plz explain cbse class 10 ncert eg 9 of chapter triangles
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given XY ||AC
find ratio ax\ab =?
hence,
angle bxy =angle bac (because corresponding angle)
angle byx =angle bca
then ,
∆ABC ~ ∆XBY
so
ar (ABC)/ar( XBY) =ab/xb whole square
(by theorem )
ar (ABC) = 2 ar (XBY) (by question ) 1
so ,
ar(ABC) /ar(XBY) = 2/1 2
from 1and 2
AB /XB whole square = 2/1
then ,
AB/XB = √2/1
so,
XB/AB =1/√2
so we find AX
then ,
we know AB
so
ab - xb =ax
√2 - 1 = ax
ratio =AX/AB = √2-1/√2
2-√2 / 2
find ratio ax\ab =?
hence,
angle bxy =angle bac (because corresponding angle)
angle byx =angle bca
then ,
∆ABC ~ ∆XBY
so
ar (ABC)/ar( XBY) =ab/xb whole square
(by theorem )
ar (ABC) = 2 ar (XBY) (by question ) 1
so ,
ar(ABC) /ar(XBY) = 2/1 2
from 1and 2
AB /XB whole square = 2/1
then ,
AB/XB = √2/1
so,
XB/AB =1/√2
so we find AX
then ,
we know AB
so
ab - xb =ax
√2 - 1 = ax
ratio =AX/AB = √2-1/√2
2-√2 / 2
Suzie786:
Ans is half... Where is the main ans??
this is full answer
Plz explain this:-ar(ABC)=2 ar
This:-ar(ABC)=2ar(XBY)
because in question xy divides the area ABC in two equal parts so we say that arABC = 2 ar XBY
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