Download Parallelogram and Rectangle Properties and more Quizzes Mathematics in PDF only on Docsity!
Q1. If the radius of a circle is tripled, the area becomes:
- 9 mes.
- 3 mes.
- 6 mes.
- 30 mes.
1 Mark
Ans: 1. 9 mes. Solu on: Let r be the radius of a circle. Area of circle If radius is tripled, then new radius will be 3r Area of new circle mes of original Hence, the area of a cirlce becomes 9 mes to the original area.
Q2. Circumference of a circle of diameter 5cm is:
- 3.14cm
- 31.4cm
- 15.7cm
- 1.57cm
1 Mark
Ans: 3. 15.7cm Solu on: Given, diameter of a circle = 5cm
Now, Circumference
Q3. In Fig., ABCD is a parallelogram in which AD = 21cm, DH = 1cm and DK = 27cm. The perimeter of the parallelogram is:
- 105cm
- 84.5cm
- 169cm
- 52.5cm
1 Mark
Ans: 1. 105cm Solu on: Area of a parallelogram = Base × Height AB × DH = AD × DK ⇒ AB × 18 = 21 × 27
Here, ABCD is a parallelogram, so AB = CD and AD = BC. Therefore, Perimeter of parallelogram ABCD = 2(AB + AD) = 2(31.5 + 21) = 105cm
∴ = π r^2
∴ = π (3r^2 ) = 9 π^2 = 9
∴ Radius = 52 cm[ ∵ radius = diameter 2 ]
= 2 π r
= 2 × 227 × 52 = 1107 = 15.7cm
⇒ AB = 21×27 18 = 632 = 31.5cm
Ans: 3. 12cm Solu on: Let d and d be the diagonals of the rhombus, where d = 16 and area of rhombus = 96cm. Area of parallelogram
Thus, the length of other diagonal is 12cm.
Q6. If the ra o of the areas of two squares is 9 : 1, then the ra o of their perimeters is:
- 2 : 1
- 3 : 1
- 3 : 2
- 4 : 1
1 Mark
Ans: 2. 3 : 1 Solu on: Let a and b be the sides of the squares, then as per the ques on
Q7. The area of a square is equal to the area of a circle. The ra o between the side of the square and the radius of the circle is:
1 Mark
Ans: 1. Solu on: Let a and r be respec vely the side of the square and radius of the circle. Here, the area of square is equal to the area of the circle. So
Hence, the correct op on is (a).
Q8. Mark (✓) against the correct answer. The length and breadth of a rectangular park are in the ra o 4 : 3 and its perimeter is 56m. The area of the field is:
1 Mark
Q4. The ra o of the areas of two squares, one having its diagonal double that of the other, is:
- 2 : 1
- 3 : 1
- 3 : 2
- 4 : 1
1 Mark
Ans: 4. 4 : 1 Solu on: Let the diagonals of two square be 2d and d Area of bigger square 2(2d) = 8d and of smaller = 2(d) = 2d Ra o in their area = 4 : 1
Q5. The length of a diagonal of a rhombus is 16cm. If its area is 96cm , then the length of other diagonal is:
- 6cm
- 8cm
- 12cm
- 18cm
1 Mark
2 2 2 2
= 8d =
2 2d^2
4 1
2
1 2 1 2
= 12 (d 1 × d 2 )
⇒ 96 = 12 (16 × d 2 )
⇒ d 2 = 96×2 16 = 12cm
a =
2 b^2
9 1
2
ab =^3
2 1
⇒ ba =^31
√ π^ : 1
a^2 = π r^2
⇒ a = π
2 r^2
⇒ ar = √ π
- 49m
- None of these
Ans: 3. 49m Solu on: Side of square = 84m Area of square = (84) = 7056m Area of rectangle = 7056m Length of rectangle = 144m Width
Q12. Area of the circle obtained in A wire is bent to form a square of side 22cm. If the wire is rebent to form a circle, its radius is:
- 196cm
- 212cm
- 616cm
- 644cm
1 Mark
Ans: 3. 616cm Solu on: Area of the circle
Q13. How many mes should a wheel of radius 7m rotate to go around the perimeter of a rectangular field of length 60m and breadth 50m?
