Question

find the area of the trapezium whose parallel sides are 7 cm 9 cm and distance between them is 8 cm​

Answer 1

Given : The Parallel sides are 7 cm 9 cm and distance between them is 8 cm .

To Find : Find the Area of the Trapezium

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SolutioN :

$\maltese$ Formula Used :

• ${\underline{\boxed{\pmb{\sf{ Area{\small_{(Trapezium)}} = \dfrac{1}{2} \times \bigg( a + b \bigg) \times Height }}}}}$

Where :

• a = 1st Parallel Side
• b = 2nd Parallel Side

$\maltese$ Calculating the Area :

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { Area = \dfrac{1}{2} \times \bigg( a + b \bigg) \times Height } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { Area = \dfrac{1}{2} \times \bigg( 7 + 9 \bigg) \times 8 } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { Area = \dfrac{1}{2} \times 16 \times 8 } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { Area = \dfrac{1}{2} \times 128 } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { Area = \dfrac{128}{2} } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { Area = \cancel\dfrac{128}{2} } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; {\underline{\boxed{\purple{\pmb{\frak{ Area = 64 \; {cm}^{2} }}}}}} \; {\orange{\bigstar}} \\ \\ \\ \end{gathered}$

$\qquad {\pink{\leadsto}}$ Area of the Trapezium is 64 cm² .

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Answer 2

Question :-

• Find the Area of the Trapezium, whose parallel sides are 7 cm || 9 cm and Distance between them is 8 cm.

Answer :-

• Area of Trapezium is 64 cm² .

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Given :-

• Parallel sides = 7 cm || 9 cm
• Distance = 8 cm

To Find :-

• Area of Trapezium = ?

Solution :-

• Here, Parallel sides are given 7 cm || 9 cm . Distance is 8 cm . And, we have to find Area of Trapezium .

➻ Formula Required :-

• $\sf{ Area \: of \: Trapezium = \frac{1}{2} \times (A + B) \times Height }$

➻ Where ,

• AB denotes to Parallel Sides .

➻ By substituting the values :-

$\dag \: \: \sf{ Area \: _{Trapezium} = \frac{1}{2} \: \times \: (A + B) \: \times \: Height }$

$\Longrightarrow \: \: \sf{Area \: _{Trapezium} = \frac{1}{2} \: \times \: (7 + 9) \: \times \: 8 }$

$\Longrightarrow \: \: \sf{Area \: _{Trapezium} = \frac{1}{2} \: \times \: 16 \: \times \: 8 }$

$\Longrightarrow \: \: \sf{Area \: _{Trapezium} = \frac{1}{1} \: \times \: 8 \: \times \: 8 }$

$\Longrightarrow \: \: \sf{Area \: _{Trapezium} = 8 \: \times \: 8}$

$\Longrightarrow \: \: \bf{Area \: _{Trapezium} = 64 \: {cm}^{2} }$

Hence :-

• Area = 64 cm² .

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$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c} \\ \underline{ { \pmb {\sf \red{ \dag \: \: More \: Formulas \: \: \dag}}}} \\ \\ \\ \footnotesize \bigstar \: \sf{Area \: of \: Square = Side \times Side} \\ \\ \\ \footnotesize\bigstar \: \sf{Area \: of \: Rectangle = Lenght \times Breadth} \\ \\ \\ \footnotesize \bigstar \: \sf{Area \: of \: Triangle = \frac{1}{2} \times Base \times Height } \\ \\ \\ \footnotesize \bigstar \: \sf{Area \: of \: Parallelogram = Base \times Height} \\ \\ \\ \footnotesize \bigstar \: \sf{Area \: of \: Trapezium = \frac{1}{2} \times [ \: A + B \: ] \times Height } \\ \\ \\ \footnotesize \bigstar \: \sf {Area \: of \: Rhombus = \frac{1}{2} \times Diagonal \: 1 \times Diagonal \: 2}\end{array}}\end{gathered}\end{gathered}$