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Perimeter And Area Test - 6

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Perimeter And Area Test - 6
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  • Question 1
    1 / -0
    Find the circumference of the circle if the area of the square is 49 cm2. (Take = )

    Solution
    Side of the square = Diameter of the circle
    Let each side of the square be x cm.
    Area of the square = Side x Side
    49 cm2 = x x x

    Side = 7 cm

    x = 7 cm

    Radius of the circle = 3.5 cm
    Circumference of the circle = 2r
    = 2 x 3.5 cm x
    = 22 cm
  • Question 2
    1 / -0
    A garden was built around a rectangular plot. Find the area of the garden.


    Solution
    Length of rectangle PQSR = 20.5 cm

    Breadth of rectangle PQSR = 14 cm

    Length of rectangle ABDC = 25.5 cm

    Breadth of rectangle ABDC = 19 cm

    Area of rectangle PQSR = 287 cm2

    Area of rectangle ABDC = 484.5 cm2

    Area of the garden = 484.5 - 287 = 197.5 cm2
  • Question 3
    1 / -0
    In the given figure, find the area of parallelogram EFHG, if the area of the shaded triangle is 9 cm2.


    Solution
    Area of the shaded triangle = 9 cm2 (given)
    Base of the triangle = 3 cm

    Area of the triangle = x base x height
    9 =

    h = 6 cm
    Area of the parallelogram = Height x Base of the parallelogram = 6 x (3+4) = 6 x 7 = 42 cm2
  • Question 4
    1 / -0
    Find the area of the shaded region in the given figure, where all of the short line segments are at right angles to each other and 1 cm long.


    Solution
    Length of the large square = 1 x 9 = 9 cm
    Breadth of the large square = 1 x 9 = 9 cm
    Area of the shaded region = Area of the large square - Area of 41 small identical squares with 1 cm as length of the side
    = 9 x 9 - 41 x 1 x 1 = 81 - 41 = 40 cm2
  • Question 5
    1 / -0
    Find the area of the given shape.

    .
    Solution
    Area of the given shape = Rectangle 1 + Rectangle 2
    = 15 × 5 + 5 × 7
    = 75 + 35 = 110 cm2

  • Question 6
    1 / -0
    Four identical squares are placed one over the other. Each square is placed at the midpoint of the other square. Each side of the square is 6 cm. Find the perimeter of the given figure.


    Solution

    Perimeter of the given figure = 6 + 6 + 3 + 6 + 3 + 6 + 3 + 6 + 6 + 6 + 3 + 6 + 3 + 6 + 3 + 6 = 78 cm
  • Question 7
    1 / -0
    The ratio of circumferences of two circles is 3 : 5. Find the ratio of their areas.


    Solution
    Circumference of a circle = 2πr
    Ratio of circumferences of two circles = =

    (where r1 and r2 are radii of the given circles)

    =
    Ratio of areas of the circles =     = =
  • Question 8
    1 / -0
    The length of the sides of a triangle are 0.7 decametres, 240 decimetres and 25 metres. Find the area of the triangle.
    Solution
    Let BC = 0.7 decametres = 7 metres
    Let AB = 240 decimetres = metres = 24 metres
    AC = 25 metres
    It is a right-angled triangle, as we can check:




    Therefore,
    Area of ABC = x BC x AB
    = x 7 x 24 = 84 m2
  • Question 9
    1 / -0
    Raghu is making a model house. He cuts a cardboard in the shape of a rectangle. He also makes two equal sized triangles and places them on either side of the rectangle, as shown in the figure below. Find the total area covered by this piece of cardboard.


    Solution
    Area covered is the area of the figure.
    Area of the rectangle =16 x 12 =192 cm2
    Area of the triangle =x 12 x 8 = 48 cm2
    Area of the triangles = 48 x 2 = 96 cm2
    Total area of the given figure = 192 + 96 = 288 cm2
  • Question 10
    1 / -0
    In a circle of radius of 38 cm, two triangles are drawn, as shown in the figure, such that . Find out the area of the shaded region.


    Solution
    Area of the circle = πr2 = x 38 x 38 = 4538.3 cm2
    Area of 1 triangle =x 38 x 38 = 722 cm2
    Area of 2 triangles = 2 x 722 = 1444 cm2
    Area of the shaded region = 4538.3 - 1444 = 3094.3 cm2
  • Question 11
    1 / -0
    In a rhombus EFGH, FH is joined and is equal to 24 m. Then, two perpendiculars are drawn from E and G to FH meeting at the same point such that both are 16 m in length. Find the area of the rhombus.
    Solution
    Area of triangle EFH = x FH x EG.
    = x 24 x 16 = 192 m2
    Area of triangle GHF = x 24 x 16 = 192 m2
    Area of the rhombus = 192 + 192 = 384 m2

  • Question 12
    1 / -0
    If the radius of a circle becomes 3 times the original radius, then the circumference of the circle will become
    Solution
    Circumference of a circle = 2πr
    New radius = 3r
    New circumference = 2π3r
    It will become 3 times the old circumference.
  • Question 13
    1 / -0
    The shape of a garden is rectangular in the middle and semi-circular at the ends, as shown in the diagram. Find the area and the perimeter of the garden.



    Solution
    Length of the rectangle = 20 - (3.5 + 3.5) = 13 m
    Breadth of the rectangle = 7 m
    Radius of the semi-circle = 3.5 m
    We know, two semi-circles = one circle
    Hence, perimeter of the circle = 2πr = 2 x x 3.5 = 22 m

    Hence, perimeter of the garden = (22 m + 13 m + 13 m) = 48 m
    Now, area of the rectangle = 13 x 7 = 91 m2
    Area of the circle (two semi-circles) = πr2 = x 3.5 x 3.5 = 38.5 m2

    Hence, area of the garden = 38.5 m2 + 91 m2 = 129.5 m2
  • Question 14
    1 / -0
    Find the area of the portion that is not shaded, if F and E are the respective midpoints of AB and CD; AC = 16 m and AB = 20 m.

    Solution


    ABDC is a rectangle and FGE is a triangle.
    Area of the rectangle = AB x AC
    ⇒ 16 x 20 = 320 m2

    As F is the midpoint of AB,
    AF = 10 m
    FE = BD = AC = 16 m
    Therefore, area of the triangle =x base x height
    Height = GH = 10 m (as GH = AF)

    ⇒ x 16 x 10 = 80 m2
    So, area of the portion that is not shaded = 320 - 80 = 240 m2
  • Question 15
    1 / -0
    What is the area of the shaded portion in the figure if the area of the rectangle ABCD is 60 m2?



    Solution
    Area of the shaded portion in the figure = Area of rectangle ABCD - Area of trapezoid AXYD
    Area of rectangle ABCD = Length x breadth
    60= BC x 4
    BC = = 15 m
    Length of side BC = 15 m
    Area of trapezoid AXYD = x height x (sum of the measures of the parallel sides)
    = x CD x (AD + XY)
    = x 4 x (15 + 3)
    = x 4 x 18
    =36 m2

    Area of the shaded portion in the figure = Area of rectangle ABCD - Area of trapezoid AXYD
    = 60 - 36
    = 24 m2
    So, the area of the shaded portion is 24 m2.
  • Question 16
    1 / -0
    A playground is 250 m long and 190 m wide. It has a platform in its centre of uniform length and width 10 m, as shown below. Find the cost of levelling the platform at Rs. 5 per square metre.




    Solution
    Area of the platform = ((190 x 10) + (250 x 10) - (10)2) m2 = (1900 + 2500 - 100) m2 = 4300 m2

    Therefore, cost of levelling the platform = Rs .(4300 x 5) = Rs. 21,500
  • Question 17
    1 / -0
    The following diagram shows a square garden. There are four square and one circular flower bed. Find the area of the garden not covered by the flower beds.

