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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, 0), (– 8, 6), (– 1, – 2) and (– 6, – 3) - Mathematics

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Question

Find the area of the quadrilateral whose vertices are at (– 9, 0), (– 8, 6), (– 1, – 2) and (– 6, – 3)

Sum
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Solution

Let the vertices A(– 9, 0), B(– 8, 6), C(– 1, –2) and D(– 6, – 3)

Plot the vertices in a graph and take them in counter-clockwise order.

Area of the Quadrilateral DCBA

= `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)]`

= `1/2[33 + 35]`

= `1/2 xx 68`

= 34 sq.units

Area of the Quadrilateral = 34 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. (ii) | Page 212
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