Question

The area of a rectangle whose vertices are (–2, 0), (5, 0), (5, 4) and (–2, 4) taken in order, is

Answer 1

Step-by-step explanation:

Let A(–4,–2), B(–3,–5), C(3,–2) and D(2,3) be the vertices of the quadrilateral ABCD.

Area of a quadrilateral ABCD= Area of △ABC+ Area of △ACD

By using a formula for the area of a triangle =

2

1

∣x

1

(y

2

−y

3

)+x

2

(y

3

−y

2

)+x

3

(y

1

−y

2

)∣

Area of △ABC

=

2

1

[−4(−5+2)+−3(−2+2)+3(−2+5)]

=

2

1

[12+9]

=

2

21

sq.units

Area of △ACD=

2

1

[−4(3+2)+−2(−2+2)+3(−2−3)]

=

2

1

[−20−15]

=

2

35

sq.units

∴Area of quadilateral=

2

21

+

2

35

=

2

56

=28 sq.unit

solution

Answer 2

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