Question

Abcd is a parallelogram n ab =5cm ,bc=4cm then what is the value of ac^2+bd^2

Answer 1

Given : Abcd is a parallelogram,  ab =5cm ,bc=4cm

To find :  the value of AC² +  BD² =

Solution:

Lets draw  DM & CN ⊥  AB

DM = CN = h  

Let say AM = BN = x    

AC²  = (AN)² + (CN)²

=> AC² = (AB + BN)² + CN²

=> AC² = (  5 + x)² + h²

=> AC² = x² + 25 + 10x  + h²

BD²  = (BM)² + (DM)²

=> BD² = (AB - BM)² + DM²

=> BD² = (  5 - x)² + h²

=> BD² = x² + 25 - 10x  + h²

Adding both

AC² +  BD² = x² + 25 + 10x  + h² + x² + 25 - 10x  + h²

=> AC² +  BD² =   50 + 2(x² + h²)

now x² + h² = 4² = 16

=> AC² +  BD² =   50 + 2(16)

=> AC² +  BD² =   50 + 32

=> AC² +  BD² =   82

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Answer 2

The value of (AC)² + (BD)² = 82 cm

Given:

AB = 5 cm = DC

BC = 4 cm = AD

Step-by-step explanation:

AC = √(AB + BC)

AC = √(25 + 16)

∴ AC = √(41) cm

Now, the length of the diagonal BD is:

BD = √(BC + DC)

BD = √(25 + 16)

∴ BD = √(41) cm

Now,

(AC)² + (BD)² = (√(41))² + (√(41))²

(AC)² + (BD)² = 41 + 41

∴ (AC)² + (BD)² = 82 cm