The diagram of the given figure is:
To find the area we can start by dividing the quadrilateral into two triangles:
Now, we can find the areas of Triangle 1 and Triangle 2.
We can use the law of cosines to find the third side of the Triangle 1.
Given a=3.1, b=6.5 and the angle C=95, use the formula:
![h^2=a^2+b^2-2ab\cdot\cos C](https://img.qammunity.org/2023/formulas/mathematics/college/l817h1qwgyo4gvq876brey8ts2u5h5p1jh.png)
By replacing the values we obtain:
![\begin{gathered} h^2=3.1^2+6.5^2-2(3.1)(6.5)\cdot\cos 95 \\ h^2=9.61+42.25-40.3\cdot(-0.0872) \\ h^2=55.3724 \\ h=\sqrt[]{55.3724} \\ h=7.4413 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/gywz5l9bl4777ns4rpwf51v5ez7kholw18.png)
And the Area of triangle 1 can be calculated using the Heron's formula as:
![\begin{gathered} A1=\sqrt[]{s(s-a)(s-b)(s-h)} \\ s=(a+b+h)/(2)=(3.1+6.5+7.4413)/(2)=8.52065 \\ A1=\sqrt[]{8.52065(8.52065-3.1)(8.52065-6.5)(8.52065-7.4413)} \\ A1=10.0368 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/l0pu4k5wpm71m5gm3h2m3j1rxr4r5hjsh0.png)
Now, to calculate the Area of Triangle 2, knowing the 3 sides, we can use the same formula:
![\begin{gathered} s=(c+d+h)/(2)=(7.1+8.5+7.4413)/(2)=11.5207 \\ A2=\sqrt[]{11.5207(11.5207-7.1)(11.5207-8.5)(11.5207-7.4413)} \\ A2=25.0511 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/1e64lxjm9r0c3vuz88efm120naju2z5mp3.png)
Then, the total area is the sum of the area1 and area2:
![\begin{gathered} At=A1+A2=10.0368+25.0511 \\ At=35.0879 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/v92fbcev08l3vu3datw7zbmt2wxcn8bs8a.png)
Answer: The area of this figure is 35.0879 yd^2