Question

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Answer 1

Answer:-

ABCD=RHOMBUS

In ∆ABC:-

P & Q are mid points of AB & BC respectively

So, PQ=1/2AC (Mid Point Theorem) ... (1)

And, PQ || AC (Mid Point Theorem) ... (2)

Similarly:-

In ∆ADC:-

S&R are the mid points of AD & CD respectively

So, RS=1/2AC (Mid Point Theorem) ... (3)

And, RS || AC (Mid points theorem) ... (4)

From (1), (2), (3) & (4)

PQ=RS and PQ || RS.

:. PQRS is a Parallelogram (Opposite sides are equal and parallel)

Now,

In ∆QRC:-

QC=RC ( CB=CD and Q & R are mid points of BC&CD)

So, angleR =angleQ (Angles opposite to equal sides are equal) ... (5)

Also,

In ∆DRS & ∆BPQ:-

DR=BP (:.DC=AB and P & R are mid points of AB & CD)

RS=PQ (:.PQRS = PARALLELOGRAM)

SD=QB (:.AD=BC and S & Q are mid points of AD and BC)

∆DRS is congruent to ∆BPQ (By SSS rule)

angle SRD = angle PQB (C.P.C.T.) ... (6)

From (5) & (6)

:. angle SRD+ angleQRC=angle PQB+angle RQC

angleSRD + angleQRC + angleR=180° (Linear pair)

:. angle R=180°-(angleSRD + angleQRC) ... (7)

anglePQB + angleRQC +angleQ=180° (Linear pair)

:. angle Q=180°-(anglePQB + angleRQC) ... (8)

180°-(angleSRD + angleQRC)=180°-(anglePQB + angleRQC) (because angle SRD+ angleQRC=angle PQB+angle RQC) .. (9)

From(7), (8) & (9)

:. angle R = angle Q ... (10)

Sum of adjacent angles of parallelogram=180°

angleR+angleQ = 180°

2× angle R=180° (angle R = angle Q)

angleR =180°/2

angle R=90°

PQRS is a parallelogram with one angle=90°

So, PQRS=Rectangle

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Answer 2

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