Class 9 Exam > Class 9 Questions > In the adjoining figure, ABCD and PQRC are re...
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In the adjoining figure, ABCD and PQRC are rectangles, where Q is the midpoint of AC. Then DP is equal to
a)DP =1/3 DC
b)DP < PC
c)DP = PC
d)DP > PC
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
In the adjoining figure, ABCD and PQRC are rectangles, where Q is the ...
PQ Is parallel AD.. By corresponding angle axiom...ang.CPQ=ang.CDA=90degree.. also, Q is the midpoint. .so, P is also the midpoint. .so,Dp=PC..
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Community Answer
In the adjoining figure, ABCD and PQRC are rectangles, where Q is the ...
Given that PQRS and ABCD are rectangle and Q is the mid pt of AC.
Since PQRS and ABCD are rectangle, every angle would be of 90°.
Consider AD and PQ.
angle ADP = 90°
angle QPC = 90°
Now, angle ADP lies corresponding to angle QPC and are equal.
If corresponding angles are equal, then the lines are parallel
=> AD || PQ
Now consider ∆ADC.
Q is the midpoint of AC (Given)
AD || PQ (shown above)
We know that,
line drawn parallel from the midpoint of a side, to other side, bisects the third side.
Hence, P would be the midpoint of DC
Hence, DP = PC
Since PQRS and ABCD are rectangle, every angle would be of 90°.
Consider AD and PQ.
angle ADP = 90°
angle QPC = 90°
Now, angle ADP lies corresponding to angle QPC and are equal.
If corresponding angles are equal, then the lines are parallel
=> AD || PQ
Now consider ∆ADC.
Q is the midpoint of AC (Given)
AD || PQ (shown above)
We know that,
line drawn parallel from the midpoint of a side, to other side, bisects the third side.
Hence, P would be the midpoint of DC
Hence, DP = PC
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In the adjoining figure, ABCD and PQRC are rectangles, where Q is the midpoint of AC. Then DP is equal toa)DP =1/3 DCb)DP < PCc)DP = PCd)DP > PCCorrect answer is option 'C'. Can you explain this answer?
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