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Perimeter and Area Test - 19

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Perimeter and Area Test - 19
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  • Question 1
    1 / -0
    A floor which measures 15m×8m15m\, \times\, 8m is to be laid with tiles measuring 50cm×25cm50cm\, \times\, 25cm. Find the number of tiles required.
    Further, if a carpet is laid on the floor so that a space of 1 m exists between its edges and the edges of the floor, what fraction of the floor is uncovered.
    Solution
    We know, 50cm=0.5m50 cm = 0.5m and 
    25cm=0.25m25 cm = 0.25m
    Tiles along the length of the room=150.5\dfrac{15}{0.5} = 3030 tiles
    Tiles along the length of the room=80.25\dfrac{8}{0.25} =3232 tiles
    Total number of tiles required =30×3230\times 32 =960960 tiles
    To leave 1m1 m between the carpet and wall on all sides, the carpet needs to be 2m2 m shorter in each dimension.
    Therefore,
      (152)m(82)m(15-2)m(8-2)m
    =(13×6)m=(13\times 6)m
    =78m2=78 m^2
    Therefore, the room area is 15×815\times 8 = 120m2120 m^2
    The carpet area is 78m278 m^2
    The fraction of the area uncovered =12078120\dfrac{120-78}{120}= 42120\dfrac{42}{120}=720\dfrac{7}{20}
                                 
                                 
                                                           
  • Question 2
    1 / -0
    The shaded portion of the figure, given along side,shows two concentric circles.
    If the circumference of the two circles are 396cm396 \:cm and 374cm374\: cm, find the area of the shaded portion.

    Solution
    Given,
    2πr=374cm2\pi r=374cm
    2πR=396cm2\pi R=396cm
    Therefore,
    2πr=374cm2\pi r=374cm
    =>r=374×72×22=>r=\cfrac{374\times7}{2\times 22}
           =59.5=59.5
    2πR=396cm2\pi R=396cm
    =>R=396×72×22=>R=\cfrac{396\times7}{2\times 22}
           =63=63
    Area of shaded portion=(πR2πr2)cm2(\pi R^2-\pi r^2)cm^2
                                        =π[(63)2(59.5)2]cm2\pi [(63)^2-(59.5)^2]cm^2
                                        =π(39693540.25)\pi (3969-3540.25)
                                        =1347.5cm21347.5cm^2 
  • Question 3
    1 / -0
    The area of a circle is 394.24394.24 cm2cm^2.  What is  the radius of the circle ?
    Solution
    Area of the circle =394.24 cm2=394.24 \ cm^2
        πr2=394.24\implies \pi r^2=394.24
        r2=394.24×722\implies r^2=\dfrac{394.24\times 7}{22}
        r2=125.44\implies r^2=125.44
        r=11.2 cm\implies r=11.2\ cm 
  • Question 4
    1 / -0
    The circumference of a circular field is 308m308 m, Find its Area.
    Solution
    Let the radius of the circle be rr cm
    Circumference of the circle=308308
    Therefore,
    2πr=3082\pi r=308

    r=308×72×22\Rightarrow r=\dfrac{308\times 7}{2\times 22}

    r=49mr=49m

    Therefore,
    Area of the circle=π×(49)2=\pi \times (49)^2

                               =227×2401=\dfrac{22}{7}\times 2401

                               =7546m2=7546m^2
  • Question 5
    1 / -0
    The diameter of a circle is 28cm28 cm. Find its circumference
    Solution
    Diameter of  a circle =28=28 cm
    Thus, circumference =π×28=\pi \times 28 cm
                                        =227×28=\dfrac{22}{7} \times 28 cm
                                        =22×4=22 \times 4 cm
                                        =88=88 cm
  • Question 6
    1 / -0
    The area enclosed by the circumferences of two concentric circles is 346.5cm2346.5\, cm^2. If the circumference of the inner circle is 8888 cm, then the radius of the outer circle is
    Solution
     GivenOisthecentreoftwoconcentriccircleswithouter Given-\\ O\quad is\quad the\quad centre\quad of\quad two\quad concentric\quad circles\quad with\quad outer

    radiusOA=r2&innercicumference=C=88cm. radius\quad OA={ r }_{ 2 }\quad \& \quad inner\quad cicumference=C=88cm.

    ar.ringbetweenthecircles=346.5cm2. ar.ring\quad between\quad the\quad circles=346.5{ cm }^{ 2 }.

