Question

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Answer 1

$\sf{ Solution} \\ \\ \sf{Let \: ABCD \: be \: the \: given \: trapezium \: in \: which} \\ \sf{AD=BC=CD=10 cm} \\ \\ \sf{AP= x \: cm} \\ \sf{∆APD=∆BQC} \\ \sf{QB= x \: cm ---①⑴} \\ \\ \sf{In ∆APD,} \\\sf{ DP = \sqrt{ {10}^{2} - {x}^{2} } \: (by \: Pythagoras \: theorem) } \\ \\ \sf \red{area \: of \: trapezium} \\\sf \green{a = \frac{1}{2} \times (sum \: of \: parallel \: sides) \times height} \\ \sf{= \frac{1}{2} \times (2x + 10 + 10) \times \sqrt{100 - {x}^{2} }} \: ーーー①ⅰ \\ \\ \sf{on \: differentiating \: both \: sides \: of \: eq \: ①} \\ \sf{we \: get} \\ \\ \sf{ \frac{dA}{dx} = (x + 10) \frac{( - 2x)}{2 \sqrt{100 - {x}^{2} } } + \sqrt{100 - {x}^{2} } } \\ \\ \sf{ \frac{ - 2 {x}^{2} - 10x + 100 }{ \sqrt{100 - {x}^{2}} } } \: \: ーーー①⑵\\ \\ \sf{for \: maximum \: put} \\ \sf{ \frac{dA}{dx} = 0} \\ \\ \sf{ \frac{ - 2 {x}^{2} - 10x + 100 }{ \sqrt{100 - {x}^{2} } } = 0} \\ \\ \sf{2(x + 10)(x - 5) = 0} \\ \\ \sf{x = 5 \: or \: x = - 10} \: ーーー① \\ \sf{Since, \: X \: represents \: distance \: cannot \: be \: negative.} \\ \sf{ \: So, \: we \: take \: x=5} \\ \\ \sf{On \: differentiating \: both \: sides \: of \: equation (ii)}\\ \\ \sf{by \: quotient \: rule} \\ \ \sf{ = \frac{ {\huge{[}}\frac{(100 - {x}^{2})( - 4x - 10)}{ - ( - 2 {x}^{2} - 10x + 100)( - x)} {\huge{]}} }{(100 - {x}^{2}) {}^{ \frac{3}{2} } } } \\ \\ \sf{ = \frac{2 {x}^{3} - 300x - 1000 }{(100 - {x}^{2}) {}^{ \frac{3}{2} } }} \:ーーー① \\ \sf{at \: x = 5} \\ \\ \sf{ \frac{ {d}^{2}a }{d {x}^{2} } = \frac{2(5) {}^{3} - 300(5) - 1000 }{((100 - (5) {}^{2}) {}^{ \frac{3}{2} } } } \\ \\ \sf {= \frac{250 - 1500 - 1000}{(100 - 25) {}^{ \frac{3}{2} } } = \frac{ - 2250}{75 \sqrt{75} } < 0} \\ \\ \sf{Thus, area \: of trapezium \: is \: maximum \: at \: x=5 } \\ \sf{and \: minimum \: value \: is:} \\ \\ \sf{A_{max}} = (5 + 10) \sqrt{100 - (5) {}^{2} } \\ \sf{put \: x = 5 \: in \: eq. \: i} \\ \sf{(15 \sqrt{100 - 25)} = 15 \sqrt{75} } \\ \sf{75 \sqrt{3} {cm}^{2} }ーーー①$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \huge\sf{@WildCat7083 }$