Questions & Answers

Question

Answers

$\left( a \right)\sqrt 3 :1$

$\left( b \right)2:\sqrt 5 $

$\left( c \right)2:\sqrt 3 $

$\left( d \right)\sqrt 5 :2$

Answer
Verified

Given data:

Area of rhombus = Area of square = 64 sq. cm.

$ \Rightarrow {A_1} = {A_2} = 64$ Sq. cm.

It is also given that a diagonal of a rhombus is twice of its other diagonal.

Let the length of one diagonal is p cm, so the length of the other diagonal is 2p cm.

Now as we all know that the area $\left( {{A_1}} \right)$ of the rhombus is half times the product of the diagonals of the rhombus.

$ \Rightarrow {A_1} = \dfrac{{p\left( {2p} \right)}}{2}$ sq.cm.

Now substitute the value and simplify we have,

$ \Rightarrow 64 = {p^2}$

$ \Rightarrow p = \sqrt {64} = 8$ Cm.

So the length of one diagonal is 8 cm and the length of the other diagonal is 16 cm.

Now consider the rhombus as shown in above diagram the diagonal of the rhombus is always bisect and perpendicular to each other as shown in the above figure, therefore, AO = OC = $\dfrac{{AC}}{2}$ and BO = OD = $\dfrac{{BD}}{2}$

Now let, AC = p, and BD = 2p.

Therefore, AC = 8cm, and BD = 16cm.

Therefore, AO = OC = $\dfrac{{AC}}{2} = \dfrac{8}{2} = 4$cm, and BO = OD = $\dfrac{{BD}}{2} = \dfrac{{16}}{2} = 8$ cm.

Now as we all know that in a rhombus all sides are equal therefore,

AB = BC = CD = DA.

Now apply Pythagoras theorem in triangle AOB we have,

$ \Rightarrow {\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{perpendicular}}} \right)^2} + {\left( {{\text{base}}} \right)^2}$

$ \Rightarrow {\left( {{\text{AB}}} \right)^2} = {\left( {{\text{OA}}} \right)^2} + {\left( {{\text{OB}}} \right)^2}$

$ \Rightarrow {\left( {{\text{AB}}} \right)^2} = {\left( {\text{4}} \right)^2} + {\left( {\text{8}} \right)^2}$

$ \Rightarrow {\left( {{\text{AB}}} \right)^2} = 16 + 64 = 80$

$ \Rightarrow AB = \sqrt {80} = 4\sqrt 5 $ Cm.

Therefore, AB = BC = CD = DA = $4\sqrt 5 $ Cm.

Now as we know that the area of the square is the square of its side.

Let the side of the square is a cm as shown in the below figure.

So the area of the square is ${A_2} = 64 = {a^2}$ sq. cm.

$ \Rightarrow a = \sqrt {64} = 8$ cm.

Now as we know that the perimeter of any shape is the sum of all the sides.

So the perimeter of the rhombus is, ${P_1} = 4\sqrt 5 + 4\sqrt 5 + 4\sqrt 5 + 4\sqrt 5 = 16\sqrt 5 $ cm.

And the perimeter of the square is, ${P_2} = 8 + 8 + 8 + 8 = 32$ cm.

So the ratio of perimeter of a square and rhombus is,

$ \Rightarrow \dfrac{{{P_2}}}{{{P_1}}} = \dfrac{{32}}{{16\sqrt 5 }} = \dfrac{2}{{\sqrt 5 }}$.

So this is the required ratio of perimeter of a square and rhombus.