Conditions for the SSS  Side Side Side congruence
Two triangles are said to be congruent if three sides of one triangle are respectively equal to the three sides of the other triangle.
Experiment to prove Congruence with SSS:
Draw ∆LMN with LM = 3 cm, LN = 4 cm, MN = 5 cm.
Also, draw another ∆XYZ with XY = 3cm, XZ = 4cm, YZ= 5cm.
We see that LM = XY, LN = XZ and MN = YZ.
Make a trace copy of ∆XYZ and try to make it cover ∆LMN with X on L, Y on M and Z on N.
We observe that: two triangles cover each other exactly.
Therefore ∆LMN ≅ ∆XYZ
Workedout problems on side side side congruence triangles (SSS postulate):
1. LM = NO and LO = MN. Show that ∆ LON ≅ ∆ NML.
Solution:
In ∆LON and ∆NML
LM = NO → given
LO = MN → given
LN = NL → common
Therefore, ∆ LON ≅ ∆ NML, by sidesideside (SSS) congruence condition
2. In the given figure, apply SSS congruence condition and state the result in the symbolic form.
Solution:
In ∆LMN and ∆LON
LM = LO = 8.9cm
MN = NO = 4cm
LN = NL = 4.5 cm
Therefore, ∆LMN ≅ ∆LON, by side side side (SSS) congruence condition
3. In the adjoining figure, apply SSS congruence condition and state the result in the symbolic form.
Solution:
In ∆LNM and ∆OQP
LN = OQ = 3 cm
NM = PQ = 5cm
LM = PO = 8.5cm
Therefore, ∆LNM ≅ ∆OQP, by Side Side Side (SSS) congruence condition
4. ∆OLM and ∆NML have common base LM, LO = MN and OM = NL. Which of the following are true?
(i) ∆LMN ≅ ∆LMO
(ii) ∆LMO ≅ ∆LNM
(iii) ∆LMO ≅ ∆MLN
Solution:
LO = MN and OM = NL → given
LM = LM → common
Thus, ∆MLN ≅ ∆LMO, by SSS congruence condition
Therefore, statement (iii) is true. So, (i) and (ii) statements are false.
5. By Side Side Side congruence prove that 'Diagonal of the rhombus bisects each other at right angles'.
Solution: Diagonal LN and MP of the rhombus LMNP intersect each other at O.
It is required to prove that LM ⊥ NP and LO = ON and MO = OP.
Proof: LMNP is a rhombus.
Therefore, LMNP is a parallelogram.
Therefore, LO = ON and MO = OP.
In ∆LOP and ∆LOM; LP = LM, [Since, sides of a rhombus are equal]
Side LO is common
PO = OM, [Since diagonal of a parallelogram bisects each other]
Therefore, ∆LOP ≅ ∆LOM, [by SSS congruence condition]
But, ∠LOP + ∠MOL = 2 rt. angle
Therefore, 2∠LOP = 2 rt. angle
or, ∠LOP = 1 rt. angle
Therefore, LO ⊥ MP
i.e., LN ⊥ MP (Proved)
[Note: Diagonals of a square are perpendicular to each other]
6. In a quadrilateral LMNP, LM = LP and MN = NP.
Prove that LN ⊥ MP and MO = OP [O is the point of intersection of MP and LN]
Proof:
In ∆LMN and ∆LPN,
LM = LP,
MN = NP,
LN = NL
Therefore, ∆LMN ≅ ∆LPN, [by SSS congruence condition]
Therefore, ∠MLN = ∠PLN  (i)
Now in ∆LMO and ∆LPO,
LM = LP;
LO is common and
∠MLO = ∠PLO
∆LMO ≅ ∆LPO, [by SAS congruence condition]
Therefore, ∠LOM = ∠LOP and
MO = OP, [Proved]
But ∠LOM + ∠LOP = 2 rt. angles.
Therefore, ∠LOM = ∠LOP = 1 rt. angles.
Therefore, LO ⊥ MP
i.e., LN ⊥ MP, [Proved]
7. If the opposite sides of a quadrilateral are equal, prove that the quadrilateral will be parallelogram.
LMNO is a parallelogram quadrilateral, whose sides LM = ON and LO = MN. It is required to prove that LMNO is a parallelogram.
Construction: Diagonal LN is drawn.
Proof: In ∆LMN and ∆NOL,
LM = ON and MN = LO, [By hypothesis]
LN is common side.
Therefore, ∆LMN ≅ ∆NOL, [by Side Side Side congruence condition]
Therefore, ∠MLN = ∠LNO, [Corresponding angles of congruent triangles]
Since, LN cuts LM and ON and the both alternate angles are equal.
Therefore, LM ∥ ON
Again, ∠MNL = ∠OLN [Corresponding angles of congruent triangles]
But LN cuts LO and MN, and the alternate angles are equal.
Therefore, LO ∥ MN
Therefore, In quadrilateral LMNO,
LM ∥ ON and
LO ∥ MN.
Therefore, LMNO is a parallelogram. [Proved]
[Note: Rhombus is parallelogram.]
Conditions for the Congruence of Triangles
Right Angle Hypotenuse Side congruence
Converse of Pythagorean Theorem
7th Grade Math Problems
8th Grade Math Practice
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