Problem Set on Triangles : 2 - Geometry of Quadrilaterals, Quantitative Aptitude Video Lecture | Study UPSC Prelims Paper 2 CSAT - Quant, Verbal & Decision Making - UPSC | Best Video for UPSC

the height of an equilateral triangle is
10 centimeters what is its area an
equilateral triangle is a triangle in
which all the sides are of equal length
so let the length be a so all of the
sides are of equal length now we have
been given the height of this triangle
as 10 centimeter now in an equilateral
triangle this altitude divides the base
so this becomes a by 2
and this is 8 by 2 it actually divides
the base in two halves now this is a
right angle triangle we can calculate a
by Pythagoras theorem so hypotenuse
square that's a square is equals to
a-square plus height square so a by 2
square plus height square so a square is
equal to a square by 4 plus 100 okay so
for a square is equals to a-square plus
400 ok now for a square minus a square
is actually 3 a square is equals to 400
so therefore a is equals to you know
root of 400 that's 20 by root 3 so we
have the value of a we can calculate the
area now so a is the base we know the
height so area is equal to half into
base is a that's 20 by root 3 into
height that's 10
so 2 times 10 is 20 so this is basically
100 by root 3 centimeter square so this
is the area option D if the area of a
square with side a is equal to the area
of a triangle with base B then what is
the altitude of the triangle
so the area of the square is equal to
the area of the triangle so I can write
a s equals to a T what is the area of a
square it is side square and we have
been given the side as a so it is a
square is equals to what is the area of
a triangle well it is half into base
into height we have been given the base
as a so times a times let the altitude
or the height of the triangle be H okay
so this a and one of this a gets
cancelled out what we have here is a is
equals to 1 by 2 H so therefore height
is equals to twice of a so option C is
the correct answer the ratio of the
basis of two triangles is X is 2 y and
that of their areas is a is 2 B then
what will be the ratio of their
corresponding altitudes so the ratio of
the basis is X is 2 y so I can write B 1
by B 2 is equals to X by Y and the ratio
of their areas is a is 2 B so a 1 by a 2
is a by B and we want the ratio of their
altitudes that we want H 1 by H 2 okay
now we can write a 1 by a 2 as you know
we have it as a by B it can be written
as a 1 is half into base into height of
the first triangle upon area of the
second triangle is half into B 2 into H
2 half and half gets cancelled out a by
B is equal to we have V 1 by V 2 times H
1 by H 2 okay so B 1 by B 2 is X by Y so
we have ay by B is equal to X by Y times
H 1 by H 2 so therefore H 1 by H 2 is
equals to a y by
B X so this is equal to a y is to B X so
option C is the correct answer if every
side of a triangle is doubled the area
of the new triangle is K times the area
of the old one what is the value of K so
for simplicity purposes let us consider
right-angled triangles so this is the
initial triangle let its base b b and
height be H now we are doubling every
side of the triangle so the base will be
doubled that will be 2b and the height
will also be doubled that will be 2 H
okay let us also say that this has area
a1 and this has area a2
now the question says that if every side
of the triangle is doubled the area of
the new triangle is K times the area of
the old one so area of the new triangle
that's a 2 is equals to K times the area
of the old one now what is a 2 a 2 is
half into base into height so half into
base is 2 B into height is 2 H it is
equals to K times what is a 1 well it is
half into base into height okay so this
half and this half goes out B H and B H
goes out what remains is K is equals to
2 times 2 that's 4 so the value of K is
for option D