Question

will mark brainliest who will first answer​

Answer 1

$\sf Given \begin{cases} & \sf{Area\:of\: trapezium\:shaped\:field = \bf{480\:m^2}} \\ & \sf{One\: parallel\: side\:of\:field = \bf{20\:m}} \\ & \sf{Distance\:between\:two\: parallel\:sides = \bf{15\:m}}\end{cases}$

To find: Length of other parallel side?

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☯ Let the length of other parallel side of trapezium shaped field be x m.

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$\setlength{\unitlength}{1.3cm}\begin{picture}(0,0)\thicklines\qbezier(0,0)(0,0)(1,2.2)\qbezier(0,0)(0,0)(4,0)\qbezier(3,2.2)(4,0)(4,0)\qbezier(1.5,2.2)(0,2.2)(3,2.2)\put(0.8,2.4){ \bf A }\put(3,2.4){ \bf D }\put(-0.3,-0.3){ \bf B }\put(4,-0.3){ \bf C }\put(4.4,0){\vector(0,0){2.2}}\put( 4.4, 0){\vector(0,-1){0.1}}\put(4.6,1){ \bf 15\ m }\put(0, -0.5){\vector(1,0){4}}\put(0, -0.5){\vector( - 1, 0){0.1}}\put(1.7, - 0.9){ \bf x\ m }\put(0.8, 2.8){\vector(1,0){2.5}}\put(0.8, 2.8){\vector( - 1, 0){0.1}}\put(1.7, 3){ \bf 20\ m }\end{picture}$

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Now,

$\dag\;{\underline{\frak{As\;we\;know\;that,}}}$

⋆ Area of trapezium is given by,

$\star\;{\boxed{\sf{\pink{Area_{\;(trapezium)} = \dfrac{1}{2} \times (a + b) \times h}}}}$

where,

a and b are length of two parallel sides and h is distance between two parallel sides or height of trapezium.

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$\dag\;{\underline{\frak{Now,\: Putting\:values\:in\;formula,}}}$

$:\implies\sf \dfrac{1}{2} \times (20 + x) \times 15 = 480\\ \\ \\ :\implies\sf (20 + x) \times 15 = 480 \times 2\\ \\ \\ :\implies\sf (20 + x) \times 15 = 960\\ \\ \\ :\implies\sf (20 + x) = \cancel{\dfrac{960}{15}}\\ \\ \\ :\implies\sf (20 + x) = 64\\ \\ \\ :\implies\sf x = 64 - 20\\ \\ \\ :\implies{\underline{\boxed{\frak{\purple{x = 44\:m}}}}}\;\bigstar$

$\therefore\:{\underline{\sf{The\: length\:of\:other\: parallel\:side\:is\: {\textsf{\textbf{44\:m}}}.}}}$