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
Answer :-
Let ABCD be a rhombus of side 10 cm and O is the intersection point of diagnols.
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
Trignometric table :-
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