Question

find the area of the rectangle whose sides are 17/3 CM long and 13/4 CM wide.​

Answer 1

Answer:

18. 41 cm²

Step-by-step explanation:

$\purple\bigstar\huge\tt{GIVEN:} \:$

❒ Length of the rectangle = 17/3 cm.

❒ Width (breadth) of the rectangle = 13/4 cm.

$\purple\bigstar\huge\tt{TO \: FIND:} \:$

❒ Area of the rectangle

$\purple\bigstar\huge\tt{SOLUTION:} \:$

➛ Area of a rectangle = Length X Breadth

$\displaystyle{ \implies \: \frac{17}{3} \times \frac{13}{4} }$

$\displaystyle{ \implies \: \frac{221}{12} }$

$\displaystyle{ \implies \: 18.41}$

∴ The area of the rectangle is 18.41 cm².

$\purple\bigstar\huge\tt{MORE \: INFORMATION :} \:$

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