Given the Quadrilateral A B C D with vertices A(-3,-8), B(6,-6), C(4,2), D(-8,2). Find the area of triangle ABC. Find the area of triangle A C D. Calculate the area of triangle ADC + the area of triangle A C D .Find the area of quadrilateral A B C D. Compare the answers obtained in 3 and 4.
Answers
Step-by-step explanation:
Let, the points be A(−2,−2),B(5,1),C(2,4) and D(−1,5)
The quadrilateral ABCD can be divided into triangles ABC and ACD and hence the area of the quadrilateral is the sum of the areas of the two triangles.
Area of a triangle with vertices (x
1
,y
1
) ; (x
2
,y
2
) and (x
3
,y
3
)
=
∣
∣
∣
∣
∣
∣
2
x
1
(y
2
−y
3
)+x
2
(y
3
−y
1
)+x
3
(y
1
−y
2
)
∣
∣
∣
∣
∣
∣
.
Hence, substituting the points (x
1
,y
1
)=(−2,−2) ; (x
2
,y
2
)=(5,1) and (x
3
,y
3
)=(2,4) in the area formula,
we get area of triangle ABC
=
∣
∣
∣
∣
∣
∣
2
(−2(1−4)+(5)(2−2)+2(−2−4)
∣
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
∣
2
6+0−12
∣
∣
∣
∣
∣
∣
=
2
6
=3 sq. units
And, substituting the points (x
1
,y
1
)=(−2,−2) ; (x
2
,y
2
)=(5,1) and (x
3
,y
3
)=(−1,5) in the area formula, we get
area of triangle ACD =
∣
∣
∣
∣
∣
∣
2
(−2)(1−5)+(5)(5+2)+(−1)(−2−1
∣
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
∣
2
8+35+3
∣
∣
∣
∣
∣
∣
=
2
46
=23 sq. units
Hence, Area of quadrilateral =3+23=26 sq. units
I didn't understand the compare alone. please forgive me
