Category: Class 10 Maths Objective Questions

[CLASS 10] Mathematics MCQs on Perimeter and Area of a Circle

Mathematics Multiple Choice Questions & Answers on “Perimeter and Area of a Circle”.

1. What is the formula for the circumference of a circle?
a) C = 2πd
b) C = 2πr
c) C = 2πa
d) C = 2πs
Answer: b
Clarification: The circumference of a circle (C) = 2πr.
Where C is the circumference of the circle and r is the radius of the given circle.
The value of π is taken as (frac {22}{7}).

2. What is the formula for the area of a circle?
a) A = πd2
b) A = πs2
c) A = πr2
d) A = πa2
Answer: c
Clarification: The area of a circle (A) = πr2.
Where A is the area of the circle and r is the radius of the given circle.
The value of π is taken as (frac {22}{7}).

3. Find the area of the circle if the radius is 3.14 cm.
a) 30.98 cm
b) 30.48 cm
c) 30.68 cm
d) 30.58 cm
Answer: a
Clarification: The area of a circle (A) = πr2
A = (frac {22}{7}) × (3.14)2
A = 30.98 cm

4. The perimeter of a circle is also called the circumference.
a) False
b) True
Answer: b
Clarification: The path that surrounds the outline of a two-dimensional shape is called the perimeter. The perimeter of a circle is also called the circumference.

5. The area of a semicircle is _____
a) (frac {2pi r}{2})
b) (frac {2pi r}{2})
c) (frac {pi r}{2})
d) (frac {pi r^2}{2})
Answer: d
Clarification: A semicircle is half of a full circle. Hence, the area of a semicircle is also half of a circle. Half of πr2 is (frac {pi r^2}{2}). Therefore, the area of a semicircle is (frac {pi r^2}{2}).

6. What is the circumference of a circle if the radius is 7 m?
a) 8 m
b) 2 m
c) 44 m
d) 22 m
Answer: c
Clarification: The circumference of a circle (C) = 2πr.
C = 2 × (frac {22}{7}) × 7
C = 44 m

7. Find the area of a semicircle if the radius is 6 cm.
a) 1.35 m
b) 6.54 m
c) 18.00 m
d) 8.05 m
Answer: b
Clarification: The area of the semicircle = πr2/2
= ((frac {22}{7}) × 62)/2
= 56.54 m

8. Find the radius of the circle if the circumference is 12 m.
a) 1.90 m
b) 1.09 m
c) 7.90 m
d) 1.40 m
Answer: a
Clarification: The circumference of a circle (C) = 2πr.
12 = 2 × (frac {22}{7}) × r
r = (frac {12}{2} times frac {7}{22})
r = 1.90 m

9. Find the radius of a circle if 2 m is the area of the circle.
a) √0.83 m
b) 5 m
c) √0.63 m
d) √38 m
Answer: c
Clarification: The area of the circle (A) = πr2
2 = (frac {22}{7}) × r2
r2 = (frac {7}{22}) × 2
r = √0.63 m

10. Find the radius of the circle if the area of the circle is 22 cm.
a) 1521.14 m
b) 1511.14 m
c) 1021.14 m
d) 1520.14 m
Answer: a
Clarification: The area of the circle (A) = πr2
A = (frac {22}{7}) (22)2
A = 1521.14 m

11. What is the circumference of the circle if the radius is 121 cm?
a) 760.00 cm
b) 765.57 cm
c) 750.57 cm
d) 760.57 cm
Answer: d
Clarification: The circumference of the circle (C) = 2πr
C = 2 × (frac {22}{7}) × 121
A = 760.57 cm

12. Find the radius of the wheel if the wheel rotates 100 times to cover 500 m.
a) 0.07 m
b) 0.47 cm
c) 0.79 m
d) 0.57 cm
Answer: c
Clarification: One rotation of the wheel = circumference of the wheel
100 rotations = 500 m
1 rotation = 5 m
So, circumference = 5 m
2πr = 5 m
r = 5 × (frac {7}{22} times frac {1}{2})
r = 0.79m

