Triangles and their properties

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Triangles andtheir properties
Area of triangle
1. When base and corresponding height is known.
½ * base * height = ½*c*h
2. When all sides are given (Heron’s formulae)
{s (s – a) (s – b) (s – c)}1/2
where, s = (a+b+c)/2
3. When two sides and corresponding angle is given.
½ a*c*sinθ
4. When all the median are given (Median is a line joining the vertex to the opposite
side at midpoint)
4/3 * {s(s – m1) (s – m2) (s –m3)}1/2
Where, s = (m1+m2+m3)/2,
m1,m2,m3 are three medians of the Triangle.
5. When all the heights are given
1/Area of ∆ = 4 {G (G – 1/h1) (G – 1/h2) (G – 1/h3),
Where, G = ½ (1/h1 + 1/h2 + 1/h3)
Sine formulae of triangle
a/SinA = b/sinB = c/SinC = 2R
Where, R is Circumradius
Cosine Formulae of triangle

CosA = b2
+c2
– a2
/ 2bc
CosB= a2
+c2
– b2
/ 2ac
CosC = b2
+a2
– c2
/ 2ba
An interesting result based of cosine formulae
If in a triangle CosA = b2
+c2
– a2
/ 2bc, Whereas b & c are the smaller sides then
Case I, b2
+c2
is greater than a2
then angle A is acute.
Case II, b2
+c2
is smaller than a2
then angle A is obtuse.
Case III, b2
+c2
is equals to a2
then angle A is Right Angle.
For example, Ques: Find the type of the triangle ABC whose 3 sides are of length 11, 3 & 9.
Solution: Sum of square of two smaller sides is 32
+ 92
= 9 + 81 = 90
Square of largest side is 112
= 121.
Hence, this triangle is obtuse angled triangle.
TRIANGLES AND ITS CERVICES
Medians of a Triangle
The medians of a triangle are line segments joining each vertex to the midpoint of the opposite
side. The medians always intersect in a single point, called the centroid.
In the adjoining figure D, E, & F are midpoints of the side of the triangle while Line Segment
AD, BE & CF are median. They are meeting at a common point I which is the centroid.
Properties:
1. Centroid divides the Median in the ratio 2:1. i.e AI/ID = 2/1

2. Apollonius Theorem:- To find the length of median when all sides are given.
4 * AD2
= 2(AC2
+ AB2
) – BC2
3. All 3 median of a triangle divides the triangle in 6 equal parts.
i.e. area of ∆AFI = area of ∆AEI = area of ∆BFI = area of ∆BDI= area of ∆CDI= area of ∆CEI=
area of ∆ABC/6
Angle Bisectors:
Angle bisectors are the line segment which bisects the internal angles of the triangle. All the
three angle bisectors meet at a common point called as Incentre.
In the adjoining figure Line Segment A1T1, A2T2, & A3T3 are Angle Bisector. While point I is
Incentre of the triangle.
Incentre is the only point from which we can draw a circle inside the triangle which will touch all
the sides of the triangle at exactly one point & this circle has a special name known as Incircle.
And the radius of this circle is known as Inradius.
Properties:
1. Inradius = Area of ∆ A1A2A3 / Semi perimeter of ∆ A1A2A3
2. Angle formed at the incentre by any two angle bisector.
Angle A1IA3 = Angle A1A2A3 /2 + 90o

Perpendicular Bisectors of Sides of the Triangle::
When the perpendicular bisectors of the side of the triangle is drawn they meet at a common
point known as circumcentre.
This circumcentre is a special point as from this point we can draw a circle which will enclose
the triangle in a way that all the vertex of the triangle lie on the circle. The radius of this circle is
known as circumradius.
Properties:In the adjoining figure DP, EP, & FP are perpendicular bisector of sides of the
triangle. Point P is circumcentre.
1. Circumradius = length of side AC * CB * BA / 4 * Area of triangle
2. Angle formed at the circumcentre by any two line segment joining circumcentre to
the vertex.
Angle APC = 2 * Angle ABC
Altitude of the Triangle:

The orthocenter of a triangle is the point where the three altitudes meet. This point may be
inside, outside, or on the triangle.
In the adjoining figure AD, BE, & CF are three altitudes of a triangle. And point O is the
orthocenter.
Properties:
1. Angle BOC + Angle BAC = 180o
Relation Between Circumradius & Inradius:
Circumradius * Inradius = a*b*c/2*(a+b+c). Where a, b, & c are the length of the
sides of the triangle.
If two triangles are similar, then the corresponding sides we have, shows the following relation:
o The ratio of sides of triangle is proportional to each other.
Like AB/DE = BC/EF = CA/FD
For example: If AB = 6cm, BC = 10cm & DE = 2cm. Find EF
Solution: AB/DE = BC/EF, 6/2 = 10/EF, Therefore EF = 2*10/6 = 10/3

o The height, angle bisector, inradius & circumradius are proportional to the sides of
triangle.
Median ∆ABC / Median ∆DEF = Height ∆ABC/ Height ∆DEF = AB/DE
o The areas of the triangles are proportional to the square of the sides of the
corresponding triangle.
Area ∆ABC/ Area ∆DEF = (AB/DE) 2
For example: If AB = 6cm, DE = 10cm & Area of ∆ABC is 135 sqcm. Find Area of ∆DEF.
Solution: Area ∆ABC/ Area ∆DEF = (AB/DE) 2
135/ Area ∆DEF = (6/10)2
, Area ∆DEF = 135*(5/3)2
= 375 sqcm.
 In a right angled triangle, the triangles on each sides of the altitude drawn from
the vertex of the right angle to the hypotenuse are similar to each other and to the
parent triangle.
o ∆ ABC ≈ ∆ ABD
o ∆ ABC ≈ ∆ CBD
o ∆ DBC ≈ ∆ ABD
Proof: In ∆ ABC & ∆ ABD. Angle B and Angle D is 90o
. & Angle A is common in both. Hence
by AA rule Both triangles are similar.

Triangles and their properties

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