Question

The area of a rectangle is 181cm. if the length of the rectangle is 20.5cm,find the width of the rectangle

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Answer 1

Answer:

hey there here's your answer

Answer :-

=> (L × B) = ½ × A

=> (20.5 × B) = ½ (2 cut) 90.5

=> (20.5 × B) = 90.5

=> B = 90.5 ÷ 20.5

=> 5

=> Breadth = 5 sq.cm

hope this helps

Answer 2

Answer :

Width of the rectangle is 8.82 cm.

Solution :

Given :

• Area of a rectangle is 181 cm.
• Length of the rectangle is 20.5 cm.

To Find :

• Width of the rectangle.

$\red\bigstar\boxed{\sf Area_{Rectangle} = Length \times width }$

Substitute the values,

$\implies\sf 181 = 20.5 \times width \\ \\ \implies\sf width = \frac{181}{20.5} \\ \\ \implies \sf width = 8.829 cm$

Hence,

Width of the rectangle is 8.82 cm.

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More to know :

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

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