13. The difference between circumference and diameter of a ring is 10 cm. Find the radius of the ring.
a) 3.07 cm
b) 0.37 cm
c) 2.33 cm
d) 4.57 cm
Answer: c
Clarification: Circumference – Diameter = 10 cm (∵ Diameter = 2 × radius)
2πr – 2r = 10 cm
2r(π – 1) = 10 cm
2r((frac {22}{7}) – 1) = 10 cm
2r((frac {15}{7})) = 10 cm
r = 10 × (frac {7}{15} times frac {1}{2})
r = 2.33 cm

14. Find the diameter of the circle if the area of the circle is 6 m.
a) 3.07 m
b) 2.74 m
c) 2.33 m
d) 4.57 m
Answer: b
Clarification: The area of the circle (A) = πr2
6 = (frac {22}{7}) × r2
r2 = 1.90
r = √1.90
r = 1.37 m
Diameter of the circle = 2 × radius
= 2 × 1.37
= 2.74 m

15. Find the diameter of the circle if the circumference of the circle is (frac {11}{5})m.
a) 23.07 cm
b) 20.74 cm
c) 72.33 cm
d) 70 cm
Answer: d
Clarification: The circumference of the circle = (frac {11}{5})m
2πr = (frac {11}{5})m
πd = (frac {11}{5})m     (∵ Diameter = 2 × radius)
d = (frac {11}{5} times frac {7}{22})m
d = 0.7 m
d = 0.7 × 100 cm
d = 70 cm

[CLASS 10] Mathematics MCQs on Area of Similar Triangle

Mathematics Question Bank for Class 10 on “Area of Similar Triangle”.

1. If the areas of two similar triangles are in the ratio 361:529. What would be the ratio of the corresponding sides?
a) 19 : 23
b) 23 : 19
c) 361 : 529
d) 15 : 23
Answer: a

2. What will be the value of BC if the area of two similar triangles ∆ABC and ∆PQR is 64cm2 and 81cm2 respectively and the length of QR is 15cm.
a) (frac {40}{9})
b) (frac {40}{6})
c) (frac {4}{3})
d) (frac {40}{3})
Answer: d

3. The areas of two similar triangles are 100cm2 and 64cm2. If the altitude of the smaller triangle is 5.5 cm, then what will be the altitude of the corresponding bigger triangle?
a) 4.4 cm
b) 4.5 cm
c) 2.4 cm
d) 2.5 cm
Answer: a

4. The area of two similar triangles is 25cm2 and 121cm2. If the median of the bigger triangle is 10 cm, then what will be the corresponding median of the smaller triangle?
a) (frac {51}{11})
b) (frac {10}{11})
c) (frac {50}{11})
d) (frac {5}{11})
Answer: d

5. ∆ABC ∼ ∆PQR, AD & PS are the angle bisectors of respectively. If AD = 1.5cm and PS = 2.3 cm then, what will be the ratio of the areas of ∆ABC and ∆PQR?
a) 19 : 15
b) 225 : 529
c) 529 : 225
d) 15 : 17
Answer: b

6. In the given figure, AC is the angle bisector and AB = 5.2 cm, AD = 4.5 cm. What will be ratios of areas of ∆ABC and ∆ACD?
a) 131 : 90
b) 131 : 91
c) 121 : 81
d) 120 : 80
Answer: c

7. What will be the ratios of areas of similar ∆ABC and ∆PQR, if the altitudes of the triangles are 10cm and 5.5cm respectively?
a) 150 : 4
b) 300 : 7
c) 100 : 3
d) 400 : 9
Answer: d

8. If two triangles are similar, then the ratio of their areas will equal to ratio of the corresponding sides.
a) True
b) False

Answer: b

[CLASS 10] Mathematics MCQs on Euclid’s Division Lemma

Mathematics Multiple Choice Questions & Answers on “Euclid’s Division Lemma”.

