Question

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Answer 1

$\large\boxed{\mathtt\red{Given:-}}$

◑ Area of the Rhombus = 600 cm²

◑ Ratios of Diagonals = 3:4

$\large\boxed{\mathtt\blue{Assumption:-}}$

◑ Let the length of Diagonals be 3x and 4x

$\large\boxed{\mathtt\green{Solution:-}}$

we know that :-

✏ Area of Rhombus = $\large\frac{Diagonal ₁ × Diagonal₂}{2}$

➽ Now, Substituting the value:-

✒ 600 cm² = $\large\frac{3x×4x}{2}$

✒ 600 cm² = 6x²

✒ x² = $\large\frac{600\:cm²}{6}$

✒ x = $\sqrt{100\:cm²}$

∴ x = 10 cm

◑ Diagonal ₁ = 3x = 30 cm

◑ Diagonal ₂ = 4x = 40 cm

✏ We know that :-

☞ Perimeter of Rhombus = ${2{\sqrt{D₁+D₂}}}$

✒ Perimetre = 2$\sqrt{2500\:cm²}$

✒ Perimetre = 2×50cm

$\small\boxed{\mathtt\red{∴ Perimeter \:of \:Rhombus=100cm}}$

∴ The perimeter of the Rhombus is 100 cm

Answer 2

Answer:

${\large{\underline{\underline{\bf{Question\: : - }}}}}$

• ⟶ The area of rhombus is 600 cm². The diognals of the rhombus are in the ratio 3:4. Find the perimeter of the rhombus.

$\begin{gathered}\end{gathered}$

${\large{\underline{\underline{\bf{Given\: : - }}}}}$

• ⟶ Area of rhombus = 600 cm².
• ⟶ Ratio of rhombus diognals = 3:4

$\begin{gathered}\end{gathered}$

${\large{\underline{\underline{\bf{To \: Find\: : - }}}}}$

• ⟶ Perimeter of rhombus

$\begin{gathered}\end{gathered}$

${\large{\underline{\underline{\bf{Using \: Formulae\: : - }}}}}$

$\bigstar{\red{\underline{\boxed{\bf{Area \: of \: rhombus = \dfrac{1}{2} \times d_1 \times d_2}}}}}$

$\bigstar{\red{\underline{\boxed{\bf{Perimeter \: of \: rhombus=2\sqrt{{(d_1)}^{2} + {(d_2)}^{2} } }}}}}$

$\green\bigstar$ Where

• ➱ $\rm{d_1}$ = length of diagonal 1
• ➱ $\rm{d_2}$ = length of diagonal 2

$\begin{gathered}\end{gathered}$

${\large{\underline{\underline{\bf{Diagram\: : - }}}}}$

$\green\bigstar$ Rhombus with diagonal

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\put(0,0){\line(1,3){1.5}}\put(0,0){\line(1,0){5}}\put(5,0){\line(1,3){1.5}}\put(1.5,4.5){\line(1,0){5}}\qbezier(1.56,4.5)(1.56,4.5)(5,0)\qbezier(6.45,4.5)(6.45,4.5)(0,0)\put(-0.5,-0.5){\sf B}\put(1,4.8){\sf A}\put(5.2,-0.5){\sf C}\put(6.7,4.75){\sf D}\put(3,1.6){\sf O}\end{picture}$

$\begin{gathered}\end{gathered}$

${\large{\underline{\underline{\bf{Solution\: : - }}}}}$

$\green\bigstar$ Let the,

• ⟶ $\rm{d_1}$ = 3x
• ⟶ $\rm{d_2}$ = 4x

$\begin{gathered}\end{gathered}$

$\green\bigstar$ Finding the diognals of rhombus

${\dashrightarrow{\sf{Area_{(Rhombus)} = \dfrac{1}{2} × d_1 × d_2}}}$

• Substuting the values

${\dashrightarrow{\sf{600 = \dfrac{1}{2} \times 3x \times 4x}}}$

${\dashrightarrow{\sf{600 = \dfrac{1}{2} \times 12 \: {x}^{2} }}}$

${\dashrightarrow{\sf{600 \times 2 = 12 \: {x}^{2} }}}$

${\dashrightarrow{\sf{1200= 12 \: {x}^{2} }}}$

${\dashrightarrow{\sf{\dfrac{1200}{12} = {x}^{2}}}}$

${\dashrightarrow{\sf{ \cancel{\dfrac{1200}{12}} = {x}^{2}}}}$

${\dashrightarrow{\sf{100= {x}^{2}}}}$

${\dashrightarrow{\sf{x = \sqrt{100}}}}$

${\dashrightarrow{\sf{x = \sqrt{10 \times 10}}}}$

${\dashrightarrow{\sf{x = 10}}}$

$\bigstar{\purple{\underline{\boxed{\bf{x = 10}}}}}$

$\begin{gathered}\end{gathered}$

$\green\bigstar$ Thus,

• Length of 1 diagonal = 3 × 10 = 30 cm
• Length of 2 diagonal = 4 × 10 = 40 cm.

∴ The diognals of rhombus are 30 cm and 40 cm.

$\begin{gathered}\end{gathered}$

$\green\bigstar$ Now, Finding the perimeter of rhombus

${\dashrightarrow{\sf{Perimeter_ {(Rhombus)} = 2\sqrt{{(d_1)}^{2} + {(d_2)}^{2}}}}}$

• Substuting the values

${\dashrightarrow{\sf{Perimeter_ {(Rhombus)} = 2\sqrt{{(30)}^{2} + {(40)}^{2}}}}}$

${\dashrightarrow{\sf{Perimeter_ {(Rhombus)} = 2\sqrt{{(30 \times 30)} + {(40 \times 40)}}}}}$

${\dashrightarrow{\sf{Perimeter_ {(Rhombus)} = 2\sqrt{{(900)} + {(1600)}}}}}$

${\dashrightarrow{\sf{Perimeter_ {(Rhombus)} = 2\sqrt{(900+1600)}}}}$

${\dashrightarrow{\sf{Perimeter_ {(Rhombus)} = 2\sqrt{(2500)}}}}$

${\dashrightarrow{\sf{Perimeter_ {(Rhombus)} = 2\sqrt{(50 \times 50)}}}}$

${\dashrightarrow{\sf{Perimeter_ {(Rhombus)} = 2 \times 50}}}$

${\dashrightarrow{\sf{Perimeter_ {(Rhombus)} = 100 \: cm }}}$

$\bigstar{\purple{\underline{\boxed{\bf{Perimeter \: of \: rhombus = 100 \: cm }}}}}$

∴ The perimeter of rhombus is 100 cm.

$\begin{gathered}\end{gathered}$

${\large{\underline{\underline{\bf{Learn \: More\: : - }}}}}$

$\green\bigstar$ Formulas of area

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

$\begin{gathered}\end{gathered}$

${\large{\underline{\underline{\bf{Request\: : - }}}}}$

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