Question

If the co- ordinate of the vertices of a quadrilateral ABCD are A (1,1),B (7,-3) C (12,2) and D (7,21), then the area of the quadrilateral is

• A. 32 Sq. units
• B. 66 sq.units
• C. 33 sq.units
• D. 132 sq.units

Answer 1

The correct option is D 132 sq.units

R.E.F image

solution : -

Let $A(1,1),B(7,-3),C(7,21)$

Area of quadrilateral ABCD = Area of triangle ABC +Area of triangle ACD

Area of triangle ABC $=\frac{1}{2}\left[x_1\right(y_2-y_2)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$

$=\frac{1}{2}\left[1\right(-3-2)+7(2-1)+12(1+3\left)\right]$

$=25sq.units$

Area of triangle ADC $=\frac{1}{2}\left[1\right(2.21)+12(21-1)+7(1-2\left)\right]$

$=107sq.units$

Area of quadrilateral ABCD = 25+107 = 132 sq units.