Math, asked by harshitha1430, 8 months ago

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

Answered by Anonymous
0

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

Attachments:
Answered by AnjuVarshithaS
0

I didn't understand the compare alone. please forgive me

Attachments:
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