Question

Second Program: triples.c 8. Create a program named triples.c "C How to Program", 8th edition, problem 4.27 on page 154. In other editions, it may be on a different page Here's the problem... A right triangle can have sides that are all integers. The set of 3 integers for the sides of a right triangle is called a Pythagorean triple Write a C program to find all of the triples with hypotenuse less than or equal to 1000 Use brute force with triple-nested for loops Include documentation just like you did in calls.c. In the output, print column headings, neatly aligned columns, and right-justify the numbers as shown below Side 1 Side 2 Hypotenuse 3 4 5 5 12 13 9. Compile. Debug if needed and recompile. Run and test the program. You can check your output for this one by doing a web search on "Pythagorean Triples" Submit calls.c and triples.c on Canvas.

to be done in c language. follow prompt

Answer 1

Dear Learner,

Here is the code for your solution,

#include<stdio.h>
#include<conio.h>
main()
{
   int side1,side2,hypotenuse;                                        //Declaring the variables side1 side2 and hypotenuse
   int i,j,k;                                                            //Declaring the iterating variables
   printf("side1 side2 hypotenuse\n");                                // These two lines
   printf("----- ----- ----------\n");                                // are printed for formatting the output
   for(side1=1;side1<10;i++)                                           // First loop for the first side
   {  
       for(side2=side1;side2<10;j++)                                   //The second nested loop is for side2. NOTE that it starts from the value
       {                                                               // of side1 in order to reduce the number of cycles
           for(hypotenuse=side2;k<10;j++)                           //This loop is to find the hypotenuse
           {
               if(hypotenuse*hypotenuse == side1*side1 + side2*side2)   //Checking if the third side satisfies the hypotenuse relation
               {
               printf("%d\t\t%d\t\t%d\n",side1,side2,hypotenuse);           // printing the values if true
               }

           }
       }
   }
}

CORRESONDING OUTPUT:

I have shown first and ending part of the printed values.

side1 side2 hypotenuse 113 ad 65 14 145 3e 18e 181 52 ១១ 181 35 15 128 264 122 15 6e 65 312 168 313 178 45 123 365 - ៖ – ១៩ 188 197 421 ៩

580 609 841 580 741 941 582 776 970 585 648 873 585 780 975 588 784 980 591 788 985 594 608 850 594 792 990 595 600 845 597 796 995 600 630 870 612 759 975 615 728 953 616 663 905 616 735 959 620 651 899 621 672 915 624 715 949 638 720 962 640 928 970 660 957 680 714 986 696 697 985 Program ended with exit code: @ 65€ 720 693

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