A) To find the length AC, we must use the trigonometric ratio
[tex]\cos \text{ 31 =}\frac{adjacent}{hypothenuse}[/tex]adjacent = 5cm
hypothenuse = AC
[tex]\begin{gathered} \cos 31\text{ = }\frac{5}{AC} \\ AC\text{ = }\frac{5}{\cos 31} \\ AC\text{ =}5.8332\operatorname{cm} \end{gathered}[/tex]B) To find the length AB, we will use the value of AC just obtained to get it
since triangle ABC is a right-angled triangle, we will use Pythagoras theorem
so that
|AC|^2 = |AB|^2 +|BC|^2
AC = 5.8332cm
BC = 4cm
|AB|^2 = |AC|^2 - |BC|^2
[tex]\begin{gathered} AB\text{ = }\sqrt[]{5.8332^2-4^2} \\ AB\text{ = }\sqrt[]{18.0262224} \\ AB\text{ = 4.245729883cm} \end{gathered}[/tex]C) The perimeter of the quadrilateral can be found by adding the length of all the sides around its edges.
Perimeter = AD +CD + BC + AB
We do not have CD and we must find CD
[tex]\begin{gathered} CD\text{ = }\sqrt[]{|AC|^2-|AD|^2} \\ CD\text{ =}\sqrt[]{5.8332^2-5^2} \\ CD\text{ =}\sqrt[]{9.02622224} \\ CD\text{ = 3.004366195cm} \end{gathered}[/tex]Perimeter of ABCD = 5 + 3.004366195 + 4 +4.245729883 = 16.25009708cm
D) To find the area of the quadrilateral we must find the area of triangle ADC and triangle ABC.
Area of triangle ADC = 1/2 x base x height= 1/2 x 3.004366195 x 5 = 7.510915488 square centimeter
Area of triangle ABC = 1/2 X base x height = 1/2 x 4 x 4.245729883 =8.491459766 square centimeter
Area of quadrilateral ABCD = 7.510915488 + 8.491459766 = 16.00237525 square centimeter
