MathematicsFigure shows a regular pentagon with all its vertices on the sides of a rectangle . If BC = 14, calculate the perimeter of the pentagon .

Figure shows a regular pentagon with all its vertices on the sides of a rectangle . If BC = 14, calculate the perimeter of the pentagon .


C:UsersuserAppDataLocalMicrosoftWindowsINetCacheContent.Worded1b6301-4eed-4f9c-b6ec-6570aa31ab033573014692444065861.jpeg

  1. A
     0.31
  2. B
     0.62
  3. C
     0.77
  4. D
    1.00  

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    Solution:

    Concept- Using the sine rule
    asin A =bsin B=csin C , when a , b and c are the length of sides of the triangle and A, B,and C are the angles opposite to that sides a,b and c respectively .assume the length of one the sides of pentagon as x. we will use the sine rule to determine the value of x  and we will multiply this value with 5 to find  the perimeter.
    All sides of the regular pentagon are equal. Assuming each side as x. name the pentagon MNOPQ. Now join the vertices N and Q to form a triangle . MNQ . also there , a rectangle  BNQC formed.
    C:UsersuserAppDataLocalMicrosoftWindowsINetCacheContent.Worda46cc23b-3ccf-446c-bc2c-0f0cc7d3c3151661398243084875650.jpeg NQ=BC=14  since they are the opposite sides of rectangle BNQC
    In triangle MNQ
    Angle NMQ=108°, all internal angles of a regular pentagon is 108°.
    MN=MQ,
    angle MNQ=Angle MQN=36°
    Apply sine rule in triangle in MNQ
    MNsin 36° =NQsin 108° 
    NQ=BC=14 we get
    MNsin 36°=NQsin 108° 
    xsin 36° =14sin 108° 
    Using the conclusion,
    sin 108°=sin 180°-108°=sin 72° and   
    sin 2x=2sin xcos x , we get   
    x=sin 36° 4×2sin 36°cos 36°  
    x=18×cos 36° 
    Substituting , Cos 36°=5+14, we get
    x=18×5+14
    x=12×15+1
    Rationalise denominator.
    x=12×15+×5-15-1
    =12×5-15-1
    =5-18
    5=2.236
    x=2.236-18
    x=1.2368
    x=0.1545
    Perimeter of the pentagon =5x because all 5 sides are equal.
    perimetre=5×0.1545=0.77
    Hence, option (3) is the correct answer.
     
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