y = the width
L = the length
A = the area of original rectangle
A = y * L = 140
y * L = 140 Divide both sides by y
L = 140 / y
New area :
( y - 2 ) ( L + 3 ) = 136
( y - 2 ) ( 140 / y + 3 ) = 136
y * 140 / y - 2 * 140 / y + 3 * y - 3 * 2 = 136
140 - 280 / y + 3 y - 6 = 136
- 280 / y + 3 y + 134 = 136 Subtract 134 to both sides
- 280 / y + 3 y + 134 - 134 = 136 - 134
- 280 / y + 3 y = 2 Multiply both sides by y
- 280 * y / y + 3 y * y = 2 * y
- 280 + 3 y ^ 2 = 2 y
3 y ^ 2 - 280 = 2 y Subtract 2 y to both sides
3 y ^ 2 - 280 - 2 y = 2 y - 2 y
3 y ^ 2 - 2 y - 280 = 0
The solutions are :
y = - 28 / 3 and y = 10
The width can't be negative so y = 10 cm
L = 140 / y = 140 / 10 = 14
L = 14 cm
The diagonal of the original rectangle :
d = sqroot ( y ^ 2 + L ^ 2 ) =
sqroot ( 10 ^ 2 + 14 ^ 2 ) =
sqroot ( 100 + 196 ) = sqrt ( 296 ) = 17.20465 cm
The area of a rectangle of width ycm is 140cm^2. If the width is reduced by 2cm, the length increases by 3cm, and the area decreases to 136cm^2 to form an equation that enables you to determine the value of Y and hence, find the diagonal of the original rectangle
workings.
Area area= length length*width width
140-136=x 3*y-2
4=x 3*y-2
collect like terms
4-3 2=x*y
3=x*y
divide both sides by x
y=3/x.. Please am not sure am correct
1 answer
1 answer