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Perimeter and Area Test - 21

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Perimeter and Area Test - 21
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  • Question 1
    1 / -0
    The circumference of a circle is 4444 m then the area of the circle is 
    Solution
    Circumference of circle: 4444 m
    Let the radius of circle be rr.
    Therefore, 2πr=442\pi r = 44
    r=7\Rightarrow r = 7 m
    Area of circle:
    πr2=227×72=154 m2\pi r^2 = \dfrac{22}7\times7^2 = 154\ m^2
  • Question 2
    1 / -0
    A circular park has a path of uniform width around it. The difference between outer and inner circumference of the circular path is 132 m132\ m. Its width is _____ (π=227)\displaystyle \left(\pi=\frac{22}{7}\right)
    Solution
    Given 2πr22πr1=132\displaystyle 2\pi r_{2}-2\pi r_{1}=132
    r2r1=1322π=132×744=21 m\displaystyle \Rightarrow r_{2}-r_{1}=\frac{132}{2\pi }=\frac{132\times 7}{44}=21\ m

  • Question 3
    1 / -0
    Find the area of the parallelogram whose base is 17 cm17\ cm and height 0.8 m0.8\ m?
    Solution
    Base =17 cm= 17\ cm
    Height =0.8 m=0.8×100=80 cm= 0.8\ m =0.8 \times 100 = 80\ cm
    Area of parallelogram 
    =b×h= b \times h
    =17×80= 17\times 80
    =1360 cm2= 1360\ cm^2
  • Question 4
    1 / -0
    If the diameter of a circle is 1414 cm, then its circumference is
    Solution
    Since, diameter =d=14  cm= d=14\;cm
    C=πd=227×14=44  cm\displaystyle C=\pi d=\frac{22}{7}\times 14=44\;cm
  • Question 5
    1 / -0
    The base of a parallelogram is three times its height. If the area of the parallelogram is 7575 sq cm, then its height is
    Solution
    let height of ||g be 'b', then base of ||g be 3b.
     area of ||g==base×\times height
    75cm2=3b×bb2=2575c{ m }^{ 2 }=3b\times b\\ { b }^{ 2 }=25
     b=5b=5cm, therefore height =5=5cm
  • Question 6
    1 / -0
    The length and breadth of a rectangle are in the ratio 3:23:2. If the sides of the rectangle are extended on each side by 1m1 m, the ratio of length to breadth becomes 10:710:7. Find the area of the original rectangle in square meters.
    Solution
    Let length of rectangle=3x=3x
     breadth of rectangle   =2x=2x
     Now,
    3x+12x+1=10721x+7=20x+10x=3 \cfrac { 3x+1 }{ 2x+1 } =\cfrac { 10 }{ 7 } \\ 21x+7=20x+10\\ x=3
    \therefore length=3×3=9m,=3\times 3=9m, breadth=2×3=6m=2\times 3=6m
     Area of original rectangle
     =(9×6)m2=54 m2=(9\times 6){ m }^{ 2 }\\ =54\ { m }^{ 2 }
  • Question 7
    1 / -0
    In the given figure, PP is the center of the circle. The area and perimeter respectively, of the figure are

    Solution
    Area of figure==area of circle + area of rectangular - area of quarter circle.
    =πr2+l×b14×πr2=34πr2+l×b=34×3.14×20×20+50×40=942+2000=2942cm2 =\pi { r }^{ 2 }+l\times b-\cfrac { 1 }{ 4 } \times \pi { r }^{ 2 }\\ =\cfrac { 3 }{ 4 } \pi { r }^{ 2 }+l\times b=\cfrac { 3 }{ 4 } \times 3.14\times 20\times 20+50\times 40\\ =942+2000\\ =2942c{ m }^{ 2 }

    Perimeter of figure
    =2πr2πr4+2(l+b)2r=3πr2+2(l+b)2r=3×3.14×202+2(50+40)2×20=94.2+18040=234.2cm=2\pi r-\cfrac { 2\pi r }{ 4 } +2(l+b)-2r\\ =\cfrac { 3\pi r }{ 2 } +2(l+b)-2r\\ =\cfrac { 3\times 3.14\times 20 }{ 2 } +2(50+40)-2\times 20\\ =94.2 +180-40\\ =234.2cm
  • Question 8
    1 / -0
    A field is in the form of a parallelogram whose base is 420 m420\ m and altitude is 36m36 m. Find the cost of watering at 1010 paise per sq. m.
    Solution
    Base  =420 m= 420\ m
    Height = 36 m=  36\ m
    Area =b×h=420×36= b \times h = 420 \times 36
    =15,120 m2= 15,120 \displaystyle \ m^{2}
    The cost of watering per sq m
    =10= 10 paise
    Cost  of watering the field =15120×0.1= 15120 \times 0.1
    Rs. 15121512
  • Question 9
    1 / -0
    The ratio between the length and breadth of a rectangular garden is 5:35:3. If the perimeter of the garden is 160160 meters, what will be the area of 5 m5\text{ m} wide road around its outside?
    Solution
    Let the length and breadth of the rectangular garden be 5x5x meter and 3x3x meters.
    Given 2(5x+3x)= 1602(5x + 3x) =  160
     16x=160 \Rightarrow  16x = 160
    x=10 \Rightarrow x = 10
     \therefore  Length of garden =50 m= 50 \text{ m} and Breadth of garden =30 m= 30 \text{ m}
    Area of 5 m5 \text{ m} wide road around its outside == Area of outer rectangle - Area of inner rectangle
    [[\because length of outer rectangle =50+2×5= 50+2\times 5 and breadth of outer rectangle =30+2×5]=30+2\times 5 ]
    Area =(50+10)×(30+10)50×30= (50 + 10) \times (30 + 10) - 50 \times 30  
    =60×4050×30= 60 \times 40 - 50 \times 30
    =(24001500)= (2400 - 1500)
    =900 m2 = 900 \text{ m}^{2}
    \therefore  Area of road is 900 m2900\text{ m}^2

  • Question 10
    1 / -0
    Circumference of a circle is equal to 
    Solution
    c=πd=2πrc=\pi d=2\pi r
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