ABCD is rohmbus &its two diagonal meet at O.show that
AD^2+DC^2+CB^2+BA^2=4DO^2+4AO^2
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LET AC AND BD INTERSECT AT A POINT O
IN TRIANGLE AOD
BY PYTHAGOROUS
AB^2 = AO^2 + DO^2
ON MULTIPLYING BY 4 ON BOTH SIDES
4AB^2.=4AO^2 + 4DO^2
ALL SIDES OF RHOMBUS ARE EQUAL SO---
AB^2 + BC^2 + CD^2 + DA^2 = 4AO^2 +4 DO^2
This is your answer.
IN TRIANGLE AOD
BY PYTHAGOROUS
AB^2 = AO^2 + DO^2
ON MULTIPLYING BY 4 ON BOTH SIDES
4AB^2.=4AO^2 + 4DO^2
ALL SIDES OF RHOMBUS ARE EQUAL SO---
AB^2 + BC^2 + CD^2 + DA^2 = 4AO^2 +4 DO^2
This is your answer.
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