Figure 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 = $\frac{1}{4}$, calculate the perimeter of the pentagon .

1. A
2. B
3. C
4. D
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### Solution:

Concept- Using the sine rule
, 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.
$\mathit{NQ}=\mathit{BC}=\frac{1}{4}$  since they are the opposite sides of rectangle BNQC
In triangle MNQ
Angle NMQ=108°, all internal angles of a regular pentagon is 108°.
$\mathit{MN}=\mathit{MQ},$
$\mathit{angle MNQ}=\mathit{Angle MQN}=36°$
Apply sine rule in triangle in MNQ

Using the conclusion,

Substituting , Cos 36
$x=\frac{1}{8×\frac{\sqrt{5+1}}{4}}$
$x=\frac{1}{2}×\frac{1}{\sqrt{5+1}}$
Rationalise denominator.
$x=\frac{1}{2}×\frac{1}{\sqrt{5+}}×\frac{\sqrt{5-1}}{\sqrt{5-1}}$
$=\frac{1}{2}×\frac{\sqrt{5-1}}{5-1}$
$=\frac{\sqrt{5-1}}{8}$
$\sqrt{5}=2.236$
$x=\frac{2.236-1}{8}$
$x=\frac{1.236}{8}$
$x=0.1545$
Perimeter of the pentagon =5x because all 5 sides are equal.
$\mathit{perimetre}=5×0.1545=0.77$
Hence, option (3) is the correct answer.

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