ML Aggarwal Mensuration Exe-16.3 Class 9 ICSE Maths Solutions Ch-16. Step by Step Solutions of Exe-16.3 Questions on Mensuration of ML Aggarwal for ICSE Class 9th Mathematics. Visit official website CISCE for detail information about ICSE Board Class-9.
ML Aggarwal Mensuration Exe-16.3 Class 9 ICSE Maths Solutions
| Board | ICSE |
| Subject | Maths |
| Class | 9th |
| Chapter-16 | Mensuration |
| Topics | Solution of Exe-16.3 Questions |
| Academic Session | 2024-2025 |
Solution of Exe-16.3 Questions on Mensuration
ML Aggarwal Mensuration Exe-16.3 Class 9 ICSE Maths Solutions Ch-16.
Question 1. Find the length of the diameter of a circle whose circumference is 44 cm.
Answer :
Consider radius of the circle = r cm
Circumference = 2 πr
2 πr = 44
So,
(2 × 22)/7 r = 44
r = (44 × 7)/(2 × 22) = 7 cm
Diameter = 2r = 2 × 7 = 14 cm
Question 2. Find the radius and area of a circle if its circumference is 18π cm.
Answer :
Consider the radius of the circle = r
Circumference = 2 πr
2 πr = 18π
So,
2r = 18
⇒ r = 18/2 = 9 cm
Here
Area = πr2
= π × 9 × 9
= 81π cm2
Question 3. Find the perimeter of a semicircular plate of radius 3.85 cm.
Answer :
Radius of semicircular plate = 3.85 cm
Length of semicircular plat = πr
Perimeter = πr + 2r = r (π + 2)
= 3.85 (22/7 + 2)
= 3.85 × 36/7
= 0.55 × 36
= 19.8 cm
(ML Aggarwal Mensuration Exe-16.3 Class 9 ICSE Maths Solutions)
Question 4. Find the radius and circumference of a circle whose area is 144π cm2.
Answer :
Area of a circle = 144 π cm2
Consider radius = r
πr2 = 144 π
⇒ r2 = 144
r = √144 = 12 cm
Here,
Circumference = 2 πr
= 2 ×12×π
= 24π cm
(ML Aggarwal Mensuration Exe-16.3 Class 9 ICSE Maths Solutions)
Question 5. A sheet is 11 cm long and 2 cm wide. Circular pieces 0.5 cm in diameter are cut from it to prepare discs. Calculate the number of discs that can be prepared.
Answer :
Length of sheet = 11 cm
Width of sheet = 2 cm
We have to cut the sheet to a square of side 0.5 cm
Here
Number of squares = 11/0.5 × 2/0.5
Multiply and divide by 10
= (11×10)/5 × (2×10)/5
= 22 × 4
= 88
Therefore, the number of discs will be equal to number of squares cut out = 88.
Mensuration Exercise-16.3
ML Aggarwal Class 9 ICSE Maths Solutions
Page 379
Question 6.
If the area of a semicircular region is 77cm2, find its perimeter.
Answer 6
Area of a semicircular region = 77 cm2
Consider r as the radius of the region
Area = ½ πr2
½ πr2 = 77
½ × 22/7 r2 = 77
So we get
r2 = (77×2×7)/22 = 49 = 72
⇒ r = 7 cm
Here
Perimeter of the region = πr + 2r
= 22/7 × 7 + 2 × 7
= 22 + 14
= 36 cm
Question 7.
(a) In the figure (i) given below, AC and BD are two perpendicular diameters of a circle ABCD. Given that the ara of the shaded portion is 308 cm2, calculate
(i) the length of AC and
(ii) the circumference of the circle.
(b) In the figure (ii) given below, AC and BD are two perpendicular diameters of a circle with centre O. If AC = 16 cm, calculate the area and perimeter of the shaded part. (Take π = 3.14)
Answer :
(a) Area of shaded portion = Area of semicircle = 308 cm2
Consider r as the radius of circle
½ πr2 = 308
½ × 22/7 r2 = 308
⇒ r2 = (308 × 2 × 7)/ 22
⇒ r2 = 196 = 142
So, r = 14 cm
(i) AC = 2r = 2 × 14 = 28 cm
(ii) Circumference of the circle = 2πr
= 28 × 22/7
So we get
= 4 × 22
= 88 cm
(b) Diameter of circle = 16 cm
Radius of circle = 16/2 = 8 cm
Here,
Area of shaded part = 2 × area of one quadrant
= ½ πr2
= ½ × 3.14 × 8 × 8
= 100.48 cm2
Perimeter of shaded part = ½ of circumference + 4r
It can be written as
= ½ × 2πr + 4r
= πr + 4r
Taking r as common
= r (π + 4)
= 8 (3.14 + 4)
= 8 × 7.14
= 57.12 cm
Question 8. A bucket is raised from a well by means of a rope which is wound round a wheel of diameter 77 cm. Given that the bucket ascends in 1 minute 28 seconds with a uniform speed of 1.1 m/sec, calculate the number of complete revolutions the wheel makes in raising the bucket.
Answer :
Diameter of wheel = 77 cm
So radius of wheel = 77/2 cm
Circumference of wheel = 2πr
= 2 × 22/7 × 77/2
= 242 cm
Question 9. The wheel of a cart is making 5 revolutions per second. If the diameter of the wheel is 84 cm, find its speed in km/hr. Give your answer correct to the nearest km.
