Examples- Triangles Video Lecture - Class 10

hello friends welcome to rising pearl
calm we are talking about series to will
be learning about crying ghouls this is
lesson number 21 friends today we are
going to focus on solving questions
related to areas of similar triangles so
let's get started now friends in the
last webisode for the first time we
looked at one special theorem regarding
the areas of similar triangles so the
theorem went like this so the theorem
about areas it was something like this
the ratio of areas of two similar
triangles is equal to the square of
ratios of their corresponding sides that
means friend what we have is if we are
given triangle ABC similar to triangle
PQR if this is what is given to us then
what we can write about their areas is
that the ratio of areas the ratio of
area so area ABC divided by area PQR
that is the ratio of the areas of
similar triangle is equal to square is
equal to square of the ratios of
corresponding sides so because these two
triangles are similar ad by PQ is equal
to AC by PR is equal to BC by QR because
these two triangles are similar but the
ratio of areas is equal to PC by QR
whole square is equal to AV by PQ whole
square is equal to EC by PR whole square
so this is a theorem that we had seen at
great length in the last video I am
going to provide a link below here as
well for your quick reference let's take
a look at some questions that may use
this theorem so first question is if the
areas of two similar triangles are equal
so we have it is given that areas of two
similar triangles if they are equal
prove that the triangles are congruent
so how do we go about proving this so
first friends let's draw two triangles
we have drawn triangle ABC and PQR and
these two triangles are similar and what
is given is their areas are equal so we
just now we learned that this theorem we
know that in similar triangles the ratio
of areas of two triangles is equal to
the square of ratios of corresponding
sides it is given that ABC is similar to
PQ R so we can write so we can actually
write area of triangle ABC divided by
area of triangle PQR is equal to AV by
PQ whole square equal to AC by PR whole
square equals to BC by Q R whole square
but it is given that these two triangles
have same area that is area of ABC is
equal to area of PQ R that means each of
these ratios is equal to 1 so each of
these ratios is equal to 1 because area
of triangle ABC divided by area triangle
PQR is equal to 1 right so what we have
here if you look at AV by PQ whole
square is equal to 1 from here we get a
be equals P Q so we get a B equals PQ if
AC by P R whole square equals 1 from
here we get AC equals P R similarly we
will have a we will have DC equal to Q R
so friends if you look at these two
triangles now a B is equal to PQ AC is
equal to PR and BC is equal to Q R so we
say these two triangles are congruent as
per s s s side-side-side con ruin rule
so friends so we have proved the two
triangles are indeed Condren let's take
a look at our second question
second question states P Q R P Q and R
are respectively midpoints of sites ABB
CCA of a triangle ABC find the ratio of
areas of triangle PQR and ABC so here
friends we don't have the diploid given
to us so let's draw it ourselves so we
have drawn a simple triangle ABC and we
have found out the midpoints P Q R
respectively that is piece the midpoint
of ad Q is midpoint of BC and R is
midpoint of AC right if you look at this
big triangle ABC we have a P equal to P
B and a are equal to R C so we can say a
B by P B is equal to a R by R C because
both ratios are equal to 1 right so if
that is the case then it means P R is
parallel to BC why because basic
proportionality theorem in a triangle
ABC if PR is such a line that divides a
B and is in same ratio then the line P R
is parallel to the third side PC right
so similarly friends we can say that Q R
is parallel to a B Y again if you look
at the triangle this with C a B C R by a
R is equal to C Q by B Q that means this
line Q R is parallel to a B right so now
offense if you look at if you look at
this quadrilateral P r QB what we have
is we have both pairs of opposite sides
parallel that means this is a
parallelogram that means we have
P r BQ is a parallelogram now in a
parallelogram we know opposite angles
are equal that means this angle must be
equal to this angle right also this side
this side must be same as this side now
let's clean this up a little bit right
so what we have done is we have found
out this angle is equal to this angle
and this side is equal to this side
also this side is equal to this side
because opposite sides of a
parallelogram are equal and parallel now
similarly friends if you look at the
quadrilateral P r CQ P R is parallel to
this and CR is parallel to P Q so
quadrilateral P r CQ is also a
parallelogram and so this angle will be
equal to this angle right now since so
similarly we can also prove that a R QP
is also a parallelogram but I think we
would most likely not need this so let's
let's keep going now if you look at the
small triangle friends P Q R and ABC if
you look at the small triangle PQR and
big triangle KBC you will notice that
angle B is equal to angle R angle P is
equal to angle C right so we can say
that these two triangles are actually
similar the small triangle and the big
triangle because angle D is equal to
angle R and angle C is equal to angle P
by a a we don't even need the third one
by a a similarity criteria or similarity