- 3
- 4
- 5
- 6
1 Mark
Ans: 3. 5 Solu on: Here, Radius (r) = 7m, Length (l) = 60m and Breadth (b) = 50m. Perimeter of circle Perimeter of rectangle Therefore Number of turns
Hence, the correct op on is (c).
Q14. If the area of a square is 225m , then its perimeter is:
- 15m
- 60m
- 225m
- 30m
1 Mark
Ans: 2. 60m Solu on: Let a be the side of the square. Then area of square = a ⇒ 225 = a ⇒ a = 15 ⇒ a = 15m Perimeter of the square = 4a = 4 × 15 = 60m
Q15. The area of a circle is 24.64m. The circumference of the circle is:
- 14.64m
- 16.36m
- 17.60m
- 18.40m
1 Mark
2 2 2
= LengthArea
= 7056144 = 49m
2 2 2 2 2
= π r^2 = 227 × 14 × 14 = 616cm^2 [ ∵ r = 14cm, find above]
= 2 π r = 2 × 227 × 7 = 44m
= 2(l + b) = 2(60 + 50) = 220m
= Perimeter of rectanglePerimeter of circle
2
2 2 2 2
2
Ans: 3. 17.60m Solu on: Area of a circle
Circumference
Q16. The area of a square is 50cm. The length of its diagonal is:
1 Mark
Ans: 2. Solu on: Let a be the side of the square, Then, area of the square
Now Diagonal of the square
Q17. Circumference of a circle is always:
- More than three mes of its diameter.
- Three mes of its diameter.
- Less than three mes of its diameter.
- Three mes of its radius.
1 Mark
Ans: 1. More than three mes of its diameter. Solu on: We know that, Circumference of a circle Circumference ⇒ Circumference So, circumference of circle is always more than three mes of its diameter.
Q18. The circumference of a circle is 44cm. Its area is:
- 77cm
- 154cm
- 208cm
- 144cm
1 Mark
Ans: 2. 154cm Solu on: Let r be the radius of the circle. Then,
Therefore Area of circle Hence, the correct op on is (b).
Q19. Area of triangle MNO of is: 1 Mark
= 24.64m^2
∴ Radius = √^ Area π
= √ 24.64×7 22 = √1.12 × 7
= √7.84 = 2.8m
∴ = 2 π r
= 2 × 227 × 2.8m
= 17.60m
2
5√2cm
10cm
10√2cm
10cm
= a^2 = 50cm^2
⇒ a = √50 = 5√2cm
= a√2 = 5√2 × √
= 5 × 2 = 10cm
= 2 π r
∴ = 2 × 3.14 × r [ ∵ π = 3.14]
= 3.14 × d [ ∵ d = 2r]
2 2 2 2 2
44 = 2 π r ⇒ r = 2×^44 22 = 7cm
7
= π r^2 = 227 × 7^2 = 154cm^2
Hence, the correct op on is (a).
Q24. The area of a semicircle of radius 4r is:
1 Mark
Ans: 1. Solu on: Given, radius of semi-circle = 4r Area of semi-circle
Q25. The radius of a circular wheel is 1.75m. How many revolu ons will it make in travelling 11km?
- 10
- 100
- 1000
- 10000
1 Mark
Ans: 3. 1000 Solu on: Radius of a circular wheel (r) Circumference
Total distance covered No. of revolu ons
Q26. A wire is in the form of a square of side 18m. It is bent in the form of a rectangle, whose length and breadth are in the ra o 3 : 1. The area of the rectangle is:
- 81m
- 243m
- 144m
- 324m
1 Mark
Ans: 2. 243m Solu on: Side of square (a) = 18m Let l = 3x and b = x be the length and breadth of the rectangle. Then Perimeter of rectangle = Perimeter of square 2(l + b) = 4a 2(3x + x) = 4 × 18 8x = 72 x = 9m Thus Length (l) = 3x = 3 × 9 = 27m Breadth (b) = x = 9m Therefore Area of the rectangle = l × b = 27 × 9 = 243m Thus, the area of the rectangle is 243m.
