Examples: Applications of Areas and Perimeters (Part- 2) Video Lecture - Class 7

hello friends this video on perimeter
and area but 18 is brought to you by
example calm no more pure Brahmin Sam
question number three plug adapt a cord
around the circular pipe of radius four
centimeters so this is a circular pipe
on which a wire was wrapped so this is
the wire and cut off the length required
of the cordon so whatever length was
required to wrap the circle that much
length was cut then she'd acted around
the square box of site for centimeters
now then the same wire which she had cut
she took the same wire and she wrapped
it around the square like this so did
she have any cord left that means let's
see that by wrapping it here she cut a
specific length of the wire the same
length of the wire when she'd acted on
the square so was that while so big that
some part of the wire was left even
after covering the square or not so that
is what we have to answer so first of
all let us try to find out the length of
the wire required to wrap the circular
pipe so length required to wrap the
circular pipe so what would be the
length required to wrap the circular
pipe that would be nothing but the
circumference of this circular pipe so
the circumference is given as 2 PI R now
radius is given as 4 centimeters so 2
into pi is 3.14 into R is 4 centimeters
this is equal to 25 point 1 2
centimeters so this much length of the
wire is needed to wrap this circular
pipe so basically he has cut 25 point
1/2 centimeter of the wire now he takes
this 25 point 1/2 centimeters of wire
and wraps it around a square now the
length of each side of the square is
given as 4 centimeters so if the length
of each side is 4 centimeter then what
would be the perimeter of the square
that is what would be the boundary of
the square so perimeter
square would be equal to 4 into lengths
of each side that is 4 so 16 centimeter
so this would be the perimeter of the
square that is the boundary of the
square and how much wire do you have you
have a wire of length 25 point 1/2
centimeter that means you have more wire
so even after covering the square you
will still be left with some amount of
while because you only need 16
centimeters and you already have 25
point 1/2 centimeters so how much cord
will be left even after covering the
square so the cord left would be 25
point 1/2 minus 16 which is equal to 9
point 1/2 centimeter so this much amount
of wire or this much length of the wire
would be left even after wrapping it
around the square box question number 4
in the following figures find the area
of the shaded portion so here what do
you see so the image that you see is
actually a square because this side is
equal to 20 this is 20 this is also 20
all the sides are equal to 20
centimeters so you have a square and
inside the square you have a triangle so
you have to find out the shaded region
so for that let us first find out the
total area of the square so area of the
square is given as side into side that
is length of each side multiplied by
length of the other side so this would
be 20 into 20 which is equal to 400
centimeter square because length of each
side of the square is equal to 20
centimeters now let us try to find out
these areas let us call these as
triangle 1 2 & 3 so let us call this
triangle 1 this is triangle - and this
is strangle 3 so let us try to find out
the areas of these triangles so if we
are able to find these areas then we
will subtract these
of the square and then we will find out
the shaded region so the area of
triangle 1 will be equal to half into
base into height so for triangle 1 we
can consider base as any of the sides
let us say we consider base as P 2 so if
P P is the base then what would be the
height the perpendicular which falls on
P D that it is PQ so what is the length
of P P so if this is 10 centimeters this
is also 10 centimeter and since this is
this is 20 centimeter so the opposite
side of the square would also be 20
centimeters so this would be 1/2 into PT
that is 10 and P Q that is 20 so two
things the 20 so this would be 100
centimeter square so that is the a of
triangle one in a similar way let us try
to find out the area of triangle 2 so
for triangle 2 also this would be half
into base into height so here let us say
that the base is you are that means this
part so the height would be nothing but
the perpendicular on u bar which is Q R
so this would be Q 1 so u R is given as
10 centimeters and Q R is given as 20 so
again to pfizer
n this is also equal to 100 centimeter
square now let's focus on area of
triangle 3 so there also it would be
half into base into height let's see if
I assume the base as su then the height
would be the perpendicular on su from
the opposite vertex which is PS so this
is half into 10 into 10 so this is 2/5
is a 10 so this is 50 centimeter square
so we found out the ears of all the
three triangles 1 2 & 3 therefore total
area of triangle 1 2 & 3 all of them
together would be hundred plus 100 plus
50 centimeter square that is 250
centimeter square so all these area 3
areas taken together
- 50 centimeter square and the total
area of the square is 400 centimeter
square therefore the area of shaded
region will be equal to the area of the
square which is 400 centimeter square so
this area of the square - the total area
of the three triangles which is 250
centimeter square that is equal to 150
centimeter square so this is the area of
the shaded region question number five
find the area of the quadrilateral ABCD
here ac is equal to 22 centimeters BM is
equal to 3 centimeters that means this
length is given as 3 centimeters this
total length of AC is given as 922
centimeters DN is also given as 3
centimeters so we have to find out the
area of the quadrilateral so when you
look at this quadrilateral you see that
the quadrilateral is made up of two
triangles pectus triangle ABC and
triangle ADC so if you find out the area
of triangle ABC and tell your crank and
ABC and then add the two you would get
the area of the quadrilateral so let us
find out the area of triangle a B C so
area of a triangle is half into base
into height so for triangle ABC base
could be AC so if AC is the base then
what would be the height height would be
BM because BL is perpendicular on AC now
AC is given as 22 centimeters BM is
given as 3 centimeters so this is 33
centimeter square in a similar way let
us try to find out the area of triangle
a DC so in this case also this would be
happened to base into height so we can
again fancy
you see as the base so if AC has the
base than the perpendicular for tangled
a DC would be DN because D n is
perpendicular on AC so this would be the
end so this is half into AC that is 22
into DN which is three so this would
again be 33 centimeter square therefore
area of the quadrilateral would be equal
to AE of triangle ABC plus area of
triangle ABC so 33 plus 33 that is 66
centimeter square so this would be the
area of the quadrilateral ABCD so with
this we have reached towards the end of
this lesson on parameter and area and I
hope that this lesson would have helped
you so this is a very interesting topic
where once you understand the concepts
you can create the diagrams on your own
looking at the problems and you can
solve them and this is going to be
really interesting so try to solve more
questions get your concept scared so I
hope this lesson would have helped you
see you all in the next lesson thank you
please visit exam fee or come for free
quality education you can learn with a
simple four-step learning process
wherein you can watch video lessons you
can ask your questions you can refer
notes and you can take a free online
test we have content for class 6 to 12
on physics chemistry mathematics and
biology along with practical videos so
please subscribe to our channel for
daily updates thank you