CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

In the given figure, AB = 6 and BC = 4. BC is extended to E such that C is the midpoint of BE and join AE. It cuts CD in P. Now consider the two points:
i) P is the midpoint of AE
ii) P is the midpoint of CD


A

Both (i) and (ii) are true

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B

Only (i) is true

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

Only (ii) is true

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

Neither (i) nor (ii) is true

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B

Both (i) and (ii) are true


Consider AEB and (\triangle PEC\) . Since PC is part of DC and DC is parallel to AB.
EAB=EPC [corresponding angles]
AEB=PEC [common angle]
By AA similarity criterion,AEB ~ PEC
So, APPE=BCCE --------------------------(I)
Since C is the mid point of BE, BC=CE. -------------------------(II)
From (II) and (I), APPE = BCCE = 1,
So AP = PE and hence P is the mid point of AE.
Now in right triangle ABE , AB = 6, BE = 2BC = 8.
So by Pythagoras theorem,
AE2=AB2+BE2
AE2=36+64=100
AE=10
Hence PE = AE2 = 102 = 5.
In right triangle PCE, CE = 4 = BC, PE = 5, by Pythagoras theorem,
PE2 = PC2 + CE2
PC2=2516=9
PC=3 .
Since AB = CD (rectangle), CD = 6 and hence DP=CDPC=63=3.
That is CD = PC. So P is the midpoint of BC also.


flag
Suggest Corrections
thumbs-up
0
BNAT
mid-banner-image