Q27. If A is the area of an equilateral triangle of height h, then:
1 Mark
8 π r^2
4 π r^2
12 π r^2
2 π r^2
8 π r^2
∵ = 12 × π r^2
= 121 × π × (4r)^2
= 162 π r^2
= 8 π r^2
= 1.75m
= 2 π r = 2 × 227 × 1.75m
= 44 × 0.25m = 11m
= 11km
∴ = 11km11m
= 11×1000 11 = 1000
2 2 2 2 2 2 2
A = √3 h^2
√3A = h
√3A = h^2
3A = h^2
Ans: 3. Solu on: Let a and h be the side and height of the equilateral triangle respec vely. Then
Therefore, Area of equilateral triangle (A)
Q28. The length of a rectangle is 8cm and its area is 48cm. The perimeter of the rectangle is:
- 14cm
- 24cm
- 12cm
- 28cm
1 Mark
Ans: 4. 28cm Solu on: Let a and b be the length and breadth of the rectangle respec vely. Then, area of the rectangle = ab
Perimeter of the rectangle = 2(a + b) = 2( 6 + 8) = 28cm
Q29. The difference between the circumference and radius of a circle is 37cm. The area of the circle is:
- 111cm
- 148cm
- 154cm
- 258cm
1 Mark
Ans: 3. 154cm Solu on: Let r and r be the radius of the two circles. Then,
Now Area of circle Hence, the correct op on is (c).
Q30. The length of the diagonal of a square is d. the area of the square is:
1 Mark
Ans: 2. Solu on: Let a be the side of the square. Then, diagonal of the square Therefore
Area of the square
Q31. What will be the area of the largest square that can be cut out of a circle of radius 10cm?
- 100cm
1 Mark
√3A = h^2
⇒ a = × a × h
(^2) √ 4
1 2
⇒ a√3 4 = 12 × h
⇒ a = √32h
= a
(^2) √ 4
⇒ A = =
( (^) √32h)^2 √ 4
h^2 √
⇒ √3A = h^2
2
⇒ 48 = a × 8 (∵ b = 8cm)
⇒ a = 6cm
2 2 2 2 2
1 2
2 π r − r = 37
⇒ 2 × 227 × r − r = 37
⇒ 44r−7r 7 = 37
⇒ r = 37×7 37 = 7cm
= π r^2 = 227 × 7 × 7 = 154cm^2
d^2
12 d 2
14 d 2
2d^2
12 d 2
= a√
= a√2 = d ⇒ a = √2d
2
√2d^ = d
2 2
2
Q33. If the diagonal of a rectangle is 17cm long and its perimeter is 46cm, the area of the rectangle is:
- 100cm
- 110cm
- 120cm
- 150cm
1 Mark
Ans: 3. 120cm Solu on: Perimeter = 46cm
and l + b = 17 = 289 Now (l + b) = (23) ⇒ l + b + 2lb = 529 ⇒ 289 + 2lb = 529 ⇒ 2lb = 529 - 289 ⇒ 21b = 240 ⇒ lb
Now area of rectangle = lb = 120cm
Q34. The length of and breadth of a rectangle are (3x + 4)cm and (4x - 13)cm. If the perimeter of the rectangle is 94cm, then x =
- 4
- 8
- 12
- 6
1 Mark
Ans: 2. 8 Solu on: Here, l = (3x + 4)cm and b = (4x - 13)cm Perimeter of rectangle = 2(l + b) = 2[(3x + 4) + (4x - 13)] = 2(7x - 9) = 14x - 18 Now, as per the ques on Perimeter of rectangle = 94cm
Q35. Area of parallelogram ABCD is not equal to:
- DE × DC
- BE × AD
- BF × DC
- BE × BC
1 Mark
⇒ 16√5 = (16√5 − 24) + Area of △PQR
⇒ Area of △PQR = 24cm^2
⇒ 12 × PR × QM = 24
⇒ 12 × 12 × QM = 24
∴ QM = 4cm
2 2 2 2 2
= l + b = 462 = 23cm
2 2 2 2 2 2 2
2
∴ 14x − 18 = 94
⇒ 14x = 94 + 18 = 112
⇒ x = 11214 = 8
Ans: 1. DE × DC Solu on: We know that, Area of parallelogram = Base × Corresponding Height So, area of parallelogram ABCD = AD × BE = BC × BE or area of parallelogram ABCD = DC × BF
Q36. The area of a rectangle is 11.6m. If its breadth is 46.4cm, then the perimeter is:
- 25.464m
- 50.928m
- 101.856m
- None of these.