1. For any integer m, square of the number is of the form ______ or ______
a) 3m + 3, 3m – 2
b) 3m – 2, 3m + 2
c) 3m + 2, 3m – 3
d) 3m, 3m + 1
Answer: d
Clarification: Let a be any arbitrary number.
Then, by Euclid’s division lemma,
a = 3q + r where 0 ≤ r ≤ 3
a2 = (3q + r)2 = 9q2 + r2 + 6qr
When r = 0,
Then, a2 = 9q2 + 02 + 6q(0)
= 9q2 = 3(3q2) = 3m where m = 3q2
Now, r = 1
a2 = (3q+r)2
= 9q2 + (1)2 + 6q(1)
= 9q2 + 1 + 6q
= 3(3q2 + 2q) + 1 = 3m + 1 where m = 3q2 + 2q
When r = 2,
a2 = (3q+r)2
= 9q2 + (2)2 + 6q(2)
= 9q2 + 12q + 4
= 3(3q2) + 3(4)q + 3 × 1 + 1
= 3[(3q2) + (4)q + 1] + 1
= 3m + 1 where m = (3q2) + (4)q + 1
Hence, square of number n is of the form 3m or 3m + 1

2. A number when divided by 60 gives 35 as quotient and leaves 81 as remainder. What is the number?
a) 2121
b) 4151
c) 2181
d) 3171
Answer: c
Clarification: According to Euclid’s Division Lemma,
Number = (quotient × divisor) + remainder = 60 × 35 + 81 = 2181

3. For any two given positive integers a and b, there exists unique whole numbers q and r such that a = bq + r, where 0 ≤ r ≤ b.
a) True
b) False
Answer: a
Clarification: According to Euclid’s Division lemma, any two numbers can be written in the form of a = bq + r where a and b are integers and q and r whole numbers.
For Example: 28 when divided by 7 give 4 as quotient and 0 as remainder.
So, according to Euclid’s Division Lemma,
28 = 7 × 4 + 0

4. When a number is divided by 3 it leaves remainder as 5. What will be the remainder when 3n + 3 are divided by 3?
a) 0
b) 3
c) 9
d) 6
Answer: a
Clarification: Let the number be n.
If the number is divided by 3 it leaves 5 as remainder.
By Euclid’s division lemma,
n = 3q + 5 where q is quotient
3n = 3(3q+5)
3n = 9q + 15
3n + 3 = 9q + 18 = 3 × 3q + 3 × 6 = 3 (3q + 6)
Hence, the remainder is 0.

5. The HCF of 80 and 567 is ___________
a) 5
b) 4
c) 1
d) 6
Answer: c
Clarification: 80 = 1×2×2×2×2×5
567 = 1×3×3×3×3×7. The largest common factor between the two numbers is 1.

6. A number in the form of 6n, where n belongs to natural numbers, can never end with the digit
a) 0
b) 1
c) 2
d) 3
Answer: a
Clarification: If 6n ends with 0, then it should have 5 as a factor.
In case of 6 only 3 and 2 are factors of 6.
Also, from the fundamental theorem of arithmetic, prime factorisation of each number is unique.
Hence, 6n can never end with 0.

7. If the HCF of two numbers is 1 and the LCM is 3395. What is the other number if one of them is 97?
a) 61
b) 57
c) 43
d) 35
Answer: d
Clarification: For two numbers a and b, we know that
(a × b) = HCF of (a, b) × LCM of (a, b)
Here a = 97, HCF is 1 and LCM is 3395
97 × b = 1 × 3395
B = (frac {3395}{97}) = 35

[CLASS 10] Mathematics MCQs on Area of Sector and Segment of a Circle

Mathematics Multiple Choice Questions & Answers on “Area of Sector and Segment of a Circle”.

1. What is the name of the sector with a larger area?
a) Large
b) Major
c) Big
d) Wide
Answer: b
Clarification: A sector is a part of a circle that is enclosed by two radii and an arc. The sector having a larger area is called a major sector and the sector having a smaller area is called a minor sector.

2. What is the name of the sector with a smaller area?
a) Small
b) Narrow
c) Minor
d) Tiny
Answer: c
Clarification: A sector is a part of a circle that is enclosed by two radii and an arc. The sector having a larger area is called a major sector and the sector having a smaller area is called a minor sector.

3. What is the formula to calculate the area of a sector?
a) (frac {x^{circ }}{360^{circ }}) × πr2
b) (frac {x^{circ }}{360^{circ }}) + πr2
c) (frac {x^{circ }}{360^{circ }}) – πr2
d) (frac {x^{circ }}{360^{circ }}) × πr3
Answer: a
Clarification: The area of the sector is (frac {x^{circ }}{360^{circ }}) × πr2
Where x° is the degree measure of the angle at the center and r is the radius of the circle.

