Question

The diagonal of a rohambus are 12cmand 16cm.find altitude

Answer 1

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

Given that diagonals of a rhombus are 12 cm and 16 cm.

It means

$\rm \: d_1 = 12 \: cm$

$\rm \: d_2 = 16 \: cm$

Let assume that side of a rhombus be a cm and its altitude is h cm.

We know, diagonals and side of a rhombus are connected by the relationship

$\rm \: {4a}^{2} \: = \: {d_1}^{2} + {d_2}^{2}$

So, on substituting the values, we get

$\rm \: {4a}^{2} \: = \: {12}^{2} + {16}^{2}$

$\rm \: {4a}^{2} \: = \: 144 + 256$

$\rm \: {4a}^{2} \: = \: 400$

$\rm \: {a}^{2} \: = \: 100$

$\rm \: {a}^{2} \: = \: {10}^{2}$

$\rm\implies \:a \: = \: 10 \: cm$

Now, we know that

$\color{green}\boxed{ \rm{ \:Area_{(rhombus)} = \frac{1}{2} \times d_1 \times d_2 = a \times h \: }}$

So,

$\rm \: \dfrac{1}{2} \times d_1 \times d_2 = a \: \times \: h$

On substituting the values, we get

$\rm \: \dfrac{1}{2} \times 12 \times 16 = 10 \: \times \: h$

$\rm \: 6 \times 16 = 10 \: \times \: h$

$\rm \: 10h = 96$

$\rm\implies \:h \: = \: 9.6 \: cm$

$\rule{190pt}{2pt}$

Additional Information :-

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