In a rhombus ABCD prove that 4AB^2=AC^2+BD^2.
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because side of rhombus are equal
, therefore
AB=BC=CD=AD
JOIN THE DIAGONALS of rhombus
we know that diagonals of rhombus bisect each other at 90 degree. therefore,
AO=HALF OF AC AND BO=HALF OF BD
now , in triangle AOB , by Pythagoras theorem
AB^2=AO^2+BO^2
AB^2=(1/2AC)^2+(1/2BD^2), BY taking lcm . 4AB^2= AC^2+BD^2
, therefore
AB=BC=CD=AD
JOIN THE DIAGONALS of rhombus
we know that diagonals of rhombus bisect each other at 90 degree. therefore,
AO=HALF OF AC AND BO=HALF OF BD
now , in triangle AOB , by Pythagoras theorem
AB^2=AO^2+BO^2
AB^2=(1/2AC)^2+(1/2BD^2), BY taking lcm . 4AB^2= AC^2+BD^2
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