Question

(i) A footpath of uniform width runs all around the inside of a rectangular field 54 m long and 35 m wide. If the area of the path is 420$m^2$, find the width of the path.

(ii) A carpet is laid on the floor of a room 8 m by 5m. There is a border of constant width all around the carpet. If the area of the border is 12 $m^2$, find its width.

Answer 1

(i)

Let width = x m

Area of path = ar(EFGH) - ar(ABCD)

$420=54\times 35-(54-2x)(35-2x)420=1890-[54\times 34-108x-70x+4x^2]420=1890-1890+178x+4x^{2}4x^2-178x+420=02x^2-89x+210=0D=b^2-4ac=(-89{)}^2-4(2\left)\right(210)=7921-1680=6241x=\frac{-b\pm \sqrt{D}}{2a}=\frac{-(-89)\pm \sqrt{6241}}{2\times 2}=\frac{89\pm 79}{4}=\frac{89+79}{4}or\frac{89-79}{4}=\frac{168}{4}or\frac{10}{4}=42or2.5$

Width of path cannot be 42,

So width of path = 2.5 m

(ii)

Let width of carpet be 'x'm
Therefore, inclusive of border:
Length of room = 8m
Breadth of room = 5m
Length of carpet = 8-x-x = (8-2x) m
Breadth of carpet = 5-x-x = (5-2x) m

Area of room (carpet + border) = 8 $\times$ 5 = 40 sq.m

Area of just carpet \)= (8-2x)(5-2x) \\ = (40-16x-10x+4x^2)\\= (4x^2-26x+40)~m^2\)

Area of just border = 12sq.m (given)

Therefore,
Area of ground = Area of just border + Area of just carpet
$\Rightarrow 40=12+(4x^2-26x+40)\Rightarrow 0=4x^2-26x+12\Rightarrow 0=2x^2-13x+6\text{(dividing equation by 2)}\Rightarrow 0=2x^2-12x-x+6\Rightarrow 0=2x(x-6)-1(x-6)\Rightarrow 0=(2x-1)(x-6)\Rightarrow x=\frac{1}{2}=0.5mor6m$

discarding 6m since it is longer than room then

Therefore, width of border = 0.5m