Draw a quadrilateral ABCD in a cartesian plane, whose vertices are(-5,5)(5'5)(5,-5) and (-5,-5). Name the quadrilateral ABCD
Answers
Answer:
Draw a Quadrilateral ABCD whose vertices are A (–4, 5), B (0, 7), C (5, –5), and D (–4, –2).
Step-by-step explanation:
Draw a Quadrilateral ABCD whose vertices are A (–4, 5), B (0, 7), C (5, –5), and D (–4, –2).
Draw a quadrilateral in cartesian
Area of quadrilateral ABCD = area of ∆ABC + area of ∆ACD
Relate the coordinate of ∆ ABC, A (–4, 5), B (0, 7), C (5, –5) with (x1, y1), (x2, y2), and (x3, y3) and use formula
\[ = \frac{1}{2}|{x_1}({y_2} - {y_3}+{x_2}({y_3} –{y_1} )+ +{x_3}({y_1} –{y_2} |\]
Plug the values, we get
=1/2|-4(7+5) + 0(-5 ,-5) + 5(5-7)|
=1/2|-4(12 +0 + 5(-2)|
=1/2|-48 -10|
=1/2 (58)
=29 unit2
Relate the coordinate of ∆ ABC, A (–4, 5), C (5, –5) and D (–4, –2). with (x1, y1), (x2, y2), and (x3, y3) and use formula
\[ = \frac{1}{2}|{x_1}({y_2} - {y_3}+{x_2}({y_3} –{y_1} )+ +{x_3}({y_1} –{y_2} |\]
Area of ∆ACD
=1/2|-4(-5+ 2) +5(-2-5) + (-4)(5 +5)|
= ½ |-4(-3) +5(-7) +(-4)(10)|
=1/2|12 -35 -40|
=1/2|-63|
= 63/2 unit2
Total area of quadrilateral (ABCD) = 29+ 63/2 = 121/2 unit2