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Perimeter And Area Test - 7

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Perimeter And Area Test - 7
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  • Question 1
    1 / -0
    A circular pit is dug in a square-shaped garden. The perimeter of the garden is 28 m. What can be the maximum circumference of the pit?
    (Take =)
    Solution
    For maximum circumference:
    Side of the square = Diameter of the circle
    Let each side of the square be x m.
    Perimeter of the garden = 28 m
    4x = 28
    x = 7 m

    Radius of the circle = 3.5 m
    Circumference of the circle = Diameter x
    = 7 ×
    = 22 m
  • Question 2
    1 / -0
    A 2.5 m wide path is built along the sides of a rectangular garden, as shown below. Find the total area of the path.

    Solution

    Length of rectangle EFGH = 22.5 m
    Breadth of rectangle EFGH = 15 m
    Length of rectangle ABCD = 27.5 m
    Breadth of rectangle ABCD = 20 m
    Area of rectangle EFGH = 22.5 x 15 = 337.5 m2
    Area of rectangle ABCD = 27.5 x 20 = 550 m2
    Area of the path = 550 - 337.5 = 212.5 m2
  • Question 3
    1 / -0
    In the given figure, find the area of the shaded region of rectangle ABCD.

    Solution


    Area of rectangle EBCR = 1 4 = 4 cm2
    Area of triangle EDR = = 4 cm2
    Area of the shaded region = Area of rectangle EBCR + Area of triangle EDR = (4 + 4) cm2 = 8 cm2
  • Question 4
    1 / -0
    The following figure shows a rectangular floor made using square-shaped tiles of side 2 m. Some of the tiles among them are broken. The broken tiles have been shaded.

    What is the area of unshaded portion of rectangular floor?

    Solution
    Length of rectangular floor = 6 × 2 = 12 m (where 2 m is the side of each tile)
    Breadth of rectangular floor = 4 × 2 = 8 m (where 2 m is the side of each tile)
    Area of rectangular floor = 12 × 8 = 96 m2
    Now,
    Number of broken tiles = 8
    Area of 1 tile = 2 × 2 = 4 m2
    Area of 8 tiles = 8 × 4 = 32 m2
    Area of unshaded portion = 96 - 32 = 64 m2
  • Question 5
    1 / -0
    Find the perimeter of the given figure comprising a triangle, a rectangle and a semicircle.

    Solution


    Perimeter = (7 + 7) + (15 + 15) +
    = 14 + 30 + 11 = 55 cm
  • Question 6
    1 / -0
    Three identical squares, with side 4 cm, are placed one above another. Each square placed on top of another is moved to the right by half the length of the side of the square, as shown below. Find the perimeter of the figure.
    Solution


    Perimeter of the given figure = (4 x 8) + (2 x 4) = 40 cm
  • Question 7
    1 / -0
    The ratio of circumferences of two circles is 6 : 5. Find the respective ratio of their areas.
    Solution
    Circumference of a circle = 2πr
    Ratio of the circumferences of two circles ==
    (where r1 and r2 are the radii of the given circles)
    =
    Ratio of their areas ===
  • Question 8
    1 / -0
    The perpendicular of a right-angled triangle measures 3 m. The area of the triangle 6 m2. What are the lengths of the other two sides of the triangle?
    Solution


    It is a right-angled triangle, so
    6 =× 3 × BC
    BC = 4 m

    By Pythagoras theorem:
    AC2 = AB2 + BC2
    AC2 = 9 + 16
    AC2 = 25
    AC = 5 m
  • Question 9
    1 / -0
    Radha takes a rectangular cardboard. She also takes two identical triangles and place them along the two vertical sides of the rectangle, as shown in the figure below. Find the area of the combined shape.
    Solution
    Area of the rectangle = 16 × 14 = 224 cm2
    Area of each triangle =× 14 × 8 = 56 cm2
    Area of two such triangles = 56 × 2 = 112 cm2

    Total area of the combined shape = 224 + 112 = 336 cm2
  • Question 10
    1 / -0
    Two triangles are drawn inside a circle of radius 28 cm, as shown in the figure below. Calculate the area of the shaded region.

