Question

show that the quadrilateral formed by joining the mid point the sides of a rhombus taken in order form a rectangle .​

Answer 1

Answer:

From triangle ABD, we have SP = 1/2 BD and SP || BD (Because S and P are mid-points)

Similarly, RQ = 1/2 BD and RQ || BD Therefore, SP = RQ and SP || RQ

So, PQRS is a parallelogram.

(1) Also,AC ⊥ BD (Diagonals of a rhombus are perpendicular) Further PQ || AC (From ∆BAC)

As SP || BD, PQ || AC and AC ⊥ BD,

therefore, we have SP ⊥ PQ, i.e. ∠SPQ = 90º.

(2) Therefore, PQRS is a rectangle[From (1) and (2)]

Answer 2

Answer:

Let ABCD be a rhombus and P, Q, R and S be the mid-points of sides AB, BC, CD and DA, respectively (Fig.). Join AC and BD. From triangle ABD, we have SP = 1/2 BD and SP || BD (Because S and P are mid-points) Similarly, RQ = 1/2 BD and RQ || BD Therefore, SP = RQ and SP || RQ So, PQRS is a parallelogram. (1) Also,AC ⊥ BD (Diagonals of a rhombus are perpendicular) Further PQ || AC (From ∆BAC) As SP || BD, PQ || AC and AC ⊥ BD, therefore, we have SP ⊥ PQ, i.e. ∠SPQ = 90º. (2) Therefore, PQRS is a rectangle[From (1) and (2)]Read more on Sarthaks.com - https://www.sarthaks.com/872093/show-that-quadrilateral-formed-joining-points-sides-rhombus-taken-order-form-rectangle