Congruence Of Triangles

 Mind Maps

Class VII - maths: Congruence of Triangles
Q) Definition of congruence triangle?

Q) Two line segments are congruence ?

Q) Given two real- life examples for congruent shapes?

Q) Among two congruent angles, one has a measure of 80° the measure of the other angle is?

Q) Congruence object are exact copies of one another?

Q) Two squares are congruent if both of them have the same ?

Q) Two circles are congruent if both of them have the same?

Q) Two rectangles are congruent if their opposite sides are?

Q) Two circles having equal areas are ?

Q) Two circles are congruent if they have the?

Q) The side of a triangle are of length 6.4 cm, 7.9 cm and 8 cm. Is this triangle a right triangle? If so, which side is the hypotenuse?

Q) The side of a triangle are of length 9 cm, 13 cm and 19 cm. Is this triangle a right triangle? If so, which side is the hypotenuse?

Q) Proof of Pythagorean Theorem using Algebra?

Q) Explain the Congruent or Similar?

Q) Two plane figure, say,F1 and F2 are congruent if the trace-copy of F1 fits exactly on that of F2. We write this as F1~= F2?

Q) When we write ∠A=∠B,we actually mean?

Q) Two line segments are congruent if?

Q) Name the theorem in which the square of the hypotenuse is equal to the sum of squares of other two sides?

Q) Converse of Pythagorean Theorem?

Q) Two equal line-segments, lying in the same straight line and sharing a common vertex?

Q) Two angles, say, ∠ABC and ∠PQR, are congruent if their measures are equal. We write this as ∠ABC~= ∠PQR or as m∠ABC = m∠PQR. However, in practice, it is common to write it as ∠ABC = ∠PQR?

Q) SAS Congruence of two triangles: Under a given correspondence, two triangles are congruent if two sides and the angle included between them in one of the triangles are equal to the corresponding sides and the angle included between them of the other triangle?

Q) RHS Congruence of two right angle triangles. Under a given correspondence two right angled triangles are congruent if the hypotenuse and a leg of one of the triangle are equal to the hypotenuse and the corresponding leg of the other triangle?

Q) The side of a triangle are of length 4.5 cm, 7.5 cm and 6 cm. Is this triangle a right triangle? If so, which side is the hypotenuse?

Q) Among two congruent angles,one has a measure of 70°then the measure of the other angle is?

Q) If ΔDEF ≅ ΔBCA, write the parts of ΔBCA that correspond to ∠E.

Q) The side of a triangle are of length 8 cm, 15 cm and 17 cm. Is this triangle a right triangle? If so, which side is the hypotenuse?

Content
• Congruence of plane figures
• Congruence of line segments
• Congruent Angles
• Congruence of Triangles
• Conditions for the Congruence of Triangles
• Side Side Side Congruence
• Side Angle Side Congruence
• Angle Side Angle Congruence
• Angle Angle Side Congruence
• Right Angle Hypotenuse Side congruence
• Congruent Shapes
• Proof of Pythagorean Theorem
• Converse of Pythagorean Theorem
Formulae
1. Congruent objects are exact copies of one another.
2. The method of superposition examines the congruence of plane figures.
3. Two plane figures, say, F_1and F_2are congruent if the trace-copy of F_1fits exactly on that of F_2.
We write this as F_1 ~= F_2.
4. Two line segments, say, AB and CD , are congruent if they have equal lengths. We write this as AB CD . However, it is common to write it as AB = CD .
5. Two angles, say, /_ABC and /_PQR, are congruent if their measures are equal. We write this as /_ABC ~= /_PQR or as m/_ABC = m/_PQR. However, in practice, it is common to write it as /_ABC = /_PQR.
6. SSS Congruence of two triangles:
Under a given correspondence, two triangles are congruent if the three sides of the one are equal to the three corresponding sides of the other.
7. SAS Congruence of two triangles:
Under a given correspondence, two triangles are congruent if two sides and the angle included between them in one of the triangles are equal to the corresponding sides and the angle included between them of the other triangle.
8. ASA Congruence of two triangles:
Under a given correspondence, two triangles are congruent if two angles and the side included between them in one of the triangles are equal to the corresponding angles and the side included between them of the other triangle.
9. RHS Congruence of two right-angled triangles:
Under a given correspondence, two right-angled triangles are congruent if the hypotenuse and a leg of one of the triangles are equal to the hypotenuse and the corresponding leg of the other triangle.
10. There is no such thing as AAA Congruence of two triangles:
Two triangles with equal corresponding angles need not be congruent. In such a correspondence, one of them can be an enlarged copy of the other. (They would be congruent only if they are exact copies of one another)
Congruence of plane figures
Congruence of plane

Look at the two figures given here (Fig). Are they congruent? You can use the method of superposition. Take a trace-copy of one of them and place it over the other. If the figures cover each other completely, they are congruent. Alternatively, you may cut out one of them and place it over the other. Beware! You are not allowed to bend, twist or stretch the figure that is cut out (or traced out). In Fig, if figure F_1 is congruent to figure F_2 , we write F_1 ≅ F_2 .

