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Tamil Nadu Board of Secondary EducationSSLC (English Medium) Class 10

Find the area of the quadrilateral whose vertices are at (– 9, – 2), (– 8, – 4), (2, 2) and (1, – 3)

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Question

Find the area of the quadrilateral whose vertices are at (– 9, – 2), (– 8, – 4), (2, 2) and (1, – 3)

Sum
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Solution

Let the vertices A(– 9, – 2), B(– 8, – 4), C(2, 2) and D(1, – 3).
Plot the vertices in a graph.

Note: Consider the points in counter clockwise order

Area of the Quadrilateral = `1/2[(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (x_2y_1 + x_3y_2 + x_4y_3 + x_1y_4)]`

Area of the Quadrilateral ABDC = `1/2[36 + 24 + 2 - 4 - (16 - 4 - 6 - 18)]`

= `1/2[58 - (- 12)] - 1/2 [58 + 12]`

= `1/2 xx 70`

= 35 sq. units

∴ The area of the quadrilateral is 35 sq. units.

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Chapter 5: Coordinate Geometry - Exercise 5.1 [Page 212]

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Samacheer Kalvi Mathematics [English] Class 10 SSLC TN Board
Chapter 5 Coordinate Geometry
Exercise 5.1 | Q 5. (i) | Page 212

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