Question

A building is 25 m long and 10 m wide , all round the building 2m wide cement concrete foot path has to be provided. Find the area of foot path

Answer 1

AnswEr :

⋆ Refrence of Image is in the Diagram :

$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\thicklines\put(7.3,2){\large{14 m}}\put(9.2,0.7){\large{29 m}}\put(8,1){\line(1,0){3}}\put(8,1){\line(0,2){2}}\put(11,1){\line(0,3){2}}\put(8,3){\line(3,0){3}}\setlength{\unitlength}{1.5cm}\put(8.4,2){\large{10 m}}\put(9.2,1.4){\large{25 m}}\put(8,1.26){\line(1,0){2.7}}\put(8.3,1){\line(0,2){1.76}}\put(10.7,1.25){\line(0,3){1.5}}\put(8.3,2.74){\line(3,0){2.4}}\put(8.02,1.07){\scriptsize{2m}}\end{picture}$

$\rule{150}{2}$

$\bullet\:\textsf{Length of Building ( l ) = 25 m}\\\\\bullet\:\textsf{Breadth of Building ( b ) = 10 m}\\\\\bullet\:\textsf{Outer Length ( L ) = [25 + 2(2)] m}\\\qquad\qquad\qquad\qquad\textsf{= [25 + 4] m = 29 m} \\\\\bullet\:\textsf{Outer Breadth ( B ) = [10 + 2(2)] m}\\ \qquad\qquad\qquad\qquad\textsf{= [10 + 4] m = 14 m}$

$\rule{200}{1}$

$\underline{\bigstar\:\textsf{According to the Question Now :}}$

$:\implies\texttt{Area of Path = Outer Area - Inner Area}\\\\\\:\implies\tt Area\:of\:Path = (Length \times Breadth)-(length \times breadth)\\\\\\:\implies\tt Area\:of\:Path = (L \times B)- (l \times b)\\\\\\:\implies\tt Area\:of\:Path =(29 \: m\times14 \:m)-(25 \:m\times 10 \:m)\\\\\\:\implies\tt Area\:of\:Path =406 \:m^2- 250\:m^2\\\\\\:\implies\underline{\boxed{\red{\tt Area\:of\:Path = 156\:m^2}}}$

⠀

$\therefore\:\underline{\textsf{Area of the Path around Building is \textbf{156 m ^2 }.}}$

Answer 2

Answer:

Inner Area = 25 m × 10 m

= 250 m²

Outer length = 25 m + ( 2m + 2m )

= 29 m

Outer breadth = 10 m + ( 2m + 2m )

= 14m

Outer Area = 29m × 14m

= 406 m²

• Area of the path = Outer Area - Inner Area

= 406 m² - 250 m²

= 156 m²