Question

Find the area of a parallelogram ABCD, three of whose vertices are A(0, 0), B(3, 5) and C(4, 2).

Answer 1

$\boxed{\bf{\red{\implies Answer}}}$

• First we need To find co-ordinates of D
• Suppose D(x, y)
• x+3/2 = 0+4/2
• x = 1
• y+5/2 = 0+2/2
• y = -3
• D(1, -3)

• Area of Parallelogram = Area of Two Triangles
• Area of Triangle ABC + Triangle ACD = Area of ||gm ABCD
• 2(Area of ABC) = Area of ABCD
• $2( \frac{1}{2} (x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))$
• $= 0(5 - 2) + 3(2 - 0) + 4(0 - 5)$
• $= 3 + 6 - 20$
• $= - 11 \\ but \: area \: cant \: be \: negative \: so \: area \: is \: 11 \: sq.units$