Question

Find the area of a rectangle whose length and breadth are 10 cm and 8 cm respectively.

Answer 1

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Question :-

• Find the area of rectangle whose length is 10 cm and breadth is 8 cm.

To Find :-

• Area of rectangle.

Solution :-

Given :

• Length = 10 cm
• Breadth = 8 cm

By using the formula,

${\longrightarrow{\sf Area\ of\ rectangle\ =\ l \times b}}$

Where,

• l = length of the rectangle
• b = breadth of the rectangle

According to the question, by using the formula, we get :

${\longrightarrow{\sf Area\ of\ rectangle\ =\ l \times b}}$

${\longrightarrow{\sf 10 \times 8}}$

${\longrightarrow{\sf 80\ cm^2}}$

∴ Hence, area of rectangle is 80 cm².

More To Know :-

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