Question

28. In a rhombus ABCD, diagonals intersect at O. The sum of AC^2 + BD^2 is
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Answer 1

Step-by-step explanation:

$Side of Rhombus are  equal.</p><p></p><p>AB = BC = CD = DA</p><p></p><p>∴AC=4×(side)2−(DiagonalBD)2</p><p></p><p>AC=4×(10)2−(16)2</p><p></p><p>=400−256</p><p></p><p>=144</p><p></p><p>=12cm</p><p></p><p></p><p>$