Congruence of line segments

In congruent line-segments we will learn how to recognize that two line-segments are congruent.
Two equal line-segments, lying in the same straight line and sharing a common vertex.

Here, two line-segments XY and YZ lying in the same straight line are equal. This is to be verified that they are congruent.
ab is perpendicular drawn at Y on XZ. With respect to ab, the image of ZY is YZ'. XY = YZ
Hence, Z' lies on X
Therefore, XY ≅ YZ Taking Y as the centre of rotation and rotating YZ through an angle 180° in anticlockwise direction, the image YZ' is obtained, where Z' lies on X
Therefore, XY ≅ YZ

Example #### Two line segments lie on the same plane but at different positions.

example
(1) (2) PQ and RS are two equal line segments on the same plane but on different positions. It is verified that they are congruent line-segments. Perpendicular bisector XY of PR is drawn. Taking XY as the axis of reflection, the image of RS and PS'. Now taking P as the centre of rotation and rotating PS' through such an angle (in anti-clock wise direction), so that PS' coincides with PQ. Since PS' that is RS = PQ.
Hence S' lies on Q and its new name is D”.
Thus PQ ≅ R'S” i.e., RS
Therefore, PQ ≅ RS.

Congruent Angles

In congruent angles we will learn how to recognize that when two angles are congruent.

example #### In case of two angles having the same vertex: Suppose /_LOM = /_PON having same vertex O. Now we will verify their congruence.
A straight line QOR is drawn through point O and a perpendicular xy is drawn on QOR at O.
With respect to the axis of reflection xy, /_P'ON' is the image of /_PON. Now with centre of rotation O, OP' is rotated through such an angle in anticlockwise direction, so that OP' coincides with OM. Since /_P'ON' is rigid figure and equal to /_LOM, ON' falls on OL.
Therefore, /_LOM ≅ /_N'OP' ≅ /_NOP = /_PON With O as center of rotation, OP is rotated through such an angle (in anticlockwise direction) so that OP lies on OL. In the same manner ON being rotated equally, falls on OM.
Therefore, /_LOM ≅ /_PON

Example #### If two equal angles are at different positions but lie on the same plane. /_LMN and /_PQR are two equal angles at different positions but lie on the same plane. • Taking xy, the perpendicular bisector of MQ as the axis of reflection, the image of /_PQR is /_P'MR'.
• Therefore, /_P'MR' ≅ /_PQR
• Now observe that /_LMN and /_P'MR' are two equal angles sharing common vertex M.
Example Congruence of Triangles

Congruent triangles are those two triangles which are said to be congruent if and only if one of them can be made to superpose on the other so as to cover it exactly. Example

Construction

Two triangles are given as follows,where
/_ABC = /_DEF and /_ACB = /_DEF.Sides AB = DE;
To Prove:DeltaABC = DeltaDEF
proof : /_ABC = /_DEF (given)
AB = DE
AC = DF
(Sides opposite to corresponding angles are in the same ratio as ratio of angles)
Hence by S.A.S congruence rule /_ABC = /_DEF is proved.

Let ΔABC and ΔDEF be two congruent triangles, then we can superpose ΔABC on ΔDEF so as to cover it exactly. The vertices of ΔABC fall on the vertices of ΔDEF in the following order A ↔ D, B ↔ E, C ↔ F.
Thus, the order in which vertices match automatically determines the correspondence between the sides and angles of the triangle. Corresponding parts are also called matching parts of triangles.
So, we have six equalities

Corresponding sides are congruent:      AB = DE       BC = EF       CA = FD
Corresponding angles are congruent:      /_A = /_D       /_B = /_E       /_C = /_F

In the congruent triangles we will observe six correspondences between their verities. The symbol used to denote correspondence is '↔' Therefore, ΔABC ≅ ΔDEF

Solution Note

The order of letters in the name of two triangles will indicated the correspondence between the vertices of two triangles. Thus, two triangles are congruent only if there exist a correspondence between their vertices such that the correspondence sides and correspondence angles of two triangles are equal.

