So far, you have been mainly studying the equality of sides and angles of a triangle or triangles. Sometimes, we do come across unequal objects, we need to compare them. For example, line-segment AB is greater in length as compared to line segment CD in
Fig. 7.41 (i) and ∠ A is greater than ∠ B in Fig 7.41 (ii).
Let us now examine whether there is any relation between unequal sides and unequal angles of a triangle. For this, let us perform the following activity:
Activity : Fix two pins on a drawing board say at B and C and tie a thread to mark a side BC of a triangle.
Fix one end of another thread at C and tie a pencil
at the other (free) end . Mark a point A with the
pencil and draw Δ ABC (see Fig 7.42). Now, shift
the pencil and mark another point A′ on CA beyond
A (new position of it)
So, A′C > AC (Comparing the lengths)
Join A′ to B and complete the triangle A′BC.
What can you say about ∠ A′BC and ∠ ABC?
Compare them. What do you observe?
Clearly, ∠ A′BC > ∠ ABC
Continue to mark more points on CA (extended) and draw the triangles with the side BC and the points marked.
You will observe that as the length of the side AC is increased (by taking different positions of A), the angle opposite to it, that is, ∠ B also increases.
Let us now perform another activity :
Activity : Construct a scalene triangle (that is a triangle in which all sides are of
different lengths). Measure the lengths of the sides.
Now, measure the angles. What do you
In Δ ABC of Fig 7.43, BC is the longest side
and AC is the shortest side.
Also, ∠ A is the largest and ∠ B is the smallest.
Repeat this activity with some other triangles.
We arrive at a very important result of inequalities in a triangle. It is stated in the form of a theorem as shown below:
Theorem 7.6 :
If two sides of a triangle are unequal, the angle opposite to the longer side is larger (or greater).
You may prove this theorem by taking a point P
on BC such that CA = CP in Fig. 7.43.
Now, let us perform another activity :
Activity : Draw a line-segment AB. With A as centre and some radius, draw an arc and mark different points say P, Q, R, S, T on it.
Join each of these points with A as well as with B (see Fig. 7.44). Observe that as
we move from P to T, ∠ A is becoming larger and larger. What is happening to the length of the side opposite to it? Observe that the length of the side is also increasing; that is ∠ TAB > ∠ SAB > ∠ RAB > ∠ QAB > ∠ PAB and TB > SB > RB > QB > PB.
Now, draw any triangle with all angles unequal to each other. Measure the lengths of the sides (see Fig. 7.45).
Observe that the side opposite to the largest angle is the longest. In Fig. 7.45, ∠ B is the largest angle and AC is the longest side.
Repeat this activity for some more triangles and we see that the converse of Theorem 7.6 is also true. In this way, we arrive at the following theorem:
Theorem 7.7 :
In any triangle, the side opposite to the larger (greater) angle is longer. This theorem can be proved by the method of contradiction.
Now take a triangle ABC and in it, find AB + BC, BC + AC and AC + AB. What
do you observe?
You will observe that AB + BC > AC,
BC + AC > AB and AC + AB > BC.
Repeat this activity with other triangles and with this you can arrive at the following heorem :
Theorem 7.8 :
The sum of any two sides of a triangle is greater than the third side.
In Fig. 7.46, observe that the side BA of Δ ABC has been produced to a point D such that AD = AC. Can you
show that ∠ BCD > ∠ BDC and BA + AC > BC? Have
you arrived at the proof of the above theorem.
Let us now take some examples based on these results.
Example 9 :
D is a point on side BC of Δ ABC such that AD = AC (see Fig. 7.47).Show that AB > AD.
In Δ DAC,
AD = AC (Given)
So, ∠ ADC = ∠ ACD
(Angles opposite to equal sides)
Now, ∠ ADC is an exterior angle for ΔABD.
So, ∠ ADC > ∠ ABD
or, ∠ ACD > ∠ ABD
or, ∠ ACB > ∠ ABC
So, AB > AC (Side opposite to larger angle in Δ ABC)
or, AB > AD (AD = AC)
In this chapter, you have studied the following points :
Cite this Simulator: