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  • How to solve this problem- - Co-ordinate geometry - JEE Main-8

The two adjacent sides of a cyclic quadrilateral are 2 and 5 and the angle between them is 600.  If the area of the
quadrilateral is 4\sqrt{3} then the permeter of the quadrilateral is :

  • Option 1)

    12.5

  • Option 2)

    13.2

  • Option 3)

    12

  • Option 4)

    13

 

Answers (1)

best_answer

Using concept PQRS , Cyclic quadrilateral \left | opposite\: angles \: of\: a\: cyclic\:quadrilateral\:are\: supplementary \right | and concept of properties of triangle

d=\sqrt{2^2+5^2-2\times2\times5cos60^{\circ}}

(Using cosinc's rule)

Area of \Delta ABC = \frac{1}{2}\times 2\times5sin60^{\circ}

\Delta ABC = 5\frac{\sqrt{3}}{2}

Area of \Delta BCD = \frac{1}{2}xy \:sin 120^{\circ}

=4\sqrt{3}-5\frac{\sqrt{3}}{2}

\frac{1}{2}xy \times \frac{\sqrt{3}}{2}= 3\frac{\sqrt{3}}{2}

xy=6

Also (\sqrt{19})^2 = x^2+y^2-2 \:xy\:cos120^{\circ}

Hence x=3
           y=2
     Permeter =12


Option 1)

12.5

Option 2)

13.2

Option 3)

12

Option 4)

13

Posted by

Himanshu

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