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Question: A cyclic quadrilateral ABCD of area \(\dfrac{{3\sqrt 3 }}{4}\) is inscribed in a unit circle. If one...

A cyclic quadrilateral ABCD of area 334\dfrac{{3\sqrt 3 }}{4} is inscribed in a unit circle. If one of its sides AB =11 and diagonal BD=3\sqrt 3 , find the lengths of the other sides.

Explanation

Solution

Area of the quadrilateral is the sum of the areas of the two triangles which are separated by diagonal BD. Use cosine rule and sine rule of the triangle in both the triangles to evaluate the sides of the quadrilateral.

Complete step-by-step answer:
First, we draw the figure using the given information.

We are given that one of its sides AB =11 and diagonal BD=3\sqrt 3 .
We have to find the length of other sides.
Since, O is the centre of the circles therefore radii OA=OB=OC=OD =11.
First, we find the measure of A\angle A.
Use the formula of sine rule for circumcircle radius that is,
BDsinA=2R\dfrac{{BD}}{{\sin A}} = 2R
here RR is the radius of the circumcircle.
Substitute all the values and find A\angle A.
3sinA=1 sinA=13 A=60  \dfrac{{\sqrt 3 }}{{\sin A}} = 1 \\\ \Rightarrow \sin A = \dfrac{1}{{\sqrt 3 }} \\\ \therefore A = 60^\circ \\\
Since the sum of opposite angles of cyclic quadrilateral is 180180^\circ therefore, C=120\angle C = 120^\circ
Now, we use cosine rule in ΔABD\Delta ABD
cosA=AB2+AD2BD22ABAD\cos A = \dfrac{{A{B^2} + A{D^2} - B{D^2}}}{{2ABAD}}
Substitute all the values.
cos60=1+AD232AD\cos 60^\circ = \dfrac{{1 + A{D^2} - 3}}{{2AD}}
Substitute cos60=12\cos 60^\circ = \dfrac{1}{2} and cross multiply to solve the equation.

12=1+AD232AD AD2AD2=0  \Rightarrow \dfrac{1}{2} = \dfrac{{1 + A{D^2} - 3}}{{2AD}} \\\ \Rightarrow A{D^2} - AD - 2 = 0 \\\

Solve the quadratic equation and discard the negative value of AD.
AD22AD+AD2=0 AD(AD2)+1(AD2)=0 (AD2)(AD+1)=0 AD=2   A{D^2} - 2AD + AD - 2 = 0 \\\ \Rightarrow AD(AD - 2) + 1(AD - 2) = 0 \\\ \Rightarrow (AD - 2)(AD + 1) = 0 \\\ \therefore AD = 2 \\\ \\\
Now, we use cosine rule in ΔBCD\Delta BCD
cosC=BC2+CD2BD22BCDC\cos C = \dfrac{{B{C^2} + C{D^2} - B{D^2}}}{{2BCDC}}
Substitute all the values.
cos120=BC2+CD232(BC)(CD)\cos 120^\circ = \dfrac{{B{C^2} + C{D^2} - 3}}{{2(BC)(CD)}}
Substitute cos120=12\cos 120^\circ = \dfrac{{ - 1}}{2} and cross multiply to solve the equation.

12=BC2+CD232(BC)(CD) BC2+CD2+(BC)(CD)3=0.....(1)  \Rightarrow - \dfrac{1}{2} = \dfrac{{B{C^2} + C{D^2} - 3}}{{2(BC)(CD)}} \\\ \Rightarrow B{C^2} + C{D^2} + (BC)(CD) - 3 = 0.....(1) \\\

We are given a cyclic quadrilateral ABCD of area 334\dfrac{{3\sqrt 3 }}{4}.
Therefore,
Area of cyclic quadrilateral = Area of ΔABC\Delta ABC+ Area of ΔBCD\Delta BCD
Write the area of the triangle using the formula 12absinθ\dfrac{1}{2}ab\sin \theta , here θ\theta is the angle between the side lengths aaand bb.
Write the areas in the mathematical form using the above result.
334=12×1×2×sin60+12×BC×CD×sin120\dfrac{{3\sqrt 3 }}{4} = \dfrac{1}{2} \times 1 \times 2 \times \sin 60^\circ + \dfrac{1}{2} \times BC \times CD \times \sin 120^\circ
Substitute sin60=sin120=12\sin \,60^\circ = \sin 120^\circ = \dfrac{1}{2}and cross multiply to solve the equation.

334=12×1×2×13+12×BC×CD×13 2+BC×CD=3 BC×CD=1.........(2)  \Rightarrow \dfrac{{3\sqrt 3 }}{4} = \dfrac{1}{2} \times 1 \times 2 \times \dfrac{1}{{\sqrt 3 }} + \dfrac{1}{2} \times BC \times CD \times \dfrac{1}{{\sqrt 3 }} \\\ \Rightarrow 2 + BC \times CD = 3 \\\ \Rightarrow BC \times CD = 1.........(2) \\\

Substitute the value of BC×CDBC \times CD in the equation (1)(1).

BC2+CD2+13=0 BC2+CD2=2..............(3)  \therefore B{C^2} + C{D^2} + 1 - 3 = 0 \\\ \Rightarrow B{C^2} + C{D^2} = 2..............(3) \\\

From equation (2)(2)and (3)(3),the only possible value of BCBCand CDCD is 11.
Therefore, the length of the other sides is AD=1AD = 1 and BC=CD=1BC = CD = 1

Note: Use the sine and cosine rule of the triangle by analysing the angle and the side which includes in the rule properly.
If the length of the sides of the cyclic quadrilateral is a,b,ca,b,c and dd then the area of the cyclic quadrilateral will be
(sa)(sb)(sc)(sd)\sqrt {(s - a)(s - b)(s - c)(s - d)}
Here, s=a+b+c+d2s = \dfrac{{a + b + c + d}}{2}