Congruent Triangles

Congruent Triangles


Congruent Triangles

 

Two triangles are said to be congruent triangles if they are exactly the same by their shapes and sizes.


********************


10 Math Problems officially announces the release of Quick Math Solver and 10 Math ProblemsApps on Google Play Store for students around the world.


Install Quick Math Solver

********************


Install 10 Math Problems


********************


Two triangles of the same shape are meant three angles of one triangle are equal to the corresponding three angles of another triangle. And, two triangles of the same size are meant three sides of one triangle are equal to the corresponding three sides of another triangle.

 

The symbol is used for congruent triangles.

 

For example:


Examples of congruent triangles

In the given triangles ABC and PQR, AB = PQ, BC = QR, AC = PR. Also, A = P, B = Q and C = R. So the triangles ABC and PQR are congruent. It is written as ΔABC ΔPQR.

 

 

Conditions for Congruency

 

The following conditions are necessary for the congruency of two triangles.


Condition - 1: Side-Side-Side (SSS) axiom

 

When three sides of a triangle are equal to three corresponding sides of another triangle, they are congruent by Side-Side-Side condition. It is called SSS axiom.


Side-Side-Side (SSS) axiom

In the given triangles, AB = PQ (S), BC = QR (S), AC = PR (S). Therefore, ΔABC ΔPQR by SSS axiom.


Condition - 2: Side-Angle-Side (SAS) axiom

 

When Side-Angle-Side serially of a triangle are equal to the corresponding Side-Angle-Side serially of another triangle, then the triangles are congruent by Side-Angle-Side condition. It is called SAS axiom.


Side-Angle-Side (SAS) axiom

In the given triangles, AB = PQ (S), B = Q (A), BC = QR (S). Therefore, ΔABC ΔPQR by SAS axiom.



Condition - 3: Angle-Side-Angle (ASA) axiom

 

When Angle-Side-Angle serially of a triangle are equal to the corresponding Angle-Side-Angle serially of another triangle, then the triangles are congruent by Angle-Side-Angle condition. It is called ASA axiom.


Angle-Side-Angle (ASA) axiom

In the given triangles, B = Q (A), BC = QR (S), C = R (A). Therefore, ΔABC ΔPQR by ASA axiom.



Condition - 4: Right angle-Hypotenuse-Side (RHS) axiom

 

In two right-angled triangles, if the hypotenuse and a side of a triangle are equal to the hypotenuse and a side of another triangle, then the triangles are congruent by Right angle-Hypotenuse-Side condition. It is called RHS axiom.


Right angle-Hypotenuse-Side (RHS) axiom

In the given triangles, B = Q = 90° (R), AC = PR (H), BC = QR (S). Therefore, ΔABC ΔPQR by RHS axiom.


Condition - 5: Side-Angle-Angle (SAA) axiom

 

When Side-Angle-Angle serially of a triangle are equal to the corresponding Side-Angle-Angle serially of another triangle, then the triangles are congruent by Side-Angle-Angle condition. It is called SAA axiom.


Side-Angle-Angle (SAA) axiom

In the given triangles, BC = QR (S), C = R (A), A = P (A). Therefore, ΔABC ΔPQR by SAA axiom.


 

Worked Out Examples

 

Example 1: In the given figure, state the condition of congruency and write their corresponding sides and angles.


Example 1: Triangles ABC and PQR

Solution: Here,

In ΔPQR and ΔABC

i.       B = Q (A) -------> both 60°

ii.    BC = QR (S) -------> both 5cm

iii. C = R (A) -------> both 45°

  ΔABC ΔPQR -------> by ASA axiom

  AB = PQ, AC = PR -------> corresponding sides of congruent triangles

  A = P -------> corresponding angles of congruent triangles


 

Example 2: In the given figure, BP = PC, AP BC, prove that ΔABP ΔAPC, also find the value of a.


Example 2: ΔABP and ΔAPC

Solution: Here,

In ΔABP and ΔAPC

i.       BP = PC (S) -------> given

ii.    APB = APC (A) -------> both right angles

iii. AP = AP (S) -------> common side of both triangles

  ΔABP  ΔAPC -------> by SAS axiom

  AB = AC -------> corresponding sides of congruent triangles

or,     2a + 3 = 5

or,     2a = 5 – 3

or,     2a = 2

or,     a = 1  Ans.

 

 

Example 3: In the given figure, AP = DP, CP = BP, PAC = 30° and PDB = 40° then,

a.     Prove that ΔAPC ΔDPB

b.     Find the values of a, b and y.


Example 3: ΔAPC and ΔDPB

Solution: Here,

In ΔAPC and ΔBPD

i.       AP = PD (S) -------> given

ii.    APC = DPB (A) -------> vertically opposite angles

iii. PC = PB (S) -------> given

  ΔAPC   ΔDPB --------> by SAS axiom

  a  = 40° -------> corresponding angles of congruent triangles

  b  = 30° -------> corresponding angles of congruent triangles

  AC = DB -------> corresponding sides of congruent triangles

or,     y + 3 = 2y

or,     3 = 2y – y

or,     y = 3  Ans.

 

 

If you have any questions or problems regarding the Congruent Triangles, you can ask here, in the comment section below.

 

Was this article helpful? LIKE and SHARE with your friends…

0 comments: