Question

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Answer 1

Question :

A footpath of uniform width runs round the inside of a rectangular field 32 m long and 24 m wide. If the path occupies 208 m² , find the width of the footpath.

Answer :

The required width of the footpath is 2 m

Step-by-step explanation :

Given :

• Length of the rectangular field = 32 m
• Width of the rectangular field = 24 m
• Area of the footpath = 208 m²

To find :

the width of the footpath

Solution :

Let "a" be the width of the footpath

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\multiput(2.1,-0.7)(0,4.2){2}{\sf\large 32 m}\put(-1.1,1.5){\sf\large 24 m}\put(1,1){\line(1,0){3}}\put(1,1){\line(0,1){1}}\put(1,2){\line(1,0){3}}\put(4,2){\line(0,-1){1}}\put(0.1,1.2){\Large \sf a m}\put(0.1,1){\vector(1,0){1}}\put(1,1){\vector(-1,0){1}}\put(0.8,0.5){\Large \sf A}\put(3.8,0.5){\Large \sf B}\put(3.8,2.1){\Large \sf C}\put(0.8,2.1){\Large \sf D}\put(1,1.2){\vector(1,0){3}}\put(4,1.2){\vector(-1,0){3}}\put(1.7,1.4){\sf (32-2a)m}\put(4.2,2){\vector(0,-1){1}}\put(4.2,1.2){\vector(0,1){1}}\put(4.4,1.5){\sf (24-2a)m}\end{picture}$

Area of the rectangular field = Area of the footpath + Area of the rectangle ABCD

Let's find the area of rectangular field first.

➙ Area of the rectangular field = length × breadth

➙ Area of the rectangular field = 32 m × 24 m

➙ Area of the rectangular field = 768 m²

 

Now, Area of the rectangle ABCD :

The length of the rectangle ABCD = (32 - 2a) m

The breadth of the rectangle ABCD = (24 - 2a) m

     

➙ Area of the rectangle ABCD = (32 - 2a) (24 - 2a)

➙ Area of the rectangle ABCD = 32(24 - 2a) - 2a(24 - 2a)

➙ Area of the rectangle ABCD = 768 - 64a - 48a + 4a²

➙ Area of the rectangle ABCD = 4a² - 112a + 768

So, Area of the rectangular field = Area of the footpath + Area of the rectangle ABCD

➙ 768 = 208 + 4a² - 112a + 768

➙  768 - 768 = 4a² - 112a + 208

➙   4a² - 112a + 208 = 0

➙  4(a² - 28a + 52) = 0

➙   a² - 28a + 52 = 0/4

➙   a² - 28a + 52 = 0

 

Solving this equation by splitting middle term,

➙  a² - 2a - 26a + 52 = 0

➙  a(a - 2) - 26(a - 2) = 0

➙   (a - 2) (a - 26) = 0

 ⇒ a - 2 = 0 ; a = 2 m

 ⇒ a - 26 = 0 ; a = 26 m

   

The width of the footpath cannot be more than the width of the rectangular field.

Hence, the width of the footpath is 2 m

Answer 2

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