Examples: Area of Quadrilaterals and Polygons Video Lecture - Class 8

hello friends this video on mensuration
part six is brought to you by
example.com normal tio cram exam so with
this we have learned a lot about the
area and trading unit of many different
shapes so let us try out a few questions
question number one the diagonal of a
quadrilateral shaped feet is 24 meters
and the perpendicular
dropped on it from the remaining
opposite vertices at 8 meters and 30
meters find the area of the feet so in
this case it is a general quadrilateral
it nothing has been spoken about whether
it is the square or a rectangle or
Adobe's integers to general
quadrilateral and what are the values
which are given here so if you look at
this if let's name this quadrilateral
ABCD so this gives you see the length of
the diagonal is given and the length of
the perpendicular drop down the
diagonals are also given now we already
know that the area of a general
quadrilateral is given as half into
diagonal into some of the perpendiculars
drops on the diagonal at all these three
versions are already given in this
problem so this would be how into
diagonal which is 24 meters into h1 and
h2 between 8 and 13 respectively so this
would be 1/2 into 24 into 21 so this is
12 so 12 into 21 is 252 meter square so
this is the area question number 2 the
floor of the building consists of 3,000
times with a normal shape and each of
its diagonals and 45 centimeters and 30
centimeters in length
find the total cost of polishing the
floor if the cost per meter square is
useful selection assume that this is the
floor and the floor is made up of tiles
which are in
shape of rhombus so here if you look at
Siam not joined and acting and just
trying to give you an idea so if you
focus at each side it is a red shape of
a rhombus
each side in the shape of a rhombus
please note that so how many talents do
we have we have a total of 3000 pedals
now if we want to find out the total
cost to polishing the floor for that
what do we need to know we need to find
out that area of the floor so how do we
find out the area of the floor so area
of the floor would be equal to the area
of the 3,000 times and each time is a
rhombus so basically if you are able to
find out the area of the 3,000 rhombus
this area of 3,000 rhombuses is the area
of the floor that simple
so for that first of all our first step
would be to find out the area of one
Robert now we have found out for
ourselves that area of orbital give
entire ass into product of the diagonals
now in this problem the length of the
diagonals are given as 45 and 30 so this
would be half into 45 into 30 so 2 into
15 is 30 so what inside into 15 is equal
to 675 centimeter square this is the
area of one rhombus so therefore the
area of 3,000 rhombuses would be equal
to 3000 into 675 which is equal to two
zero two five zero zero zero centimeter
square so we can convert it into meter
square so we know that one meter is
equal to hundred centimeters therefore
one meter squared will be equal to ten
thousand centimeter square therefore one
centimeter square will be equal to one
divided by 10,000 meter square right so
therefore this much centimeter square
the equal to two zero two five zero zero
zero divided by 10,000 meter square so
this here is zero zero zero will cancel
out so it is two zero two point five
meter square so this is the total area
of the floss so now let us try to find
out the part when the problem it is
given that the cost per meter square is
rupees 4 that means cost for one meter
square so if the area is one meter
square then the cost is rupees 4 so
therefore if the area is 200 two point
five meter square then the cost would be
rupees 4 into two hundred two point five
which is equal to rupees 810 so Ellie
Hendricks ten would be the total cost of
polishing the floor thank you please
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