Question

The are a of trapezium is 204cm² and its height is 12cm.

If one of the parallel sides is 15cm, find the length of other

parallel side​

Answer 1

Answer:

Let b be the length of the other parallel side

Then

a=15 cm h=12cm

area of trapezium=204

1/2(a+b)*h=204

1/2(15+b)*12=204

180+12b/2=204

180+12b=204*2

180+12b=408

12b=408-180

12b=228

b=228/12

b=19 cm

Answer 2

Answer:

${\large{\underline{\underline{\bf{\green{Given\:} : -}}}}}$

• ⟶ Area of trapezium = 204 cm².
• ⟶ Height of trapezium = 12 cm.
• ⟶ One parallel sides of trapezium = 15 cm

$\begin{gathered}\end{gathered}$

${\large{\underline{\underline{\bf{\green{To \: Find}\: : -}}}}}$

• ⟶ Length of other parallel side of trapezium.

$\begin{gathered}\end{gathered}$

${\large{\underline{\underline{\bf{\green{Concept}\: : - }}}}}$

⟶ Here the concept of Area of Trapezium has been used. We are given that Area of trapezium is 204 cm², height of cylinder is 12 cm and one parallel side if trapezium is 15 cm. We need to find the length of Rectangle.

⟶ So, we'll find the other parallel side of trapezium by insert the values in the required  formula.

$\begin{gathered}\end{gathered}$

${\large{\underline{\underline{\bf{\green{Using \: Formula}\: : -}}}}}$

• ⟶ Area of trapezium = ½(a + b)h

$\pink\bigstar$ Where

• ➱ a and b = bases of trapezium
• ➱ h = height (the perpendicular distance between a and b)

$\begin{gathered}\end{gathered}$

${\large{\underline{\underline{\bf{\green{Solution}\: : - }}}}}$

$\pink\bigstar$ Finding other parallel side of trapezium

${\dashrightarrow{\sf{Area \: of \: trapezium = \dfrac{1}{2} \bigg(a + b \bigg)h}}}$

• Substuting the values

${\dashrightarrow{\sf{204 \: {cm}^{2} = \dfrac{1}{2} \bigg(15 + b \bigg)12}}}$

${\dashrightarrow{\sf{204 \times 2 = \bigg(15 + b \bigg)12}}}$

${\dashrightarrow{\sf{408 = \bigg(15 + b \bigg)12}}}$

${\dashrightarrow{\sf{408 = \bigg(15 \times 12+ b \times 12 \bigg)}}}$

${\dashrightarrow{\sf{408 = \bigg(180 + 12b \bigg)}}}$

${\dashrightarrow{\sf{ 12b = 408 - 180}}}$

${\dashrightarrow{\sf{ 12b = 228}}}$

${\dashrightarrow{\sf{b = \dfrac{228}{12}}}}$

${\dashrightarrow{\sf{b = \cancel{\dfrac{228}{12}}}}}$

${\dashrightarrow{\sf{b = 19 \: cm}}}$

$\bigstar\red{\underline{\boxed{\bf{Other \: parallel \: side \: of \: trapezium = 19 \: cm}}}}$

∴ The lenght of other parallel side of trapezium is 19 cm.

$\begin{gathered}\end{gathered}$

${\large{\underline{\underline{\bf{Verification\: : - }}}}}$

$\pink\bigstar$ Checking our answer

${\dashrightarrow{\sf{Area \: of \: trapezium = \dfrac{1}{2} \bigg(a + b \bigg)h}}}$

• Substuting the values

${\dashrightarrow{\sf{204= \dfrac{1}{2} \bigg(15+ 19 \bigg)12}}}$

${\dashrightarrow{\sf{204= \dfrac{1}{2} \bigg( \: 34 \: \bigg)12}}}$

${\dashrightarrow{\sf{204= \dfrac{1}{2} \times 34 \times 12}}}$

${\dashrightarrow{\sf{204= \dfrac{1}{\cancel{2}} \times 34 \times \cancel{12}}}}$

${\dashrightarrow{\sf{204=34 \times 6}}}$

${\dashrightarrow{\sf{204 \: {cm}^{2} =204 \: {cm}^{2} }}}$

$\bigstar\red{\underline{\boxed{\bf{LHS = RHS}}}}$

∴ Hence Verified!

$\begin{gathered}\end{gathered}$

${\large{\underline{\underline{\bf{\green{Diagram } \: : - }}}}}$

$\setlength{\unitlength}{1.5cm}\begin{picture}\thicklines\qbezier(0,0)(0,0)(1,2.2)\qbezier(0,0)(0,0)(4,0)\qbezier(3,2.2)(4,0)(4,0)\qbezier(1.5,2.2)(0,2.2)(3,2.2)\put(0.8,2.4){ \bf A }\put(3,2.4){ \bf D }\put(-0.3,-0.3){ \bf B }\put(4,-0.3){ \bf C }\put(4.4,0){\vector(0,0){2.2}}\put( 4.4, 0){\vector(0,-1){0.1}}\put(4.6,1){ \bf 12\ cm }\put(0, -0.5){\vector(1,0){4}}\put(0, -0.5){\vector( - 1, 0){0.1}}\put(1.7, - 0.9){ \bf 19\ cm  }\put(0.8, 2.8){\vector(1,0){2.5}}\put(0.8, 2.8){\vector( - 1, 0){0.1}}\put(1.7, 3){ \bf 15\ cm  }\end{picture}$

$\begin{gathered}\end{gathered}$

${\large{\underline{\underline{\bf{\green{Learn \: More}\: : - }}}}}$

$\pink\bigstar$ Formulas of area :

$\boxed{\begin {minipage}{9cm}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {minipage}}$

$\begin{gathered}\end{gathered}$

${\large{\underline{\underline{\bf{\green{Request}\: : - }}}}}$

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