Question

show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a rhombus is a rectangle.???​

Answer 1

Answer:

Let ABCD be a rhombus and P, Q, R and S be the mid-points of sides AB, BC, CD and DA respectively.

In ABD and BDC we have

From (i) and (ii) we get

PQRS is a ||gm

As diagonals of a rhombus bisect each other at right angles.

AC BD

Since SP||BD, PQ||AC and AC BD

From above results, we have

||gm PQRS is a rectangle.

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

$</p><p>Let ABCD be the rectangle and P,Q,R,S be the midpoints of AB,BC,CD,DA respectively.</p><p></p><p>Join AC, a diagonal of the rectangle.</p><p>In ΔABC, we have:</p><p></p><p>PQ∣∣AC and PQ=21AC [By midpoint theorem]</p><p></p><p>Again, in ΔDAC, the points S and R are the mid points of AD and DC, respectively.</p><p></p><p>SR∣∣AC and SR=21AC [By midpoint theorem]</p><p></p><p>Now, PQ∣∣AC and SR∣∣AC and PQ∣∣SR</p><p></p><p>Also, PQ=SR [Each equal to 21AC] . . . . . . . (i)</p><p></p><p>So, PQRS is a parallelogram.</p><p></p><p>Now, in ΔSAP and ΔQBP, we have:</p><p></p><p>AS=BQ,∠A=∠B=90∘andAP=BP</p><p></p><p>i.e.,ΔSAP∼ΔQBP</p><p></p><p>PS=PQ . . . . . . . . . (ii)</p><p></p><p>Similarly, ΔSDR∼ΔQCR</p><p></p><p>SR=RQ . . . . . . . . (iii)</p><p></p><p>From (i), (ii) and (iii), we have:</p><p></p><p>PQ=PS=SR=RQ</p><p></p><p>Hence, PQRS is a rhombus.</p><p></p><p>$

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