Question

Who can solve the 25. Question number

Answer 1

GIVEN: BD=AD,PD parallel to CQ

TO PROVE: ar(tr.BPQ)=1/2 ar(tr.ABC)

CONSTRUCTION: join CD

PROOF:since D is the is point of AB,ar(tr.BDC)=ar(tr.ADC)=1/2 ar(tr.ABC)

but ar(tr.BDC)=ar(tr.BDP)+ar(tr.DPC) - eq. 1
so 1/2 ar(tr.ABC)=ar(tr.BDP)+ar(tr.DPC)

now ar(tr.DPC)=ar(tr. DPQ) - eq.2
[because both the triangles lie on the same base DP and between same parallels DP parallel to QC]

from eq. 1 and eq. 2
1/2 ar(tr.ABC)=ar(tr. BDP)+ar(tr.DPQ)
1/2 ar(tr.ABC)=ar(tr. BQP)

hence proved

[here tr. stands for triangle which is to treplaced with the symbol of a triangle,ar. stands for area and eq. for equation]


hope this might help.