SOLUTION: The sides of a parallelogram ABCD are AB=424 mm and AD=348 mm. If angle ABC is 130 degrees: A. Find the lengths of the long and short diagonals. B. Find the area and perimeter of

B. Let me start with the perimeter. It is 

     P = 2*(424 + 348) mm = . . .   Please calculate it yourself.

   Now with the area. The area of a parallelogram (of any parallelogram) is

     S = , 

   where a and b are measures of two its adjacent sides and  is the angle between them 
   (see the lesson Area of a parallelogram in this site).
   
   So in your case  S = 424*348*sin(130°) = 424*348*sin(180°-130°) = 424*348*sin(50°) = . . . Please calculate it yourself.


A. Now let us calculate the long diagonal of the parallelogram.

   Make a sketch of the parallelogram and its elements.

   Use the Cosines Law (see the lesson Proof of the Law of Cosines revisited in this site or everywhere else). 
   
   It says that if you know the measures of two sides a and b of a triangle and the angle  between them
   then the third side of the triangle is 

    = .

   So in your case c =  = .

   Notice that I replaced the angle 130° by the supplementary angle 50° and changed the sign of cosines.

   Now calculate it yourself please.


   Next, let us calculate the short diagonal of the parallelogram.

   Make a sketch again. 

   Use the same Law of Cosines. This time you have the same measures of the sides. The only difference is in that 
   the angle between the sides is not 130°. It is the supplementary angle 180° - 130° = 50°.
   
   Thus the formula for the third side in this case is 

     c = .

   Please calculate it yourself.

   From my side, the problem is solved.