Question

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Perimeter of rhombus is 96 cm and obtuse angle of it is 120° .Find length of it's Digonals

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Answer 1

Answer:

The lengths of the diagonals of the rhombus are 24 √3 cm and 24 cm.

Step-by-step-explanation:

NOTE: Refer to the attachment for the diagram.

In figure, □ABCD is a rhombus.

∴ AB = BC = CD = AD - - - ( 1 ) [ Sides of a rhombus ]

m∠BCD = 120° - - - [ Given ]

Diagonals AC & BD intersect each other at point O.

Now, we know that,

Perimeter of rhombus = 4 * side

⇒ P ( □ABCD ) = 4 * AB

⇒ 96 = 4 * AB

⇒ AB = 96 ÷ 4

⇒ AB = 24 cm

∴ AB = BC = CD = AD = 24 cm - - - ( 2 ) [ From ( 1 ) ]

Now, in □ABCD,

m∠ABC + m∠BCD = 180° - - - [ Adjacent angles ]

⇒ m∠ABC + 120° = 180° - - - [ Given ]

⇒ m∠ABC = 180° - 120°

⇒ m∠ABC = 60° - - - ( 3 )

Now, we know that,

Diagonals of a rhombus bisect the opposite angles.

∴ m∠ACB = m∠ACD = ½ * m∠BCD

⇒ m∠ACB = ½ * 120 - - - [ Given ]

⇒ m∠ACB = 60°

Also,

m∠DBC = m∠DBA = ½ * m∠ABC

⇒ m∠DBC = ½ * 60 - - - [ From ( 3 ) ]

⇒ m∠DBC = 30°

We know that,

Diagonals of a rhombus are perpendicular bisectors of each other.

∴ In △BOC, m∟BOC = 90°

m∠ACB = m∠OCB = 60° - - - [ A - O - C ]

m∠DBC = m∠OBC = 30° - - - [ D - O - C ]

∴ △BOC is a 30°-60°-90° triangle.

Now,

BO = ( √3 / 2 ) * BC - - - [ Side opposite to 60° ]

⇒ BO = ( √3 / 2 ) * 24 - - - [ From ( 2 ) ]

⇒ BO = √3 * 24 ÷ 2

⇒ BO = √3 * 12

⇒ BO = 12 √3 cm - - - ( 4 )

Now,

CO = ½ * BC - - - [ Side opposite to 30° ]

⇒ CO = ½ * 24 - - - [ From ( 2 ) ]

⇒ CO = 24 ÷ 2

⇒ CO = 12 cm - - - ( 5 )

Now, we know that,

Diagonals of a rhombus bisect each other.

∴ BO = OD = ½ * BD

⇒ BO = ½ * BD

⇒ 12 √3 = ½ * BD - - - [ From ( 4 ) ]

⇒ BD = 12 √3 * 2

⇒ BD = 12 * 2 √3

⇒ BD = 24 √3 cm

Also,

CO = OA = ½ * AC

⇒ CO = ½ * AC

⇒ 12 = ½ * AC - - - [ From ( 5 ) ]

⇒ AC = 12 * 2

⇒ AC = 24 cm

∴ The lengths of the diagonals of the rhombus are 24 √3 cm and 24 cm.

Answer 2

Let's denote the length of one side of the rhombus by $a$. Since the rhombus is symmetric, we can divide it into four congruent right triangles, each with one acute angle of 60 degrees and hypotenuse equal to $a$. The other two sides of each triangle are equal and can be found using the Pythagorean theorem: $$\begin{aligned} b &= a\sin 60^\circ = \frac{a\sqrt{3}}{2}\ c &= a\cos 60^\circ = \frac{a}{2} \end{aligned}$$ The perimeter of the rhombus is four times the length of one side, so we have $4a=96$ or $a=24$.

The diagonals of a rhombus are perpendicular bisectors of each other. Let's denote the lengths of the two diagonals by $d_1$ and $d_2$, where $d_1$ is the longer diagonal. Using the fact that $d_1$ and $d_2$ are perpendicular bisectors of each other, we can divide the rhombus into four congruent right triangles, each with one acute angle of 30 degrees and hypotenuse equal to $d_1/2$ or $d_2/2$. The other two sides of each triangle are equal and can be found using the sine and cosine of 30 degrees: $$\begin{aligned} e &= \frac{d_1}{2} \sin 30^\circ = \frac{d_1}{4}\ f &= \frac{d_1}{2} \cos 30^\circ = \frac{d_1\sqrt{3}}{4} \end{aligned}$$ We also know that $e^2+f^2=(a/2)^2=(24/2)^2=144$. Substituting the expressions for $e$ and $f$, we get $$\frac{d_1^2}{16}+\frac{3d_1^2}{16}=\frac{144}{1}$$ Simplifying and solving for $d_1$, we get $$d_1=\sqrt{\frac{256}{4}}=16$$ Therefore, the length of the longer diagonal is $d_1=16$ cm, and the length of the shorter diagonal is $d_2=d_1/\sqrt{3}=16/\sqrt{3}$ cm (since the diagonals of a rhombus are perpendicular bisectors of each other, we could have used the length of the shorter diagonal to find the length of the longer one).