4. Find the area of the sector if the radius is 5 cm and with an angle of 50°.
a) 11.90 cm
b) 10.90 cm
c) 12.90 cm
d) 13.90 cm
Answer: b
Clarification: The area of the sector = (frac {x^{circ }}{360^{circ }}) × πr2
= (frac {50^{circ }}{360^{circ }} times frac {22}{7}) × 52
= 10.90 cm

5. Find the area of the sector if the radius is 12 cm and with an angle of 134°.
a) 158.38 cm
b) 168.00 cm
c) 167.38 cm
d) 168.38 cm
Answer: d
Clarification: The area of the sector = (frac {x^{circ }}{360^{circ }}) × πr2
= (frac {134^{circ }}{360^{circ }} times frac {22}{7}) × 122
= 168.38 cm

6. A man goes to a walking track twice a day in the shape of a sector with an angle of 123° and a radius of 138 m. Find the area covered by the man of the walking track in a day.
a) 20441.4 m
b) 20882.8 m
c) 40882.8 m
d) 81765.6 m
Answer: c
Clarification: The area of the sector = (frac {x^{circ }}{360^{circ }}) × πr2
= (frac {123^{circ }}{360^{circ }} times frac {22}{7}) × 1382
= 20441.4 m
Area covered by the man of the walking track in a day = 20441.4 + 20441.4
= 40882.8 m

7. A horse is grazing in a field. It is tied to a pole with a rope of length 6 m. The horse moves from point A to point B making an arch with an angle of 70°. Find the area of the sector grazed by the horse.
a) 20.99 m
b) 21.99 m
c) 22.99 m
d) 23.99 m
Answer: b
Clarification: The area of the sector = (frac {x^{circ }}{360^{circ }}) × πr2
= (frac {70^{circ }}{360^{circ }} times frac {22}{7}) × 62
= 21.99 m

8. Number of sectors in a circle are ____
a) 2
b) 3
c) 4
d) 1
Answer: a
Clarification: A circle contains two sectors. The sector having a larger area is called a major sector and the sector having a smaller area is called a minor sector.

9. A segment is a part of a circle that is enclosed by two radii and an arc.
a) False
b) True
Answer: a
Clarification: A segment is a part of a circle that is obtained by subtracting the triangle from the sector whereas, a sector is part of a circle that is enclosed by two radii and an arc.

10. Find the area of the segment if the area of the sector is 44 m and the part of a triangle in the sector is 12 m.
a) 39 m
b) 22 m
c) 32 m
d) 31 m
Answer: c
Clarification: The area of the segment = ((frac {x^{circ }}{360^{circ }}) × πr2 ) – (frac {bh}{2})
= Area of the sector – Area of the triangle
= 44 – 12
= 32 m

[CLASS 10] Mathematics MCQs on Criteria for Similarity of Triangle

Mathematics Multiple Choice Questions & Answers on “Criteria for Similarity of Triangle”.

1. The perimeters of two similar triangles ABC, PQR is 64 cm and 24 cm respectively. If PQ is 12 cm what will be the length of AB?
a) 30 cm
b) 32 cm
c) 12 cm
d) 16 cm

Answer: b
Clarification: We know that the ratio of the perimeters of similar triangles is the same as the ratio of their corresponding sides.
∴ (frac {Perimeter , of , triangle ABC}{Perimeter , of , triangle PQR} = frac {AB}{PQ})
(frac {64}{24}=frac {AB}{12})
AB = (frac {64times 12}{24}) = 32 cm

2. If ∠D = ∠L, ∠E = ∠M then, ∆DEF & ∆LMN are similar according to which test?

a) AAA test
b) AA test
c) SAS test
d) SSS test
Answer: b

3. If (frac {AB}{XY} = frac {BC}{YZ} = frac {AC}{XZ}) then, ∆ABC & ∆XYZ are similar according to which test?
a) AAA test
b) AA test
c) SAS test
d) SSS test
Answer: d

[CLASS 10] Mathematics MCQs on Surface Area and Volume of Different Solid Shapes

Mathematics Quiz for Schools on “Surface Area and Volume of Different Solid Shapes”.

