Question

The area of a rhombus is 60 cm². One diagonal is 10 cm. The other diagonal is
(a) 6cm
(b) 12cm
(c) 3cm
(d)24cm

Answer 1

Given :

• Area = 60 cm²
• Diagonal 1 = 10 cm

To Find :

• Diagonal 2 = ?

$\\ \qquad{\rule{200pt}{2pt}}$

SolutioN :

$\maltese$ Formula Used :

• ${\underline{\boxed{\pmb{\sf{ Area{\small_{(Rhombus)}} = \dfrac{1}{2} \times \bigg| Diagonal \; 1 \bigg| \times \bigg| Diagonal \; 2 \bigg| }}}}}$

$\maltese$ Calculating the Diagonal 2 :

$\begin{gathered} \qquad \; \longrightarrow \; \sf { Area = \dfrac{1}{2} \times \bigg| Diagonal \; 1 \bigg| \times \bigg| Diagonal \; 2 \bigg| } \\ \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \sf { 60 = \dfrac{1}{2} \times 10 \times Diagonal \; 2 } \\ \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \sf { 60 = \dfrac{10}{2} \times Diagonal \; 2 } \\ \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \sf { 60 = \cancel\dfrac{10}{2} \times Diagonal \; 2 } \\ \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \sf { 60 = 5 \times Diagonal \; 2 } \\ \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \sf { \dfrac{60}{5} = Diagonal \; 2 } \\ \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \sf { \cancel\dfrac{60}{5} = Diagonal \; 2 } \\ \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; {\pmb{\underline{\boxed{\red{\sf{ Diagonal \; 2 = 12 \; cm }}}}}} \\ \\ \\ \\ \end{gathered}$

$\therefore \;$ Second Diagonal of the Rhombus is 12 cm .

$\\ \qquad{\rule{200pt}{2pt}}$

Answer 2

Step-by-step explanation:

According to question:-

Given that:-

• $\large\underline{\sf{Area \: of \: rhombus=60cm²}}$
• $\large\underline{\sf{Diagonal_{1}=10cm}}$

To find:-

• $\large\underline{\sf{ Diagonal_{2} = ?}}$

Formula used:-

$\large \underline{\boxed{\sf{{ \red{\implies \frac{1}{2} \times d_{1} \times d_{2} }}}}}$

Solution:-

$\large \underline{\sf{ \implies \frac{1}{2} \times d_{1} \times d_{2} = area}}$

On substituting values we get:-

$\large\underline{\sf{ \implies \frac{1}{2} \times 10cm \times d_{2} = {60cm}^{2} }}$

$\large\underline{\sf{ \implies 5cm \times d_{2} = {60cm}^{2} }}$

$\large\underline{\sf{ \implies d_{2} = \frac{60 {cm}^{2} }{5cm} }}$

$\large \underline{\boxed{\sf{{ \red{ \dag d_{2} = 12cm \dag}}}}}$

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Additional Information !!

$\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \quad \footnotesize\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered} \end{gathered} </p><p>FormulasofAreas:−</p><p></p><p> </p><p>⋆Square=(side) </p><p>2</p><p> </p><p>⋆Rectangle=Length×Breadth</p><p>⋆Triangle= </p><p>2</p><p>1</p><p></p><p> ×Breadth×Height</p><p>⋆Scalene△= </p><p>s(s−a)(s−b)(s−c)</p><p></p><p> </p><p>⋆Rhombus= </p><p>2</p><p>1</p><p></p><p> ×d </p><p>1</p><p></p><p> ×d </p><p>2</p><p></p><p> </p><p>⋆Rhombus= </p><p>2</p><p>1</p><p></p><p> d </p><p>4a </p><p>2</p><p> −d </p><p>2</p><p> </p><p></p><p> </p><p>⋆Parallelogram=Breadth×Height</p><p>⋆Trapezium= </p><p>2</p><p>1</p><p></p><p> (a+b)×Height</p><p>⋆EquilateralTriangle= </p><p>4</p><p>3</p><p></p><p> </p><p></p><p> (side) </p><p>2</p><p> </p><p></p><p> </p><p>$