Question

1. A rectangular field is of dimensions 25 m x 16.4 m. Two paths run parallel to the sides of the
rectangle through the centre of the field. The width of the longer path is 1.7 m and that of the
shorter path is 2 m. Find :
(i) the area of the paths.
(ii) the area of the remaining portion of the field.​

Answer 1

Answer:

x x  

Step-by-step explanation:

Answer 2

$\large\underline\purple{\bold{Solution :- }}$

$\tt \: Length \: of \: rectangular \: field,  \: l=25 \: m</p><p>$

$\tt \: Width \: of \: rectangular \: field, b=16.4 \: m$

$\tt \: \red{Length \: of \: \: shorter \: path, l_1=16.4 \: m }$

$\tt \: \red{</p><p>Width \: of \: the \: shorter \: path, b_1=2 \: m}$

$\tt \: \green{Length \: of \: the \: longer \: path, l_2=25 \: m}$

$\tt \: \green{Width \: of \: the \: longer \: path,  \: b_2=1.7 \: m \: }$

$\tt \: \purple{Area \: of \: shorter \: path, A_1=l_1×b_1}$

$\tt \: = 16.4 \times 2$

$\tt \: = 32.8 \: {m}^{2}$

$\tt \: \pink{ Area \: of \: longer \: path, A_2=l_2×b_2}$

$\tt \:  = 25 \times 1.7$

$\tt \:  = 42.5 \: {m}^{2}$

$\tt \:  Thus, \: Area \: of \: the \: path, \:  P=A_1+A_2−Area \: of \: common \: path$

$\tt \:  = 32.8 + 42.5 - 2 \times 1.7$

$\tt \:  = 75.3 - 3.4$

$\tt \:  = 71.9 \: {m}^{2}$

$\tt \:  \red{Area \: of \: the \: rectangular \: field,  \: A=l×b}$

$\tt \:  = 25 \times 16.4$

$\tt \:  = 410 \: {m}^{2}$

$\tt \:  \orange{Hence \: area \: of \: the \: remaining \: portion \: of \: field}$

$\tt\implies \boxed{ \red{ \bf \: \:Area = \: 410 - 71.9 = 338.1 \: {m}^{2} }}$