Answer :
Diameter of wheel = 84 cm
Radius of wheel = 84/2 = 42 cm
Here,
Circumference of wheel = 2πr
= 2× 22/7 ×42
= 264 cm
So the distance covered in 5 reductions = 264 × 5 = 1320 cm
Time = 1 second
Speed of wheel = 1320/1 × (60×60)/(100×1000)
= 47.25 km/hr
= 48 km/hr
Question 10. The circumference of a circle is 123.2 cm. Calculate :
(i) the radius of the circle in cm.
(ii) the area of the circle in cm2, correct to the nearest cm2.
(iii) the effect on the area of the circle if the radius is doubled.
Answer :
Circumference of a circle = 123.2 cm
Consider radius = r cm
(i) 2πr = 123.2
(2×22)/7 r = 1232/10
So,
r = (1232 ×7)/(10×2×22)
⇒ r = 19.6 cm
Hence, radius of the circle is 19.6 cm
(ii) Here, Area of the circle = πr2
= 22/7 × 19.6 × 19.6
= 1207.36
= 1207 cm2
(iii) If radius is doubled = 19.6 × 2 = 39.2 cm
So the area of circle = πr2
= 22/7 × 39.2 × 39.2
= 4829.44 cm2
Effect on area = 4829.44/1207 = 4 times
(ML Aggarwal Mensuration Exe-16.3 Class 9 ICSE Maths Solutions)
Question 11.
(a) In the figure (i) given below, the area enclosed between the concentric circles is 770 cm2. Given that the radius of the outer circle is 21 cm, calculate the radius of the inner circle.
(b) In the figure (ii) given below, the area enclosed between the circumferences of two concentric circles is 346.5 cm2. The circumference of the inner circle is 88 cm. Calculate the radius of the outer circle.
Answer :
(a) Radius of outer circle (R) = 21 cm
Consider r cm as the radius of inner circle
Area of the ring = π (R2 – r2)
= 22/7 ×(212 – r2)
= 22/7 ×(441 – r2)
Area of the ring = 770 cm2
22/7 ×(441 – r2) = 770
441 – r2 = (770×7)/22 = 245
⇒ r2 = 441 – 245 = 196
⇒ r = √196 = 14
Therefore, the radius of inner circle is 14 cm.
(b) It is given that, Area of ring = 346.5 cm2
Circumference of inner circle = 88 cm
Radius = (88 × 7)/(2×22) = 14 cm
Consider R cm as the radius of outer circle
Area of ring = π (R2 – r2)
= 22/7 ×(R2 – 142)
= 22/7 ×(R2 – 196) cm2
Area of ring = 346.5 cm2
By equating we get
22/7 ×(R2 – 196) = 346.5
R2 – 196 = (346.5×7)/22 = 110.25
⇒ R2 = 110.25 + 196 = 306.25
So,
R = √306.25 = 17.5
Therefore, the radius of outer circle is 17.5 cm.
Question 12. A road 3.5 m wide surrounds a circular plot whose circumference is 44 m. Find the cost of paving the road at ₹50 per m2.
Answer :
Circumference of circular plot = 44 m
Radius of circular plot = (44×7)/(22×2) = 7 m
Width of the road = 3.5 m
So the radius of outer circle = 7 + 3.5 = 10.5 m
Area of road = π (R2 – r2)
= 22/7 ×(10.52 – 72)
We can write,
= 22/7 ×(10.5 + 7) (10 – 7)
= 22/7 × 17.5 × 3.5
= 192.5 m2
Rate of paving the road = Rs 50 per m2
Total cost = 192.5 × 50
= Rs 9625
(ML Aggarwal Mensuration Exe-16.3 Class 9 ICSE Maths Solutions)
Question 13. The sum of diameters of two circles is 14 cm and the difference of their circumferences is 8 cm. Find the circumference of the two circles.
Answer :
Sum of the diameters of two circles = 14 cm
Consider R and r as the radii of two circles
2R + 2r = 14
Dividing by 2
R + r = 7 …(1)
Difference of their circumferences = 8 cm
2 πR – 2 πr = 8
2π ×(R – r) = 8
⇒ (2 × 22)/7(R – r) = 8
R – r = (8×7)/(2×22) = 14/11 …(2)
By adding both the equations
2R = 7 + 14/11
By taking LCM
2R = (77 + 14)/11 = 91/11
R = 91/(11×2) = 91/22
From equation (1)
R + r = 7
91/22 + r = 7
⇒ r = 7 – 91/22
Taking LCM
r = (154 – 91)/22 = 63/22
Circumference of first circle = 2 πr
= 2× 22/7 × 91/22
= 26 cm
Circumference of second circle = 2 π R
= 2× 22/7 × 63/22
= 18 cm
Question 14. Find the circumference of the circle whose area is equal to the sum of the areas of three circles with radius 2 cm, 3 cm and 6 cm.
