Question

Find the area of quadrilateral ABCD whose vertices are A(-5 , 7) , B (−4 , −5) , C (−1 ,
−6) and D (4 , 5).

Answer 1

Given to find  the area of quadrilateral ABCD whose vertices are

• A= (-5,  7)
• B = ( -4 , -5)
• C = (-1 , -6)
• D = ( 4, 5 )

Formula to know :-

$\dfrac{1}{2} \left|x_1(y_2-y_3) +x_2(y_3-y_1) +x_3(y_1-y_2)\right|$

Diagram :-

$\setlength{\unitlength}{1 cm}\begin{picture}(20,15)\thicklines\qbezier(1,1)(1,1)(6,1)\qbezier(1,1)(1,1)(1.6,4)\qbezier(6,1)(6,1)(6.6,4)\qbezier(1.6,4)(1.6,4)(6.6,4)\qbezier(6.6,4)(6.6,4)(1,1)\put(0.2,0.6){\sf A(-5,7)}\put(5.5,0.6){\sf B(-4,-5)}\put(0.9,4.2){\sf D(4,5)}\put(6.1,4.2){\sf C(-1,-6)}\end{picture}$

SOLUTION :-

Area of Quadrilateral ABCD = ar(△ABC) + ar(△CDA)

In △ABC ,

$x_1= -5 \\x_2 = -4 \\x_3 = -1$

$y_1 = 7 \\y_2 = -5 \\y_3 = -6$

Now we can find area of △ ABC

Substituting the values in formula

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

$\dfrac{1}{2} \bigg|-5[-5-(-6)] -4[-6-(7)] -1[7-(-5)]\bigg|$

$\dfrac{1}{2} \bigg|-5 (-5+6)-4(-6-7)-1(7+5)\bigg|$

$\dfrac{1}{2} \bigg|-5(1) -4(-13) -1(12)\bigg|$

$\dfrac{1}{2} \bigg|-5+52-12\bigg|$

$\dfrac{1}{2}\bigg|35\bigg|$

$\dfrac{35}{2}$

So, the area of triangle ABBC is 35/2

Now finding Area of triangle ACD

$x_1 = -5\\x_2 = -1 \\x_3 = 4$

$y_1 = 7 \\y_2 = -6 \\y_3 = 5$

Now finding area by using the formula

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

$\dfrac{1}{2} \bigg|-5(-6-5) -1(5-7) +4(7+6)\bigg|$

$\dfrac{1}{2} \bigg| -5(-11) -1(-2) +4(13)\bigg|$

$\dfrac{1}{2} \bigg|55+2+52\bigg|$

$\dfrac{109}{2}$

So, area of quadrilateral = ar△(ABC) +ar△(ACD)

$\dfrac{35}{2} +\dfrac{109}{2}$

$\dfrac{144}{2}$

$72 units^2$

So, area of quadrilateral ABCD is 72 units²

Note:-

If you are confusing you to see calculation refer attachment