anyone pls solve this qustn
Answers
||✪✪ QUESTION ✪✪||
Prove that the sum of squares of the sides of a rhombus is equal to the sum of the square of its diagonals ?
or,
To Prove :- ( AB² + BC² + CD² + DA² ) = ( AC² + BD² ).
|| ✰✰ ANSWER ✰✰ ||
❁❁ Refer To Image First .. ❁❁
➡ We know that the diagonals of a rhombus bisect each other at right angles.
So, we can say that :- ∠AOB = ∠BOC = ∠COD = ∠DOA = 90°..
Also, Diagonals Bisect Each other In Equal Parts..
So , OA = (1/2) AC and OB = (1/2) BD.
Now, using Pythagoras' theorem in Right ∆AOB , we get ,
AB² = OA² + OB²
putting values we get,
→ AB² = [(1/2)AC]² + [(1/2)BD]²
→ AB² = (1/4) [ AC² + BD²]
→ 4AB² = (AC² + BD²)
or,
→ AB² + AB² + AB² + AB² = ( AC² + BD² ) .
Now, we know That, All sides of Rhombus are Equal ..
So,
•°• AB²+BC²+CD²+DA² = (AC² + BD²)
✪✪ Hence Proved ✪✪
❖❖❖❖❖❖QUESTION ❖❖❖❖❖❖
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
❖❖❖❖❖❖SOLUTION ❖❖❖❖❖❖
Refer to the attachment for the diagram.
➷➷➷➷➷➷GIVEN➷➷➷➷➷➷
ABCD is a rhombus whose diagonals AC and BD intersect at O.
➷➷➷➷➷➷TO PROVE➷➷➷➷➷➷
AB^2 + BC^2 + CD^2
+ AD^2 = AC^2 + BD^2
➷➷➷➷➷➷➷PROOF➷➷➷➷➷➷
Since,
diagonals of rhombus bisect each other perpendicularly
Therefore,
Angle AOD = angle AOB = angle BOC = angle COD = 90°
And, OA= OC ; OB = OD
↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓
By Pythagoras in ∆ AOB
AB^2 = OA^2 + OB^2 →→→eqn (1)
By Pythagoras in ∆ BOC
BC^2 = OB^2 + OC^2 →→→eqn (2)
By Pythagoras in ∆ COD
CD^2 = OC^2 + OD^2 →→→eqn (3)
By Pythagoras in ∆ AOD
AD^2 = OA^2 + OD^2 →→→eqn (4)
Adding eqn (1), (2) , (3) and (4)
AB^2+ BC^2+ CD^2+ AD^2 = 2 ( OA^2 +
OB^2 + OC^2 + OD^2 )
*(since, OA= OC and OB = OD hence,)
AB^2+BC^2 + CD^2 + AD^2 =
2( 2 OA^2+ 2OB^2)
AB^2+BC^2+CD^2+AD^2=
4(OA^2+OB^2)
*(Since, OA = AC/2 and OB =
BC/2 ,hence)
AB^2+BC^2+CD^2+AD^2 =
4 [ (AC/2) ^2 + (BC/2)^2 ]
AB^2+BC^2+CD^2+AD^2=
4 ( AC^2 / 4 + BC^2 /4 )
AB^2+BC^2+CD^2+AD^2 =
(4/4) (AC^2 + BC^2)
AB^2 +BC^2 +CD^2 +AD^2 =
AC^2+BC^2
☆☆☆☆☆☆PROVED☆☆☆☆☆☆☆