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**Hint:**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

${\text{area}} = \sqrt {(s - a)(s - b)(s - c)(s - d)} $

Where $s$is the semi-perimeter of the quadrilateral which means $s = \dfrac{{a + b + c + d}}{2}$

**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 $ABCD$ be the quadrilateral.

Let us assume that $AB = a,BC = b,CD = c,AD = d$

Upon joining $AC$ let us assume that $AC = x$

$\angle ABC = \theta $

And we know that in the cyclic quadrilateral sum of the opposite angle sis equal to $180^\circ $

So $\angle ABC + \angle ADC = 180^\circ $

$\theta + \angle ADC = 180^\circ $

$\angle ADC = 180^\circ - \theta $

Now we can apply cosine formula in $\Delta ADC$ and $\Delta ABC$

Now in $\Delta ABC$

$\cos B = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}$

And we know that $\angle B = \theta $

$\cos \theta = \dfrac{{{a^2} + {b^2} - {x^2}}}{{2ab}}$$ - - - - - (1)$

And in $\Delta ADC$

$\cos D = \dfrac{{{c^2} + {d^2} - {x^2}}}{{2cd}}$ and we know that $\angle D = 180 - \theta $

And $\cos (180 - \theta ) = - \cos \theta $

$ - \cos \theta = \dfrac{{{c^2} + {d^2} - {x^2}}}{{2cd}}$$ - - - - (2)$

Now we can write equation (1) as

${x^2} = {a^2} + {b^2} - 2ab\cos \theta $

And from equation (2), we get

${x^2} = {c^2} + {d^2} + 2cd\cos \theta $

Equating both we get that

${a^2} + {b^2} - 2ab\cos \theta $$ = {c^2} + {d^2} + 2cd\cos \theta $

So we get

$\cos \theta (2cd + 2ab) = {a^2} + {b^2} - {c^2} - {d^2}$

$\cos \theta = \dfrac{{{a^2} + {b^2} - {c^2} - {d^2}}}{{2(ab + cd)}}$

And we know that the area of the quadrilateral is given as

Area of quadrilateral $ABCD$$ = $ area of$\Delta ABC$+area of $\Delta ADC$

And if we are given two sides and the included angle, then the area is given by $\dfrac{1}{2}ab\sin \theta $ where $\theta $ is the included angle

So we get that

Area of quadrilateral $ABCD$$ = $ area of$\Delta ABC$+area of $\Delta ADC$

$ = \dfrac{1}{2}ab\sin \theta + \dfrac{1}{2}cd\sin (180 - \theta)$

And we know $\sin (180 - \theta ) = \sin \theta $

So Area of quadrilateral $ABCD$$ = \dfrac{1}{2}ab\sin \theta + \dfrac{1}{2}cd\sin \theta $

$A = = \dfrac{1}{2}(ab + cd)\sin \theta $

Upon squaring both the sides we get

${A^2} = \dfrac{1}{4}{(ab + cd)^2}{\sin ^2}\theta $

And we know that ${\sin ^2}\theta = 1 - {\cos ^2}\theta $

${A^2} = \dfrac{1}{4}{(ab + cd)^2}(1 - {\cos ^2}\theta )$

Now we know that $\cos \theta = \dfrac{{{a^2} + {b^2} - {c^2} - {d^2}}}{{2(ab + cd)}}$

So on putting this value we get that

${A^2} = \dfrac{1}{4}{(ab + cd)^2}(1 - \dfrac{{{{({a^2} + {b^2} - {c^2} - {d^2})}^2}}}{{4{{(ab + cd)}^2}}})$

${A^2} = \dfrac{1}{4}\left( {\dfrac{{4{{(ab + cd)}^2} - {{({a^2} + {b^2} - {c^2} - {d^2})}^2}}}{4}} \right)$

We can write it as

$16{A^2} = {(2(ab + cd))^2} - {({a^2} + {b^2} - {c^2} - {d^2})^2}$

On applying ${a^2} - {b^2} = (a + b)(a - b)$

$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})$$16{A^2} = ({a^2} + {b^2} + 2ab - ({c^2} + {d^2} - 2cd)( - ({a^2} + {b^2} - 2ab) + {c^2} + {d^2} + 2cd)$

$16{A^2} = ({(a + b)^2} - {(c + d)^2})({(c + d)^2} - {(a + b)^2})$

Again solving we get that

$16{A^2} = ((a + b + c + d)(a + b - c - d)(c + d + a + b)(c + d - a - b))$

And we know that $s$ is the semi-perimeter and

${\text{area}} = \sqrt {(s - a)(s - b)(s - c)(s - d)} $

$s = \dfrac{{a + b + c + d}}{2}$

So as we know that $s = \dfrac{{a + b + c + d}}{2}$

$2s = a + b + c + d$

$2s - 2d = a + b + c - d$

$2s - 2c = a + b + d - c$

$2s - 2b = a + c + d - b$

$2s - 2a = c + b + d - a$

We get that

$16{A^2} = 2(s - a)2(s - c)2(s - b)2(s - d)$

${A^2} = (s - a)(s - c)(s - b)(s - d)$

So for the cyclic quadrilateral the area is given by

${\text{area}} = \sqrt {(s - a)(s - b)(s - c)(s - d)} $

Here sides are given as $a = 36m,b = 77m,c = 75m,d = 40m$

$s = \dfrac{{a + b + c + d}}{2}$

$s = \dfrac{{36 + 77 + 75 + 40}}{2} = 114$

${\text{area}} = \sqrt {(114 - 36)(114 - 77)(114 - 75)(114 - 40)} $

${\text{area}} = \sqrt {78.37.39.74} $

On solving this we get that

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

**Note:**for the triangle if all the sides are given then the area of the triangle is given by the formula

${\text{area}} = \sqrt {s(s - a)(s - b)(s - c)} $

Where $s = \dfrac{{a + b + c}}{2}$

And this is called the Heron’s formula.

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