Triangle Properties and Exterior Angle Property Video Lecture | Mathematics (Maths) Class 7 (Old NCERT)

0:0:0: hello dear
0:0:1: students in today's video we will learn
0:0:4: about
0:0:5: triangles the classification of
0:0:7: triangles on the basis of its sides and
0:0:9: its angles and then we will learn about
0:0:12: the median and altitude of a
0:0:15: triangle and finally we will learn about
0:0:17: the exterior angle property of a
0:0:20: triangle in the earlier classes you have
0:0:23: learned about Lines line segments and
0:0:26: angles and also a little about triangles
0:0:30: but in this chapter we will put our
0:0:31: Focus entirely on triangles and its
0:0:35: properties so let's start with what is a
0:0:38: triangle a triangle is a two-dimensional
0:0:41: shape which is made by joining any three
0:0:43: points a b and c on a sheet of paper
0:0:46: that do not fall on the same straight
0:0:48: line we can call this closed figure
0:0:52: triangle ABC or
0:0:55: triangle B CA or
0:0:57: triangle c a b
0:1:0: the segments AB B C and C A are called
0:1:3: the sides of the triangle
0:1:5: ABC and the points a B and C are called
0:1:10: the vertices of the triangle
0:1:12: ABC vertices is the plural of
0:1:15: vertex we can classify the triangles
0:1:18: according to the sides and angles if we
0:1:21: classify the triangle on the basis of
0:1:23: its sides we have one equilateral
0:1:27: triangle a triangle that has all the
0:1:30: three sides of the same length two
0:1:33: isoceles
0:1:34: triangle a triangle in which only two of
0:1:37: the sides are of the same length the
0:1:39: third one is of different length three
0:1:42: scaling
0:1:43: triangle a triangle that has all the
0:1:46: three sides of different
0:1:48: lenss if we classify the triangle on the
0:1:51: basis of its angles we have one acute
0:1:55: angle
0:1:56: triangle a triangle that has all angles
0:1:59: smaller than than
0:2:0: 90° two right angle
0:2:3: triangle a triangle that has one angle
0:2:6: equal to
0:2:8: 90° three obtuse angle
0:2:11: triangle a triangle that has one angle
0:2:14: greater than
0:2:17: 90°
0:2:19: now understanding more about the
0:2:21: triangles let's learn about the median
0:2:23: of a
0:2:24: triangle here we have a triangle made of
0:2:28: paper suppose we have to find the
0:2:30: midpoint of this side can you tell us
0:2:33: how to go about it to find the midpoint
0:2:36: of this side of a
0:2:38: triangle we fold the paper in such a way
0:2:41: that the two vertices of this side cover
0:2:43: each other we will draw the
0:2:46: crease then the point where the crease
0:2:48: intersects this side of the
0:2:50: triangle is the midpoint of this side
0:2:54: now if we connect the midpoint of this
0:2:56: side with the opposite vertex then we
0:2:58: will get a median of the tri triangle
0:3:1: the median is a line segment that
0:3:3: connects a vertex of a triangle to the
0:3:5: midpoint of its opposite
0:3:7: side now let's learn about the altitude
0:3:11: of a
0:3:12: triangle sometimes in a triangle it is
0:3:15: necessary to find the shortest distance
0:3:17: between a vertex and its opposite side
0:3:20: but how do we do so suppose we are given
0:3:24: a triangle
0:3:25: pqr and we have to find its height that
0:3:29: is is we have to find the shortest
0:3:31: distance between the vertex p and the
0:3:34: side QR to find this
0:3:37: distance we can make line segments in
0:3:40: different ways but do you think all
0:3:43: these lines give us the correct
0:3:45: information about the distance between
0:3:48: the vertex p and its opposite side QR or
0:3:52: say the height of this
0:3:53: triangle think about it no not all of
0:3:58: them give us the correct
0:4:0: information from the vertex P to find
0:4:3: the height of a triangle we draw a
0:4:5: perpendicular to its opposite side QR
0:4:9: which intersects the side QR at Point g
0:4:12: a perpendicular drawing from a Vertex to
