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Perimeter and Area Test - 25

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Perimeter and Area Test - 25
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  • Question 1
    1 / -0
    A bicycle wheel has diameter 1m. If the bicycle travels one kilometer, then the number of revolutions the wheel make is.
    Solution
    Let the number of revolution of the wheel is nn.
    Since, Radius=12\text{Radius} = \dfrac{1}{2}
    Then,
    n ×n\ \times Circumference of wheel = Distance travelled by bicycle
    n× 2π ×12=1 kmn \times  2\pi  \times \frac {1}{2}=1\ \text{km} 
    n× π=1000 mn \times  \pi =1000\ \text{m}
    n=1000π mn=\frac {1000}{\pi }\ \text{m}
  • Question 2
    1 / -0
    In the figure shown here, BDECBDEC is a rectangle. ABCABC and DEFDEF are straight lines. The area, in cm2^2 , of the whole figure is ______ .

    Solution
     Area of the whole figure= area of rectangle BDEC+ area of triangle ABD+ area of triangle ECF
    =DE×EC+12×AB×BD+12×CE×EF=DE\times EC+\dfrac { 1 }{ 2 } \times AB\times BD+\dfrac { 1 }{ 2 } \times CE\times EF
    =8×10+12×8×3+12×8×11=8\times 10+\dfrac { 1 }{ 2 } \times 8\times 3+\dfrac { 1 }{ 2 } \times 8\times 11
    =80+12+44=80+12+44
    =136cm2=136{ cm }^{ 2 }
  • Question 3
    1 / -0
    The areas of two concentric circles forming a ring are 154 sq. cm and 616 sq. cm. The breadth of the ring is
    Solution
    Breadth of the ring is equal to the difference between the radius of the outer circle and the radius of the inner circle
    Let the radius of the outer circle be r2r_2 and that of the inner circle be r1r_1

    We know that, the area of a circle is πr2\pi r^2

    Given that the area of outer circle is 616 cm2616 \ \displaystyle cm^{2}
    $$\displaystyle \therefore \   \pi r_{2}^{2}=616 \ cm^{2}$$

    r22=616×722[ π=227]\displaystyle \Rightarrow r_{2}^{2}=\frac{616\times 7}{22}\quad \quad \left[\because \ \pi=\dfrac{22}{7}\right]

    =196 cm2=196 \ cm^2

     r2=14 cm\displaystyle \therefore \ r_{2}=14 \ cm

    Also, the area of the inner circle is 154 cm2\displaystyle 154 \ cm^{2}

    $$\displaystyle \therefore \  \pi r_{1}^{2}= 154 \ cm^2$$

    r12=154×722[ π=227]\displaystyle \Rightarrow r_{1}^{2}=\frac{154\times 7}{22}\quad \quad \left[\because \ \pi=\dfrac{22}{7}\right]

    =49 cm2=49 \ cm^2

    r1=7 cm.\displaystyle \therefore r_{1}=7 \ cm.

     \displaystyle \therefore \ The required breadth =r2r1\displaystyle = r_{2}-r_{1}

    =147=14-7

    =7 cm=7 \ cm

    Hence, the breadth of the ring is 7 cm7 \ cm.

  • Question 4
    1 / -0
    The perimeter of circle is π\displaystyle \pi cm, then the area
    Solution
    Given, Circumference

    of a circle of radius r =2πr =π  = 2\pi r  = \pi 


    r=12 r = \dfrac {1}{2}

    Area of a circle of radius r $$ =

    \pi {r}^{2} $$



    Area of circle of radius  12=π(12)2=π4  \dfrac {1}{2} = \pi { ( \dfrac {1}{2} )}^{ 2 } = \dfrac { \pi }{ 4 }

  • Question 5
    1 / -0
    The cost of fencing a circular field at the rate of Rs. 12 per meter is Rs. 1320. The field is to be ploughed at Rs. 2 per m2\mathrm {m^{2}}, then cost of ploughing is: (use π=227)\displaystyle \left ( \text{use }\pi =\frac{22}{7} \right )
    Solution

    Given that at a rate of Rs. 12Rs.\ 12 per meter, the cost of fencing the circular field is Rs.1320Rs. 1320

