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Question: In a circular grassy plot, a quadrilateral shape with its corners touching the boundary of the plot ...

In a circular grassy plot, a quadrilateral shape with its corners touching the boundary of the plot is to be paved with the bricks. Find the area of the quadrilateral when the sides of the quadrilateral are 36m,77m,75m and 40m36{\text{m,77m,75m and 40m}}.
A) 4686 square meter4686{\text{ square meter}}
B) 2886 square meter{\text{2886 square meter}}
C) 3856 square meter{\text{3856 square meter}}
D) none of these{\text{none of these}}

Explanation

Solution

As we know that it forms a cyclic quadrilateral and we know that if sides of the cyclic quadrilateral is given for example a, b, c, d then the area is given by
area=(s−a)(s−b)(s−c)(s−d){\text{area}} = \sqrt {(s - a)(s - b)(s - c)(s - d)}
Where ssis the semi-perimeter of the quadrilateral which means s=a+b+c+d2s = \dfrac{{a + b + c + d}}{2}

Complete step by step solution:
Here as we are given that in a circular grassy plot, a quadrilateral shape with its corners touching the boundary of the plot is to be paved with the bricks. So it will form the cyclic quadrilateral.

Let ABCDABCD be the quadrilateral.
Let us assume that AB=a,BC=b,CD=c,AD=dAB = a,BC = b,CD = c,AD = d
Upon joining ACAC let us assume that AC=xAC = x
∠ABC=θ\angle ABC = \theta
And we know that in the cyclic quadrilateral sum of the opposite angle sis equal to 180∘180^\circ
So ∠ABC+∠ADC=180∘\angle ABC + \angle ADC = 180^\circ
θ+∠ADC=180∘\theta + \angle ADC = 180^\circ
∠ADC=180∘−θ\angle ADC = 180^\circ - \theta
Now we can apply cosine formula in ΔADC\Delta ADC and ΔABC\Delta ABC
Now in ΔABC\Delta ABC
cos⁡B=a2+b2−c22ab\cos B = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}
And we know that ∠B=θ\angle B = \theta
cos⁡θ=a2+b2−x22ab\cos \theta = \dfrac{{{a^2} + {b^2} - {x^2}}}{{2ab}} −−−−−(1) - - - - - (1)
And in ΔADC\Delta ADC
cos⁡D=c2+d2−x22cd\cos D = \dfrac{{{c^2} + {d^2} - {x^2}}}{{2cd}} and we know that ∠D=180−θ\angle D = 180 - \theta
And cos⁡(180−θ)=−cos⁡θ\cos (180 - \theta ) = - \cos \theta
−cos⁡θ=c2+d2−x22cd- \cos \theta = \dfrac{{{c^2} + {d^2} - {x^2}}}{{2cd}} −−−−(2) - - - - (2)
Now we can write equation (1) as
x2=a2+b2−2abcos⁡θ{x^2} = {a^2} + {b^2} - 2ab\cos \theta
And from equation (2), we get
x2=c2+d2+2cdcos⁡θ{x^2} = {c^2} + {d^2} + 2cd\cos \theta
Equating both we get that
a2+b2−2abcos⁡θ{a^2} + {b^2} - 2ab\cos \theta =c2+d2+2cdcos⁡θ= {c^2} + {d^2} + 2cd\cos \theta
So we get
cos⁡θ(2cd+2ab)=a2+b2−c2−d2\cos \theta (2cd + 2ab) = {a^2} + {b^2} - {c^2} - {d^2}
cos⁡θ=a2+b2−c2−d22(ab+cd)\cos \theta = \dfrac{{{a^2} + {b^2} - {c^2} - {d^2}}}{{2(ab + cd)}}
And we know that the area of the quadrilateral is given as
Area of quadrilateral ABCDABCD == area ofΔABC\Delta ABC+area of ΔADC\Delta ADC
And if we are given two sides and the included angle, then the area is given by 12absin⁡θ\dfrac{1}{2}ab\sin \theta where θ\theta is the included angle
So we get that
Area of quadrilateral ABCDABCD == area ofΔABC\Delta ABC+area of ΔADC\Delta ADC
