English, asked by anreesakunju, 1 month ago

Prove that the quadrilateral formed by the joining the midpoint of a rectangle is a rhombus​

Answers

Answered by mathdude500
13

\large\underline{\sf{Solution-}}

Let assume that ABCD be a rectangle such that P, Q, R, S are the midpoints of the sides of a rectangle AB, BC, CD, DA respectively.

Construction: Join PQ, QR, RS, SP and AC

Now, In triangle ABC

\sf P \: is \: midpoint \: of \: AB \\  \\

\sf Q \: is \: midpoint \: of \: BC \\  \\

So, By Midpoint Theorem, we get

\implies\sf \: PQ \:  \parallel \: AC \: and \: PQ \:  =  \: \dfrac{1}{2}AC  -  -  - (1)\\  \\

Now, In triangle ADC

\sf S \: is \: midpoint \: of \: AD \\  \\

\sf R \: is \: midpoint \: of \: DC \\  \\

So, By Midpoint Theorem, we have

\implies\sf \: SR \:  \parallel \: AC \: and \: SR \:  =  \: \dfrac{1}{2}AC  -  -  - (2)\\  \\

From equation (1) and (2), we concluded that

\implies\sf \: PQRS \: is \: a \: parallelogram. \\  \\

[ ∵ If in a quadrilateral, one pair of opposite sides are equal and parallel, then quadrilateral is a parallelogram ]

\implies\sf \: PQ = SR \:  \: and \:  \: PS = QR \\  \\

Now, As it is given that ABCD is a rectangle.

\sf \: AD = BC \\  \\

\sf \: \dfrac{1}{2} AD = \dfrac{1}{2} BC \\  \\

\implies\sf \: AS = BQ \\  \\

Further also, P is the midpoint of AB

\implies\sf \: AP = PB \\  \\

Now,

\sf \: In \:  \triangle \: ASP \: and \:   \triangle \: QBP \\  \\

\boxed{\begin{aligned}& \sf \:AS = BQ \:  \{proved \: above \} \\ \\& \sf \: AP = PB \:  \:\{proved \: above \}   \\ \\& \sf \:  \angle \: SAP= \angle \:PBQ \:  \: \{each\: {90}^{ \circ}  \} \end{aligned}} \implies\sf \:  \triangle \: ASP \:  \cong \: \triangle \: QBP \\  \\

[ By SAS Congruency ]

\implies\sf \: SP = PQ \\  \\

\implies\sf \: PQ = QR = RS = SP \\  \\

\implies\sf \: PQRS \: is \: a \: rhombus \\  \\

Hence, the line segments joining the midpoints of the consecutive sides of a rectangle form a rhombus.

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Answered by sujal34negi
0

Explanation:

So we know that OERF is a parallelogram. Therefore, it is proved that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a rhombus is a rectangle. P, Q, R and S are respectively the midpoints of the sides AB, BC, CD and DA of a quadrilateral ABCD....

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