    Solution


    When we combine the two unshaded triangles, we get a square of side 28 cm, as shown in the above figure.
    Area of the two triangles combined = Area of the square = 28 x 28 = 784 cm2

    Area of the circle = πr2 =× 28 × 28 = 2464 cm2
    Now, area of the shaded portion can be calculated by subtracting the total area of the two unshaded triangles from the total area of the circle.

    Area of the shaded portion = 2464 - 784 = 1680 cm2
  • Question 11
    1 / -0
    In a rhombus ABCD, BD is one of the diagonals and it measures 24 m. Two perpendiculars are drawn from A and C to BD, each 16 m in length. Find the area of the rhombus.
    Solution


    Area of triangle ABD =× BD × AE
    =× 24 × 16 = 192 m2
    Area of triangle CBD =× 24 × 16 = 192 m2
    Area of the rhombus = 192 + 192 = 384 m2
  • Question 12
    1 / -0
    If the radius of a circle increases to 4 times its initial value, then the circumference of the circle will become __________ its initial value.
    Solution
    Initial radius = r
    Initial circumference of the circle = 2πr
    New radius = 4r
    New circumference of the circle = 2π(4r) = 8πr
    The circumference becomes 4 times its initial value.
  • Question 13
    1 / -0
    What is the value of (x + y) in the given figure, if the perimeter is 34 cm?

    Solution
    Perimeter = Sum of all sides
    34 = 3 + x + y + 3 + 3 + 3 + 3 + 3 + 4
    34 = 22 + x + y
    x + y = 12 cm
  • Question 14
    1 / -0
    Find the area of the shaded reason in the following figure.

    Solution
    Area of the shaded region = Area of rectangle ABCD - (Area of triangle DEF + Area of triangle ECG)
    = 20 × 16 - (× 4 × 4 + × 15 × 12)
    = 320 - (8 + 90) = 222 cm2
  • Question 15
    1 / -0
    The area of a trapezium is 384 sq. cm. If the lengths of its parallel sides are in the ratio of 3 : 5 and the perpendicular distance between them is 12 cm, then length of the smaller of the parallel sides is
    Solution
    Let the lengths of parallel sides be 3x and 5x.
    Now, according to given information:
    Area of trapezium = 384 cm2

    8x = 384/6
    x = 8
    Lengths of parallel sides of trapezium = 3x and 5x = 3(8) and 5(8)
    = 24 cm and 40 cm
    So, length of the smaller side is 24 cm.
  • Question 16
    1 / -0
    A park is 300 m long and 150 m wide. It has two roads in its centre of uniform width of 5 m, one parallel to its length and the other parallel to its breadth. The municipal corporation wants to cover that path by tiles leaving the centre portion of both the paths. Find the cost of covering the paths at the rate of Rs. 5 per square metre.
    Solution


    Area to be covered with tiles = Area of both the paths - Area common between both the paths
    = (300 × 5) + (150 × 5) - 2(5 × 5)
    = 1500 + 750 - 50
    = 2200 m2

    Cost of covering 1 m2 = Rs. 5
    Cost of covering 2200 m2 = Rs. [5 × 2200] = Rs. 11,000
  • Question 17
    1 / -0
    In a park, there are some flower beds made in the shapes of square and circle. The following figure shows the dimensions of the same. Find the area of the park not covered by the flower beds.


    Solution
    Total area of the park = 50 x 50 = 2500 m2
    Area of the flower beds = [ x 5 x 5]m2 + 4 x [5 x 5] m2
    = 78.5 m2 + 100 m2 = 178.5 m2
    Area of the park not covered by the flower beds = (2500 – 178.5) m2 = 2321.5 m2
  • Question 18
    1 / -0
    A pet dog is tied with a leash to a wall outside a house. If the length of the leash is 20 m, find the area of land on which the dog can move freely.
    Solution


    Radius of the semicircle = 20 m
    Required area =
    == 628.5 m2

  • Question 19
    1 / -0
    From a square-shaped wooden piece of side 30 cm, a triangle was cut as shown in the figure below. Find out the area of the remaining wooden piece.

