wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The given figure shows a trapezium in which AB || DC and diagonal AC and BD intersect at point P. If AP : CP = 3:5 and area of APB is 27 m2, then the area of CPD is ___.


A

72 m2

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

75 m2

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

81 m2

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

84 m2

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

The correct option is B

75 m2


In ΔAPB and ΔCPD,

APB=CPD (vertically opposite angles)

PDC=PBA (alternate angles)

ΔAPB ΔCPD (by AA similarity criterion)

AP2CP2=ar(ΔAPB)ar(ΔCPD)

(Since the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.)

(APCP)2=ar(ΔAPB)ar(ΔCPD)

(35)2=27 m2ar(ΔCPD)

ar(ΔCPD)=27×259=75 m2


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Areas of Similar Triangles
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon