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Perimeter and Area Test - 24

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Perimeter and Area Test - 24
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  • Question 1
    1 / -0
    ABCDABCD is a square field of side 80  m80\;m. If EFEF is perpendicular to DCDC, find the area of ΔEDC\Delta EDC and the area of the shaded portion respectively.

    Solution
    ar ΔEDC=12DC×EF\Delta EDC=\dfrac { 1 }{ 2 } DC\times EF

                        == 12×80×80\dfrac { 1 }{ 2 } \times 80\times 80 

                        == 32003200 m2{ m }^{ 2 }

    area of shaded portion == Area of square - 32003200 m2{ m }^{ 2 }

                                            == (80×80) 3200{ \left( 80\times 80 \right)  }-3200

                                            == 32003200 m2{ m }^{ 2 }

  • Question 2
    1 / -0
    Mrs. Kaushik has a square plot with the measurement as shown in the figure. she wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of Rs.  55  per  m2Rs.\;55\;per\;m^2.

    Solution
    Area of the garden = Area of the outer square - Area of the inner rectangle
                                                              =25×25  m220×15  m2\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=25\times25\;m^2-20\times15\;m^2
                                                              =(625300)  m2=325  m2\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=(625-300)\;m^2=325\;m^2
    Cost of developing a garden @ Rs.  55  per  m2=Rs.  (55×325)=Rs.  17,875Rs.\;55\;per\;m^2=Rs.\;(55\times325)=Rs.\;17,875.
  • Question 3
    1 / -0
    In figure, ABAB and CDCD are two perpendicular diameters of a circle with centre OO. If OAOA==7cm7cm, find the area of the shaded region. [Use π=227\pi=\displaystyle\frac{22}{7}]

    Solution
    Area of shaded region == area of larger circle - area of smaller circle - area of the triangle
                                           
                                           =πR2πr212×DC×BO=\pi { R }^{ 2 }-\pi { r }^{ 2 }-\dfrac { 1 }{ 2 } \times DC\times BO

                                            =πR2πr212×DC×BO=\pi { R }^{ 2 }-\pi { r }^{ 2 }-\dfrac { 1 }{ 2 } \times DC\times BO

                                            =227(72(3.5) 2)12×14×7=\dfrac { 22 }{ 7 } \left( { 7 }^{ 2 }-{ \left( 3.5 \right)  }^{ 2 } \right) -\dfrac { 1 }{ 2 } \times 14\times 7

                                            =227(4912.25)49=\dfrac { 22 }{ 7 } \left( 49-12.25 \right) -49

                                           =(227×36.75)49=(\dfrac { 22 }{ 7 } \times 36.75)-49

                                           =115.549= 115.5 - 49

                                           =66.5= 66.5 cm2{ cm }^{ 2 }
  • Question 4
    1 / -0
    If the diameter of a semicircular protractor is 14cm14cm then find its perimeter.
    Solution
    Diameter =14cm=14cm
    Radius=Diameter2=14cm2=7cm\displaystyle\frac{Diameter}{2}=\frac{14cm}{2}=7cm
    Length of the semicircular part =πr=227×(7)=22cm=\pi r=\displaystyle\frac{22}{7}\times (7)=22cm
    Total perimeter=length of semicircular part +Diameter
    =22cm+14cm=36cm=22cm+14cm=36cm
    Thus, the perimeter of the protractor is 36cm36cm

  • Question 5
    1 / -0
    The area of the shaded portion in the given figure is

    Solution
    Step 1 : Find the required area\textbf{Step 1 : Find the required area}
                     Area of the outer semi-circle =π2×52\text{Area of the outer semi-circle =} \dfrac{\pi}{2}\times 5^2

                     Area of the inner semi-circle =π2×(51)2\text{Area of the inner semi-circle =} \dfrac{\pi}{2}\times (5-1)^2
                   
                      Area of shaded region = Area of outer semicircle -  Area of inner semicircle\therefore \text{ Area of shaded region = Area of outer semicircle -  Area of inner semicircle}
                                                                   =π2× 52π2× 42=\dfrac{\pi}{2}\times 5^2-\dfrac{\pi}{2}\times 4^2
                    
                                                                   = π2(2516)=  \dfrac{\pi}{2}(25-16)
                                                                   =4.5  π  unit2= \text{4.5  $\pi$  unit}^2

    Hence,  Area of shaded portion is 4.5 π sq. unit\boldsymbol{\textbf{Hence,  Area of shaded portion is 4.5 $\pi$} \textbf{ sq. unit}}
  • Question 6
    1 / -0
    A circular ground whose diameter is 140140 meters is to be fenced by wire three times around its circumference. Find the length of wire needed.

    [[use π=227\displaystyle \pi = \frac {22}{7}]]
    Solution
    Diameter of circular ground =140m=140m.
    Radius of circular ground =70m=70m.
    Circumference of circular ground =2πr=2\pi r

    =2×227×70\displaystyle =2\times \frac {22}{7}\times 70
    =440m=440m
    Length of wire needed to fenced three times
    =440×3=440\times 3
    =1320m=1320m
  • Question 7
    1 / -0
    The inner circumference of a circular tracks is 220 m220\ m. The track is 77 mm wide everywhere. Calculate the cost of putting up a fence along the outer circle at the rate of Rs. 22 per metre. Use π=227\pi=\displaystyle\frac{22}{7}
    Solution
    Circumference of inner side = 220m220 m

    2πr=220r=220×744=35m\Rightarrow 2\pi r=220\Rightarrow r=\dfrac { 220\times 7 }{ 44 } =35m

    Now, width of track  =7m=7m

    \therefore Outer radius =35+7=42m=35+7 = 42 m

    Therefore outer circumference  =2πR=2×227×42=264m=2\pi R=2\times \dfrac { 22 }{ 7 } \times 42=264m

    \therefore Cost of fencing == Rs. (264×2)\left( 264\times 2 \right) == Rs. 528528

  • Question 8
    1 / -0
    The radii of two circles are 88cm and 66cm respectively. Find the radius of the circle having its area equal to the sum of the areas of the two circles.
    Solution
    Let the radius of required circle is rr cm. 
    So, πr2=πr12+πr22\pi r^2=\pi r^2_1+\pi r^2_2
    r2=r12+r22r^2=r^2_1+r^2_2
    r2=(8)2+(6)2r^2=(8)^2+(6)^2
    r2=100r^2=100
    r=10cmr=10cm
  • Question 9
    1 / -0
    If the radius of circle is π \displaystyle \pi , then its area will be:
    Solution
    Area of circle == πr2=π×(π ) 2=π 3\pi { r }^{ 2 }=\pi \times { \left( \pi  \right)  }^{ 2 }={ \pi  }^{ 3 }
  • Question 10
    1 / -0
    Find the area of a circle whose circumference is 22cm22cm.
    Solution
    Let rr be the radius of the circle.

    Then, Circumference=22cm=22cm

    2πr=22\displaystyle \Rightarrow 2\pi r=22

    2×227×r=22\displaystyle\Rightarrow 2\times \frac{22}{7}\times r=22

    r=72cm\Rightarrow r=\displaystyle\frac{7}{2}cm

    \therefore Area of the circle=πr2=\displaystyle \pi r^2

    =227×72×72cm2=\displaystyle\frac{22}{7}\times\frac{7}{2}\times \frac{7}{2}cm^2

    =38.5cm2=38.5 \, cm^2
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