CBSE BOARD X, asked by Anonymous, 5 months ago

 \huge \bf Question

A rhombus of side of 10 cm has two angles 60° each. Find the length of diagonals and also find its area

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Answers

Answered by Anonymous
61

Answer :-

Let ABCD be a rhombus of side 10 cm and O is the intersection point of diagnols.

\sf \angle BAD = \angle BCD = 60 \degree

\setlength{\unitlength}{1 cm}\begin{picture}(20,15)\thicklines\qbezier(1,1)(1,1)(6,1)\qbezier(1,1)(1,1)(1.6,4)\qbezier(1.6,4)(1.6,4)(6.6,4)\qbezier(6,1)(6,1)(6.6,4)\qbezier(6.6,4)(6.6,4)(1,1)\qbezier(1.6,4)(1.6,4)(6,1)\put(0.7,0.5){\sf A}\put(6,0.5){\sf B}\put(1.4,4.3){\sf D}\put(6.6,4.3){\sf C}\end{picture}

According to the properties of rhombus, it bisects the diagnol into two equal parts :-

→ AO = OC

→ BO = OD

In the right triangle AOB :-

∠ AOB = 30°

cos 30° = OA/OB

→ √3/2 = OA/10

→ OA = 5√3

OA is half of the diagonal AC -

→ AC = 2 ( OB )

AC = 10 √3 cm

sin30° = OB/AB

→ 1/2 = OB/10

→ OB = 5 cm

OB is half of the diagonal BD -

→ BD = 2 ( OB )

BD = 10 cm

Diagnols of rhombus =

  • BD = 10 cm
  • BD = 10 cm AC = 10√3 cm

Calculating Area :-

Area of rhombus = 1/2 × Product of diagnols

= 1/2 × 10 × 10√3 cm²

= 50√3 cm²

\boxed{\rm Area\: of\: rhombus = 50\sqrt{3} cm^2 }

Trignometric table :-

\bullet\:\sf Trigonometric\:Values :\\\\\boxed{\begin{tabular}{c|c|c|c|c|c}Radians/Angle & 0 & 30 & 45 & 60 & 90\\\cline{1-6}Sin \theta & 0 & $\dfrac{1}{2} &$\dfrac{1}{\sqrt{2}} & $\dfrac{\sqrt{3}}{2} & 1\\\cline{1-6}Cos \theta & 1 & $\dfrac{\sqrt{3}}{2}&$\dfrac{1}{\sqrt{2}}&$\dfrac{1}{2}&0\\\cline{1-6}Tan \theta&0&$\dfrac{1}{\sqrt{3}}&1&\sqrt{3}&Not D{e}fined\end{tabular}}


Anonymous: Great!
EliteSoul: Nice
BrainlyPopularman: Awesome
Answered by Anonymous
40

Given :-

A rhombus of side of 10 cm has two angles 60° each.

To find :-

  • Length of diagonals
  • Area of Rhombus

Explanation :-

Properties of Rhombus

  • The diagonals of a rhombus bisect each other at 90°

  • A pair of adjacent sides are equal

Solution :-

  • WXYZ is a Rhombus
  • ∠XYZ = 60°
  • ∠XWZ = 60°
  • WX = 10 cm
  • ∠XWO = 30°
  • WO = OY and XO = OZ

In WOX

→ sin 30° = OX/WX

→ ½ = OX/10

→ 2 × OX = 10

→ OX = 10/2 = 5cm

Now,

→ cos 30° = WO/WX

→ √3/2 = WO/10

→ √3 × 10 = 2 × WO

→ 10√3 = 2 × WO

→ WO = 10√3/2 = 5√3 cm

Hence,

  • Required diagonals
  • XZ = 2OX = 2 × 5 = 10cm
  • WY = 2WO = 2 × 5√3 = 10√3cm

Area of rhombus

→ ½ × product of diagonals

→ ½ × XZ × WY

→ ½ × 10 × 10√3

→ 5 × 10√3

→ 50√3 cm²

→ 50 × 1.73

→ 86.5 cm²

•°• Area of rhombus is 86.5 cm²

Attachments:

amitkumar44481: Perfect :-)
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Anonymous: Thankieee ♡♡
EliteSoul: Great
Anonymous: Thanks!
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