Question

plz give me answer plz
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Answer 1

Answer:

1 ) on further calculation D2 = 20cm

2) therefore, the length of each side of rhombus is 26cm

3) perimeter of rhombus is 104cm

Answer 2

$\large\underline{\sf{Solution-43}}$

Given that

• Side of rhombus, a = 20 cm

• Length of one diagonal, x = 24 cm

Let assume that

• Length of second diagonal be y cm

We know,

In rhombus, side a, and diagonals x and y are connected by the relationship

$\boxed{ \rm{ \: {4a}^{2} = {x}^{2} + {y}^{2} \: }}$

So, on substituting the values, we get

$\rm \: 4 \times {20}^{2} = {24}^{2} + {y}^{2}$

$\rm \: 4 \times 400 = 576 + {y}^{2}$

$\rm \: 1600 = 576 + {y}^{2}$

$\rm \: 1600 - 576 = {y}^{2}$

$\rm \: {y}^{2} = 1024$

$\rm\implies \:\boxed{ \rm{ \:y \: = \: 24 \: cm \: }}$

Now, Area of rhombus is given by

$\rm \: Area_{(rhombus)} \: = \: \dfrac{1}{2} \times x \times y$

$\rm \: Area_{(rhombus)} \: = \: \dfrac{1}{2} \times 20 \times 24$

$\rm\implies \:\boxed{ \rm{ \:Area_{(rhombus)} \: = \: 240 \: {cm}^{2} \: }}$

$\red{\large\underline{\sf{Solution-44}}}$

Given that,

• Area of rhombus = 480 sq. cm

• Length of one diagonal, x = 48 cm

Let assume that

• Length of other diagonal is y cm

We know,

$\rm \: Area_{(rhombus)} \: = \: \dfrac{1}{2} \times x \times y$

$\rm \: 480 \: = \: \dfrac{1}{2} \times 48 \times y$

$\rm\implies \:\boxed{ \rm{ \:y \: = \: 20 \: cm \: \: }}$

Let assume that

• Side of rhombus = a cm

We know,

In rhombus, side a, and diagonals x and y are connected by the relationship

$\boxed{ \rm{ \: {4a}^{2} = {x}^{2} + {y}^{2} \: }}$

On substituting the values of x and y, we get

$\rm \: {4a}^{2} = {48}^{2} + {20}^{2}$

$\rm \: {4a}^{2} = 2304 + 400$

$\rm \: {4a}^{2} = 2704$

$\rm \: {a}^{2} = 676$

$\rm \: {a}^{2} = {26}^{2}$

$\rm\implies \:\boxed{ \rm{ \:a \: = \: 26 \: cm \: }}$

Now,

$\rm \: Perimeter_{(rhombus)} \: = \: 4 \times a$

$\rm \: Perimeter_{(rhombus)} \: = \: 4 \times 26$

$\rm\implies \:\boxed{ \rm{ \:\rm \: Perimeter_{(rhombus)} \: = \: 104 \: cm \: \: }}$

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

Additional information :-

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