Question

the area of trapezium is 900cmsq the distance between the parallel sides is 24cm if one of the parellel sides is double the other find the length of 100 parellel sides​

Answer 1

☯︎ Aɴsᴡᴇʀ

we have,

• area of trapezium = 900 cm²
• height of trapezium = 24 cm
• one of the parallel side is double the other

• let one parallel side of the trapezium be x

then, other parallel side is 2x

we know that,if

• two parallel sides are ‘a’ and ‘b’ and distance between them is ‘h’ then,

$\underline{ \boxed{ \bold{area \: of \: trapezium = \frac{1}{2}(a + b) \times h }}}$

here,

• a = x
• b = 2x
• h = 24 cm
• Area of trapezium = 900 cm²

$\bold{: \implies 900 = \frac{1}{ \cancel{2} } (x + 2x) \times \cancel{ 24}} \\ \\ \bold{: \implies 900 = 3x \times 12 } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ : \implies \: 36x = 900}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ : \implies x = \cancel{ \frac{900}{36}} }\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ : \implies \boxed{ \bold{\: x = 25 } }} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

so, length of one parallel side is 25 cm

then, length of other side is 2 × 25 = 50 cm

therefore, length of two parallel sides of the trapezium are 25 cm and 50 cm

Answer 2

$\sf Given \begin{cases} & \sf{Area\:of\:trapezium = \bf{900\;cm^2}} \\ & \sf{Distance\: between\: parallel\:sides\:(Height) = \bf{24\;cm}} \end{cases}$

To find: Length of parallel sides.

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀⠀

☯ Let one parallel side of trapezium be x cm.

Then, another parallel side will be 2x cm.

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⋆ Reference to image is shown in diagram

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$\setlength{\unitlength}{1.25cm}\begin{picture}(0,0)\thicklines\qbezier(0,0)(0,0)(1,2.2)\qbezier(0,0)(0,0)(4,0)\qbezier(3,2.2)(4,0)(4,0)\qbezier(1.5,2.2)(0,2.2)(3,2.2)\put(0.8,2.4){ \bf A }\put(3,2.4){ \bf D }\put(-0.3,-0.3){ \bf B }\put(4,-0.3){ \bf C }\put(4.4,0){\vector(0,0){2.2}}\put( 4.4, 0){\vector(0,-1){0.1}}\put(4.6,1){ \bf 24 \:cm }\put(0, -0.5){\vector(1,0){4}}\put(0, -0.5){\vector( - 1, 0){0.1}}\put(1.7, - 0.9){ \bf 2x \:cm }\put(0.8, 2.8){\vector(1,0){2.5}}\put(0.8, 2.8){\vector( - 1, 0){0.1}}\put(1.7, 3){ \bf x \: cm }\end{picture}$

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

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

where,

• "a" and "b" are two parallel sides and "h" is height or distance between two parallel sides.

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$:\implies\sf \dfrac{1}{2} \times (x + 2x) \times 24 = 900\\ \\ \\ :\implies\sf \dfrac{1}{ \cancel{2}} \times 3x \times \cancel{24} = 900\\ \\ \\ :\implies\sf 3x \times 12 = 900\\ \\ \\ :\implies\sf 3x = \cancel{ \dfrac{900}{12}}\\ \\ \\ :\implies\sf 3x = 75\\ \\ \\ :\implies\sf x = \cancel{ \dfrac{75}{3}}\\ \\ \\ :\implies{\underline{\boxed{\frak{\purple{x = 25}}}}}\;\bigstar$

Therefore,

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• $\sf First\: side\: of\: trapezium,\: x = \bf{25\:cm}$
• $\sf Second\: side\: of \:trapezium, \:2x = 2 \times 25 = \bf{50\;cm}$

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$\therefore\:{\underline{\sf{Hence,\; length\:of\;two\: parallel\:sides\;are\: \bf{25\;cm}\;and\;\bf{50\:cm}}}}.$