Question

Points A, B, C, D, and O have coordinates (0, 3), (4, 4), (4, y), (4, 0), and (0, 0), respectively. The area of ACDO is 5 times greater than the area of ABDO. Find y.

Answer 1

Area of Quadrilateral ACDO= Area (ΔAOD)+Area(ΔACD)-----(1)

Area of Quadrilateral ABDO= Area (ΔAOD)+Area(ΔABD)------(2)

⇒ (1) = 5 ×(2)→→→→GIven

→Area (ΔAOD)+Area(ΔACD)=5 Area (ΔAOD)+5 Area(ΔABD),

→5 Area (ΔABD)+4 Area (ΔAOD)-Area (ΔACD)=0-----(3)

As, Area of a Triangle
`={\text{having vertices}} (x_(1),y_(1)),(x_(2),y_(2)) {\text{and}} (x_(3),y_(3))=(1)/(2)*[x_(1)(y_(2)-y_(3))-x_2(y_(3)-y_(1))+x_(3)(y_(1)-y_(2))]`

Area (ΔACD)

`=(1)/(2)[0(y-0)-4(0-3)+4(3-y)]\\\\= (1)/(2)[12+12-4y]\\\\=12-2 y`

Area (ΔABD)

`=(1)/(2)[0(4-0)-4(0-3)+4(3-4)]\\\\ =(1)/(2)[12-4]=4`

Area (ΔAOD)
`=(1)/(2)[0(3-0)-0(0-0)+4(0-3)]=6`

Putting these values in equation (3)

→20+24-12+2 y=0

→ 2 y= -32

Dividing both sides by , 2 we get

→y= -16