Question

If A(15,14), B(30,15),C(27,5) and D(10,5) are vertices of a quadrilateral then what will be area of Quadrilateral ABCD?

Answer 1

Answer:

Ar ΔABD=1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)

1/2[15(5-5)+30(5-15)+10(14-15)

1/2[0-300+10

1/2*290

145 sq unit

Ar ΔBDC=1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]

1/2[30(5-5)+27(5-15)+10(15-5)

1/2[0+270+100

1/2+370

185 sq unit

Ar of quadrilateral=145+185=330 sq unit

Answer 2

Answer:

150

Hello.  I hope this helps you.

Step-by-step explanation:

The easy way to deal with something like this is with the surveyor's area formula.  Writing O for the origin, the idea is that

area(ABCD) = area(OAB) + area(OBC) + area(OCD) + area(ODA)

where these "areas" are really "signed areas", so when the labels are anticlockwise around the triangle the signed area is positive, and when the labells are clockwise, the signed area is negative.  So for example, area(OAB) = -area(OBA).  Once you've got that idea, the formula above is easy to confirm.

Next, the signed area of triangle OAB is half the determinant made up from the coordinates of A and B.  Now to calculation...

$\displaystyle2\times\mbox{signed area}(ABCD)=\\ \\\left|\begin{array}{cc}15&14\\30&15\end{array}\right|+\left|\begin{array}{cc}30&15\\27&5\end{array}\right|+\left|\begin{array}{cc}27&5\\10&5\end{array}\right|+\left|\begin{array}{cc}10&5\\15&14\end{array}\right|\\ \\=(15)(15)-(14)(30)+(30)(5)-(15)(27)+(27)(5)-(5)(10)+(10)(14)-(5)(15)\\ \\=225 - 420 + 150 - 405 + 135 - 50 + 140 - 75\\ \\=-300\\ \\\Rightarrow\mbox{signed area}(ABCD) = -150$

Of course, we don't really want the signed area, so the (normal) area is just 150.

Incidentally, the fact that the signed area was negative tells us that the vertices ABCD in that order are going clockwise around the quadrilateral!