Question

The area of a trapezium shaped field is 600m2 . If the height of the trapezium is 24m and one its parallel sides is of length 20m, find the length of the other parallel side.

Answer 1

Given:-

• Area of trapezium shaped field is 600m².

• Height of trapezium is 24 m.

• Length of one of its parallel side is 20 m.

To Find:-

• Length of the other parallel side.

Solution:-

Here, Area of Trapezium = $\dfrac{1}{2}\times height \times ( sum\ of \ parallel \ sides)$

Let the another parallel side be x m.

So, Putting values we get,

⇒ 600 m² =  $\dfrac{1}{2}\times 24 \ m \times ( 20 \ m + x \ m )$

⇒ $\dfrac{600 \ m^{2} \times 2}{24 \ m} = ( 20 \ m + x \ m )$

⇒ $50 \ m - 20 \ m = x$

⇒ $x = 30 \ m$

Hence, Other parallel side of trapezium is 30 m.

Some Important Terms:-

• Area of Triangle = $\dfrac{1}{2}\times base \times height$

• Area of square = $Side^{2}$

• Area of Rectangle = $Length \times Breadth$

Answer 2

$\frak{Given} \begin{cases} \sf Area\:of\:trapezium\:shaped\:field\: = \frak{600\:m^2} & \\ \\ \sf Height\:of\:trapezium\:shaped\:field\: = \frak{24\:m}& \\ \\ \sf One\:of\:the\:parallel\:side\:of\:field\: = \frak{20\:m}&\end{cases}$

$\frak{To\:find\::}$ Length of other parallel side of trapezium shaped field.

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☯ Let's consider the other parallel side of trapezium shaped field be x m.

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⋆ DIAGRAM

$\setlength{\unitlength}{1.1cm}\begin{picture}(0,0)\thicklines\qbezier(0,0)(0,0)(1,2.2)\qbezier(0,0)(0,0)(4,0)\qbezier(3,2.2)(4,0)(4,0)\qbezier(1.5,2.2)(0,2.2)(3,2.2)\put(0.8,2.4){ \bf A }\put(3,2.4){ \bf D }\put(-0.3,-0.3){ \bf B }\put(4,-0.3){ \bf C }\put(4.4,0){\vector(0,0){2.2}}\put( 4.4, 0){\vector(0,-1){0.1}}\put(4.6,1){ \bf 24\ m }\put(0, -0.5){\vector(1,0){4}}\put(0, -0.5){\vector( - 1, 0){0.1}}\put(1.7, - 0.9){ \bf x\ m }\put(0.8, 2.8){\vector(1,0){2.5}}\put(0.8, 2.8){\vector( - 1, 0){0.1}}\put(1.7, 3){ \bf 20\ m }\end{picture}$

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

The Area of trapezium is given by,

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

Where,

• a and b are two parallel sides of trapezium.
• And, h is the height or distance between two parallel sides of trapezium.

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$\dag\:{\underline{\frak{Substituting\:given\:values\:in\:formula,}}}\\\\\\ :\implies\sf 600 = \dfrac{1}{\cancel{2}} \times (20 + x) \times \cancel{24}\\\\\\ :\implies\sf 600 = (20 + x) \times 12\\\\\\ :\implies\sf (20 + x) = \cancel{\dfrac{600}{12}}\\\\\\ :\implies\sf (20 + x) = 50\\\\\\ :\implies\sf x = 50 - 20\\\\\\ :\implies{\boxed{\frak{\pink{x = 30}}}}\:\bigstar$

$\therefore{\underline{\sf{Hence,\:the \:length\: of\:other\:parallel\:side\:is\:{\pmb{30\:m}}.}}}$