Advertisement Remove all ads

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

Answer in Brief

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

Advertisement Remove all ads

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?
Advertisement Remove all ads

APPEARS IN

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

Video TutorialsVIEW ALL [2]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×