0:4:15: its opposite side is called the altitude
0:4:17: of the triangle our line segment PJ is
0:4:20: the altitude of the triangle
0:4:23: pqr that forms a right angle on the side
0:4:26: QR this gives us the distance between
0:4:29: the vert X and its opposite side that is
0:4:32: the height of a
0:4:33: triangle note a triangle has three
0:4:37: medians from each of its vertices to its
0:4:40: signs a triangle has three altitudes
0:4:43: from each of its vertices to its
0:4:46: signs moving forward let us now learn
0:4:49: about the exterior angle of a triangle
0:4:52: and its property consider any
0:4:55: triangle and let's call it triangle AB b
0:4:59: c we know what the interior angles are
0:5:4: but what are exterior angles of a
0:5:6: triangle it's simple we just have to
0:5:10: extend any one side say let us produce
0:5:14: side BC to point D like this since we
0:5:18: have just extended BC to D it means that
0:5:21: BD is also a straight line that is the
0:5:24: angle ACD is in the exterior of the
0:5:26: triangle which is why it's called the
0:5:28: exterior
0:5:30: angle we know these three angles are in
0:5:33: the
0:5:34: interior but this time we will look at
0:5:36: them with respect to their exterior
0:5:39: angle first let's look at this angle we
0:5:42: know that it lies in the interior of the
0:5:45: triangle and it is adjacent to this
0:5:47: exterior angle so with respect to this
0:5:50: exterior angle it is the interior
0:5:52: adjacent angle what about the other two
0:5:56: angles angle a and angle B A and B both
0:6:0: are in the interior region of the
0:6:2: triangle and are not adjacent to the
0:6:4: exterior angle
0:6:6: ACD they are called as interior opposite
0:6:10: angles now that we have learned about
0:6:12: the exterior and interior opposite
0:6:14: angles of the
0:6:16: triangle let's try to derive an
0:6:18: interesting relationship between them
0:6:21: let's name the interior angles of this
0:6:23: triangle as angle one angle 2 and angle
0:6:27: three and the interior angle as angle 4
0:6:32: we already know that the sum of the
0:6:35: interior angles of the triangle is
0:6:38: 180° that means angle 1 + angle 2 +
0:6:43: angle 3 is equal to
0:6:46: 180° let's name this equation as
0:6:49: equation 1 now look at the angles 3 and
0:6:53: four we can clearly see that those two
0:6:56: form angles on the same straight line
0:6:59: and we also know that angles on a
0:7:1: straight line sum up to
0:7:4: 180° so we can write this as angle 3 +
0:7:9: angle 4 is equal to
0:7:12: 180° or we can write it down as angle 3
0:7:16: is equal to
0:7:19: 180° minus angle 4 now name this
0:7:23: equation as equation two now we will be
0:7:27: substituting the value of angle three
0:7:29: from equation 2 and using it in equation
0:7:31: 1 on
0:7:33: solving we get that angle 1 plus angle 2
0:7:37: is equal to angle 4 great we have
0:7:42: therefore derived the relation between
0:7:44: these three
0:7:45: angles this derivation proves that the
0:7:49: exterior angle of a triangle is always
0:7:51: equal to the sum of its two interior
0:7:54: opposite
0:7:56: angles now that you have learned about
0:7:58: the exterior angle property let us see a
0:8:2: solved example to use the property we
0:8:4: have just
0:8:5: learned in the given figure find the
0:8:8: value of x let us first Mark these
0:8:11: angles as angle one angle 2 and angle
0:8:14: three for our
0:8:16: convenience now we can use the property
0:8:18: of exterior angle and write that angle 1
0:8:23: plus angle 2 is equal to angle 3 here
0:8:26: angle 1 is 70° and angle 2 is unknown
0:8:31: therefore X and angle three that is our
0:8:34: exterior angle is
0:8:37: 120° by putting in the values 70° + x is
0:8:42: equal to
0:8:44: 120° that is X is equal to
0:8:48: 120° -
0:8:50: 70° thus X comes out to be equal to
0:8:54: 50° easy
0:8:57: right to learn about more such topics
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