 3.5π + 3.5π
= 7π + 7π
= 14π
= 14 × 22/7
= 2 × 22
= 44 cm
Hence, the length of the boundary = 44 cm
(ii) Now we find the area of the shaded region = Area of semicircle with AB as diameter –
Area of the semicircle with radius AE –
Area of the semicircle with radius BC +
Area of semicircle with diameter 7 cm
= 1/2 × π(14/2)2 – 1/2 × π(3.5/2)2 – 1/2 × π(3.5/2)2 + 1/2 × π(7/2)2
= 1/2π [72 – 1.752 – 1.752 + 3.52]
= 1/2 π[49 – 3.0625 – 3.0625 + 12.25]
= 1/2 π[49 – 6.125 + 12.25]
= 1/2 π [42.875 + 12.25]
= 1/2 π [55.125]
= 1/2 × 22/7 × 55.125
= 11 × 7.875
= 86.625 cm2
Hence, the required area of the shaded region is 86.625 cm2
Вопрос 43. На рисунке АВ = 36 см, а М — середина АВ. На AB, AM и MB в качестве диаметров начерчены полуокружности. Окружность с центром С касается всех трех окружностей. Найдите площадь заштрихованной области.
Решение:
Given that,
AB = 36 cm
AM = BM = 1/2 × AB = 1/2 × 36 = 18 cm [M is mid-point of AB]
AM = BM = 18 cm
AP = PM = MQ = QB = 9 cm
Let us considered the radius of circle with centre C be ‘r’ i.e CR = r
Join P to C and M to C, MC ⊥ AB
MR = AM = 18 cm
CM = MR – CR
CM = (18 – r )………(1)
PC = PE + CE
PC = (9 + r)…….(2)
Now In ∆ PCM,
By using Pythagoras theorem, we get
PC2 = PM2 + MC2
(9 + r)2 = 92 + (18 – r)2
81 + r2 + 18r = 81 + 324 + r2 – 36r [From eq (1) and (2)]
54r = 324
r = 324/54
r = 6
Radius of circle with C as a centre = 6 cm
Now we find the area of shaded region = Area of semicircle with diameter AB –
Area to semicircles with diameter AM and MB –
Area of circle with C as a centre
= 1/2 π(36/2)2 – 2 × 1/2 π(18/2)2 – π(6)2
= 1/2 π(18)2 – π(9)2 – π(6)2
= 1/2 π × 324 – 81π – 36π
= 162π – 81π – 36π
= 162π – 117π
= 45π cm2
Hence, the area of required shaded region is 45π cm2.
Вопрос 44. На рисунке ABC — прямоугольный треугольник, в котором ∠A = 90°, AB = 21 см и AC = 28 см. Полуокружности описаны на AB, BC и AC как диаметры. Найдите площадь заштрихованной области.
Решение:
Given that,
ABC is a right-angled triangle in which ∠A = 90°, AB = 21 cm and AC = 28 cm.
To find: the area of the shaded region.
Now In right ΔABC,
By using Pythagoras theorem, we get
BC2 = AB2 + AC2
BC2 = 212 + 282
BC2 = 1225
BC = √1225
BC = 35 cm
Diameter BC = 35 cm
Now we find the area of shaded region, A = Area of semicircle with AC as a diameter +
Area of right angle ∆ ABC +
Area of semicircle with AB as a diameter –
Area of semicircle with BC as diameter
= 1/2 π(21/2)2 + 1/2 π(28/2)2 + 1/2 × 21 × 28 – 1/2 π(35/2)2
= 1/2 π(21/2)2 + 1/2 π(28/2)2 – 1/2 π(35/2)² + 1/2 × 21 × 28
= 1/2 π [10.52 + 142 – 17.52] + 14 × 21
= 1/2 π [110.25 + 196 – 306.25] × 294
= 1/2 π [306.25 – 306.25] + 294
= 1/2 π × 0 + 294
= 0 + 294
= 294 cm2
Hence, the area of required shaded region is 294 cm2
Вопрос 45. На рисунке изображен поперечный разрез железнодорожного тоннеля. Радиус ОА круглой части равен 2 м. Если ∠AOB = 90°, рассчитайте:
(i) высота туннеля
(ii) периметр поперечного сечения
(iii) площадь поперечного сечения.
Решение:
Given that,
The radius OA of the circular part = 2 m
∠AOB = 90°
Let OM ⊥ AB.
(i) Now In ∆OAB,
By using Pythagoras Theorem, we get
AB2 = OA2 + OB2
AB2 = 22 + 22
AB2 = 8
AB = √8
AB = √4×2
AB = 2√2 cm
Here, D b e the mid point so, AD = BD = √2
So, OD2 = OA2 – AD2
= 22 – (√22)
= √2
Let the height of the tunnel to be h.
