lets see who can solve question no. 9
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join BD.
through mid point theorem SR||BD and SR is half of BD........1
pf||BD pf is half of BD......2
from 1 and 2 SR||pf and SR=pf
srpf is ||grm
since s and f are midpoints join them.
asfd is ||grm and sfcb is ||grm
∆rsf and|| grm adfs are on same base bet same ||s
area of rsf=1\2 area of adfs ....1
similarly area of PSF =1\2area of bsfc....2
add 1 and 2
area of rsf±psf=1\2area of adfs+bcfs
take 1\2as common
hence area of ||grm psrf=1\2ar(ABCD)
through mid point theorem SR||BD and SR is half of BD........1
pf||BD pf is half of BD......2
from 1 and 2 SR||pf and SR=pf
srpf is ||grm
since s and f are midpoints join them.
asfd is ||grm and sfcb is ||grm
∆rsf and|| grm adfs are on same base bet same ||s
area of rsf=1\2 area of adfs ....1
similarly area of PSF =1\2area of bsfc....2
add 1 and 2
area of rsf±psf=1\2area of adfs+bcfs
take 1\2as common
hence area of ||grm psrf=1\2ar(ABCD)
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