prove that the sum of the square of the sides of a rhombus is equal to the sum of the squares of its diagonal.
Answers
Given:
- ABCD is a rhombus.
- AC & BD are the diagonals.
To proof:
- Sum of square of all sides = Sum of the squares of it's diagonals.
→ AB^2 + BC^2 + CD^2 + AD^2 = AC^2 + BD^2
Solution:-
Since sides of rhombus are equal.
OA = OC = OB = OD
• In ∆AOB, ∠AOB = 90°
→ AB^2 = AO^2 + BO^2 --(1)
• In ∆BOC, ∠BOC = 90°
→ BC^2 = OB^2 + OC^2 --(2)
• In ∆COD, ∠COD = 90°
→ CD^2 = OC^2 + OD^2 --(3)
• In ∆ AOD, ∠AOD = 90°
→ AD^2 = OA^2 + OD ^2 --(4)
Adding (1), (2), (3), & (4).
→ AB^2 + BC^2 + CD^2 + AD^2 = AO^2 + BO^2 OB^2 + OC^2 + OD^2 + OA^2 + OD^2.
→ 2OA^2 + 2OB^2 + 2OC^2 + 2OD^2
→ 2OA ^2 + 2OB^2 + 2OA^2 + 2OB^2
→ 4OA^2 + 4OB^2
→ (2OA)^2 + (2OB)^2
→ AC^2 + BD^2
Hence,
- Sum of square of all sides = Sum of the squares of it's diagonals.
Answer:
In rhombus ABCD, AB = BC = CD = DA
We know that diagonals of a rhombus bisect each other perpendicularly.
That is AC ⊥ BD, ∠AOB=∠BOC=∠COD=∠AOD=90° and
Consider right angled triangle AOB
AB2 = OA2 + OB2 [By Pythagoras theorem]
⇒ 4AB2 = AC2+ BD2
⇒ AB2 + AB2 + AB2 + AB2 = AC2+ BD2
∴ AB2 + BC2 + CD2 + DA2 = AC2+ BD2
Thus the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.