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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:

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, $A B C+A D C$
The area of the triangle formed by ABC will be,
Area $A B C =\frac{1}{2}[(-4)((-5)-(-2))+(-3)((-2)-(-2))+3((-2)-(-2))]$
$=\frac{1}{2}[12+0+9]=\frac{21}{2}$ Square units.

The area of the triangle formed by $ADC$ will be,

Area $A D C=\frac{1}{2}[(-4)((-2)-(-3))+3(3-(-2))+2((-2)-(-2))]$
$=\frac{1}{2}[20+15+0]=\frac{35}{2}$ 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 the base AC will be $(3-(-4))=7$ units.
Therefore,
Area of triangle ABC:
$=\frac{1}{2} \times(\text { Base }) \times(\text { Height })=\frac{1}{2} \times(7)(3)$
Area of triangle ADC:
$=\frac{1}{2} \times(\text { Base }) \times(\text { Height })=\frac{1}{2} \times(7)(5)$

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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