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  • Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and (2, 3).

4.   Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and (2, 3).

Answers (1)

best_answer

From the figure:

quadrilateral area 2

The coordinates are  A(-4,-2),\ B(-3,-5),\ C(3,-2)\ and\ D(2,3)

Divide the quadrilateral into 2 parts of triangles.

Then the area will be, ABC + ADC

Area of the triangle formed by ABC will be,

Area_{(ABC)} = \frac{1}{2}\left [ (-4)((-5)-(-2))+(-3)((-2)-(-2))+3((-2)-(-2)) \right ]                       = \frac{1}{2}\left [ 12+0+9 \right ] = \frac{21}{2}\ Square\ units.

Area of the triangle formed by ADC will be,

Area_{(ADC)} = \frac{1}{2}\left [ (-4)((-2)-(-3))+3(3-(-2))+2((-2)-(-2)) \right ]                        = \frac{1}{2}\left [ 20+15+0 \right ] = \frac{35}{2}\ Square\ units.

Therefore, the area of the quadrilateral will be: 

= \frac{21}{2}+\frac{35}{2} = 28\ square\ units.

 

Alternatively,

The points A and C are in the same ordinates.

Hence, the length of base AC will be (3-(-4)) = 7\ units.

Therefore,

Area of triangle ABC:

= \frac{1}{2} \times (Base) \times (Height) = \frac{1}{2}\times(7)(3)

Area of triangle ADC:

= \frac{1}{2} \times (Base) \times (Height) = \frac{1}{2}\times(7)(5)

Therefore, the area will be, \frac{1}{2}\times(7)\times(5+3) =28\ square\ units.

Posted by

Divya Prakash Singh

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