ABCD and PQRC are rectangles and Q us the mid-point of AC show that P is the mid-point of DC and R is the mid- point of BC.also, find the ratio of ar(ABCD) and ar(PQRC)
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Answered by
40
hope u got the figure as said in the quesn..
ok so, join diagonal BD
we know that in a rectangle diagonals are equal
now consider triangle BQR and triangle CQR
BQ=CQ (1/2AC=1/2BD)
QR=QR (common)
by RHS similarity rule the two triangles are congruent.
by CPCT, BR=CR i.e, R is the midpoint of BC
BR=CR=1/2BC-(1)
Similarly, prove triangle QDP and triangle CDP are congruent
and by CPCT we get that DP=CP i.e, P is the midpoint of CD
DP=CP=1/2CD-(2)
first part of the question is done!!
ar(ABCD) = length × breadth = lb = CD×BC
ar(PQRC) = length ×breadth = lb = PC× RC
=1/2CD ×1/2BC
[from (1) and (2)..]
by transposing the 1/2 to the other side,
2 ar (PQRC) = CD×BC
=2 ar (PQRC) = ar (ABCD)
=ar (PQRC)= 1/2 ar (ABCD).
hence the proof.
sorry for taking time. hope its clear!
ok so, join diagonal BD
we know that in a rectangle diagonals are equal
now consider triangle BQR and triangle CQR
BQ=CQ (1/2AC=1/2BD)
QR=QR (common)
by RHS similarity rule the two triangles are congruent.
by CPCT, BR=CR i.e, R is the midpoint of BC
BR=CR=1/2BC-(1)
Similarly, prove triangle QDP and triangle CDP are congruent
and by CPCT we get that DP=CP i.e, P is the midpoint of CD
DP=CP=1/2CD-(2)
first part of the question is done!!
ar(ABCD) = length × breadth = lb = CD×BC
ar(PQRC) = length ×breadth = lb = PC× RC
=1/2CD ×1/2BC
[from (1) and (2)..]
by transposing the 1/2 to the other side,
2 ar (PQRC) = CD×BC
=2 ar (PQRC) = ar (ABCD)
=ar (PQRC)= 1/2 ar (ABCD).
hence the proof.
sorry for taking time. hope its clear!
Answered by
4
Answer:
By mid pt. therom
PB=1/2AD and PB is parallel to AD
therefore in triangle DAC
by converse of mpt
DA is parallel to PB and B is half of AC
therefore P is the mid pt. of DC
similarly in triangle CAB
by converse of mpt
R is the mid pt. of BC
Ratio=1:4 since P is mpt of DC and PB is half of AD
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