So,
The area of ∆ OAB = 1/2 × Base × height
= 1/2 × OA × OB
1/2 × 2 × 2
= 2
(i) Height of the tunnel (h) = OC + OD
h = √2 + 2
h = (2 + √2)m
(ii) Central angle of major arc, θ = 360° – 90° = 270°
Perimeter of cross-section,
= length of the major Arc AB + AB
= θ/360° × 2πr + 2√2
= 270°/360° × 2π × 2 + 2√2
= 3/4 × 4π + 2√2
= (3π + 2√2) m
(iii) Area of cross-section, A = θ/360° × Area of circle + area of ∆AOB
= θ/360° × πr2 + 1/2 × base × height
= 270°/360° × π× 22 + 1/2 × 2 × 2
= 3/4 × π × 4 + 2
= (3π + 2)m
Hence, the height of the tunnel is (2 + √2)m,
Perimeter of cross-section is (3π + 2√2) m and
Area of cross section is (3π + 2)m.
Вопрос 46. На рисунке изображен воздушный змей, у которого BCD имеет форму квадранта окружности радиусом 42 см. ABCD — квадрат, а ΔCEF — равнобедренный прямоугольный треугольник, равные стороны которого равны 6 см. Найдите площадь заштрихованной области.
Решение:
Given :
Radius of a quadrant of a circle, r = 42 cm.
Equal sides of an isosceles right-angled ∆ = 6 cm
To find: the area of the shaded region.
Now we find the area of shaded region(A) = Area of quadrant + Area of isosceles ∆
A = 1/4 πr2 + 1/2 × base × height
A = 1/4 × 22/7 × 422 + 1/2 × 6 × 6
A = 1/2 × 11 × 6 × 42 + 18
A = 11 × 3 × 42 + 18
A = 33 × 42 + 18
A = 1386 + 18
A = 1404 cm2
Hence, the area of shaded region is 1404 cm2
Вопрос 47. На рисунке ABCD — трапеция площадью 24,5 см2. В нем AD || ВС, ∠DAB = 90°, AD = 10 см и ВС = 4 см. Если АВЕ — квадрант круга, найдите площадь заштрихованной области. (Возьмите π = (22/7).
Решение:
Given,
Area of trapezium ABCD, A = 24.5 cm2
AD || BC, ∠DAB = 90°, AD = 10 cm, BC = 4 cm and ABE is quadrant of a circle.
Now in trapezium ABCD,
Area of the trapezium, A = 1/2 (sum of parallel sides) × perpendicular distance between the parallel sides(h)
A = 1/2 (AD + BC) × AB
24.5 = 1/2 (10 + 4) × AB
24.5 × 2 = 14 AB
AB = 49/14
AB = 7/2
AB = 3.5 cm
So, the radius of the quadrant of the circle, r = AB = 3.5 cm
Area of the quadrant of the circle = 1/4 ×πr2
= (1/4) (22/7 x 3.5 x 3.5)
= 9.625 cm2
Now we find the area of the shaded region = Area of the trapezium – Area of the quadrant of the circle
= 24.5 – 9.625
= 14.875 cm2
Hence, the area of the shaded region is 14.875 cm2
Вопрос 48. На рисунке ABCD — трапеция с AB || DC, AB = 18 см, DC = 32 см, а расстояние между AB и DC равно 14 см. Нарисованы окружности равного радиуса 7 см с центрами А, В, С и D. Затем найдите площадь заштрихованной части рисунка. (Используйте π = 22/7).
Решение:
Given,
AB = 18 cm, DC = 32 cm,
Distance between AB and DC(h)= 14 cm and radius of each circle(r) = 7cm
Since, AB ||DC
So, ∠A + ∠D = 180° & ∠B + ∠C = 180°
Area of sector = (θ /360) × πr2
Area of sector with ∠A and ∠D = (180 /360) × 22/7 × 72
= 1/2 × 22 × 7 = 11 × 7 = 77 cm2
Similarly, Area of sector with ∠B & ∠C = (180 /360) × 22/7 × 72
= 1/2 × 22 × 7 = 11 × 7 = 77 cm2
Now in trapezium ABCD,
Area of trapezium = 1/2 (sum of parallel sides) × perpendicular distance between Parallel sides(h)
= 1/2 (AB + DC) × (h)
= 1/2(18 + 32) × 14
= 1/2(50)× 14
= 25 × 14 = 350 cm2
Now we find the area of shaded region = Area of trapezium –
(Area of sector with ∠B and ∠C +
Area of sector with ∠A and ∠D )
= 350 -(77+77) = 350 – 154 = 196 cm2
Hence, the Area of shaded region is 196 cm2
Вопрос 49. Из тонкого куска металла в форме трапеции ABCD, в которой AB || CD и ∠BCD = 90°, удаляется четверть окружности BEFC (см. рисунок). Учитывая AB = BC = 3,5 см и DE = 2 см, рассчитайте площадь оставшегося куска металлического листа.
