Question

A rectangular lawn is 2 m longer than it is wide.
The area of the lawn is 21 m². The gardener wants to edge the lawn with edging strips
which are sold in lengths of 13 m. How many will she need to buy?

Answer 1

Let's breadth be x

• Length be x+2

ATQ

$\\ \sf\longmapsto x(x+2)=21$

$\\ \sf\longmapsto x^2+2x=21$

• Use completing the square method

$\\ \sf\longmapsto 4(x^2+2x)=4(21)$

$\\ \sf\longmapsto 4x^2+8x=84$

$\\ \sf\longmapsto (2x)^2+2(2x)(2)=84$

$\\ \sf\longmapsto (2x)^2+2(2x)(2)+2^2=2^2+84$

$\\ \sf\longmapsto (2x+2)^2=88$

$\\ \sf\longmapsto 2x+2=\pm\sqrt{88}$

• As it's dimensions we will take positive

$\\ \sf\longmapsto 2x+2\approx 9.7$

$\\ \sf\longmapsto 2x=9.7-2=7.7$

$\\ \sf\longmapsto x=3.8$

• x+2=3.8+2=5.8

Now

$\\ \sf\longmapsto Perimeter=2(L+B)$

$\\ \sf\longmapsto perimeter=2(3.8+5.8)$

$\\ \sf\longmapsto perimeter=2(9.6)$

$\\ \sf\longmapsto Perimeter=19.2m$

• Length of 1 strip=13m

Total strips

$\\ \sf\longmapsto \dfrac{19.2}{13}$

$\\ \sf\longmapsto 1.47strips$

• She needs 2 strips

Answer 2

Answer:

• width is y metres

• length is (2 + y) metres

${ \tt{area = length \times width}} \\ \\ { \rm{21 = (2 + y)(y)}} \\ \\ { \rm{21 = 2y + {y}^{2} }} \\ \\ { \rm{ {y}^{2} + 2y - 21 = 0 }} \\ \\ { \rm{ \dashrightarrow \: by \: factorisation : }} \\ \\ { \boxed{ \rm{y \approx4 \: and \: - 6}}}$

• Therefore, length = 4 + 2 = 6 meters

• The gardener will need;

$= { \rm{ \frac{13}{6} }} \\ \\ = { \underline{ \underline{ \rm{ \: 3 \: strips \: \: }}}}$