1 Mark
Ans: 2. 50.928m Solu on: Here, area of rectangle(A) = 11.6m , breadth(b) = 46.4cm = 0.464m. Let l be the length of the rectangle Then area of rectangle = l × b = l × 0. Area of rectangle = 11. ⇒ l × 0.464 = 11.
Now Perimeter = 2(l + b) = 2(25 + 0.464) = 2(25.464) = 50.928m
Q37. Area of circular garden with diameter 8m is:
- 12.56m
- 25.12m
- 50.24m
- 200.96m
1 Mark
Ans: 3. 50.24m Solu on: Given, diameter = 8m So, radius Area of circle garden
Q38. Each side of an equilateral triangle is 8cm long. Its area is:
1 Mark
Ans: 3. Solu on: Side of an equilateral triangle
Q39. Each diagonal of a square is 12cm long. Its area is:
- 144cm
1 Mark
[ ∵ AD = BC]
2
2
⇒ l = 0.46411.6 = 11600464 = 25m
2 2 2 2
2
= 82 m = 4m[ ∵ radius = diameter 2 ]
∴ = π r^2 = 227 × 4 × 4 = 50.24m^2
32cm^2
64cm^2
16√3cm^2
16√2cm^2
16√3cm^2
= 8cm
∴ Area = √3 4 a^2 = √3 4 × 8 × 8cm^2
= 16√3cm^2
2
Circumference of the circle
Q42. In fig., ABCD is a parallelogram of area 144cm , the value of x is:
1 Mark
Ans: 3. 9 Solu on: Area of parallelogram = Base × Height
Q43. The length of a rectangle 8cm more than the breadth. If the perimeter of the rectangle is 80cm, then the area of the rectangle is:
- 192cm
- 364cm
- 384cm
- 382cm
1 Mark
Ans: 3. 384cm Solu on: Let l and b be the length and breadth of the rectangle, then l = b + 8. Perimeter of rectangle = 2(l + b) = 2(l + l - 8) = 4l - 16 ⇒ 80 = 4l - 16 ⇒ 4l = 80 + 16 = 96 ⇒ l = 24cm ⇒ b = l - 8 = 24 - 8 = 16cm Area of rectangle = l × b = 24 × 16 = 384cm
Q44. The radii of two circles are in the ra o 2 : 3. The ra o of their areas is:
- 2 : 3
- 4 : 9
- 3 : 2
- 9 : 4
1 Mark
Ans: 2. 4 : 9 Solu on: Let r and r be the radius of the two circles. So
Now Ra o of areas Thus, the required ra o is 4 : 9. Hence, the correct op on is (b).
Q45. If the diameter of a circle is equal to the diagonal of a square, then the ra o of their areas is:
- 7 : 1
- 1 : 1
- 11 : 7
1 Mark
⇒ 2r = 10cm [ ∵ diameter = 2 × radius]
⇒ r = 5cm
∴ = 2 π r = 2 × π × 5 = 10 π cm
2
⇒ 144 = 16 × x
⇒ x = 14416 = 9
2 2 2 2 2 2
1 2
r r 12 = 23
2
2
π r^21 π r^22
r 1 r 2 23
4 9
Ans: 3. 11 : 7 Solu on: Let r and a be the diameter of the circle and side of the square respec vely. Then, Diameter of circle = 2r Diagonal of square Now, as per the ques on Diameter of circle = Diagonal of square
Therefore
Hence, the correct op on is (c).
Q46. The area of a parallelogram is 100cm. If the base is 25cm, then the corresponding height is:
- 4cm
- 6cm
- 10cm
- 5cm
1 Mark
Ans: 1. 4cm Solu on: Let b = 25cm and h be the base and the corresponding height of the parallelogram. Then Area of parallelogram = b × h ⇒ 100 = 25 × h ⇒ h = 4cm
Q47. If radius of a circle is increased to twice its original length, how much will the area of the circle increase?
