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

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