Answer :
Radius of first circle = 2 cm
Area of first circle = πr2
= π (2)2
= 4 π cm2
Radius of second circle = 3 cm
Area of first circle = πr2
= π (3)2
= 9 π cm2
Radius of second circle = 6 cm
Area of first circle = πr2
= π (6)2
= 36 π cm2
So the total area of the three circles = 4 π + 9 π + 36 π = 49 π cm2
Area of the given circle = 49 π cm2
Radius = √(49 π/ π) = √49 = 7 cm
Circumference = 2 πr = 2 × 22/7 × 7 = 44 cm
Mensuration Exercise-16.3
ML Aggarwal Class 9 ICSE Maths Solutions
Page 380
Question 15. A copper wire when bent in the form of a square encloses an area of 121 cm2. If the same wire is bent into the form of a circle, find the area of the circle.
Answer :
Area of square = 121 cm2
So side = √121 = 11 cm
Perimeter = 4a = 4 × 11 = 44 cm
Circumference of circle = 44 cm
Radius of circle = (44×7)/(2×22) = 7 cm
Area of the circle = πr2
= 22/7 (7)2
So,
= 22/7 × 7× 7
= 154 cm2
Question 16. A copper wire when bent into an equilateral triangle has area 121√3 cm2. If the same wire is bent into the form of a circle, find the area enclosed by the wire.
Answer :
Area of equilateral triangle = 121 √3 cm2
Consider a as the side of triangle
Area = √3/4 a2
√3/4 a2 = 121√3
a2 = (121×√3×4)/√3
⇒ a2 = 484
So,
a = √484 = 22 cm
Here
Length of the wire = 66 cm
Radius of the circle = 66/2π
= (66×7)/(2×22)
= 21/2 cm
Area of the circle = πr2
By calculation
= 22/7 × (21/2)2
= 22/7 × 21/2 × 21/2
= 693/2
= 346.5 cm2
Question 17. (a) Find the circumference of the circle whose area is 16 times the area of the circle with diameter 7 cm.
(b) In the given figure, find the area of the unshaded portion within the rectangle. (Take π = 3.14)
Answer :
(a) It is given that, Diameter of the circle = 7 cm
Radius of the circle = 7/2 cm
Area of the circle = πr2
= 22/7 × 7/2 × 7/2
= 77/2 cm2
Area of bigger circle = 77/2 × 16 = 616 cm2
Consider r as the radius
πr2 = 616
We can write,
22/7 r2 = 616
r2 = (616 × 7)/22
⇒ r2 = 196 cm2
r = √196 = 14 cm
Circumference = 2πr
= 2 × 22/7 × 14
= 88 cm
(b) It is given that, Radius of each circle = 3 cm
Diameter of each circle = 2 × 3 = 6 cm
Here
Length of rectangle (l) = 6 + 6 + 3 = 15 cm
Breadth of rectangle (b) = 6 cm
So the area of rectangle = l × b
= 15 × 6
= 90 cm2
Area of 2 ½ circles = 5/2 πr2
= 5/2 ×3.14 ×3×3
= 5 ×1.57×9
= 70.65 cm2
So,
the area of unshaded portion = 90 – 70.65
= 19.35 cm2
(ML Aggarwal Mensuration Exe-16.3 Class 9 ICSE Maths Solutions)
Question 18. In the adjoining figure, A6CD is a square of side 21 cm. AC and BD are two diagonals of the square. Two semicircle are drawn with AD and BC as diameters. Find the area of the shaded region. Take π = 
.
Answer :
Side of square = 21 cm
So the area of square = side2 = 212 = 441 cm2
∠AOD + ∠COD + ∠AOB + ∠BOC = 441
x + x + x + x = 441
⇒ 4x = 441
x = 441/4 = 110.25 cm2
We should find the area of shaded portion in square ABCD which is ∠AOD and ∠BOC
∠AOD + ∠BOC = 110.25 + 110.25 = 220.5 cm2
Here
Area of two semicircle = πr2
= 22/7 × 10.5 × 10.5
= 346.50 cm2
So the area of shaded portion = 220.5 + 346.5 = 567 cm2
Question 19.
(a) In the figure (i) given below, ABCD is a square of side 14 cm and APD and BPC are semicircles. Find the area and the perimeter of the shaded region.
(b) In the figure (ii) given below, ABCD is a square of side 14 cm. Find the area of the shaded region.
(c) In the figure (iii) given below, the diameter of the semicircle is equal to 14 cm. Calculate the area of the shaded region. Take π = 22/7.
Answer :
(a) ABCD is a square of each side (a) = 14 cm
APD and BPC are semi-circle with diameter = 14 cm
Radius of each semi-circle (a) = 14/2 = 7 cm
(i)
Area of square = a2 = 142 = 196 cm2
Area of two semicircles = 2 × ½ πr2
= πr2
= 22/7 × 7 × 7
= 154 cm2
So the area of shaded portion = 196 – 154 = 42 cm2
(ii)
Length of arcs of two semicircles = 2πr
= 2 × 22/7 × 7
= 44 cm
So the perimeter of shaded portion = 44 + 14 + 14 = 72 cm
(b) ABCD is a square whose each side (a) = 14 cm
4 circles are drawn which touch each other and the sides of squares
Radius of each circle (r) = 7/2 = 3.5 cm
(i)
Area of square ABCD = a2 = 142 = 196 cm2
Area of 4 circles = 4 × πr2
Substituting the values
= 4 × 22/7 × 7/2 × 7/2
= 154 cm2
So the area of shaded portion = 196 – 154 = 42 cm2
(ii) Here
Perimeter of 4 circles = 4 × 2 πr
Substituting the values
= 4 × 2 × 22/7 × 7/2
= 88 cm
So the perimeter of shaded portion = perimeter of 4 circles + perimeter of square
Substituting the values
= 88 + 4 × 14
So we get
= 88 + 56
= 144 cm
(c) Area of rectangle ACDE = ED × AE
Substituting the values
= 14 × 7
= 98 cm2
Area of semicircle DEF = πr2/2
= (22×7×7)/(7×2)
= 77 cm2
So,
the area of shaded region = 77 + (98 – 2× ¼ × 22/7 ×7×7)
= 77 + 21
= 98 cm2
(ML Aggarwal Mensuration Exe-16.3 Class 9 ICSE Maths Solutions)
Question 20.
