Question

12. In the adjoining figure, ABCD is a square grassy lawn of area
729 m2. A path of uniform width runs all around it. If the area
of the path is 295 m2, find
(i) the length of the boundary of the square field enclosing
the lawn and the path.
(ii) the width of the path.​

Answer 1

Step-by-step explanation:

Hope it is helpful to you

Answer 2

Given:

• Area of Square =729m²

so it's side $\rm{\sqrt{729} =27m}$

•Area of the path =295m²

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To find :

(1) The length of the boundary of the square field enclosing the lawn and the path =?

(2) The width of the path=?

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Solution :

•Let's take the width of the path =x m

Then, Side of out field=27+x+x=(27+2x)

and Area of Square PQRS=(27+2x)²m²

Now ,

Area of PQRS-Area of ABCD=Area of path

$\\ \implies \sf \: {(27 + 2x)}^{2} {m}^{2} - 729 {m}^{2} = 295 {m}^{2}$

$\\ \implies \sf \: 729 + 4 {x}^{2} + 108m - 729 = 295 \\ \\ \implies \sf \: 4 {x}^{2} + 108x - 295 = 0 \\ \\ \implies \sf \: 4 {x}^{2} + 108x - 295 = 0$

By using the quadratic formula method , we have

➡ a=4 , b=108 , c=-295

$\\ \implies \sf \: x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}$

$\\ \implies \sf \: x = \frac{ - 108 \pm \sqrt{ {(108)}^{2} - 4 \times 4 \times (- 295)} }{8}$

$\\ \implies \sf \: x = \frac{ - 108 \pm \sqrt{11664 + 4720} }{8}$

$\\ \implies \sf \: x = \frac{ - 108 \pm128}{8} = \frac{20}{8} = \frac{10}{4}$

$\\ \implies \sf \: x = \frac{5}{2} = 2.5$

Hence the width of the path is 2.5m

Now side of square field =27+2x=$\rm{27+2 \times 2.5=32m}$

Therefore,

$\rm{ Length \: of \: boundary = 4 \times side}$

$\rm{Length \: of \: boundary =4 \times 32=128m}$

$\rm{Length \: of \: boundary =128m}$

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Additional information :

(1) For solving any quadratic equation we use formula method first we calculate by determinant form and then we use formula method to get roots of given equation.

Formula method :

$\\ \red \bigstar \sf \: x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}$

Determinant method :

$\\ \red \bigstar \sf \delta \:d \: = {b}^{2} - 4ac$