Conditions for the Congruence of Triangles

The conditions for the congruence of triangles are defined by three angles and three sides i.e. the six measuring parts. But out of these if three are properly satisfied, then automatically the other three are also satisfied. There are four such cases known as axioms for congruency.

Conditions #### To tell that the triangles are congruent if:

• SSS (Side Side Side postulate) congruent triangles condition
• SAS (Side Angle Side postulate) congruent triangles condition
• ASA (Angle Side Angle postulate) congruent triangles condition
• AAS (Angle Angle Side postulate) congruent triangles condition
• RHS or HL (Right angle Hypotenuse Side or Hypotenuse – Leg postulate) congruent triangles condition.
Side Side Side Congruence

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 ΔLMN with LM = 3 cm, LN = 4 cm, MN = 5 cm.
Also, draw another ΔXYZ with XY = 3cm, XZ = 4cm, YZ= 5cm. We see that LM = XY, LN = XZ and MN = YZ.
Make a trace copy of ΔXYZ and try to make it cover ΔLMN with X on L, Y on M and Z on N.
We observe that: two triangles cover each other exactly.
Therefore ΔLMN ≅ ΔXYZ

#### Worked-out problems on side side side congruence triangles (SSS postulate):

1. LM = NO and LO = MN. Show that Δ LON ≅ Δ NML. Solution:

In ΔLON and ΔNML
LM = NO → given
LO = MN → given
LN = NL → common

Therefore, Δ LON ≅ Δ NML, by side-side-side (SSS) congruence condition
2. In the given figure, apply SSS congruence condition and state the result in the symbolic form. Solution:

In ΔLMN and ΔLON
LM = LO = 8.9cm
MN = NO = 4cm
LN = NL = 4.5 cm

Therefore, ΔLMN ≅ ΔLON, by side side side (SSS) congruence condition

3. In the adjoining figure, apply S-S-S congruence condition and state the result in the symbolic form. Solution:

In ΔLNM and ΔOQP
LN = OQ = 3 cm
NM = PQ = 5cm
LM = PO = 8.5cm

Therefore, ΔLNM ≅ ΔOQP, by Side Side Side (SSS) congruence condition

4. ΔOLM and ΔNML have common base LM, LO = MN and OM = NL. Which of the following are true? (i) ΔLMN ≅ ΔLMO
(ii) ΔLMO ≅ ΔLNM
(iii) ΔLMO ≅ ΔMLN

Solution:

LO = MN and OM = NL → given
LM = LM → common
Thus, ΔMLN ≅ ΔLMO, by SSS congruence condition
Therefore, statement (iii) is true. So, (i) and (ii) statements are false.

Side Angle Side Congruence

Conditions for the SAS - Side Angle Side congruence  Two triangles are said to be congruent if two sides and the included angle of one are respectively equal to the two sides and the included angle of the other.

#### Experiment to prove Congruence with SAS:

• ΔLMN with LM – 8 cm, MN – 10 cm, /_M – 60°
• Also, draw another ΔXYZ with XY = 8cm, YZ = 10cm, /_Y= 60°.
• We see that LM = XY, AC = /_M = /_Y and MN = YZ
• Make a trace copy of ΔXYZ and try to make it cover ΔLMN with X on L, Y on M and Z on N.
We observe that: two triangle cover each other exactly.

Therefore ΔLMN ≅ ΔXYZ

#### Worked-out problems on side angle side congruence triangles (SAS postulate): 1. In the kite shown, PQ = PS and /_QPR = /_SPR.

(i) Find the third pair of corresponding parts to make Δ PQR ≅ ΔPSR by SAS congruence condition.
(ii) Is /_QRP = /_SRP?

Solution:

(i) In Δ PQR and Δ PSR
PQ = PS → given
/_QPR = /_SPR → given
PR = PR → common
Therefore, ΔPQR ≅ ΔPSR by SAS congruence condition

(ii) Yes, /_QRP = /_SRP (corresponding parts of congruence triangle).