(a) Find the area and the perimeter of the shaded region in figure (i) given below. The dimensions are in centimetres.
(b) In the figure (ii) given below, area of ∆ABC = 35 cm2. Find the area of the shaded region.
Answer :
(a) There are 2 semicircles where the smaller is inside the larger
Radius of larger semicircles (R) = 14 cm
Radius of smaller circle (r) = 14/2 = 7 cm
(i)
Area of shaded portion = Area of larger semicircle – Area of smaller circle
= ½ πR2 – ½ πr2
We can write it as
= ½ π ×(R2 – r2)
= ½ × 22/7 ×(142 – 72)
By further calculation
= 11/7 ×(14 + 7)×(14 – 7)
= 11/7 ×21×7
= 231 cm2
(ii)
Perimeter of shaded portion = circumference of larger semicircle + circumference of smaller semicircle + radius of larger semicircle
= πR + πr + R
= (22/7 × 14) + (22/7 × 7) + 14
= 44 + 22 + 14
= 80 cm
(b) Area of △ABC formed in a semicircle = 3.5 cm
Altitude CD = 5 cm
So the base AB = (area×2)/altitude
= (35×2)/5
= 14 cm
Diameter of semicircle = 14 cm
Radius of semicircle (R) = 14/2 = 7 cm
So the area of semicircle = ½ πR2
= ½ × 22/7 × 7 × 7
= 77 cm2
Area of shaded portion = Area of semicircle – Area of triangle
= 77 – 35
= 42 cm2
Question 21.
(a) In the figure (i) given below, AOBC is a quadrant of a circle of radius 10 m. Calculate the area of the shaded portion. Take π = 3.14 and give your answer correct to two significant figures.
(b) In the figure, (ii) given below, OAB is a quadrant of a cirlce. The radius OA = 3.5 cm and OD = 2 cm. Calculate the area of the shaded portion.
Answer :
(a) Shaded portion = Quadrant – △AOB
Radius of the quadrant = 10 m
Area of quadrant = ¼ πr2
= ¼ × 3.14 × 10 × 10
= (3.14×100)/ 4
= 314/4
= 78.5 m2
Area of △AOB = ½ × AO × OB
= ½ × 10 × 10
= 50 m2
So the area of shaded portion = 78.5 – 50 = 28.5 m2
(b) Radius of quadrant = 3.5 cm
(i)
Area of quadrant = ¼ πr2
= ¼ × 22/7 × 3.5 × 3.5
= 9.625 cm2
(ii)
Area of △AOD = ½ × AO × OD
= ½ × 3.5 × 2
= 3.5 cm2
So the area of shaded portion = Area of quadrant – Area of △AOD
= 9.625 – 3.6
= 6.125 cm2
Mensuration Exercise-16.3
ML Aggarwal Class 9 ICSE Maths Solutions
Page 381
Question 22. A student takes a rectangular piece of paper 30 cm long and 21 cm wide. Find the area of the biggest circle that can be cut out from the paper. Also find the area of the paper left after cutting out the circle. (Take π = 22/7 )
Answer :
Length of rectangle = 30 cm
Width of rectangle = 21 cm
Area of rectangle = l × b
= 30 × 21
= 630 cm2
So the radius of the biggest circle = 21/2 cm
Area of the circle = πr2
= 22/7 × 21/2 × 21/2
So,
= 693/2
= 346.5 cm2
So,
The area of remaining part = 630 – 346.5 = 283.5 cm2
Question 23. A rectangle with one side 4 cm is inscribed in a circle of radius 2.5 cm. Find the area of the rectangle.
Answer :
ABCD as a rectangle
AB = 4 cm
Diameter of circle AC = 2.5 × 2 = 5 cm
Here,
= 3 cm
Area of rectangle = AB × BC
= 4 × 3
= 12 cm2
(ML Aggarwal Mensuration Exe-16.3 Class 9 ICSE Maths Solutions)
Question 24.
(a) In the figure (i) given below, calculate the area of the shaded region correct to two decimal places. (Take π = 3. 142).
(b) In the figure (ii) given below, ABC is an isosceles right angled triangle with ∠ABC = 90°. A semicircle is drawn with AC as diameter. If AB = BC = 7 cm, find the area of the shaded region. Take π =
Answer :
(a) ABCD is a rectangle which is inscribed in a circle of length = 12 cm
Width = 5 cm
= 13 cm
Diameter of circle = AC = 13 cm
Radius of circle = 13/2 = 6.5 cm
Area of circle = πr2
= 3.142 × (6.5)2
= 3.142 × 42.25
= 132.75 cm2
Area of rectangle = l × b
= 12 × 5
= 60 cm2
So the area of the shaded portion = 132.75 – 60 = 72.75 cm2
(b) Area of △ABC = ½ × AB × BC
= ½ × 7 × 7
= 49/2 cm2
Here
AC2 = AB2 + BC2
Substituting
= 49 + 49
So we get
AC = 7√2
Radius of semi-circle = 7√2/2 cm
Area of semi-circle = π/2 × (7√2/2)2
= ½ × 22/7 × 98/4
= 77/2 cm2
Area of the shaded region = Area of the semi-circle – Area of △ABC
= 77/2 – 49/2
= 28/2
= 14 cm2
Question 25. A circular field has perimeter 660 m. A plot in the shape of a square having its vertices on the circumference is marked in the field. Calculate the area of the square field.
Answer :
Perimeter of circular field = 660 m
Radius of the field = 660/2 π
= (660×7)/(2×22)
= 105 m
ABCD is a square which is inscribed in the circle where AC is the diagonal which is the diameter of the circular field
Consider a as the side of the square
AC = √2a
a = AC/√2
a = (105 × 2)/√2
Multiply and divide by √2
a = (105×2×√2)/(√2×√2)
By calculation
a = (105×2×√2)/2
a = 105√2 m
Area of the square = a2
It can be written,
= (105√2)2
= 105√2 × 105√2
= 22050 m2
Question 26. In the adjoining figure, ABCD is a square. Find the ratio between
(i) the circumferences
(ii) the areas of the incircle and the circumcircle of the square.
Answer :
Consider side of the square = 2a
Area = (2a)2 = 4a2
Diagonal of AC = √2AB
(i) Radius of the circumcircle = ½ AC
It can be written,
= ½ (√2 × AB)
= √2/2 × 2a = √2a
Circumference = 2 πr = 2 × π × √2a = 2√2 πa
Radius of incircle = AB = ½ × 2a = a
Circumference = 2 πr = 2 πa
Here,
Ratio between the circumference incircle and circumcircle = 2 πa: 2√2 πa
= 1: √2
(ii) Area of incircle = πr2 = πa2
Area of circumcircle = πR2
= π(√2a)2
= π×2a2
= 2×πa2
So,
The ratio = πa2: 2 πa2
= 1: 2
(ML Aggarwal Mensuration Exe-16.3 Class 9 ICSE Maths Solutions)
Question 27.
(a) The figure (i) given below shows a running track surrounding a grassed enclosure PQRSTU. The enclosure consists of a rectangle PQST with a semicircular region at each end.
PQ = 200 m ; PT = 70 m.
(i) Calculate the area of the grassed enclosure in m2.
(ii) Given that the track is of constant width 7 m, calculate the outer perimeter ABCDEF of the track.
(b) In the figure (ii) given below, the inside perimeter of a practice running track with semi-circular ends and straight parallel sides is 312 m. The length of the straight portion of the track is 90 m. If the track has a uniform width of 2 m throughout, find its area.
Answer :
(a) Length of PQ = 200 m
Width PT = 70 m
(i) We know that, Area of rectangle PQST = l×b
= 200 ×70
= 14000 m2
Radius of each semi-circular part on either side of rectangle = 70/2 = 35 m
Area of both semi-circular parts = 2×½ πr2
= 22/7 × 35 × 35
= 3850 m2
So the total area of grassed enclosure = 1400 + 3850 = 17850 m2
(ii) We know that, Width of track around the enclosure = 7 m
Outer length = 200 m
So,
the width = 70 + 7 × 2
= 70 + 14
= 84 m
Outer radius = 84/2 = 42 m
Here,
Circumference of both semi-circular part = 2πr
= 2 × 22/7 × 42
= 264 m
Outer perimeter = 264 + (200×2)
= 264 + 400
= 664 m
(b) Inside perimeter = 312 m
Total length of the parallel sides = 90 + 90 = 180 m
Circumference of two semi-circles = 312 – 180 = 132 m
Here
Radius of each semi-circle = 132/2π
= 66/3.14
= 21.02 m
Diameter of each semi-circle = 66/π × 2
So,
= 132/π
= 132/3.14
Multiply and divide by 100
= (132 × 100)/314
= 42.04 m
Width of track = 2 m
Outer diameter = 42.04 + 4 = 46.04 m
Radius = 46.04/2 = 23.02 m
We know that, Area of two semi-circles = 2 × ½ × πR2
= πR2
= 3.14 × (23.02)2
= 3.14 × 23.02 × 23.02
= 1663.95 m2
Area of rectangle = 90 × 46.04
= 4143.6 m2
Total area = 1663.95 + 4143.60 = 5807.55 m2
Area of two inner circles = 2 × ½ πr2
= 3.14 × 21.02 × 21.02
= 1387.38 m2
Area of inner rectangle = 90 × 42.04
= 3783.6 m2
Total inner area = 3783.60 + 1387.38
= 5170.98 m2
Area of path = 5807.55 – 5170.98
= 636.57 m2
Mensuration Exercise-16.3
ML Aggarwal Class 9 ICSE Maths Solutions
Page 382
Question 28.
(a) In the figure (i) given below, two circles with centres A and B touch each other at the point C. If AC = 8 cm and AB = 3 cm, find the area of the shaded region.
(b) The quadrants shown in the figure (ii) given below are each of radius 7 cm. Calculate the area of the shaded portion.
Answer :
(a) AC = 8 cm
BC = AC – AB = 8 – 3 = 5 cm
Area of big circle of radius AC = πR2
= 22/7 × 8 × 8
By calculation
= 64 × 22/7 cm2
Area of smaller circle = πr2
= 22/7 × 5 × 5
= (25×22)/7 cm2
Here
Area of shaded portion = (64×22)/7 – (25×22)/7
Taking out the common terms
= 22/7 ×(64 – 25)
= 22/7 × 39
= 122.57 cm2
(b) We know that, Radius of each quadrant = 7 cm
Here
Area of shaded region = Area of square – 4 area of the quadrant
= (side)2 – 4 × ¼ πr2
= 142 – 22/7 ×7 ×7
By calculation
= 196 – 154
= 42 cm2
Question 29.
(a) In the figure (i) given below, two circular flower beds have been shown on the two sides of a square lawn ABCD of side 56 m. If the centre of each circular flower bed is the point of intersection O of the diagonals of the square lawn, find the sum of the areas of the lawn and the flower beds.
(b) In the figure (ii) given below, a square OABC is inscribed in a quadrant OPBQ of a circle. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14)
Answer:
(a) Side of square lawn ABCD (a) = 56 cm
Area = a2 = 52 = 3136 cm2
We know that, Length of the diagonal of the square = √2a
= √2 × 56 cm
Radius of each quadrant = (√2×56)/2
= 28√2 cm
So the area of each segment = ¼ πr2 – area of △OBC
= (¼× 22/7 ×28√2 ×28√2) – (½ ×28√2 ×28√2)
Taking out the common terms
= [28√2 × 28√2 ×(1/4 × 22/7)] – 1/2
By calculation
= 784 × 2(11/14 – ½)
So ,
= 784 × 2 × 4/14
= 448 cm2
Here,
Area of two segments = 448 × 2 = 896 cm2
So the total area of the lawn and beds = 3136 + 896 = 4032 cm2
(b) In the figure, OPBQ is a quadrant and OABC is a square which is inscribed in a side of square = 20 cm
OB is joined
Here,
OB = √2 a = √2 ×20 cm
Radius of quadrant = OB = 20√2cm
Area of quadrant = ¼ πr2
= ¼ × 3.14 × (20√2)2
By calculation
= ¼ × 3.14 × 800
So,
= 314 × 2
= 628 cm2
Area of square = a2 = 202 = 400 cm2
So,
the area of shaded portion = 628 – 400 = 228 cm2
(ML Aggarwal Mensuration Exe-16.3 Class 9 ICSE Maths Solutions)
Question 30.
(a) In the figure (i) given below, ABCD is a rectangle, AB = 14 cm and BC = 7 cm. Taking DC, BC and AD as diameters, three semicircles are drawn as shown in the figure. Find the area of the shaded portion.
(b) In the figure (ii) given below, O is the centre of a circle with AC = 24 cm, AB = 7 cm and ∠BOD = 90°. Find the area of the shaded region. (Use π = 3.14).
Answer:
(a) ABCD is a rectangle
Three semicircles are drawn with AB = 14 cm and BC = 7 cm
Area of rectangle ABCD = l × b
Substituting the values
= 14 × 7
= 98 cm2
Radius of each outer semicircles = 7/2 cm
So the area = 2 × ½ πr2
= 22/7 × 7/2 × 7/2
= 77/2
= 38.5 cm2
Here
Area of semicircle drawn on CD as diameter = ½ πR2
= ½ × 22/7 × 72
By calculation
= 11/7 × 7 × 7
= 77 cm2
So the area of shaded region = 98 + 38.5 – 77
= 59.5 cm2
(b) AC = 24 cm
AB = 7 cm
∠BOD = 90°
In △ABC
Using Pythagoras theorem
BC2 = AC2 + AB2
BC2 = 242 + 72
By calculation
BC = √ (576 + 49) = √625 = 25 cm
So the radius of circle = 25/2 cm
We know that, Area of △ABC = ½ × AB × AC
= ½ × 7 × 24
= 84 cm2
Area of quadrant COD = ¼ πr2
Substituting the values
= ¼ × 3.14 × 25/2 × 25/2
By calculation
= 1962.5/16
= 122.66 cm2
Area of circle = πr2
= 3.14 × 25/2 × 25/2
= 1962.5/4
= 490.63 cm2
Area of shaded portion = Area of circle – (Area of △ABC + Area of quadrilateral COD)
= 490.63 – (84 + 122.66)
= 490.63 – 206.66
= 283.97 cm2
Mensuration Exercise-16.3
ML Aggarwal Class 9 ICSE Maths Solutions
Page 383
Question 31.
(a) In the figure given below ABCD is a square of side 14 cm. A, B, C and D are centres of the equal circle which touch externally in pairs. Find the area of the shaded region.
(b) In the figure (ii) given below, the boundary of the shaded region in the given diagram consists of three semi circular arcs, the smaller being equal. If the diameter of the larger one is 10 cm, calculate.
(i) the length of the boundary.
