A rhombus ABCD has AB = 10 and m∠A = 60°. Find the lengths of the diagonals of ABCD.
Answers
Given:
1. ∠A of rhombus is equal to 60°
2. Length of side AB = 10 cm
To find:
Length of the diagonals of ABCD.
Solution:
- As per the question, side AB is equal to 10 cm. Thus, all other sides are also equal to 10 cm.
- Now, ∠BAD=60° and AB = AD. Therefore, △ABD is an equilateral triangle.
- Similarly, △CBD is also an equilateral triangle.
- We know that height of an equilateral triangle is (√3 / 2) x side.
- Thus, AO = CO = (√3 / 2) x AB
= (√3 / 2) x 10
- Now, AC = AO + CO
=(√3/2)x10+(√3/2)x10
=10√3 cm
- Thus, the length of the diagonal is 10√3 cm.
Given :
ABCD is a rhombus with ∠A = 60°, AB = 10 cm.
To Find:
We have to find the length of the diagonals, AC and BD.
Solution :
The important properties of the diagonals of a rhombus are:
the diagonals of a rhombus bisect each other
the diagonals are perpendicular to each other
the diagonals bisect the angles of the rhombus
Let O be the point where the two diagonals intersect each other as shown in the figure.
Given that ∠BAD = 60°,
By using the diagonal properties, ∠BAO= 30° and ∠AOB= 90°
∴ By using the properties of the interior angles of a triangle,
∠ABO+∠BAO+∠AOB= 180°
∠ABO= 180°-30°-90°
⇒ ∠ABO =60°
In right-angled ΔAOB,
Similarly, we have,
By using the diagonal properties, AO= OC and OB= OD
⇒ Length of AC = 2×AO = 10√3
⇒ Length of BD = 2×OB = 10
Hence, the length of the diagonals is 10 and 10√3.