    Hence, circumference of the field $$ = \dfrac {1320}{12} = 110 \  m$$
    We know that, circumference of a circle is 2πr2\pi r

     2πr=110\therefore \ 2\pi r=110

    2×227×r=110 \Rightarrow 2 \times \dfrac {22}{7} \times r = 110   

    $$ \Rightarrow r = \dfrac {35}{2} \  m $$

    We also know that the area of a circle is πr2\pi r^2

    So, area of the field =227×352 ×352=19252  m2  = \dfrac {22}{7} \times \dfrac {35}{2}  \times \dfrac {35}{2} = \dfrac {1925}{2}  \ {m}^{2} 

    Given that the cost of ploughing one m2m^2 area is Rs. 2Rs. \ 2.

    Hence, cost of ploughing the field at Rs. 2 Rs. \ 2 per m2=19252×2=Rs. 1925 {m}^{2} = \dfrac {1925}{2} \times 2 = Rs. \ 1925 .


    Hence, option B is correct.

  • Question 6
    1 / -0
    Four circular cardboard pieces, each of radius 7 cm are placed in such a way that each piece touches the two other pieces. The area of the space enclosed by the four pieces is:
    Solution
    Area of enclosed space = Area of square - 4 x Area of quadrant 
    =side×side4×πr24 = \quad side \times \quad side - 4 \times \dfrac {πr^2}{4}
    =(14) 24\displaystyle ={ \left( 14 \right)  }^{ 2 }-4(14×227×72) \displaystyle \left( \frac { 1 }{ 4 } \times \frac { 22 }{ 7 } \times { 7 }^{ 2 } \right) 
    =196154\displaystyle =196-154
    =42cm2\displaystyle =42{ cm }^{ 2 }

  • Question 7
    1 / -0
    One side of a parallelogram is 8 cm8 \ cm. If the corresponding altitude is $$6\  cm$$, then its area is given by
    Solution
    As we know,
    Area of Parallelogram =Base×Height=Base \times Height
    Here, $$Base=8\ cm, Height=6\  cm$$
    So, area of the Parallelogram =8×6 cm2=8\times 6\ cm^2=48 cm2=48\ cm^2.
  • Question 8
    1 / -0
    The perimeter of sheet of paper in the shape of a quadrant of a circle is 25 cm , then area of the paper is
    Solution

    Let the radius of the quadrant be rr 

    Given, 
    Perimeter of the square paper =πr2+2r=25 cm  = \dfrac {\pi r}{2} + 2r =25  cm 

    227×r2+2r=25\Rightarrow \dfrac {22}{7} \times \dfrac {r}{2} + 2r = 25

    22r+28r14=25 \Rightarrow \dfrac {22r + 28r}{14} = 25

    50r14=25 \Rightarrow \dfrac {50r}{14} = 25

    r=7 cm \Rightarrow r = 7  cm


    Area of the square paper quadrant of radius rr

    =πr24=227×7×74=772 cm2  = \dfrac {\pi {r}^{2}}{4} \\ = \dfrac {\dfrac {22}{7} \times 7 \times 7}{4} = \dfrac {77}{2}  {cm}^{2} 

  • Question 9
    1 / -0
    A race track is in the form of a ring whose inner circumference is 440 m440 \ \mathrm{m} & outer circumference is 506 m506 \ \mathrm{m} The width of the track is:
    Solution
    Let the radius of the outer circle be R and inner circle be r.

    Outer circumference =506  m = 506  \ \mathrm{m}

    $$\implies 2 \times \frac {22}{7}

    \times R = 506 \ \mathrm{m} $$

    R =80.5  m R  = 80.5\ \mathrm{  m}


    Inner circumference =440 m = 440  m

    $$\implies 2 \times \frac {22}{7}

    \times r = 440 \ \mathrm{m} $$

    r =70 m r  = 70 \ \mathrm{ m}

    Hence, width of the track $$ = (80.5 - 70)\  \mathrm{m} = 10.5 \ \mathrm{ m} $$

  • Question 10
    1 / -0
    If the radius of the circle is increased by 100100%, then the area is increased by
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