How do you write a c plus plus program to find area of a triangle where base4 and height6?

Answer 1

pi x radius x radius the forula for working out the area of a circle if this is what you're asking.

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Please give the Proper Answer...

Answer 2

#include <iostream>

int main()

{

float width=0,height=0;

printf( " |.\n" );

printf( " | .\n" );

// "Answers.com formatting has removed multiple spaces from the following lines.

// Insert the indicated number of spaces to form a rectangle:

printf( " | .\n" ); // insert 1 extra space before period.

printf( " | .\n" ); // insert 2 extra spaces before period.

printf( "A | .\n" ); // insert 3 extra spaces before period.

printf( " | .\n" ); // insert 4 extra spaces before period.

printf( " | .\n" ); // insert 5 extra spaces before period.

printf( " | .\n" ); // insert 6 extra spaces before period.

printf( " ._________\n\n" ); // insert 5 extra spaces before B.

printf( " B\n\n" );

printf( "Enter size A: " );

scanf_s( "%f", &height );

printf( "Enter size B: " );

scanf_s( "%f", &width );

printf( "The area is %.2f units.\n", width * height * 0.5 );

return( 0 );

}

Answer 3

The area of a rectangle is calculated by multiplying the the length of its base by its height and then dividing by 2. Thus given the base and height of any triangle, the following function will calculate the area:

float areaof_triangle( float base, float height )

{

return( ( base * height ) / 2 );

}

Remember that a right-angled triangle is half a rectangle thus its area is half the area of the imagined rectangle, hence we divide the area of the rectangle (base times height) by two. The same principal applies to any triangle so long as the height is known. For a non-right angled triangle ABC, if the base is BC then the height is AB * sin(B). This calculates the length of an imagined line AD (the height), such that AD is perpendicular to AB, thus creating two right angled triangles of the same height. If the triangle ABC is isosceles, then AD (and BD) will be exactly half the length of AB, thus AD can then be calculated as the square root of AB2-AD2 (using Pythagoras' theorem).