- 1.4 mes.
- 2 mes.
- 3 mes.
- 4 mes.
1 Mark
Ans: 4. 4 mes. Solu on: Let r be the radius of the circle. Area of circle If radius is increased to twice its original length, then radius will be 2r. Now, area of new circle mes of original area Hence, the area of circle will be increased by 4 mes.
Q48. Circumference of a circle disc is 88cm. Its radius is:
- 8cm
- 11cm
- 14cm
- 44cm
1 Mark
Ans: 3. 14cm Solu on: We know that, circumference
Hence, the radius is 14cm
Q49. The height of an equilateral triangle is Its area is:
1 Mark
= a√
2r = a√2 ⇒ a = √2r
= Area of squareArea of circle
= π r = = =
2 a^2
(^227) ×r 2 (√2r)^2
(^227) ×r 2 2r^2
11 7
2
∴ = π^2
= π (2r)^2 = 4 π^2 = 4
= 2 π r
⇒ 88 = 2 × 227 × r
⇒ r = 88 × 72 × 22 [ ∵ circumference = 88cm, give]
⇒ r = 14cm
√6cm.
3√3cm^2
2√3cm^2
2√2cm^2
- 125cm
- 158cm
- 128cm
- 144cm
Ans: 3. 128cm Solu on: Let a be the length of the equal sides of the right isosceles triangle whose hypotenuse is Then using Pythagoras theorem in the triangle, we get
Therefore Area of the triangle
Thus, the area of the square is 128cm.
Q54. Each side of an equilateral triangle is equal to the radius of a circle whose area is 154cm. The area of the triangle is:
1 Mark
Ans: (^) 2.
Solu on: Let each side of an equilateral triangle Then area Now radius of the circle Then area
Q55. In Fig., the area of the parallelogram is:
- 516cm
- 616cm
- 416cm
- 606cm
1 Mark
Ans: 1. 516cm Solu on: Here, ABCD is a parallelogram, so AD = BC = 21cm. Therefore Area of parallelogram = Base × Height = BC × DK = 21 × 27 = 567cm
Q56. 1 Mark
2 2 2 2 2
16√22cm
a^2 + a^2 = (16√2)^2
2a^2 = 512
a^2 = 256
= 12 × Base × Height
= 12 × a × a
= 12 × a^2
= 12 × 256 = 128cm^2
2
2
7√3 4 cm 2
49√3 4 cm 2
35cm^2
49cm^2
49√3 4 cm 2
= a
= √3 4 a^2
= a
= π r^2 = π a^2
∴ π a^2 = 154 ⇒ 227 a^2 = 154
a^2 = 154×7 22 = 49 = (7)^2
∴ a = 7
∴ Area of △ = √3 4 (7)^2 = 49√3 4 cm^2
2 2 2 2 2 2
The length of a rectangle 8cm more than the breadth. If the perimeter of the rectangle is 80cm, then the length of the rectangle is:
- 16cm
- 24cm
- 28cm
- 18cm
Ans: 2. 24cm Solu on: Let l and b be the length and breadth of the rectangle, then l = b + 8. Perimeter of rectangle = 2(l + b) = 2(l + l - 8) = 4l - 16 ⇒ 80 = 4l - 16 ⇒ 4l = 80 + 16 = 96 ⇒ l = 24m
Q57. The length of the diagonals of a rhombus of a rhombus are 8cm and 14cm. The area of one of the 4 triangles formed by the diagonals is:
- 12cm
- 8cm
- 16cm
- 14cm
1 Mark
Ans: 4. 14cm Solu on: Let d = 8cm and d = 14cm. Area of parallelogram
Since, the diagonals of a rhombus divides it into 4 equal parts, so Area of the required triangle
Q58. The cost of fencing a semi-circular garden of radius 14 m at ₹ 10 per metre is:
- ₹ 1080
- ₹ 1020
- ₹ 700
- ₹ 720
1 Mark
Ans: 4. ₹ 720 Solu on: Radius of circle (r) = 14m Perimeter of semi-circular garden
Cost of fencing = 72 × ₹ 10 = ₹ 720 Hence, the correct op on is (d).
