ABCD is a rhombus and its two diagonals meet at O,then,,
AD²+CD²+BC²+AB²=.............
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AB^2 = AO^2 + DO^2 ALL SIDES OF RHOMBUS ARE EQUAL
ON MULTIPLYING BY 4 ON BOTH SIDES
4AB^2.=4A0^2 + 4DO^2
SO---
AB^2 + BC^2 + CD^2 + DA^2 = 4AO^2 +4 DO^2
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