Areas of Parallelograms and Triangles Video Lecture - Class 9

hello friends we have been learning a
lot about parallel lines triangles and
parallelogram and we know certain
properties of them but have you ever
wondered what will happen if we will
combine triangles with parallel lines or
parallelogram with parallel lines so
let's start this session by juggling
with these shapes and figuring out
certain very simple and yet very
important properties here you can see a
road and on both the sides of the road
there are straight lines
so using this situation let's see what
we can come up with consider there are
two poles on both the sides of this road
and let's draw two triangles by joining
these poles so by looking at the diagram
can you tell me what is common between
the two triangles they have a common
base and one more thing is common
between them they also have a common
height because if they see the boundary
of the road they form parallel lines and
we know the distance between two
parallel lines is always constant
therefore height and base is st. for
these triangles so when we calculate the
area we use a formula half into base
into height now since the base and the
height of both the triangles is same the
area of the triangles is also going to
be the same so from this we can conclude
then two triangles are drawn between the
same parallel lines and having the same
base their areas are also going to be
equal let's try some other combination
of triangles and see what we can learn
through it now consider one light pole
on one side and three light
on other side of the road again we have
two triangles now can you tell me what
is common in both of these triangles in
terms of base and height we can see the
height of both the triangles is same
because they lie between the same
parallel lines but what can we say about
the base here we can see that the bases
are different if we notice the diagram
we can clearly see that we have split
one big triangle into two by joining a
line even if they didn't have the
parallel lines we can still say that
both the triangles have the same height
so the conclusion is if we draw two
triangles by splitting one big triangle
the height of all the three triangles is
going to be the same and then we find
the areas of these triangles their areas
will not be equal but the ratio of their
areas will be equal to the ratio of
their bases because then we take the
ratio of area the constant value half
and the height will get cancelled out
and the remaining thing will be the
ratio of the bases here we spoke about
the basis but suppose the bases were
equal and the triangles are still
between the same parallel lines with
equal area so now what can you see about
the height of these triangles I have
already given you a hint
the height of both the triangles will be
same so what does this imply it says
that triangles with equal basis and
equal area will have equal height or we
can say that they lie between the same
parallel lines here we are done with
properties of triangles let's do a small
recap before we move to the property of
parallelogram here we move to one and
the last property of the session based
on parallelogram now if you look at the
same streetlights arrangement we can
make a parallelogram out of it now
suppose if we add two more streetlights
on the top then we can see that we have
two different parallelogram but with the
same base so can you think of something
of the area of these parallelograms we
know the formula to calculate the area
of parallelogram is base into height we
know the base of both the parallelogram
is same but then what about the height
of these parallelogram
let's draw perpendicular heights for
both the parallelogram and what we can
see now are in the heights same yes they
are so from this we can say that the
area of both the parallelogram is equal
or we can say that the parallelograms on
the same base and between the same
parallel lines are equal in area
what is the session interesting we have
learnt few important properties about
triangles and parallelogram but where
are we going to use these properties
let's have a look at one problem which
we will solve using these properties and
the given figure pqrst is a Pentagon a
line through Q padding to P our needs as
part produced at you we have to show
that area of triangle PQR is equal to
area of triangle P you are and area of
quadrilateral PT su is equal to area of
Pentagon pqrst before we start let's
name the point of intersection of PU and
Q R as Y friends if you look at a
question the question R word like areas
of triangles and parallel lines so by
looking at these keywords you can
directly say that the problem is based
on areas of parallelogram and triangle
chapter now we have to prove that area
of triangle PQR is equal to area of
triangle P you are next thing is you
don't have to be afraid by looking at
the diagram let's just try to simplify
the diagram and concentrate on the part
we are given and we need to prove first
so the most important part in such
questions is to find out the parallel
lines which are P R and Q you if we look
closely both the triangles also lie
between the same parallel lines and they
have a common base that is P R so now
can you tell me using which property we
can say that the triangles have equal
area yes you are right using property
one
now we can say that area of triangle PQR
is equal to area of triangle P you are
here we are done with the first part of
the solution
the next part says that we have to prove
that the Pentagon and the quadrilateral
are having equal areas now the second
part may look complicated but you will
see how we can easily solve it using
these first part itself here we can
split the quadrilateral into
quadrilateral PPS r and triangle p ru
and in the same way we can split the
pentagon into quadrilateral PT s r and p
RQ and as we have already proved that
both the triangles have same area so
from 1 2 & 3 we can say that the
quadrilateral and the Pentagon this also
has equal area wasn't this also simple
let me leave you all with another
question so that you can polish your
skills and strengthen your math powers
in the circle quadrilateral ABCD and C X
Y D have equal area you have to prove
that that both the quadrilaterals are
parallelogram and show that area of
triangle C XY is equal to half area of
quadrilateral ABCD now again looking at
the diagram you may feel it is way too
complicated but it's not I can give you
a hint before I leave you can solve this
question very easily using property
number 4 till then keep watching keep
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