જવાબ : Angle described by the minute hand in 60 minutes = 360°
જવાબ : Length of pendulum = radius of sector = r cm
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જવાબ : Circumference of circle = 2 πr = 39.6 cm
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જવાબ :
જવાબ : Side of the square = 2 ⨯ radius of circle = 2a cm Then, Perimeter of the square = (4 ⨯ 2a) = 8a cm
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જવાબ : Area of flower bed = (area of quadrant OPQ)-(area of the quadrant ORS)
જવાબ :
જવાબ :
જવાબ : In rhombus, AB = BC = CD = AD
⇒ AC = BD = 2r
જવાબ :
Area of shaded region = 2(Area of square) – 4(Area of semicircle) = 2 × 196 – 308 = 392 – 308 = 84 cm2જવાબ : ∠ABC = ∠BAC = ∠ACB = 60°
જવાબ : Side = 14 cm, radius, r = 14/2 = 7 cm
Area of the shaded region = ar (square) – 4 (ar of quadrant)
જવાબ : AB = BC = CA
= 2(3.5) = 7 cm
=> ∆ABC is an equilateral ∆
જવાબ : r = 28/4= 7 cm
Area of the shaded region = ar(square) – 4(circle) = (side)2 – 4 (πr2)
= (28)2 – 4 × 22/7 × 7 × 7 = 784 – 616 = 168 cm2
જવાબ : Here θ = 360/3] = 120°, r = 6 cm
Area of shaded region = 3(ar of minor segment) = 3[ar(minor sector) – ar(∆ABC)]
જવાબ : PO = 42 m …[Given in the ques.
PS = QR …[∵ side of square
OR = OP …[Radius
Let OR be the radius of circle = x
So, PR = OR + OP = 2x
Using Pythagoras’ theorem,
⇒(PR)2 = (RQ)2 + (PQ)2
⇒(2x)2 = (42)2 + (42)2
જવાબ :
જવાબ : Side = 28 cm, Radius = 28/2 cm = 14 cm
The area of the shade = Area of square + 3/4 (Area of circle) + 3/4 (Area of circle)
= Area of square + 3/2 (Area of circle)
= (28)2 + 3/2 × 22/7× 14 × 14
= 784 + 924 = 1708 cm2
જવાબ :
જવાબ :
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જવાબ :
જવાબ :
જવાબ :
જવાબ :
જવાબ :
જવાબ :
Match the radius with the resp. area of sector of angle of 45°
1 |
2 |
A |
2π |
2 |
4 |
B |
9π/2 |
3 |
6 |
C |
π/2 |
4 |
8 |
D |
8π |
જવાબ :
1-C, 2-A, 3-B, 4-D
Match the radius with the resp. area of sector of angle of 90°
1 |
2 |
A |
9 π |
2 |
4 |
B |
π |
3 |
6 |
C |
16 π |
4 |
8 |
D |
4 π |
જવાબ :
1-B, 2-D, 3-A, 4-C
Match the radius with the resp. area of semi-circle
1 |
1/√π |
A |
8 |
2 |
2/√π |
B |
½ |
3 |
3/√π |
C |
2 |
4 |
4/√π |
D |
9/2 |
જવાબ :
1-B, 2-C, 3-D, 4-A
Match the radius with the resp. area of circle
1 |
1/√π |
A |
16 |
2 |
2/√π |
B |
9 |
3 |
3/√π |
C |
4 |
4 |
4/√π |
D |
1 |
જવાબ :
1-D, 2-C, 3-B, 4-A
Match the radius with the resp. area of semi-circle
1 |
8 |
A |
8 π |
2 |
6 |
B |
4 π |
3 |
4 |
C |
16π |
4 |
2 |
D |
32π |
જવાબ :
1-D, 2-C, 3-A, 4-B
જવાબ :
1-D, 2-C, 3-A, 4-B
Match the radius with the resp. perimeters of semi-circle
1 |
12 |
A |
12(π + 2) |
2 |
14 |
B |
16(π + 2) |
3 |
16 |
C |
14(π + 2) |
જવાબ :
1-A, 2-C, 3-B
Match the radius with the resp. perimeters of circle
1 |
12 |
A |
36π |
2 |
14 |
B |
24π |
3 |
16 |
C |
28π |
4 |
18 |
D |
32π |
જવાબ :
1-B, 2-C, 3-D, 4-A
Match the radius with the resp. perimeters of semi-circle
1 |
2 |
A |
6(π + 2) |
2 |
4 |
B |
2(π + 2) |
3 |
6 |
C |
8(π + 2) |
4 |
8 |
D |
4(π + 2) |
જવાબ :
1-B, 2-D, 3-A, 4-C
Match the radius with the resp. perimeters of circle
1 |
2 |
A |
8π |
2 |
4 |
B |
12π |
3 |
6 |
C |
4π |
4 |
8 |
D |
16π |
જવાબ :
1-C, 2-A, 3-B, 4-D
Math
Chapter 12 : Areas related to Circles
The GSEB Books for class 10 are designed as per the syllabus followed Gujarat Secondary and Higher Secondary Education Board provides key detailed, and a through solutions to all the questions relating to the GSEB textbooks.
The purpose is to provide help to the students with their homework, preparing for the examinations and personal learning. These books are very helpful for the preparation of examination.
For more details about the GSEB books for Class 10, you can access the PDF which is as in the above given links for the same.