Question

Find the length of a rectangle whose area is 246 sq cm and breadth is 6cm

Answer 1

Answer:

Area = 246cm2

Breadth = 6cm

SO length = Area/Base = 246/6 = 41cm

as Area = L × B

Answer 2

Step-by-step explanation:

Question :-

• Find the length of a rectangle whose area is 246 cm² and breadth is 6 cm.

To Find :-

• Length of the rectangle.

Solution :-

Given :

• Area = 246 cm²
• Breadth = 6 cm

By using the formula,

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

Where,

• l = length
• b = breadth

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

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

${\sf{\longrightarrow 246\ =\ l \times 6}}$

${\sf{\longrightarrow \dfrac{246}{6}\ =\ l}}$

${\sf{\longrightarrow 41\ =\ l}}$

${\sf{\longrightarrow l\ =\ 41\ cm}}$

$\therefore$ Hence, length of the rectangle is 41 cm.

More To Know :-

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