Question

the four sequential sides of a quadrilateral have lengths $a = 5.1$, $b = 7.7$, $c = 8.6$, and $d = 10.1$ (all measured in yards). the angle between the two smallest sides is $\alpha = 114^{\circ}$.
what is the area of this figure?
area = \boxed{\space} $\text{yd}^2$
question help: \boxed{\text{video}}

Explanation:

Step1: Calculate the area of the triangle formed by the two smallest sides

The formula for the area of a triangle with two sides \(a\), \(b\) and included - angle \(\alpha\) is \(S_1=\frac{1}{2}ab\sin\alpha\).
Substitute \(a = 5.1\), \(b = 7.7\), and \(\alpha=114^{\circ}\) (\(\sin114^{\circ}=\sin(90^{\circ}+24^{\circ})=\cos24^{\circ}\approx0.9135\)) into the formula:
\(S_1=\frac{1}{2}\times5.1\times7.7\times\sin114^{\circ}\approx\frac{1}{2}\times5.1\times7.7\times0.9135=\frac{1}{2}\times5.1\times7.7\times0.9135\approx17.99\)

Step2: Use the Law of Cosines to find the length of the diagonal

The Law of Cosines formula is \(e^{2}=a^{2}+b^{2}-2ab\cos\alpha\).
Substitute \(a = 5.1\), \(b = 7.7\), and \(\alpha = 114^{\circ}\) (\(\cos114^{\circ}=-\cos(90^{\circ}+24^{\circ})=-\sin24^{\circ}\approx - 0.4067\)) into the formula:
\(e^{2}=5.1^{2}+7.7^{2}-2\times5.1\times7.7\times\cos114^{\circ}\)
\(e^{2}=26.01 + 59.29-2\times5.1\times7.7\times(-0.4067)\)
\(e^{2}=26.01+59.29 + 31.57\)
\(e^{2}=116.87\), so \(e\approx10.81\)

Step3: Calculate the area of the second triangle

Assume the second triangle has sides \(c = 8.6\), \(d = 10.1\), and \(e\approx10.81\). Use Heron's formula \(s=\frac{c + d+e}{2}=\frac{8.6 + 10.1+10.81}{2}=\frac{29.51}{2}=14.755\)
The area of the triangle \(S_2=\sqrt{s(s - c)(s - d)(s - e)}\)
\(S_2=\sqrt{14.755\times(14.755 - 8.6)\times(14.755 - 10.1)\times(14.755 - 10.81)}\)
\(S_2=\sqrt{14.755\times6.155\times4.655\times3.945}\)
\(S_2=\sqrt{14.755\times6.155\times4.655\times3.945}\approx42.01\)

Step4: Calculate the area of the quadrilateral

The area of the quadrilateral \(S=S_1 + S_2\)
\(S\approx17.99+42.01 = 60\)

Answer:

\(60\)