Question

Find the area of a trapezium whose sides are 12cm and 10cm and the distance between them is 5 cm

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Answer 1

Answer:

55 cm^2

pls refer the above image for stepwise solution

(the answer will be different if the sides aren't parallel sides)

Answer 2

Answer:

Diagram :

$\setlength{\unitlength}{1.5cm}\begin{picture}\thicklines\qbezier(0,0)(0,0)(1,2.2)\qbezier(0,0)(0,0)(4,0)\qbezier(3,2.2)(4,0)(4,0)\qbezier(1.5,2.2)(0,2.2)(3,2.2)\put(0.8,2.4){ \bf A }\put(3,2.4){ \bf D }\put(-0.3,-0.3){ \bf B }\put(4,-0.3){ \bf C }\put(4.4,0){\vector(0,0){2.2}}\put( 4.4, 0){\vector(0,-1){0.1}}\put(4.6,1){ \bf{5\ cm} }\put(0, -0.5){\vector(1,0){4}}\put(0, -0.5){\vector( - 1, 0){0.1}}\put(1.7, - 0.9){ \bf{12\ cm} }\put(0.8, 2.8){\vector(1,0){2.5}}\put(0.8, 2.8){\vector( - 1, 0){0.1}}\put(1.7, 3){ \bf{10\ cm} }\end{picture}$

Here's the diagram of trapezium. For clear understanding please visit to website Brainly.in.

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Solution :

As per the provided information in the question, we have :

• ✧ Sides of trapezium = 12cm and 10cm.
• ✧ Distance between the sides (Height) = 5 cm.

We need to find the area of trapezium.

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Here's the required formula to find the area of trapezium :

$\begin{gathered}{\implies{\small{\sf{Area_{(Trapezium)} = \dfrac{1}{2}\bigg\{sum \: of \: parallel \: sides \bigg\}height}}}}\end{gathered}$

Substituting all the given values in the formula to find area :

$\begin{gathered}{\implies{\small{\sf{Area_{(Trapezium)} = \dfrac{1}{2}\bigg\{sum \: of \: parallel \: sides \bigg\}height}}}}\\ \\{\implies{\small{\sf{Area_{(Trapezium)} = \dfrac{1}{2}\bigg\{ \: a + b \: \bigg\}height}}}}\\\\{\implies{\small{\sf{Area_{(Trapezium)} = \dfrac{1}{2}\bigg\{10 + 12\bigg\}5}}}}\\\\{\implies{\small{\sf{Area_{(Trapezium)} = \dfrac{1}{2}\bigg\{ \: \: 22 \: \: \bigg\}5}}}}\\\\{\implies{\small{\sf{Area_{(Trapezium)} = \dfrac{1}{2} \times 22 \times 5}}}}\\\\{\implies{\small{\sf{Area_{(Trapezium)} = \dfrac{1}{\cancel{2}} \times \cancel{22} \times 5}}}}\\\\{\implies{\small{\sf{Area_{(Trapezium)} = 11 \times 5}}}}\\\\{\implies{\small{\sf{\underline{\underline{\red{Area_{(Trapezium)} = 55 \: {cm}^{2}}}}}}}}\end{gathered}$

Hence, the area of trapezium is 55 cm².

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Learn More :

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

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