Question

Find the area of quadrilateral formed by joining the vertices (-5,-2),(0,3),(10,-2)(4,-9).

Answer 1

Answer:

divide the quadrilateral in two parts .. use the formula of coordinate geometry of area of triangle.. calculate areas seprately. add them

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Answer 2

Answer:

Formula used in solving this sum

$\frac{1}{2} (x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)$

Here,

In △ABC

$(x_1,y_1) = ( - 5_,2)$

$(x_2,y_2) = (0_,3)$

$(x_3,y_3) = (10_,2)$

∴ Area of △ABC

$= \frac{1}{2} ( - 5(3 + 2) + 0( - 2 - 2) + 10(2 - 3))$

$= \frac{1}{2} ( - 5(5) + 10( - 1))$

$= \frac{1}{2} ( - 25 - 10)$

$= \frac{1}{2} ( - 35)$

$= \frac{ - 35}{2}$

Similarly

In △ADC

$(x_1,y_1) = ( - 5_,2)$

$(x_2,y_2) = (10_,2)$

$(x_3,y_3) = (5_,0)$

Area of △ADC

$= \frac{1}{2} ( - 5( - 2 - 5) + 10(5 - 2) + 0)$

$= \frac{1}{2} ( - 5( - 7) + 10(3))$

$= \frac{1}{2} (35 + 30)$

$= \frac{1}{2} (65)$

$= \frac{65}{2}$

Now area of quadrilateral ABCD = Area of △ ABC + area of triangle

$= \frac{65}{2} + ( \frac{ - 35}{2} )$

$= \frac{65}{2} - \frac{35}{2}$

$= { \dfrac{ \cancel{30} {}^{ \: \: \: \: 15} }{ \cancel{2}_{ \: \: \: \: 1}}}$

= 15 unit²

Answer: Area of the Quadrilateral = 15 unit²