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What will be the area of the quadrilateral ABCD whose vertices are A (8, 6), B (9, 0), C (1, 2) and D (3, 4)?(a) -20 units(b) 26 units(c) 23 units(d) 3 unitsThe question was asked in an online quiz.I'm obligated to ask this question of Geometry in section Coordinate Geometry of Mathematics – Class 10

Answer»

Correct ANSWER is (b) 26 units

Easiest EXPLANATION: Area of quadrilateral = Area of ∆ABC + Area of ∆ADC

We know that, area of triangle = \(\frac {1}{2}\){x1 (y2 – y3) + x2(y3 – y1) + x3(y1 – y2)}

The coordinates of VERTICES of the triangle are (8, 6), (9, 0) and (1, 2).

The area of triangle = \(\frac {1}{2}\){8(0 – 2) + 9(2 – 6) + 1(6 – 0)} = \(\frac {1}{2}\) { – 16 – 36 + 6} = \(\frac {-46}{2}\) = – 23 units

The area of triangle cannot be zero. So, Area of ∆ABC = 23 units

We know that, area of triangle = \(\frac {1}{2}\){x1(y2 – y3) + x2(y3 – y1) + x3 (y1 – y2)}

The coordinates of vertices of the triangle are (8, 6), (3, 4) and (1, 2).

The area of triangle = \(\frac {1}{2}\) {8(4 – 2) + 3(2 – 6) + 1(6 – 4)} = \(\frac {1}{2}\) {16 – 12 + 2} = \(\frac {6}{2}\) = 3 units

So, Area of ∆ADC = 3 units

Area of quadrilateral = Area of ∆ABC + Area of ∆ADC = 23 + 3 = 26 units


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