Question

A path of uniform width surrounds a lawn of dimensions 10m by 8m. If the area of the path is 88sq.m, find the width of the path.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

A path of uniform width surrounds a lawn of dimensions 10m by 8m.

Let assume that ABCD be the lawn having dimensions 10 m by 8 m.

Length of lawn = 10 m

Breadth of lawn = 8 m

So,

$\rm \: Area_{(ABCD)} = Length \times Breadth$

$\rm \: Area_{(ABCD)} = 10 \times 8$

$\rm\implies \:Area_{(ABCD)} = 80 \: {m}^{2}$

As, it is given that lawn is surrounded by a path of uniform width.

So, Let assume that the uniform width be x m.

Let the outer region be EFGH.

So,

Length = 10 + 2x m

Breadth = 8 + 2x m

So,

$\rm \: Area_{(EFGH)} = Length \times Breadth$

$\rm \: Area_{(EFGH)} = (10 + 2x)(8 + 2x)$

$\rm \: Area_{(EFGH)} =80 + 20x + 16x + {4x}^{2}$

$\rm \: Area_{(EFGH)} =80 + 36x + {4x}^{2}$

Now, it is given that

$\rm \: Area_{(path)} = 88 \: {m}^{2}$

$\rm \: Area_{(EFGH)} - Area_{(ABCD)} = 88$

$\rm \: 80 + 36x + {4x}^{2} - 80 = 88$

$\rm \: 36x + {4x}^{2} = 88$

On dividing both sides by 4, we get

$\rm \: 9x + {x}^{2} = 22$

$\rm \: {x}^{2} + 9x - 22 = 0$

$\rm \: {x}^{2} + 11x - 2x - 22 = 0$

$\rm \: x(x + 11) - 2(x + 11) = 0$

$\rm \: (x + 11)(x - 2) = 0$

$\rm\implies \:x = 2 \: \: or \: \: x = - 11 \: \{rejected \}$

So, Width of the path is 2 m

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$