2. Identify the congruent triangle: Solution:

In ΔLMN,
65° + 45° + /_L = 180°
110° + /_L = 180°
/_L = 180° - 110°

Therefore, /_L = 70°
Now in ΔXYZ and ΔLMN
/_X = /_L (given in the picture)
XY = LM (given in the picture)
XZ = NL (given in the picture)
Therefore, ΔXYZ ≅ ΔLMN by SAS congruence axiom

3. By using SAS congruency proof that, angles opposite to equal side of an isosceles triangle are equal. Solution:

Given: ΔPQR is isosceles and PQ = PR
Construction: Draw PO, the angle bisector of /_P, PO meets QR at O.
Proof: In ΔQPO and ΔRPO
PQ = PR (given)
PO = PO (common)
/_QPO = /_RPO (by construction)
Therefore, ΔQPO ≅ ΔRPO (by SAS congruence)
Therefore, /_PQO = /_PRO (by corresponding parts of congruent triangle)

4. Show that bisector of the vertical angle of an isosceles triangle bisects the base at right angle. Solution:

Given: ΔPQR is isosceles, and PO bisects /_P
Proof: In ΔPOQ and ΔPOR
PQ = PR (isosceles triangle)
/_QPO = /_RPO (PO bisects /_P)
PO = PO (common)
Therefore, Δ POQ ≅ Δ POR (by SAS congruence axiom)
Therefore, /_POQ = /_POR (by corresponding parts of congruent triangle)

Angle Side Angle Congruence

Conditions for the ASA - Angle Side Angle congruence
Two triangles are said to be congruent if two angles and the included side of the one are respectively equal to the two angles and the included side of the other.

#### Experiment to prove Congruence with ASA:

Draw a ΔLMN with /_M = 60°, MN = 5 cm, /_N = 30°. Also, draw another ΔXYZ with /_Y = 60°, YZ = 5cm, /_Z = 30°.
We see that /_M = /_Y, MN = YZ and /_N = /_Z.
Make a trace copy of ΔXYZ and try to make it cover ΔLMN with X on L, Y on M and Z on N.
We observe that: two triangles cover each other exactly.

Therefore ΔLMN ≅ ΔXYZ

#### Worked-out problems on angle side angle congruence triangles (ASA postulate):

1. ΔPQR ≅ ΔXYZ by ASA congruence condition. Find the value of x and y. Solution:

WE know Δ PQR ≅ ΔXYZ by ASA congruence. Therefore /_Q = /_Y i.e., x + 15 = 80° and /_R = /_Z i.e., 5y + 10 = 30°.
Also, QR = YZ.
Since, x + 15 = 80°
Therefore x = 80 – 15 = 65°
Also, 5y + 10 = 30°
So, 5y = 30 – 10

Therefore, 5y = 20
⇒ y = 20/5
⇒ y = 4°

Therefore, the value of x and y are 65° and 4°.

Angle Angle Side Congruence

Conditions for the AAS – Angle Angle Side congruence
Two triangles are said to be congruent if two angles and non- included side of the one triangle is equal to the two angles and the non- included side of the other.

#### Experiment to prove Congruence with AAS:

Draw a ΔLMN with /_M = 40°, /_N = 70°, LN = 3 cm.
Also, draw another ΔXYZ with /_Y = 40°, /_Z = 70°, XZ = 3cm. We see that /_M = /_Y, /_N = /_Z and LN = XZ
Make a trace copy of ΔXYZ and try to make it cover LMN with X on L, Y on M and Z on N. Two triangles cover each other exactly.
Therefore ΔLMN ≅ ΔXYZ

Note

Angle Angle Side (AAS) and Angle Side Angle (ASA) are more or less the same congruence condition.

#### Worked-out problems on angle angle side congruence triangles (AAS postulate):

1. OB is the bisector of /_AOC, PM ⊥ OA and PN ⊥ OC. Show that ΔMPO ≅ ΔNPO. Solution:
In ΔMPO and ΔNPO
PM ⊥ OM and PN ⊥ ON
Therefore /_PMO = /_PNO = 90°
Also, OB is the bisector of /_AOC
Therefore /_MOP = /_NOP
OP = OP common
Therefore, ΔMPO ≅ ΔNPO by AAS congruence condition

Right Angle Hypotenuse Side congruence

Conditions for the RHS - Right Angle Hypotenuse Side congruence
Two triangles triangle are congruent if the hypotenuse and one side of the one triangle are respectively equal to the hypotenuse and one side of the other.

