Find the area of the quadrilateral whose vertices taken in order are

Question.

Find the area of the quadrilateral whose vertices taken in order are (–4, –2), (–3, –5), (3, –2) and (2, 3).


Solution:

Join A and C. The given points are

$\mathrm{A}(-4,-2), \mathrm{B}(-3,-5), \mathrm{C}(3,-2)$ and $\mathrm{D}(2,3)$

Find the area of the quadrilateral whose vertices taken in order are

Area of $\triangle \mathrm{ABC}$

$=\frac{\mathbf{1}}{\mathbf{2}}[(-4)(-5+2)-3(-2+2)+3(-2+5)]$

$=\frac{\mathbf{1}}{\mathbf{2}}[12+0+9]=\frac{\mathbf{2 1}}{\mathbf{2}}=10.5$ sq. units

Area of $\Delta \mathrm{ACD}$

$=\frac{1}{2}[(-4)(-2-3)+3(3+2)+2(-2+2)]$

$=\frac{\mathbf{1}}{\mathbf{2}}\left[20+15 \mid=\frac{\mathbf{3 5}}{\mathbf{2}}=17.5\right.$ sq. units.

Area of quadrilateral ABCD

$=\operatorname{ar}(\Delta \mathrm{ABC})+\operatorname{ar}(\Delta \mathrm{ACD})$

$=(10.5+17.5) \mathrm{sq} .$ units $=28 \mathrm{sq} .$ units

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now