    Solution


    Area of the wooden piece = [30 × 30] = 900 cm2
    Area of the triangle = × 20 × 10 = 100 cm2
    Are of the remaining wooden piece = 900 - 100 = 800 cm2
  • Question 20
    1 / -0
    A straight road of length 1 km and width 20 m is to be covered with concrete. One kg of concrete can cover 5 cm × 5 cm. Find the cost of covering the whole road if 1 kg of concrete costs Rs. 10.
    Solution
    Amount of concrete required =

    =

    Cost of 1 kg of concrete = Rs. 10

    Cost of 80,00,000 kg of concrete = 80,00,000 × 10 = Rs. 8,00,00,000
  • Question 21
    1 / -0
    A park is 225 m long and 175 m wide. It has a path of uniform width of 30 m along its sides. This path is to be covered by tiles. Find the area of the park covered by tiles and the cost of covering the path at Rs. 25 per m2, respectively.

    Solution
    Area of the park covered by tiles
    = Total area of the park - Area of the inner park (not covered by tiles)
    = [225 x 175] m2 - [[225 - 60] x [175 - 60]]
    = 39,375 - 18,975 = 20,400 m2
    Cost of covering the path
    = Area of the path x Cost of covering per m2
    = 20,400 x 25 = Rs. 5,10,000
  • Question 22
    1 / -0
    Which of the following figures has the maximum area?
    Solution
    Area of figure 1 =
    Area of figure 2 =
    Area of figure 3 = 5 cm 13 cm = 65 cm2
    Area of figure 4 =
    So, the figure given in option (4) has the maximum area.
  • Question 23
    1 / -0
    State 'T' for true and 'F' for false for the following statements.

    1. The length of fence required to cover the boundary of a circular park of radius 60 m is 370 m.
    2. The ratio of the area of a circle to its radius is always πr : 1.
    3. 76,500 m2 = 7.65 hectares
    4. If 1 km2 = x m2, the value of x is 1,00,000.
    Solution
    1. Radius = 60 m, so perimeter = 2πr = 2 x x 60 = 377.1 m (370 - False)
    2. = = ( - True)
    3. 1 hectare = 10,000 m2; so 76,500 m2 = hectares = 7.65 hectares (7.65 - True)
    4. 1 km2 = 1000 x 1000 = 10,00,000 m2, x = 10,00,000 (1,00,000 - False)
  • Question 24
    1 / -0
    Consider the following statements and choose the correct option accordingly.

    Statement-I: A park is 55 m long and 25 m wide. Along the length, there is a path of width 5 m on either side. However, along the width of the park, on one side there is a path of width 10 m, and on the other side, there is a path of width 15 m. Then, the area of path is 925 m2.
    Statement-II: If the area of a circular park is 2,29,11,400 cm2, then its radius is 27 m.
    Solution
    Statement-I
    Length of the park = 55 m
    Width of the park = 25 m
    The shaded region shows the path along the boundary.



    Area of the path along the length of the park = 2 x 5 x 55 = 550 m2
    Area of the path along the width of the park = 15 x 10 + 15 x 15 = 375 m2
    Hence, total area of the path = 550 + 375 = 925 m2

    Statement-II
    If the radius is 27 m, then

    Area = = x 27 x 27 = 2291.14 m2
    = 10000 x 2291.14 = 2,29,11,400 cm2
  • Question 25
    1 / -0
    Find the values of A, B, and C.

    1. In a square park of side 25 m, there is a circular pond of radius 7 m. The area not covered by the pond is [A].
    2. A square-shaped room of side m is to be covered by square-shaped floor tiles of side 0.75 m. The number of tiles required is [B].
    3. If the radius of a circle is 28 m, then the circumference of the semicircle will be [C].
    Solution
    1. Area of the park = 25 x 25 = 625 m2

    Area of the pond ==x 7 x 7 = 154 m2
    So, area not covered by the pond = 625 – 154 = 471 m2

    2. Area of the room = x = 56.25 m2

    Area of each floor tile = 0.75 x 0.75 = 0.5625 m2

    Number of floor tiles required = = 100

    3. Radius of the circle = 28 m
    Then, circumference of the semicircle = = = 36 x 4 = 144 m
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