Q59. If each side of a rhombus is doubled, how much will its area increase?
- 1.5 mes.
- 2 mes.
- 3 mes.
- 4 mes.
1 Mark
Ans: 2. 2 mes. Solu on: Let b be the side and h be the height of a rhombus. Area of rhombus = b × h area of rhombus = base x corresponding height If each side of rhombus is doubled, then side of rhombus = 2b
2 2 2 2 2
1 2
= 12 (d 1 × d 2 )
= 12 (8 × 14)
= 56cm^2
= 564 = 14cm^2
= π r + 2r
= 227 × 14 + 2 × 14
= 72m
∴ [ ∵ ]
- 10cm
Ans: 1. 4cm Solu on:
From the above it is clear that largest circle will have diameter equals smaller side i.e., 8cm So, diameter = 8cm
Q65. A path of 1m runs around and inside a square garden of side of 20m. The cost of levelling the path at the rate of Rs. 2.25 per square metre is:
- Rs. 154
- Rs. 164
- Rs. 182
- Rs. 171
1 Mark
Ans: 4. Rs. 171 Solu on: Width of the path = 1m Side of the square garden = 20m Side of the inner square = (20 - 2)m = 18m Area of the path = Area of square garden - Area of inner square = 20 - 18 = 400 - 324 = 76m Cost of levelling = Rs. 2.25 × 76 = Rs. 171 Thus, the required cost is Rs. 171
Q66. A wire is bent to form a square of side 22cm. If the wire is rebent to form a circle, its radius is:
- 22cm
- 14cm
- 11cm
- 7cm
1 Mark
Ans: 2. 14cm Solu on: Given, side of a square = 22cm Perimeter of square and circumference of circle are equal, because the wire has same length. According to the ques on, Perimeter of square = Circumference of circle
Hence, the redius is 14cm
Q67. The area of a rhombus is 36cm and the length of one of its diagonals is 6cm. The length of the second diagonal is:
- 6cm
- 12cm
- None of these.
1 Mark
Ans: 3. 12cm Solu on: Area of rhombus = 36cm
∴ Radius = Diameter 2 = 4cm
2 2
2
⇒ 4 × (Side) = 2 × π × r
⇒ 4 × 22 = 2 × 227 × r[ ∵ π = 227 ]
⇒ r = 4 × 22 × 72 × 22
⇒ r = 14cm
2
6√2cm
2
Length of one diagonal = 6cm Length of second diagonal
Q68. The sides of a triangle measure 13cm, 14cm and 15cm. Its area is:
- 84cm
- 91cm
- 168cm
- 182cm
1 Mark
Ans: 1. 84cm Solu on: Sides are 13cm, 14cm, 15cm
and area
Q69. Each diagonal of a square is 14cm. Its area is:
- 196cm
- 88cm
- 98cm
- 148
1 Mark
Ans: 3. 98cm Solu on: Let a be the side of the square. Then diagonal of the square Now Area of the square
Q70. Mark (✓) against the correct answer. The area of a circle is 154cm. Its diameter is:
- 14cm
- 11cm
- 7cm
- 22cm
1 Mark
Ans: 1. 14cm Solu on: Let the radius of the circle be r cm Then, its area will be
Diameter of the circle = 2r = (2 × 7)cm = 14cm
Q71. The ra o of the area of a square of side a and that of an equilateral triangle of side a is:
1 Mark
Ans: 4.
= One diagonalArea×
= 36×2 6 = 12cm
2 2 2 2
2
∴ s = a + b + c 2
= √s(s − a)(s − b)(s − c)
= √21 × 8 × 7 × 6
= √3 × 7 × 2 × 2 × 2 × 7 × 2 × 3
= 2 × 2 × 3 × 7 = 84cm^2
2 2 2
2
= a√2 = 14 ⇒ a = √2^14 cm
= a^2 = ( )
2
√2^14 = 98cm^2
2
( π r^2 )cm^2
π r^2 = 154
⇒ ( 227 × r × r) = 154
⇒ r^2 = ( 154×7 22 ) = 49
⇒ r = √49 = 7cm