(ii) the area of the shaded region. (Take π to be 3.14)
Answer :
(a) Side of square ABCD = 14 cm
Radius of each circle drawn from A, B, C and D and touching externally in pairs = 14/2 = 7 cm
We know that, Area of square = a2
= 14 × 14
= 196 cm2
Area of 4 sectors of 90° each = 4 × π × r2
= 4 × 22/7 × 7 × 7 × ¼
= 154 cm
Area of each sector of 270° angle = 3/4 πr2
= ¾ × 22/7 × 7 × 7
= 231/2
= 115.5 cm2
Area of 4 sectors = 115.5 × 4 = 462 cm2
= 196 + 462 – 154
= 658 – 154
= 504 cm2
(b) We know that, Radius of big semi-circle = 10/2 = 5 cm
Radius of each smaller circle = 5/2 cm
(i) Length of boundary = circumference of bigger semi-circle + 2 circumference of smaller semi-circles
It can be written,
= πR + πr + πr
= 3.14 ×(R + 2r)
= 3.14 ×[5 + (2 × 5/2)]
By calculation
= 3.14 × 10
= 31.4 cm
(ii) Here,
Area of shaded region = area of bigger semi-circle + area of one smaller semi-circle – area of other smaller semi-circle
Area of bigger semi-circle = ½ πR2
= 3.14/2 × 5 × 5
= 1.57 × 25
= 39.25 cm
Question 32.
(a) In the figure (i) given below, the points A, B and C are centres of arcs of circles of radii 5 cm, 3 cni and 2 cm respectively. Find the perimeter and the area of the shaded region. (Take π = 3.14).
(b) In the figure (ii) given below, ABCD is a square of side 4 cm. At each corner of the square a quarter circle of radius 1 cm, and at the centre a circle of diameter 2 cm are drawn. Find the perimeter and the area of the shaded region. Take π = 3.14.
Answer :
(a) Radius of bigger circle = 5 cm
Radius of small circle (r1) = 3 cm
Radius of smaller circle (r2) = 2 cm
(i) Perimeter of shaded region = circumference of bigger semi-circle + circumference of small semi-circle + circumference of smaller semi-circle
= πR + π r1 + π r2
= π (R + r1 + r2)
= π (5 + 3 + 2)
= 3.14 × 10
= 31.4 cm2
(ii) Area of shaded region = area of bigger semi-circle + area of smaller semi-circle – area of small semicircle
= ½ πR2 + ½ πr22 – ½ πr12
= ½ π (R2 + r22 – r12)
= ½ π (5 2 + 22 – 32)
By calculation
= ½ π (25 + 4 – 9)
= ½ π × 20
So,
= 10 × 3.14
= 31.4 cm2
(b) We know that, Side of square ABCD = 4 cm
Radius of each quadrant circle = 1 cm
Radius of circle in the square = 2/2 = 1 cm
(i)
Perimeter of shaded region = circumference of 4 quadrants + circumference of circle + 4 × ½ side of square
= [4 × ¼ (2 πr)] + 2 πr + (4×2)
By calculation
= 2 πr + 2 πr + 8
= 4 πr + 8
So,
= (4 × 3.14 × 1) + 8
= 12.56 + 8
= 20.56 cm
(ii) Area of shaded region = area of square – area of 4 quadrants – area of circle
= side2 – (4 × ¼ πr2) – πr2
= 42 – πr2– πr2
= 16 – 2πr2
So,
= 16 – (2 × 3.14 × 12)
= 16 – 6.28
= 9.72 cm2
(ML Aggarwal Mensuration Exe-16.3 Class 9 ICSE Maths Solutions)
Question 33.
(a) In the figure given below, ABCD is a rectangle. AB = 14 cm, BC = 7 cm. From the rectangle, a quarter circle BFEC and a semicircle DGE are removed. Calculate the area of the remaining piece of the rectangle. (Take π = 22/7)
(b) The figure (ii) given below shows a kite, in which BCD is in the shape of a quadrant of circle of radius 42 cm. ABCD is a square and ∆ CEF is an isosceles right angled triangle whose equal sides are 6 cm long. Find the area of the shaded region.
Answer :
(a) Area of remaining piece = area of rectangle ABCD – area of semicircle DGE – area of quarter BFEC
= (14×7) – [½ ×π (7/2)2] – (¼ π × 72)
= (14×7) – [½×22/7 × 7/2 × 7/2] – (¼ × 22/7 × 7×7)
So,
= 98 – 77/4 – 154/4
= 98 – 19.25 – 38.5
= 98 – 57.75
= 40.25 cm2
(b) ABCD is a square of side = radius of quadrant = 42 cm
△CEF is an isosceles right triangle with each side = 6 cm
Area of shaded portion = area of quadrant + area of isosceles right triangle
= ¼ πr2 + ½ EC × FC
= ¼ × 22/7 × 42 × 42 + ½ × 6 × 6
= 1386 + 18
= 1404 cm2
Mensuration Exercise-16.3
ML Aggarwal Class 9 ICSE Maths Solutions
Page 384
Question 34.
(a) In the figure (i) given below, the boundary of the shaded region in the given diagram 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, calculate
(i) the length of the boundary.
(ii) the area of the shaded region.
(b) In the figure (ii) given below, a piece of cardboard, in the shape of a trapezium ABCD, and AB || DC and ∠BCD = 90°, quarter circle BFEC is removed. Given AB = BC = 3.5 cm and DE = 2 cm. Calculate the area of the remaining piece of the cardboard.