=12absin⁡θ+12cdsin⁡(180−θ)= \dfrac{1}{2}ab\sin \theta + \dfrac{1}{2}cd\sin (180 - \theta)
And we know sin⁡(180−θ)=sin⁡θ\sin (180 - \theta ) = \sin \theta
So Area of quadrilateral ABCDABCD =12absin⁡θ+12cdsin⁡θ= \dfrac{1}{2}ab\sin \theta + \dfrac{1}{2}cd\sin \theta
A==12(ab+cd)sin⁡θA = = \dfrac{1}{2}(ab + cd)\sin \theta
Upon squaring both the sides we get
A2=14(ab+cd)2sin⁡2θ{A^2} = \dfrac{1}{4}{(ab + cd)^2}{\sin ^2}\theta
And we know that sin⁡2θ=1−cos⁡2θ{\sin ^2}\theta = 1 - {\cos ^2}\theta
A2=14(ab+cd)2(1−cos⁡2θ){A^2} = \dfrac{1}{4}{(ab + cd)^2}(1 - {\cos ^2}\theta )
Now we know that cos⁡θ=a2+b2−c2−d22(ab+cd)\cos \theta = \dfrac{{{a^2} + {b^2} - {c^2} - {d^2}}}{{2(ab + cd)}}
So on putting this value we get that
A2=14(ab+cd)2(1−(a2+b2−c2−d2)24(ab+cd)2){A^2} = \dfrac{1}{4}{(ab + cd)^2}(1 - \dfrac{{{{({a^2} + {b^2} - {c^2} - {d^2})}^2}}}{{4{{(ab + cd)}^2}}})
A2=14(4(ab+cd)2−(a2+b2−c2−d2)24){A^2} = \dfrac{1}{4}\left( {\dfrac{{4{{(ab + cd)}^2} - {{({a^2} + {b^2} - {c^2} - {d^2})}^2}}}{4}} \right)
We can write it as
16A2=(2(ab+cd))2−(a2+b2−c2−d2)216{A^2} = {(2(ab + cd))^2} - {({a^2} + {b^2} - {c^2} - {d^2})^2}
On applying a2−b2=(a+b)(a−b){a^2} - {b^2} = (a + b)(a - b)
16A2=(2(ab+cd)+a2+b2−c2−d2)(2(ab+cd)−a2−b2+c2+d2)16{A^2} = (2(ab + cd) + {a^2} + {b^2} - {c^2} - {d^2})(2(ab + cd) - {a^2} - {b^2} + {c^2} + {d^2}) 16A2=(a2+b2+2ab−(c2+d2−2cd)(−(a2+b2−2ab)+c2+d2+2cd)16{A^2} = ({a^2} + {b^2} + 2ab - ({c^2} + {d^2} - 2cd)( - ({a^2} + {b^2} - 2ab) + {c^2} + {d^2} + 2cd)
16A2=((a+b)2−(c+d)2)((c+d)2−(a+b)2)16{A^2} = ({(a + b)^2} - {(c + d)^2})({(c + d)^2} - {(a + b)^2})
Again solving we get that
16A2=((a+b+c+d)(a+b−c−d)(c+d+a+b)(c+d−a−b))16{A^2} = ((a + b + c + d)(a + b - c - d)(c + d + a + b)(c + d - a - b))
And we know that ss is the semi-perimeter and
area=(s−a)(s−b)(s−c)(s−d){\text{area}} = \sqrt {(s - a)(s - b)(s - c)(s - d)}
s=a+b+c+d2s = \dfrac{{a + b + c + d}}{2}
So as we know that s=a+b+c+d2s = \dfrac{{a + b + c + d}}{2}
2s=a+b+c+d2s = a + b + c + d
2s−2d=a+b+c−d2s - 2d = a + b + c - d
2s−2c=a+b+d−c2s - 2c = a + b + d - c
2s−2b=a+c+d−b2s - 2b = a + c + d - b
2s−2a=c+b+d−a2s - 2a = c + b + d - a
We get that
16A2=2(s−a)2(s−c)2(s−b)2(s−d)16{A^2} = 2(s - a)2(s - c)2(s - b)2(s - d)
A2=(s−a)(s−c)(s−b)(s−d){A^2} = (s - a)(s - c)(s - b)(s - d)
So for the cyclic quadrilateral the area is given by
area=(s−a)(s−b)(s−c)(s−d){\text{area}} = \sqrt {(s - a)(s - b)(s - c)(s - d)}
Here sides are given as a=36m,b=77m,c=75m,d=40ma = 36m,b = 77m,c = 75m,d = 40m
s=a+b+c+d2s = \dfrac{{a + b + c + d}}{2}
s=36+77+75+402=114s = \dfrac{{36 + 77 + 75 + 40}}{2} = 114
area=(114−36)(114−77)(114−75)(114−40){\text{area}} = \sqrt {(114 - 36)(114 - 77)(114 - 75)(114 - 40)}
area=78.37.39.74{\text{area}} = \sqrt {78.37.39.74}
On solving this we get that

area=39.2.37.39.37.2=(39)(37)(2)=2886m2.{\text{area}} = \sqrt {39.2.37.39.37.2} = (39)(37)(2) = 2886{m^2}.

Note: for the triangle if all the sides are given then the area of the triangle is given by the formula
area=s(s−a)(s−b)(s−c){\text{area}} = \sqrt {s(s - a)(s - b)(s - c)}
Where s=a+b+c2s = \dfrac{{a + b + c}}{2}
And this is called the Heron’s formula.