Question

-Show that the quadrilateral formed by joining the mid-points of the pair of adjacent sides of a rectangle is a rhombus.

Answer 1

Hint: In order to solve this question, we have to write what is given to us or what is mentioned in this question. This will give us a clear picture of what our approach should be. In this question we have to use the basics of midpoint theorem and congruence in triangles.

Complete step-by-step answer:

The midpoint theorem states that “The line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side.”
It is given that ABCD is a rectangle and P, Q, R, S, are the mid points AB, BC, CD, DA sides.
Let ABCD be the rectangle and P, Q, R, S be the midpoints of AB, BC, CD, DA respectively.
Join AC, a diagonal of the rectangle. In $\vartriangle ABC$ we have,
$\left. {PQ} \right\|AC$ and PQ=$\dfrac{1}{2}$ AC (By midpoint theorem)
Again, in $\vartriangle DAC$ , the points S and R are the midpoints of AD and DC, respectively.
$\left. {SR} \right\|AC$ and SR= $\dfrac{1}{2}$ AC (by midpoint theorem)
Now, $\left. {PQ} \right\|SR$ (each equal to $\dfrac{1}{2}$ AC) ……(i)
So, PQRS is a parallelogram.
Now, in $\vartriangle SAP$ and $\vartriangle QBP$, we have,
AS=BQ,$\angle A = \angle B = \angle {90^ \circ }$ and AP=BP (by AAS criterion)
i.e., $\vartriangle SAP \cong \vartriangle QBP$
PS=PQ ……(ii)
Similarly, $\vartriangle SDR \cong \vartriangle QCR$
SR=RQ ……(iii)
From (i), (ii), (iii), we have,
PQ=PS=SR=RQ
Hence, PQRS is a rhombus.

Note: In this question it should be noted that the basics are the midpoint theorem which states that the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side. Also we should know about some congruence criterion in triangles SSS (all sides are equal), SAS (the sides and the angle included in them should be equal), AAS (in this two angle and the side should be equal), RHS (the hypotenuse, the right angle and a side should be same). By these concepts the question should be easy to solve and its other types.