Question

Area of a Trapezium is
132 sq. cm. If it's parallel
sides are 10 cm and 12
cm, then the distance
between the parallel sides
is:
​

Answer 1

Given : Area of a Trapezium is 132 sq. cm and it's parallel sides are 10 cm and 12 cm .

Need To Find : The distance between parallel sides or Height.

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❍ Let's consider Height of Trapezium be x .

$\underline {\frak{\dag{ As\:We\:know\:that\::}}}$

⠀⠀⠀⠀⠀$\underline {\boxed {\sf{ Area_{(Trapezium)} = \dfrac{1}{2}\times h ( a + b) \:sq.units}}}$

⠀⠀⠀⠀⠀Here h is the Height or Distance between the parallel sides of Trapezium in cm and a and b are two parallel sides of Trapezium in cm and we have given with the Area of Trapezium is 132 sq.cm .

⠀⠀⠀⠀⠀⠀$\underline {\bf{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}$

⠀⠀⠀⠀⠀$:\implies \sf{ 132cm^{2} = \dfrac{1}{2}\times x\times ( 10 + 12) }$

⠀⠀⠀⠀⠀$:\implies \sf{ 132cm^{2} = \dfrac{1}{\cancel{2}}\times x \times \cancel {22} }$

⠀⠀⠀⠀⠀$:\implies \sf{ 132cm^{2} = x \times 11 }$

⠀⠀⠀⠀⠀$:\implies \sf{ x= \dfrac{\cancel {132}}{\cancel {11}} }$

⠀⠀⠀⠀⠀$\underline {\boxed{\pink{ \mathrm {  x = 12\: cm}}}}\:\bf{\bigstar}$

Therefore,

• Height of Trapezium = x = 12 cm

⠀⠀⠀⠀⠀$\underline {\therefore\:{ \mathrm {  Height \:or\:Distance \:Between \:Parallel \:sides\:of\:Trapezium \:is\:\bf{12\: cm}}}}$

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$\large {\boxed{\sf{\mid{\overline {\underline {\star Verification \::}}}\mid}}}$

$\underline {\frak{\dag{ As\:We\:know\:that\::}}}$

⠀⠀⠀⠀⠀$\underline {\boxed {\sf{ Area_{(Trapezium)} = \dfrac{1}{2}\times h ( a + b) \:sq.units}}}$

⠀⠀⠀⠀⠀Here h is the Height or Distance between the parallel sides of Trapezium in cm and a and b are two parallel sides of Trapezium in cm and we have given with the Area of Trapezium is 132 sq.cm .

⠀⠀⠀⠀⠀⠀$\underline {\bf{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}$

⠀⠀⠀⠀⠀$:\implies \sf{ 132cm^{2} = \dfrac{1}{2}\times 12 \times ( 10 + 12) }$

⠀⠀⠀⠀⠀$:\implies \sf{ 132cm^{2} = \dfrac{1}{\cancel{2}}\times 12 \times \cancel {22} }$

⠀⠀⠀⠀⠀$:\implies \sf{ 132cm^{2} = 12 \times 11 }$

⠀⠀⠀⠀⠀$\underline {\boxed{\pink{ \mathrm {  132 cm^{2} = 132\: cm^{2}}}}}\:\bf{\bigstar}$

⠀⠀⠀⠀⠀$\therefore \underline {\bf {Hence \:Verified \:}}$

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$\large {\boxed{\sf{\mid{\overline {\underline {\star More\:To\:know\::}}}\mid}}}$

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

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