Question

2. Write a program with three functions that will be called from the main function The functions will calculate the volume, area and perimeter of a cuboid. You should able to pass three parameters such as length, width and height, and a function will return area, another will return volume and final one will return perimeter. Define all your variables in floating point numbers 3. Modify the problem#2 to calculate the volume, area and perimeter in a sing le function and return three calculated values- volume, area and perimeter in a single return value. float calculation (float I, float w, float h); //prototype float result calculation (10, 8, 6); l/calling function printf ("Cuboid's volume= %f, area= %f, perimeter= %f", result[0], result[1] result[2); //printing the result.

Answer 1

Question 2:

Here is code:

#include <stdio.h>

#include <math.h>

float

volume(float l, float w, float h)

{

return (l * w * h);

}

float

area(float l, float w, float h)

{

return 2 * (w * l + l * h + h * w);

}

float

perimeter(float l, float w, float h)

{

return 4 * (l * w * h);

}

int

main()

{

float width, length, height;

float c_area, c_volume, c_perimeter;

printf("Enter value of width, length & height of the cuboids:\n");

scanf("%f%f%f", &width, &length, &height);

c_area = area(length, width, height);

c_volume = volume(length, width, height);

c_perimeter = perimeter(length, width, height);

printf("Area of cuboids is: %f\n", c_area);

printf("\nVolume of cuboids is : %f\n", c_volume);

printf("\nPerimeter of cuboids is : %f\n", c_perimeter);

return 0;

}

Question 3:

Here is code:

#include <stdio.h>

#include <stdlib.h>

float

volume(float l, float w, float h)

{

return (l * w * h);

}

float

area(float l, float w, float h)

{

return 2 * (w * l + l * h + h * w);

}

float

perimeter(float l, float w, float h)

{

return 4 * (l * w * h);

}

float *

calculation(float l, float w, float h)

{

float * result = (float *) malloc(3 * sizeof(float));

result[0] = volume(l,w,h);

result[1] = area(l,w,h);

result[2] = perimeter(l,w,h);

return result;

}

int

main()

{

float width, length, height;

float c_area, c_volume, c_perimeter;

printf("Enter value of width, length & height of the cuboids:\n");

scanf("%f%f%f", &width, &length, &height);

float * result = calculation(length, width, height);

printf("Cuboid's volume = %f, area = %f, perimeter=%f\n", result[0], result[1], result[2]);

return 0;

}

Output:

Enter value of width, length & height of the cuboids: 10 20 30 Cuboid's volume = 6000.000000, area = 2200.000000, perimeter=24ee0.000000