Question 1185345
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Find the area of a quadrilateral having the points A(-3, 4), B(-2, -1), C(4, 4) and D(0, 5).
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<pre>
I will assume that the vertices of the quadrilateral go in that order as the points are listed.


Then I see that the diagonal AC of the quadrilateral is horizontal y=4 and divides the quadrilateral in two triangles.


Triangle ABC has the vertex  B  BELOW  the diagonal line y= 4. 

It has the base AC of the length 4 - (-3) = 7 units and the altitude drawn to AC of the length 4 - (-1) = 5 units.

Hence, the area of the triangle ABC is  {{{(1/2)*7*5}}} = 17.5 square units.



Next, triangle ADC has the vertex  D  ABOVE  the diagonal line y= 4. 

It has the base AC of the length 4 - (-3) = 7 units and the altitude drawn to AC of the length 5 - 4 = 1 units.

Hence, the area of the triangle ADC is  {{{(1/2)*7*1}}} = 3.5 square units.



Finally, the area of the quadrilateral ABCD is the sum of the areas of the triangles ABC and ADC


    {{{Area[ABCD]}}} = 17.5 + 3.5 = 21 square units.    <U>ANSWER</U>
</pre>

Solved.



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How Alan solved the problem, &nbsp;it assumes that the reader knows &nbsp;EVERYTHING &nbsp;in &nbsp;Geometry, &nbsp;including &nbsp;Analytic &nbsp;Geometry.


In my view, &nbsp;it is absolutely unrealistic hypothesis.


Much more closer to reality is to assume that the reader knows &nbsp;NOTHING &nbsp;except the basic knowledge.


So I adapted my solution correspondingly.