Answer 34
(a) (i) Length of boundary = Circumference of bigger semi-circle + Circumference of small semi-circle + 2 × circumference of the smaller semi-circles
= πR + πr1 + (2 × πr2)
= π (R + r1) + 2πr2
= 22/7 ×(7 + 3.5) + (2× 22/7 × 3.5/2)
= (22/7 × 10.5) + 11
So,
= 33 + 11
= 44 cm
(ii)
Area of shaded region = Area of bigger semicircle + area of small semicircle – 2 × area of smaller semicircles
= [½ π (7)2 + ½ π (3.5)2] – [2 × ½ π (1.75)2]
= (½× 22/7 ×7 ×7) + (½ × 22/7 ×3.5 ×3.5) – (22/7 ×1.75 ×1.75)
So,
= 77 + 19.25 – 9.625
= 86.625 cm2
(b) Here, ABCD is a trapezium in which AB || DC and ∠C = 90°
AB = BC = 3.5 cm and DE = 2 cm
Radius of quadrant = 3.5 cm
We know that
Area of trapezium = ½ (AB + DC) × BC
Substituting the values
= ½ (3.5 + 3.5 + 2)× 3.5
By further calculation
= ½ (9×3.5)
= 4.5 × 3.5
= 15.75 cm2
So the area of quadrant = ¼ πr2
Substituting the values
= ¼ × 22/7 × 3.5 × 3.5
= 9.625 cm2
Area of shaded portions = 15.75 – 9.625 = 6.125 cm2
Question 35.
(a) In the figure (i) given below, ABC is a right angled triangle, ∠B = 90°, AB = 28 cm and BC = 21 cm. With AC as diameter a semi-circle is drawn and with BC as radius a quarter circle is drawn. Find the area of the shaded region correct to two decimal places.
(b) In the figure (ii) given below, ABC is an equilateral triangle of side 8 cm. A, B and C are the centers of circular arcs of equal radius. Find the area of the shaded region correct upto 2 decimal places. (Take π = 3.142 and √3 = 1.732).
Answer :
(a) In right △ABC
∠B = 90°
Using Pythagoras theorem
AC2 = AB2 + BC2
= 282 + 212
= 784 + 441
= 1225
AC = √1225 = 35 cm
Radius of semi-circle (R) = 35/2
Radius of quadrant (r) = 21 cm
So the area of shaded region = area of △ABC + area of semi-circle – area of quadrant
= (½ ×28 ×21) + (½ πR2 – ¼ r2)
= 294 + (½ × 22/7 × 35/2 × 35/2) – ¼ × 22/7 × 21 × 21
= 294 + 1925/4 – 693/2
= 294 + 481.25 – 346.5
= 775.25 – 346.50
= 428.75 cm2
(b) △ABC is an equilateral triangle of side 8 cm
A, B, C are the centres of three circular arcs of equal radius
Radius = 8/2 = 4 cm
Here
Area of △ABC = √3/4a2
= √3/4 ×8 ×8
= √3/4 ×64
So we get
= 16√3cm2
= 16 ×1.732
= 27.712 cm2
So the area of 3 equal sectors of 60° whose radius is 4 cm = 3 × πr2 × 60/360
= 3 ×3.142 ×4 ×4 × 1/6
= 3.142 × 8
= 25.136 cm2
Area of shaded region = 27.712 – 25.136
= 2.576 = 2.58 cm2
(ML Aggarwal Mensuration Exe-16.3 Class 9 ICSE Maths Solutions)
Question 36. A circle is inscribed in a regular hexagon of side 2√3 cm. Find
(i) the circumference of the inscribed circle
(ii) the area of the inscribed circle
Answer :
ABCDEF is a regular hexagon of side 2√3 cm. and a circle is inscribed in it with centre O.
Radius of inscribed circle = √3/2 × side of regular hexagon
= √3/2 × 2√3
= 3 cm
(i) We know that, Circumference of the circle = 2πr
= 2π × 3
By calculation
= (6 × 22)/ 7
= 132/7 cm
(ii) Area of the circle = πr2
= π ×3×3
By calculation
= (9 × 22)/7
= 198/7 cm2
Mensuration Exercise-16.3
ML Aggarwal Class 9 ICSE Maths Solutions
Page 385
Question 37. In the figure (i) given below, a chord AB of a circle of radius 10 cm subtends a right angle at the centre O. Find the area of the sector OACB and of the major segment. Take π = 3.14.
Answer :
Radius of the circle = 10 cm
Angle at the centre subtended by a chord AB = 90°
Area of sector OACB = πr2 × 90/360
= 3.14 × 10 × 10 × 90/360
So we get
= 314 × ¼
= 78.5 cm2
Here,
Area of △OAB = ½ × 10 × 10 = 50 cm2
Area of minor segment = Area of sector △ACB – Area of △OAB
= 78.5 – 50
= 28.5 cm2
Area of circle = πr2
= 3.14 × 10 × 10
= 314 cm2
Area of major segment = area of circle – area of minor segment
= 314 – 28.5
= 285.5 cm2
— : End of ML Aggarwal Mensuration Exe-16.3 Class 9 ICSE Maths Solutions :–
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