Question

The area of a rhombus is 600 sq. in. and the diagonals are in ratio 3:4. Find the diagonals and one side.

Answer 1

Given:

• The area of a rhombus is 600 sq.units
• The diagonals of the rhombus are in the ratio 3 : 4.

To find:

• The measures diagonals and a side of the rhombus.

Solution:

First let's find the lengths of diagonals of the rhombus!

Let us assume that the length first diagonal be 3x units and the length of second diagonal be 4x units. We know the area of the rhombus. So, on substituting the known values in the formula:

$\\ \longmapsto \sf \pmb{Area_{(rhombus)} = \dfrac{1}{2} (d_1d_2) \: sq.units} \\ \\ \longmapsto{ \sf{600 = \dfrac{1}{2} \times 3x \times 4x}} \\ \\ \longmapsto{ \sf{600 = \dfrac{1}{ \cancel{2}} \times \cancel{12} {x}^{2} }} \\ \\ \longmapsto{ \sf{600 = 6 {x}^{2} }} \\ \\ \longmapsto{ \sf{ \dfrac{ \cancel{600}}{ \cancel{6}} = {x}^{2} }} \\ \\ \longmapsto{ \sf{ 100= {x}^{2} }} \\ \\ \longmapsto{ \sf{\sqrt{100} = x}} \\ \\ \longmapsto \underline{ \underline{ \sf{10 = x}}}$

Since the value of x is 10, the lengths of first and second diagonal will be:

• Diagonal 1 → 3x = 3(10) = 30 units
• Diagonal 2 → 4x = 4(10) = 40 units

‎

In a rhombus, when the diagonals intersects each other, they form four right-angled triangles, where its parameters will be:

◍ Hypotenuse = Side

◍ Perpendicular = 1/2 × Diagonal 2 (i.e, 20 units)

◍ Base = 1/2 × Diagonal 1 (i.e, 15 units)

As we're asked to find the side of the rhombus, we can use Pythagorean Theorem to find it!

$\underline{ \boxed{\sf \pmb{ {(Hypotenuse)}^{2} = {(Perpendicular)}^{2} + {(Base)}^{2} }}}$

On substituting the values,

$\\ \longmapsto{ \sf{{(Hypotenuse)}^{2} = {(20)}^{2} + {(15)}^{2} }} \\ \\ \longmapsto{ \sf{{(Hypotenuse)}^{2} = 400 + 225}} \\ \\ \longmapsto{ \sf{{(Hypotenuse)}^{2} = 625}} \\ \\ \longmapsto{ \sf{Hypotenuse = \sqrt{625} }} \\ \\ \longmapsto \underline{ \underline{ \sf{Hypotenuse = 25}}}$

The length of side of the rhombus is 25 units.

_______________________________________

Answer 2

$\underline{\underline{\huge{\red{\tt{\textbf Answer :-}}}}}$

First let's find the lengths of diagonals of the rhombus!

Let us assume that the length first diagonal be 3x units and the length of second diagonal be 4x units. We know the area of the rhombus. So, on substituting the known values in the formula:

$\\ \longmapsto \sf \pmb{Area_{(rhombus)} = \dfrac{1}{2} (d_1d_2) \: sq.units} \\ \\ \longmapsto{ \sf{600 = \dfrac{1}{2} \times 3x \times 4x}} \\ \\ \longmapsto{ \sf{600 = \dfrac{1}{ \cancel{2}} \times \cancel{12} {x}^{2} }} \\ \\ \longmapsto{ \sf{600 = 6 {x}^{2} }} \\ \\ \longmapsto{ \sf{ \dfrac{ \cancel{600}}{ \cancel{6}} = {x}^{2} }} \\ \\ \longmapsto{ \sf{ 100= {x}^{2} }} \\ \\ \longmapsto{ \sf{\sqrt{100} = x}} \\ \\ \longmapsto \underline{ \underline{ \sf{10 = x}}}$

Since the value of x is 10, the lengths of first and second diagonal will be:

• Diagonal 1 → 3x = 3(10) = 30 units
• Diagonal 2 → 4x = 4(10) = 40 units

‎

In a rhombus, when the diagonals intersects each other, they form four right-angled triangles, where its parameters will be:

◍ Hypotenuse = Side

◍ Perpendicular = 1/2 × Diagonal 2 (i.e, 20 units)

◍ Base = 1/2 × Diagonal 1 (i.e, 15 units)

As we're asked to find the side of the rhombus, we can use Pythagorean Theorem to find it!

$\underline{ \boxed{\sf \pmb{ {(Hypotenuse)}^{2} = {(Perpendicular)}^{2} + {(Base)}^{2} }}}$

On substituting the values,

$\\ \longmapsto{ \sf{{(Hypotenuse)}^{2} = {(20)}^{2} + {(15)}^{2} }} \\ \\ \longmapsto{ \sf{{(Hypotenuse)}^{2} = 400 + 225}} \\ \\ \longmapsto{ \sf{{(Hypotenuse)}^{2} = 625}} \\ \\ \longmapsto{ \sf{Hypotenuse = \sqrt{625} }} \\ \\ \longmapsto \underline{ \underline{ \sf{Hypotenuse = 25}}}$

• The length of side of the rhombus is 25 units.

_______________________________________