Решение:
Given,
In trapezium ABCD
AB || CD and ∠BCD = 90°
AB = BC =3.5 cm & DE = 2 cm
CE = CB = 3.5 cm [CE and BC are the radii of quarter circle BFEC]
So, DC = DE + EC
DC = 2 cm + 3.5 cm
DC = 5.5 cm
Area of remaining piece of the metal sheet (A) = Area of trapezium ABCD – Area of quarter circle BFEC
A = 1/2(AB + DC) × BC – 1/4 x π x (BC)2
A = 1/2 (3.5 + 5.5) × 3.5 – 1/4 x π(3.5)2
A = 1/2 × 9 × 3.5 – 1/4 x π(3.5)2
A = 4.5 × 3.5 – 22/7 × 3.5 × 3.5/4
A = 15.75 – 11 × 3.5/4
A = 15.75 – 9.625
A = 6.125 cm2
Hence, the area of remaining piece of the metal sheet (Shaded region) is 6.125 cm2
Вопрос 50. На рисунке АВС — равносторонний треугольник со стороной 8 см. А, В и С — центры дуг окружности радиусом 4 см. Найдите площадь заштрихованной области с точностью до 2 знаков после запятой. (Возьмите π = 3,142 и √3 = 1,732).
Решение:
In Equilateral triangle all the angles are each 60°.
The corners form sectors of a circle.
When we join the sectors we form a major sector with the middle angle as (60 × 3) = 180°
Area of the shaded region = Area of the triangle – area of the sector.
Area of the triangle = 1/2 × base × height
As we know that,
Base = 8/2 = 4 cm
Hypotenuse = 8 cm
Height = √82 – 42
= √48
= 4√3
= 4 × 1.732
= 6.928 cm
Also, area of the triangle = 1/2 × 6.928 × 8 = 27.712 cm2
Now area of the Sector,
Radius of the sector = 8/2 = 4 cm
= 180/360 × 3.142 × 42 = 25.136 cm2
Now we find the area of the shaded region = 27.712 – 25.136 = 2.576 cm2
Hence, the area of the shaded region is 2.576 cm2
Вопрос 51. Стороны треугольного поля равны 15 м, 16 м и 17 м. По трем углам поля отдельно привязаны корова, буйвол и лошадь веревками длиной 7 м каждая, чтобы пастись в поле. Найдите площадь поля, которую не могут пасти три животных.
Решение:
Let ABC be the triangular field with sides AC = 15 m, AB = 16 m and BC = 17 m
And,
Let the place where the buffalo, the horse and the cow are tied,
are three sectors i.e. sector BFG, sector CHI and sector ADE
Area of triangular field = √s(s – a)(s – b)(s – c) [by using Heron’s formula]
s = (a + b + c)/2
s = (15 + 16 + 17)/2
s = 48/2
s = 24 m
=√24(24 – 15)(24 – 16)(24 – 17)
=√24 x 9 x 8 x 7
=√12096
=109.98 m2
Area of triangular field = 109.98 m2
Area of the grazed part = Area of the sector ADE + Area of sector BFG + Area of sector CHI
= π x 72 x ∠A/360 + π x 72 x ∠B/360 + π x 72 x ∠C/360
= π x 72(∠ A + ∠ B + ∠ C)/360
= 22/7 x (7)2 x 180/360
= 154/2
= 77 m2
So, the area of the field which cannot be grazed by these animals
= 109.98 m2 – 77 m2
= 32.98 m2
Hence, the area of the field which cannot be grazed by these animals is 32.98 m2
Вопрос 52. На данном рисунке сторона квадрата равна 28 см, а радиус каждой окружности равен половине длины стороны квадрата, где О и О' — центры окружностей. Найдите площадь заштрихованной области.
Решение:
Given that,
Side of square = 28 cm
Radius of each circle is half of the length of the side of the square
So, radius of each circle = 28/2 cm =14 cm
As we know that
Area of Square = (Side)2
Area of Circle = πr2
Now we find the area of Shaded region = Area of Square +3/4 (Area of Circle) + 3/4(Area of Circle)
= (28)2 + 3/2 x 22/7 × 14 × 14
= 784 cm2 + 924 cm2
= 1708 cm2
Hence, the area of shaded region is 1708 cm2
Вопрос 53. В больнице использованная вода собирается в цилиндрический резервуар диаметром 2 м и высотой 5 м. После повторного использования эта вода используется для орошения парка больницы длиной 25 м и шириной 20 м. Если бак заполнен полностью, то какой будет высота стоячей воды, используемой для орошения парка?
Решение:
Given that
Diameter of cylinder (d) = 2 m
Radius of cylinder (r) = 1 m
Height of cylinder (H) = 5 m
Now we know that volume of cylindrical tank is,
V = πr2H = π × (1)2 × 5 = 5π m
Length of the park (l) = 25 m
Breadth of park (b) = 20 m
Let us considered the height of standing water in the park = h
Volume of water in the park = l x b x h = 25 × 20 × h
Now for irrigation in the park water is used from the tank. So,
Volume of cylindrical tank = Volume of water in the park
⇒ 5π = 25 × 20 × h
⇒ 5π/25 × 20 = h
⇒ h = π/100 m
⇒ h = 0.0314 m
Hence, the height of standing water used for irrigating the park is 0.0314 m