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NCERT Exemplar Solutions Class 6 Mathematics Solutions for Exercise in Chapter 4 - Fractions and Decimals
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42.28 - 3.19 = 39.09
Answer:
True.
Subtracting 3.19 from 42.28,
42.28- 3.19 =39.09
Video transcript
"hello students welcome to lyra q
a video session i am sef your math tutor
and question for today is
if two equal parts of a circle
intersects within the circle
true that the segments of one of the
chord
are equal to corresponding segments of
other chord
so let a b and c d be the two equal
chords such that
a b is equal to cd
so these are the two equal chords which
is mentioned in the question
now in above question it is given that
this a b
and c d intersects at a point so let us
see in the figure
a b and c e those chords intersects
at point e over here
so
and you can see further that it is now
bit to be proven that
the line segment e equal to d e and c e
equal to b
so this is given
and this is to be proven
to proceed with the proof first of all
we will construct something
in the figure so if i show you over here
from the center of the circle we have
drawn a perpendicular a b
perpendicular to a b and that
perpendicular is
om so om over here is a perpendicular to
our chord a b so this is where
it forms 90 degree from the center
and in the step two similarly you have
to draw draw o
n which is perpendicular to cd
in the step three you have to join o e
so this is the oe so this is the extra
construction which we have done
now the diagram is there which is shown
now from the diagram it is seen that om
bisects
a b so om is perpendicular to a b
as we know that a perpendicular drop
from the center to the chord bisects
that chord
so here we can say that from figure
o m perpendicular to a b
o m by 6
a b similarly
o n also bisects cd
now as in this case it is known that a b
is equal to cd
so from this we can say that am is equal
to
nd
let this be our result number one
now another result we can obtain here is
mb
equal to cn
and let this be our result number two
now consider triangle ome and o
and e so in this triangle
o m e and o
and e you can see that
there can be rhs congruency
as per the rhs right hand side
congruency
angle ome
and angle o n e these two angles
are equal because they are perpendicular
and o
e is equal to o e that is the middle
common one
and it is a common sides that is why it
is a equal one
now om and on if you see they are also
equal
because a b and c d are equal and so
they are equidistant from the center so
c o m and o n this is om and this is o n
and because a b and c
d are equal chords
and so they are equidistant from the
center and that is the reason why we are
stating that om
is equal to o n so let us write it down
something
over here
as per rhs congruency
triangle omega
and triangle o and e
they are similar
hence we can say angle ome
will be equal to angle o n e
because perpendiculars
now you can see oe and oe are the common
side so oe equal to oe
for both the triangle
and moreover om will be equal to o n
this is because a b and c d are equal
chords
so they are equidistant from the center
now this is this has become our hs
congruency
and we can say that triangle ome is
similar to triangle one
i'll write down here
they are congruent these two triangles
are congruent
hence we can say that m e is equal to e
n
this is because they are congruent parts
of the congruent triangle
so they are
congruent parts and they are
corresponding parts of the congruent
triangle
so cp ct can be the short form for
corresponding parts of corresponding
congruent triangle now uh
let this name be as result number three
now from equations one and two we get
am plus m e is equal to nd plus en
from result 1 and 2
am plus
m e is equal to nd plus en
and hence we can say a is equal to ed
so
now from equations 2 and 3 we get
mb minus m e
is equal to c n minus n
so from this we can say that eb
is equal to ce
hence we have proved
if you have any query regarding this you
can drop it in our comment section
and subscribe to lido for more such q a
thank you for watching"
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