Question

5. Find the area of the square whose side is thrice of 'a' . A. 9a 2sq. Unit
B. 9a sq. Unit
C. a 2 sq. Unit
D. 4a2sq.Unit

Answer 1

Given : The side of square is is thrice of a .

Need To Find : Area of Square.

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❍ The side of square is is thrice of a .

Then ,

• Side of a square is 3a .

Now ,

$\frak{\underline {\dag As, \:We \:know\:that \::}}$

$\qquad \qquad \underline {\boxed {\sf{\bigstar Area_{(Square)} = Side \times Side \:or\:Side^{2}\:sq.units}}}$

⠀⠀⠀⠀⠀⠀$\underline {\bf{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}$

$\qquad \qquad :\implies \sf{ Area_{(Square)} = (3a)^{2}}$

⠀⠀⠀⠀⠀$\underline {\boxed{\pink{ \mathrm {  Area_{(Square)} = 9a^{2}\: sq.units}}}}\:\bf{\bigstar}$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm { Area \:of\:Square \:is\:\bf{9a^{2}\: sq.units}}}}$

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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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