Areas of Similar Triangles Video Lecture - Class 10

hello friends welcome to resent all calm
we are talking about Series two
triangles today's webisode number twenty
threads today we are going to learn a
special theorem on areas of similar
triangles
so let's take a look at what is this
special theorem so like we always do
friends let's first understand what the
theorem is telling us the ratio of areas
of two similar triangles is equal to the
square of ratio of their corresponding
sides so first thing that we have to
understand trends is that this is not
about the similar triangle itself but it
is about the areas of two similar
triangles so that means we start with we
have triangle ABC it is given that ABC
and P Q are these two triangles are
similar so we have been given triangle
ABC is similar to triangle PQR this is
what is already given to us so what the
theorem is telling us is that the ratio
of the areas of two similar triangles so
the area of triangle ABC can be written
as a b c we know that area we write it
as abbreviated from a R so area of
triangle ABC divided by area of triangle
PQR PQ R because that is what will be
the ratio of the areas so the actual
area of triangle ABC divided by the
actual area of PQ R what this ratio is
this is equal to square of the ratio of
corresponding sides so we know the
corresponding sides in this triangle is
a B by PQ right so this ratio think we
don't have enough space let's just
probably write it over here
so this ratio is equal to square of
ratio of corresponding sides which is a
B by a B by PQ square of this ratio so
whatever this ratio is it will be times
two to the power two and since we know
basically that in similar triangle all
these ratios are saying that is right we
have a B by PQ equal to AC by PR PR
which is equal to BC BC by Q R we know
that in two similar triangles their
corresponding sides are always in the
same ratio so the area of the triangle
ABC divided by area of triangle PQR is
equal to the the ratio of the
corresponding sides squared so a B by PQ
square we can also say the same thing is
equal to AC by PR square
which is also equal to BC by QR square
because all these ratios are same so
friends the area the ratio of the areas
of two similar triangles is equal to
square of the ratios of the
corresponding sides now that we
understand the question or the theorem
let's look at how can we solve it now
friends we will start out by simply so
we have got triangle ABC similar is
similar to triangle PQR so what we do is
from point A we drop a perpendicular on
side BC let's say it intersects here at
X similarly from point P we drop a
perpendicular on Q are intersecting Q
are at why now friends we know that area
of a triangle is equal to 1/2 times base
times height that is what how we find
out the area of a triangle so in this
case area of triangle ABC so area of
triangle a b c will be equal to let's
write it here half base is bc half bc x
ax a X this is just a multiplied
multiplication sign and similarly area
of triangle PQR PQR will be equal to
half I think let's not write the
multiplication sign like this this
market is dot because the thing this
multiplication sign will get confused
with X so half dot base is QR QR x py
p1 so if we we are concerned about the
ratio remember we want to find out the
ratio of area of triangle ABC by the
area of PQ R so we divide both so half
and half gets cancelled so what we have
here is BC times BC times ax ax divided
by Q R times QR x py P Y let's say this
is our equation 1 then friends what we
do is that now we will so we have this
90 degree we have X we have Y now
friends we are going to take a look at
this triangle and this triangle and Y
you will see that in a moment if you
look at these two triangles we know that
angle B is equal to angle Q why because
triangle ABC is similar to triangle PQR
so all corresponding angles are equal
and also this angle is 90 degree by
construction so these two triangles are
similar triangles so here friends what
we have is we can express a X a X
divided by py p y equal to a B divided
by p q PQ right because these two
triangles are similar we can say a X
divided by py is equal to a B divided by
PQ corresponding sides are in the same
ratio but now remember that
ABC is similar to triangle PQR so we
have a B by PQ equal to BC by Q R right
we actually have because AD by PQ is
equal to BC by Q R right
so from here friends what we see is that
so in this we have a B by PQ and over
here we have AV by P Q so from this two
we can say ax by p y py this ratio ax by
py is equal to BC by Q R B C by Q R so
the ratio of the altitudes so ax by py
is nothing but it is equal to BC by Q
are now friends now earlier we have seen
that the area of the the ratio of the
areas area of triangle ABC divided by
area of triangle PQR we saw this ratio
to be equal to BC times a X divided by Q
R times we should have putted by dot
notation x py but
just now saw that this ratio a X by py
we can express that as BC by QR we just
saw that right when we looked at these
two triangles we're similar from there
we saw that a X by P Y can be written as
BC by QR so we have already BC by QR
here
QR this x instead of ax by py we will
write BC by QR and so you will notice
your friends we will get PC by QR square
that is BC by QR x BC by QR is nothing
but VC by QR square so friends the ratio
of the areas of two similar triangles
ABC and PQ are we got this as the square
of corresponding side PC divided by Q R
and because we know that ABC and P Q R
our corresponding they are actually
similar triangles so each of these
ratios are same so we don't have to
really prove anything else because we
know that a B by PQ is equal to BC by Q
R is equal to AC by P R right so all
these ratios are saying so we have
proved that the area of ABC divided by
area of PQ R is equal to 1 of the ratio
square so in this case it is PC by Q R
square and because all these ratios are
same that means we can write
says a B by P Q square or BC by Q R
square or AC by P R square so since this
is how we are going to prove it now that
is just clean it up and write it how you
are going to be doing it in your tests
so we start out with seeing what is
given we draw two triangles it is given
that ABC is similar to triangle PQR we
need to prove that the area of triangle
ABC divided by area of PQ r equal to the
square of their corresponding sides we
draw a perpendicular ax and py
respectively on each of these triangles
we know the area of triangle is half
times base times altitude so we
separately write out the areas of
triangle ABC and PQ R and then when we
divide them to find out the ratio of the
areas we divide them we get a
relationship which is x times BC / py
times Q R we call it equation number 1
then we closely take a look at the
triangles a b x and p qy and here we
find out that angle b is equal to angle
Q because angles in a similar triangle
are equal and ax P equals P Y Q because
both of them are 90 degrees this and
this are 90 degrees and this and this
angles are equal so these two triangles
are similar by a a similarity rule
therefore we find out that ax by P Y can
be written as BC by QR now we replace
the X by py in equation 2 by equation 1
so here we get the ratio of the two
areas is equal to BC by Q R square so
now since we know that all the ratios
are same all these ratios are same so we
can express the area
ladies a divided by area of triangle PQR
that is we can express a ratio of areas
of two similar triangles as two similar
triangles as PC by QR whole square a B
by PQ whole square or AC by PR whole
square so friends hence we have proved
the theorem in the next video we will
take a look at some of the questions
that may come up regarding areas of
similar triangles