Question

the area of a Trapezium is 156 square CM if one of the parallel side is 9 cm and the distance between them is 12 cm find the length of the other parallel side ​

Answer 1

$\textbf{\huge{\underline{Solution:}}}$

Let ABCD be a trapezium with AB and CD as parallel sides and AM as height.

$\textbf{\underline{\underline{Given\:In\:The\:Question:}}}$

$\sf \: Area = {156cm}^{2}$

$\sf \: One \: parallel \: side = 9cm$

$\sf \: Height = 12cm$

Let one parallel side be 'b' and other as 'a'

$\textbf{\underline{\underline{Using\:Formula :}}}$

$\ \sf \: Area = \dfrac{1}{2} \times (a + b) \times h$

$\hookrightarrow \sf \: 156 = \dfrac{1}{2} \times (a + 9) \times 12$

$\hookrightarrow\sf \: 156 = \dfrac{1}{2} \times (a + 9) \times 12$

$\hookrightarrow \sf 156 = \dfrac{1}{ \cancel2} \times (a + 9) \times \cancel 12$

$\hookrightarrow \sf156 = 6(a + 9)$

$\sf \hookrightarrow 156 = \: 6a + 54$

$\sf \hookrightarrow \: 6a = 156 - 54$

$\sf \hookrightarrow \: 6a = 102$

$\sf \hookrightarrow \: a = \cancel\dfrac{102}{6}$

$\sf \hookrightarrow \: a = 17$

Hence,Length of other parallel side is 17 cm.

$\textbf{\huge{\underline{Verification:}}}$

$\textbf{\underline{\underline{Given\:In\:The\:Question:}}}$

$\sf \: One \: Parallel \: Side = 9cm$

$\sf \: Another \: Parallel \: Side = 17cm$

$\sf \: Height = 12cm$

To Check Whether Area is 156 cm^2

$\tt\: Area \: of \: Trapezium = \dfrac{1}{2} (a + b) \times h$

$\sf \: Area = \dfrac{1}{ \cancel2} (9 + 17) \times \cancel 12$

$\sf \: Area = 26 \times 6$

$\sf \: Area = 156 {cm}^{2}$

Hence,Verified!

Answer 2

$\huge{\underline{\underline{\red{\mathfrak{Answer :}}}}}$

$\displaystyle{\green {\mathsf{Given \: \begin{cases} Area \: of \: Trapezium \: = \: 156 \: cm^2 \\\\ One \: of \: Parallel \: Side \: is \: 9 \: cm \\\\ Height \: is \: 12 \: cm</p><p> \end{cases}}}}$

______________________________

To Find :

Other Parallel side

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Solution :

let one side be x and other be y,

Area of Trapezium = 156 cm²

x = 9 cm

h = 12 cm

We have formula for Area of trapezium :

$\Large \displaystyle \longrightarrow {\underline{\boxed{\blue{\sf{\frac{1}{2} \: (x \: + \: y) \: \times \: h}}}}}$

Put Values

⇒ 156 = ½ (9 + y) *12

⇒156 = (9 + y)*6

⇒156 = 54 + 6y

⇒156 - 54 = 6y

⇒6y = 102

⇒y = 102/6

⇒y = 17

$\Large \displaystyle {\boxed{\red{\sf{Other \: parallel \: side \: = \: 17 \: cm}}}}$