    Solution
    Total area of the garden = 60 x 60 = 3600 m2
    Area of the flower beds = [ x 5 x 5] m2 + 4 x [12 x 12] m2
    = 78.5 m2 + 576 m2
    Area not covered by the flower beds = 3600 - 654.5 = 2945.5 m2
  • Question 18
    1 / -0
    A sheep is tied by a rope to the fence outside a farm. If the length of the rope is 20 m, find the area of the land on which the sheep can move freely.
    Solution


    Radius of the semicircle= 20 m
    Required area =

    = = 628.5 m2
  • Question 19
    1 / -0
    From a square-shaped iron piece of side 40 cm, a triangle was cut. Find out the area of the remaining iron piece.


    Solution



    Area of the iron piece = [40 x 40] = 1600 cm2

    Area of the triangle = x 25 x 10 = 125 cm2
    Area of the remaining iron piece = 1600 - 125 = 1475 cm2
  • Question 20
    1 / -0
    A wall of length 20 m and width 30 m is to be painted. With 1 litre of paint, an area of 800 cm x 800 cm can be painted. Find out the cost of painting the whole wall, if the cost of 1 litre of paint is Rs. 200.
    Solution
    Amount of paint required =

    = = 9.37 litres

    Cost of 1 litre of paint = Rs. 200
    Cost of 9.37 litres of paint = 9.37 × 200 = Rs. 1875
  • 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.

    1. Find the area of the park not covered by tiles.
    2. Find the cost of constructing the path at Rs. 25 per m2.


    Solution
    1. Area of the park not covered by tiles = Total area - Area of the path
    = [225 x 175] - [[225 - 60] x [175 - 60]]
    = 39375 - 18975
    = 20,400 m2

     2. Cost of covering the path = Area of the path x Cost of covering per m2
    = 18975 x 25 = Rs. 4,74,375
  • Question 22
    1 / -0
    Find the area of the shaded portion in each of the given figures, respectively.







    Solution
    A.
    Area of triangle ABE = x 25 x 20 = 250
    Area of triangle ABC = x 10 x 20 = 100
    Area of the shaded portion = 250 - 100 =150

    B.
    Area of rectangle ABCD = 40 x 25 = 1000
    Area of triangle ECB = x 25 x 20 = 250
    Area of the shaded portion = 1000 - 250 = 750


    C.
    Area of square ABCD =24 x 24 = 576
    Area of circle with centre O = x 6 x 6 = 113.14
    Area of triangle ECD = x 5 x 24 = 60
    Area of the shaded portion = 576 - [113.14 + 60] = 402.86

    D.
    Area of rectangle ABCD= 25 x 15 = 375
    Area of rectangle PQRS= 3 x 15 = 45
    Area of rectangle XYZW= 25 x 3= 75
    Area of square in the middle = 3 x 3 = 9
    Area of the shaded portion is = 375 - [45 + 75] + 9 m2 = 255 + 9 m2 = 264 m2
  • Question 23
    1 / -0
    State 'T' for true and 'F' for false.

    1. The length of fence required to cover the boundary of a square park of side 46 m is 178 m.
    2. Area of a circle : radius = : 1
    3. 34,500 m2 = 3.45 hectares
    4. If 1 km2 = x m2, the value of x is 1,00,00,000.
    Solution
    1. Side = 46 m, so perimeter = 4 x 46= 184 m

    2.

    3. 1 hectare = 10,000 , so 34,500 = = 3.45 hectares

    4. 1 = 1000 x 1000 = 10,00,000, x = 10,00,000
  • 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 park of dimensions 25 m x 25 m, there is a circular pond of radius 10 m. The area not covered by the pond is [A].
    2. A square-shaped room measuring m is to be covered by square-shaped floor tiles of side 0.75 m each. The number of tiles required is [B].
    3. If the radius of a circle is 25 m, then the perimeter of the semicircle is [C].
    Solution
    1. Area of the park = 25 x 25 = 625
    Area of the pond = = x 10 x 10 = 314.28
    So, area not covered by the pond = 625 - 314.28 = 310.7

    2. Area of the room = x = 42.25
    Area of each floor tile = 0.75 x 0.75 = 0.56
    Number of tiles required = = 75.1; or rounding to nearest whole number, 75 tiles

    3. If radius = 25 m, then
    Circumference = 2πr + 2r
    = x 2 x x 25 + 2 x 25 = 128.5 m
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