Question

The four sequential sides of a quadrilateral have lengths a = 3.1, b = 6.9, c = 9.6, and d = 10.6 (all measured in yards). The angle between the two smallest sides is a = 112°. What is the area of this figure?I got 58.72 but there saying that wrong

Answer 1

Step 1

see the figure below to better understand the problem

Step 2

Applying the law of sines

Find out the area of triangle ABC

$\begin{gathered} A=\frac{1}{2}(3.1)(6.9)sin(112^o) \\ A=9.916\text{ yd}^2 \end{gathered}$

Step 3

Applying the law of cosines

Find out the length of the side AC

$\begin{gathered} AC^2=3.1^2+6.9^2-2(3.1)(6.9(cos112^o) \\ AC=8.6\text{ m} \end{gathered}$

Step 4

Applying the law of cosines

Find out the measure of angle D

$AC^2=AD^2+DC^2-2(AD)(DC)cosD$

substitute given values

$\begin{gathered} 8.6^2=10.6^2+9.6^2-2(10.6)(9.6)cosD \\ solve\text{ for cosD} \\ cosD=\frac{10.6^2+9.6^2-8.6^2}{2(10.6)(9.6)} \\ \\ angle\text{ D}=50.1^o \end{gathered}$

Step 5

Applying the law of sines

Find out the area of the triangle ADC

$\begin{gathered} A=\frac{1}{2}(10.6)(9.6)sin(50.1^o) \\ \\ A=39.03\text{ yd}^2 \end{gathered}$

The area of the quadrilateral is equal to

$\begin{gathered} A=9.92+39.03 \\ A=48.95\text{ yd}^2 \end{gathered}$

The area is 48.95 square yards (rounded to two decimal places)