# If A(−3, 5), B(−2, −7), C(1, −8) and D(6, 3) Are the Vertices of a Quadrilateral Abcd, Find Its Area. - Mathematics

If A(−3, 5), B(−2, −7), C(1, −8) and D(6, 3) are the vertices of a quadrilateral ABCD, find its area.

#### Solution

It is given that A(−3, 5), B(−2, −7), C(1, −8) and D(6, 3) are the vertices of a quadrilateral ABCD.

Area of the quadrilateral ABCD = Area of ∆ABC + Area of ∆ACD

$\text{ ar } \left( ∆ ABC \right) = \frac{1}{2}\left| x_1 \left( y_2 - y_3 \right) + x_2 \left( y_3 - y_1 \right) + x_3 \left( y_1 - y_2 \right) \right|$

$= \frac{1}{2}\left| - 3\left[ - 7 - \left( - 8 \right) \right] + \left( - 2 \right)\left( - 8 - 5 \right) + 1\left[ 5 - \left( - 7 \right) \right] \right|$

$= \frac{1}{2}\left| - 3 + 26 + 12 \right|$

$= \frac{35}{2}\text{ square units }$

$\text{ ar } \left( ∆ ACD \right) = \frac{1}{2}\left| x_1 \left( y_2 - y_3 \right) + x_2 \left( y_3 - y_1 \right) + x_3 \left( y_1 - y_2 \right) \right|$

$= \frac{1}{2}\left| - 3\left( - 8 - 3 \right) + 1\left( 3 - 5 \right) + 6\left[ 5 - \left( - 8 \right) \right] \right|$

$= \frac{1}{2}\left| 33 - 2 + 78 \right|$

$= \frac{109}{2} \text{ square units }$

∴ Area of the quadrilateral ABCD =$\frac{35}{2} + \frac{109}{2} = \frac{144}{2} = 72$  square units

Hence, the area of the given quadrilateral is 72 square units.

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 10 Maths
Chapter 6 Co-Ordinate Geometry
Exercise 6.5 | Q 10 | Page 54