Asmakan wrote:
Hi Sandy,
How we want to find the area of ABD and we are using the coordinate of C?
Hi,
Unfortunately Sandy, is not with us
However, to answer your ques. - Simple trick don't complicate with formulas
Plz see the attach diag.
Let us extended the line from co-ordinate C to point M (-7,0) and to Point O (0,7)
Extend the line from the co-ordinate B (-3,4) to Point O (0,7)
Extend the line from the co-ordinate D (4,-3) to Point M (-7,0) ( co-ordinate of point D = mirror image of the co-ordinate of Point B and hence the sign changes and value gets interchanged)
it is the best approach. Thanks!
Now you can see 4 \(\triangle\) s
Since we need the Area of the parallelogram ABCD and that can be found = Area of the square AMCO - Sum of the Areas of all the 4 \triangle s
i.e. Area of the parallelogram ABCD = Area of the square AMCO - { Area of \(\triangle\) AMD + Area of \(\triangle\) DMC + Area of \(\triangle\) COB + Area of \(\triangle\) OAB }
First :
Area of the square \(AMCO = 7 * 7 = 49\) (distance between each point =7)Now,
Area of \(\triangle\) AMD = \(\frac{1}{2} * 7 * 3\) = \(\frac{21}{2}\)( base =7 and height = distance from base to point D = 3)
Area of \(\triangle\) DMC = \(\frac{1}{2}* 7 * 3\) = \(\frac{21}{2}\)( base =7 and height = distance from base to point D = 3)
Area of \(\triangle\)COB = \(\frac{1}{2} * 7 * 3\) =\(\frac{21}{2}\)( base =7 and height = distance from base to point B = 3)
Area of \(\triangle\) OAB = \(\frac{1}{2} * 7 * 3\) =\(\frac{21}{2}\)( base =7 and height = distance from base to point B = 3)
Sum of all 4 \(\triangle\) s = \({4* \frac{21}{2}} = 42\)Therefore Area of the parallelogram ABCD = 49 - 42 =7