1. A funnel is in the shape of a right circular cone with a base radius of 3 cm and a height of 4 cm. Find the slant height of the funnel.
a) 4 cm
b) 5 cm
c) 7 cm
d) 7.5 cm
Answer: b
Clarification: Slant height = (sqrt {h^2+r^2})
= (sqrt {4^2+3^2})
= (sqrt {16+9})
= √25
= 5 cm

2. What is the total surface area of a cylinder with a radius of 7 m and a height of 8 m?
a) 609.4 m2
b) 659.4 m2
c) 650.4 m2
d) 689.4 m2
Answer: b
Clarification: Total surface area of a cylinder = 2πr(h + r)
= 2 × (frac {22}{7}) × 7 × (8 + 7)
= 659.4 m2

3. A right circular cone has a radius of 7 cm and a height of 24 cm. Find the area of the sheet required to make 7 such cones.
a) 3846.5 cm2
b) 1052 cm2
c) 1153.4 cm2
d) 3172 cm2
Answer: a
Clarification: Slant height = (sqrt {h^2+r^2})
= (sqrt {24^2+7^2})
= (sqrt {576+49})
= 25 cm
Area required to make 7 such cones = 7 × Lateral surface area of a right circular cone
= 7 × (πrl)
= 7 × 3.14 × 7 × 25
= 3846.5 cm2

4. What is the formula for the volume of a right circular cone?
a) (frac {1}{3})πr3h
b) (frac {1}{3})πr2h
c) (frac {1}{2})πr2h
d) (frac {1}{2})πr3h
Answer: b
Clarification: Volume of right circular cone = (frac {1}{3})πr2h
Where ‘r’ is the radius of the base and ‘h’ is the height of the right circular cone.

5. What is the formula for the curved surface area of a rectangular circular cylinder?
a) 2πr2h
b) 2πrh
c) 2π(r + h)
d) 2π(rh)
Answer: b
Clarification: The curved surface area of a rectangular circular cylinder = 2πrh
Where ‘r’ is the radius of the base and ‘h’ is the height of the right circular cylinder.

6. What is the total surface area of an iron sphere having a radius of 11 cm?
a) 1529.76 cm2
b) 1514.76 cm2
c) 1519.76 cm2
d) 1419.76 cm2
Answer: c
Clarification: The total surface area of an iron sphere = 4πr2
= 4 × (frac {22}{7}) × 112
= 1519.76 cm2

7. What is the volume of a hemisphere if the radius of the hemisphere is 3 m?
a) 135.4 m3
b) 56.52 m3
c) 120.23 m3
d) 105.5 m3
Answer: b
Clarification: Radius of the hemisphere = (frac {2}{3})πr3
= (frac {2}{3}) × 3.14 × 33
= 56.52 m3

8. Find the slant height of the right circular cone if the base diameter of the right circular cone is 14 cm and the height is 24 cm.
a) 11 cm
b) 13 cm
c) 28 cm
d) 25 cm
Answer: d
Clarification: Radius (r) = (frac {diameter}{2} = frac {14}{2}) = 7 cm
Slant height = (sqrt {h^2+r^2})
= (sqrt {24^2+7^2})
= (sqrt {576+49})
= 25 cm

9. The lateral surface area of a right circular cone is 2020 cm2 and its radius is 10 cm. What is its slant height?
a) 28.3 cm
b) 25.6 cm
c) 68.23 cm
d) 64.33 cm
Answer: d
Clarification: The lateral surface area of a right circular cone is 2020 cm2.
πrl = 2020
3.14 × 10 × l = 2020
l = (frac {2020}{3.14 times 10})
= 64.33 cm

10. Find the volume of the right prism with an area of base 121 m2 and a height of 23 m.
a) 4793 m3
b) 2763 m3
c) 2783 m3
d) 4783 m3
Answer: c
Clarification: The volume of the right prism = Area of base × height
= 121 × 23
= 2783 m3

11. Find the lateral surface area of a rectangular room with a height of 22 m, length 27 m and a breadth of 23 m.
a) 4793 m2
b) 2263 m2
c) 2200 m2
d) 4783 m2
Answer: c
Clarification: The lateral surface area of a cuboid = 2h(l + b)
= 2 × 22 (27 + 23)
= 2200 m2

12. Find the total surface area of a square-shaped box having a length of 2 m for its side.
a) 47 m2
b) 22 m2
c) 24 m2
d) 48 m2
Answer: c
Clarification: The total surface area of a cube = 6a2
= 6 × 22
= 24 m2

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