Question

Hi i will mark as brain list
please tell me answer of this question in easy way but Step-by-step explanation
Que no 40

Answer 1

$\underline{ \bold{ \purple{QUESTION \: \: }}}$ ➡

$\bold{A \: \: rectangular \: \: hall \: \: has \: \: length \: \: greater \: \: than \: \: } \\ \bold{its \: \: breadth \: \: by \: \: 5 \:m. \: If \: \: area \: \: of \: \: the \: \: hall \: \: is}\\ \bold{618.75 \: m {}^{2}, \: then \: \: find \: \: perimeter \: \: of \: \: the \: \: hall.}$

$\underline{ \bold{ \purple{ANSWER \: \: }}}$ ➡ $\boxed{ \bold{(d) \: \: 100 \: m{}^{2} }}$

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$\underline{\bold{ \purple{SOLUTION\: \: }}}$➡

$\bold{Let,} \\ \\ \bold{Breadth \: \: of \: \: the \: \: rectangular \: \: hall = a \: \: metre} \\ \\ \bold{So,} \\ \\ \bold{Length \: \: of \: \: the \: \: rectangular \: \: hall = (a + 5) \: \: metre}$

$\bold{Given,} \\ \\ \bold{Area \: \: of \: \: the \: \: rectangular \: \: hall = 618.75\: \: m { }^{2} }$

$\bold{A.T.Q.,} \\ \\ \bold{Length \times Breadth = Area} \\ \\ \bold{ = > (a + 5) \times a = 618.75} \\ \\ \bold{ = > a {}^{2} + 5a = 618.75} \\ \\ \bold{ = >100 a {}^{2} + 500a = 61875 } \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{[Multiplying \: \: both \: \: sides \: \: by \: \: 100]} \\ \\ \bold{ = > 100a {}^{2} + 500a - 61875 = 0} \\ \\ \bold{ = > 4a {}^{2} + 20a - 2475 = 0} \\ \\ \bold{[Dividing \: \: both \: \: sides \: \: by \: \: 25]} \\ \\ \bold{ = > 4a {}^{2} + 110 a - 90a - 2475= 0} \\ \\ \bold{ = > 2a(2a + 55) - 45(2a + 55) = 0} \\ \\ \bold{ = > (2a - 45)(2a + 55) = 0}$

$\bold{So,} \\ \\ \bold{2a - 45 = 0 \: \: \: \: \: \: \: \: \: \: or , \: \: \: \: \: \: \: \: \: \: 2a + 55 = 0} \\ \bold{ = > a = \frac{45}{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = > \: a = \frac{ - 55}{2} } \\ \bold{ = > a = 22.5 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = > a = - 27.5} \\ \\ \\ \bold{Here,} \\ \\ \bold{a \: \: not \: \: equal \: \: to \: \: - 27.5} \\ \\ \bold{Thus, \: \: \: \: \: \: \: \: \: \: \: \: \: \: a = 22.5} \\ \\ \\ \bold{Hence,} \\ \\ \bold{Breadth \: \: of \: \: the \: \: rectangular \: \: hall = 22.5 \: \: m} \\ \\ \bold{And,} \\ \\ \bold{Length \: \: of \: \: the \: \: rectangular \: \: hall = (22.5 + 5) \: m} \\ \\ \bold{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 27.5 \: \: m}$

$\bold{Now,} \\ \\ \bold{Perimeter \: \: of \: \: the \: \: rectangular \: \: hall} \\ \\ = \bold{2(Length + Breadth)} \\ \\ \bold{ = 2(27.5 + 22.5) \: \: m} \\ \\ \bold{ =( 2 \times 50) \: \: m} \\ \\ \bold{ =100\: \: m}$

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✝ $\star{ \mathfrak{ \purple{ \: \: THANKS...}} \: \: \star}$ ✝