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3. Show that the figure formed by joining the midpoints of sides of a rhombus successivelyis a rectangle. |
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Answer» Given AC,BD are diagonals of a quadrilateral ABCD are perpendicular. P,Q,R and S are the mid points of AB,BC, CD and AD respectively. Proof:In ΔABC, P and Q are mid points of AB and BC respectively. ∴ PQ|| AC and PQ = ½AC ..................(1) (Mid point theorem) Similarly in ΔACD, R and S are mid points of sides CD and AD respectively. ∴ SR||AC and SR = ½AC ...............(2) (Mid point theorem) From (1) and (2), we get PQ||SR and PQ = SR Hence, PQRS is parallelogram ( pair of opposite sides is parallel and equal)Now, RS || AC and QR || BD.Also, AC ⊥ BD (Given)∴RS ⊥ QR. Thus, PQRS is a rectangle. |
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