Question

Find the area of a trapezium whose parallel sides are 57 cm and 33 cm and the
distance between them is 13 cm.​

Answer 1

Answer:

585cm²

Explanation:

In a trapezium,

• Parallel sides - 57cm & 33cm.
• Distance between the parallel sides (height) is 13cm.

★ Formula :

• Area of a trapezium : ½(a + b)h

Where,

• (a + b) is the sum of parallel sides.
• ‘h' denotes the height.

On substituting the given values :

→ ½(57 + 33)13

→ ½ • 90 • 13

→ 45 • 13

→ 585

∴ Area of the trapezium is 585cm²

________________________

Answer 2

Given : The parallel sides are 57 cm and 33 cm and the distance between them is 13 cm.

Need to find: The Area of Trapezium.

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❍ The Formula for Area of Trapezium is given by :

⠀⠀⠀$\dag\;{\underline{\frak{As\;we\;know\;that,}}}$

$\star\;{\boxed{\sf{\pink{Area_{\;(trapezium)} = \dfrac{1}{2} \times (a + b) \times h}}}}$

Where,

• a and b are the two parallel sides and h is distance between two parallel sides or height of trapezium.⠀⠀⠀⠀

$\dag\;{\underline{\frak{Now,\: Substituting\:values\:in\;formula,}}}$

$:\implies\sf \dfrac{1}{2}(57 + 33) \times 13 \\\\\\:\implies\sf \dfrac{1}{\cancel {2}} (57 + 33 )\times \cancel {13}\\\\\\:\implies\sf (57 + 33) \times 6.5 \\\\\\:\implies\sf 90\times 6.5 \\\\\\:\implies{\underline{\boxed{\frak{\pink{Area = 585\;cm^{2}}}}}}\;\bigstar$

$\therefore{\underline{\sf{Hence, \;the\;Area\;of\: Trapezium \;\;is\;\bf{ 585\;cm^2}.}}}$..

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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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