Question

$\large{\underline{ \tt{ \widetilde{ \red{Question}}}}}$

The area of a rhombus is 16 cm² and the length of one of its diagonal is 4 cm. Calculate the length of other diagonal.
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Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

• Area of a rhombus = 16 cm²

• The length of one diagonal, $\rm \: d_1$ = 4 cm.

Let assume that

• The length of second diagonal = $\rm \: d_2$ cm

We know,

Area of rhombus having diagonals $\rm \: d_1\: and\:d_2$ is given by

$\boxed{ \rm{ \:Area_{(Rhombus)} \: = \: \frac{1}{2} \times d_1 \times d_2 \: \: }}$

So, on substituting the values, we get

$\rm \: 16 = \dfrac{1}{2} \times 4 \times d_2$

$\rm \: 16 = 2 \times d_2$

$\bf\implies \:d_2 \: = \: 8 \: cm$

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Additional Information :-

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

Answer 2

GIVEN :-

The area of a rhombus is 16 cm² and the length of one of its diagonal is 4 cm

TO FIND :-

Calculate the length of other diagonal = ?

SOLUTION :-

First diagonal (d₁) = 4 cm

Area of rhombus,

2 × (d₁ × d₂)

2 × (4 cm × d₂)

16 × 2 = 4 cm x d₂

32 = 4 cm × d₂

32/4 = 8 cm

8 cm = d₂

length of other diagonal = 8 cm