rule we can say that these two triangles
are actually similar
right we actually do not even need this
third point so we actually did not
really we didn't use this one we don't
need to have this so the small triangle
and the big triangle are actually
similar triangles so then friends what
we can say is that we know that area of
the similar triangle PQR divided by area
of similar triangle ABC can they express
SPR by BC y PR by BC because so we know
that this angle was equal to this angle
right these angles were equal similarly
this angle was equal to this angle so
the only angle left is this angle must
be equal to this angle right so if you
look at let's just do it this way area
of triangle PQR divided by area of
triangle ABC this is the ratio that we
need to actually find out this is our
original question right this is the
question find the ratio of areas of
triangle PQR and ABC so PQ our area of
triangle PQR divided by ABC is equal to
square of corresponding sides so we are
trying to find out what are the
corresponding sides so in the small
triangle angle R is equal to angle B of
big triangle so which side is opposite
of angle r PQ so let's write PQ here PQ
in this case I have actually written PR
opposite to Q is PR and opposite of a is
BC so that's okay so here we are taking
angle B and angle are so opposite to
angle R is PQ and opposite of B is AC
let's write AC now AC let's write it
here AC is equal to a R
plus our see now because our is a mid
point I can write a CS twice a R or I
can write this as twice R see the length
AC I can write this as two times AR or
two times RC now our say remember
friends p PR CQ is a parallelogram which
we have proved so this length is
actually equal to this length right so
two RC is nothing but 2 RC same as 2 PQ
2 PQ so now if you come back here and
here our numerator is PQ but instead of
AC if you put 2 PQ here 2 PQ then what
you get old square you will get P Q P Q
will cancel out it will be 1 by 2 whole
square which will be basically equal to
1 by 4 so this ratio area of small
triangle divided by area of big triangle
is equal to 1 by 4
since let's take a look at one last
question before we wrap up this
particular ribosome so here what we have
is we have a triangle ABC and b PQ such
that two equilateral triangles there two
equilateral triangles such that p is the
midpoint of BC so let's draw the
triangles because the triangles are not
given so I have drawn a equilateral
triangle ABC and again like I have said
always these questions are not
construction questions so you are not
really drawing with your protractor and
your ruler you are just roughly drawing
the shapes if it is not given to us so I
have drawn triangle ABC it is given that
ABCs
a little triangle also it is given that
p is a midpoint of BC so on BC I take a
midpoint P now in this case I have drawn
Q so I have drawn a 60 on line BC at
Point P I have drawn a 60 degree angle
that is this angle I have drawn this as
60 degrees and this will intersect a B
and this point I'm calling it is Q so b
PQ b PQ is a equilateral triangle
because angle b is 60 degrees this is 60
so this angle will automatically become
60 degree but instead of drawing the
angle here I could have actually also
drawn it this way right so I absolutely
could have done that way so in that case
this angle will be 60 degrees and Q will
be outside it will not be inside the
triangle ABC but Q will be lying outside
so both scenarios we have to consider as
we solve this question so what do we
have to solve here they find the ratio
of areas of triangle ABC the big
triangle to the area of the small
triangle B P Q so friends again the way
we are going to look at this is ABC b PQ
both are equilateral triangle which is
given to us so angle all angles in a b c
and b PQ b PQ are 60 degrees right
therefore ABC and b PQ will be basically
similar triangles because all angles of
the big triangle ABC and the small
triangle b PQ and i think let us just
see it will be little easier to visually
see it here it's exactly the same in
this case as well so here this angle is
60 this angle is 60 this angle is 60
similarly here this angle is 60 this
angle is 60 this angle is 60 because ABC
is an equilateral triangle and b QP is
also an equilateral triangle so because
all angles
60 we can say that these two triangles
are similar same this triangle and the
big triangle are similar triangles now
in similar triangles we know that the
ratio of areas is square of ratio of
corresponding sides right so what we
have here is area of triangle ABC
divided by area of triangle B PQ is BC
by BP because it's an equilateral
triangle it's a special case all the
sides are the same that is this side is
equal to this side is equal to this side
similarly in the small triangle its
equilibria triangle this side is equal
to this side is equal to this side so we
have big triangle so BC by BB whole
square right and why did I pick BC and
BB because P is a midpoint I want to
express the ratio instead of writing the
ratio in terms of b q and AC actually i
could have done that as well so
essentially here I have taken BC by BP
right and it is given to us that BC is
nothing but 2 times BP right so BP and
BP will cancel out so I have 2 times 2 2
square is 4 so the ratio of area of big
triangle to the area of small triangle
is 4 so friends this wraps up the
problems questions on the areas of
similar triangles in the next webisode
we are going to take a look at some
right angle triangle related theorems