    Tofindout To\quad find\quad out-

    r2=? { r }_{ 2 }=?

    Solution Solution-

    innercicumference=88cm. inner\quad cicumference=88cm.

    Lettheradiusoftheinnercircleber1. Let\quad the\quad radius\quad of\quad the\quad inner\quad circle\quad be\quad { r }_{ 1 }.

    r1=C2π =346.52×227 cm=14cm. \therefore \quad { r }_{ 1 }=\dfrac { C }{ 2\pi  } =\dfrac { 346.5 }{ 2\times \frac { 22 }{ 7 }  } cm=14cm.

    Soar.ring=ar.outercirclear.innercircle=π{r22r12} So\quad ar.ring=ar.outer\quad circle-ar.inner\quad circle=\pi \left\{ { { r }_{ 2 } }^{ 2 }-{ { r }_{ 1 } }^{ 2 } \right\}

    346.5=227×{r22r12} \Longrightarrow 346.5=\dfrac { 22 }{ 7 } \times \left\{ { { r }_{ 2 } }^{ 2 }-{ { r }_{ 1 } }^{ 2 } \right\}

    r22=306.25 { { \Longrightarrow r }_{ 2 } }^{ 2 }=306.25

    r2=17.5cm. \Longrightarrow r_2=17.5cm.

    Theradiusoftheoutercircle=17.5cm. \therefore \quad The\quad radius\quad of\quad the\quad outer\quad circle=17.5cm.

    AnsOptionB. Ans-\quad Option\quad B.

  • Question 7
    1 / -0
    The radii of two circles are 25 cm25\ cm and 18 cm18\ cm. Find the radius of the circle which has a circumference equal to the sum of the circumference of these two circles.
    Solution
    The circumference of a circle =2πr=2\pi r
    Let RR be the radius of the circle which has a circumference equal to the sum of the circumferences of the two circles.
    2πR=2π×25+2π×182\pi R=2\pi \times 25+2\pi \times 18
        R=25+18=43 cm\implies R=25+18=43\ cm
  • Question 8
    1 / -0
    The diameter of a circle is 28cm28 cm. Find the one-fourth of its area.
    Solution
    The diameter of a circle is 28cm28cm
    Therefore,
    Radius of the circle=282=14cm\dfrac{28}{2}=14cm
    Therefore,
    Area of the circle=π(14)2=\pi (14)^2

                              =227×196=\dfrac{22}{7} \times 196

                              =616cm2=616cm^2

    So, 14×616=154cm2\dfrac 14 \times 616 = 154 cm^2
  • Question 9
    1 / -0

    Directions For Questions

    The cross-section of a piece of metal 44 m in length is shown below. Calculate :

    ...view full instructions

    The area of the cross-section;

    Solution
    The given diagram can be divided in two parts, rectangle ABCD and triangle CEF.
    AB=12AB=12cm
    BE=16BE=16cm
    FD=7.5FD=7.5cm
    AD=BC=10AD=BC=10cm
    Now,
    AB=CDAB=CD  (Opposite sides of a rectangle)
    =>AB=CF+FD=>AB=CF+FD
    =>12=CF+7.5=>12=CF+7.5
    =>CF=4.5=>CF=4.5cm
    Again,
    BE=BC+CEBE=BC+CE
    =>16=10+CE=>16=10+CE
    =>CE=6=>CE=6cm
    Area of cross section=Area of rectangle ABCD+\triangleCEF 
                                       =AB×AD+12×CE×EF=AB\times AD+\dfrac{1}{2}\times CE\times EF
                                       =10×12+12×6×4.5=10\times 12+\dfrac{1}{2}\times 6\times 4.5
                                       =133.5cm2=133.5cm^2                 

  • Question 10
    1 / -0
    Find the area of a ring shaped region enclosed between two concentric circles of radii 2020 cm and 1515 cm.
    Solution
    Let us given R=20R = 20 cmcm & r=15r = 15 cmcm
    The area of a ring shaped region=π(20)2π(15)2=\pi(20)^2-\pi(15)^2
                                                        =(400225)×227=(400-225)\times \dfrac{22}{7}
                                                        =(175)×227=(175)\times \dfrac{22}{7}
                                                       =550 =550 sq. cm

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