Elitmus
Exam
Numerical Ability
Area and Volume
Q. ABCD is a square. P is the mid point of AB. the line passing through A and perpendicular to DP intersects the diagonal at Q and BC at R, If AB =2 then PR=?
Read Solution (Total 5)
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- Let the line passing thru A meet DP in E.
So In triangle DAP
DP= Sqroot(5)
AE=2/Sqroot(5)
triangle AEP is similar to triangle ABR
So AR= sqroot(5)
=>RB=1
=> PR=Sqroot(2) - 11 years agoHelpfull: Yes(11) No(17)
- let E is point of intersection of line AR.
IN TRIANGLE ADP;
WE GET,
OP=SQRT(5).
now in similar triangles AEP AND ADP (By angle p in both triangles is common and one side AP is also common)
therefore
AP/DP=EP/AP
AP^2= EP*DP
1=EP*SQRT(5)
SO, EP=1/SQRT(5)
AGAIN IN SIMILAR TRIANGLES APE AND PER(BY ONE SIDE EP IS COMMON)
AP/PE=PE/PR
AP*PR=PE^2
1*PR={1/SQRT(5)}^2
HENCE, PR=1/5.
WHICH IS REQUIRED VALUE OF PR. - 11 years agoHelpfull: Yes(4) No(4)
- Ans : square root of 2. if u draw the diagram it will give u an triangle where u can apply pythagoras theorem..
- 11 years agoHelpfull: Yes(1) No(2)
- Plesse give options
- 11 years agoHelpfull: Yes(0) No(3)
- underroot(29) / 5.
- 7 years agoHelpfull: Yes(0) No(0)
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