#### Experiment to prove Congruence with RHS: Draw a ΔLMN with /_M = 90°, LM = 3cm LN = 5 cm,
Also, draw another ΔXYZ with /_Y = 90°, XY = 3cm and XZ = 5cm.
We see that /_M = /_Y, LM = XY and LN = XZ.
Make a trace copy of ΔXYZ and try to make it cover ΔLMN with X on L, Y on M and Z on N.
We observe that: Two triangles cover each other exactly.
Therefore, ΔLMN ≅ ΔXYZ

#### Worked-out problems on right angle hypotenuse side congruence triangles (HL postulate):

1. ΔPQR is an isosceles triangle such that PQ = PR, prove that the altitude PO from P on QR bisects PQ. Solution:
In the right triangles POQ and POR,
/_POQ = /_POR = 90°
PQ = PR [since, ΔPQR is an isosceles. Given PQ = PR]
PO = OP [common]
Therefore Δ POQ ≅ Δ POR by RHS congruence condition
So, QO = RO (by corresponding parts of congruence triangles)

2. ΔXYZ is an isosceles triangle such that XY = XZ, prove that the altitude XO from X on YZ bisects YZ. Solution:

In the right triangles XOY and XOZ,
/_XOY = /_XOZ = 90°
XY = XZ [since, ΔXYZ is an isosceles. Given XY = XZ]
XO = OX [common]
Therefore Δ XOY ≅ Δ XOZ by RHS congruence condition
So, YO = ZO (by corresponding parts of congruence triangles)

3. In the adjoining figure, given that AB = BC, YB = BZ, BA ⊥ XY and BC ⊥ XZ. Prove that XY = XZ Solution:
In right triangles YAB and BCZ we get,
YB = BZ [given]
AB = BC [given]
So, by RHS congruence condition
Δ YAB ≅ Δ BCZ
/_Y = /_Z (since by corresponding parts of congruence triangles are equal)
XZ = XY (since sides opposite to equal angles are equal)

Congruent Shapes

The two figures which are same shape and same size are called congruent figures and the relation of two figures being congruent is called congruence.
The symbol to denote congruence is .
Two line segments are congruent if both have the same length. Two circles are congruent if both of them have the same radii. Two triangles are congruent if both of them have the same length of sides. Two rectangles are congruent if both of them have the opposite sides are equal. Two squares are congruent if both of them have the same edges. Thus, the method of comparing two figures is known as the method of superposition.

Congruent or Similar

Congruent shapes: Similar shapes:
Two geometrical shapes which are identical in shape and size are said to be congruent. We use the symbol '≅' to denote congruence.Geometrical shape which have same shape but different size are called similar shape or figure. We use the symbol '∼' to denote similarly.
For example:
(i) Two circle of same radii.
(ii) Two line segments of same length.
For example:
(i) Two circle of different radii.
(ii) Two line segments of different measure.

Note: Two congruent figures are always similar but two similar figures may not be congruent.

Proof of Pythagorean Theorem
• The proof of Pythagorean Theorem in mathematics is very important.
• In a right angle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
• States that in a right triangle that, the square of a (a2) plus the square of b (b2) is equal to the square of c (c2).
• In short it is written as: a2 + b2 = c2 • Let QR = a, RP = b and PQ = c. Now, draw a square WXYZ of side (b + c). Take points E, F, G, H on sides WX, XY, YZ and ZW respectively such that WE = XF = YG = ZH = b.
• Then, we will get 4 right-angled triangle, hypotenuse of each of them is 'a': remaining sides of each of them are band c. Remaining part of the figure is the
• square EFGH, each of whose side is a, so area of the square EFGH is a2.
• Now, we are sure that square WXYZ = square EFGH + 4 Δ GYF

or, (b + c)2 = a2 + 4 • 1/2 b • c
or, b2 + c2 + 2bc = a2 + 2bc
or, b2 + c2 = a2 Proof of Pythagorean Theorem using Algebra: Given: A Δ XYZ in which /_XYZ = 90°.

To prove: XZ2 = XY2 + YZ2

Construction: Draw YO ⊥ XZ

Proof: In ΔXOY and ΔXYZ, we have,
/_X = /_X → common
/_XOY = /_XYZ → each equal to 90°
Therefore, Δ XOY ∼ Δ XYZ → by AA-similarity
⇒ XO/XY = XY/XZ
⇒ XO * XZ = XY2 ............ (i)

In ΔYOZ and ΔXYZ, we have,
/_Z = /_Z → common
/_YOZ = /_XYZ → each equal to 90°
Therefore, Δ YOZ ∼ Δ XYZ → by AA-similarity
⇒ OZ/YZ = YZ/XZ
⇒ OZ * XZ = YZ2 .............. (ii)

From (i) and (ii) we get,
XO * XZ + OZ * XZ = (XY2 + YZ2)
⇒ (XO + OZ) * XZ = (XY2 + YZ2)
⇒ XZ * XZ = (XY2 + YZ2)
⇒ XZ2 = (XY2 + YZ2)

Converse of Pythagorean Theorem

#### Converse of Pythagorean Theorem states that:

In a triangle, if the square of one side is equal to the sum of the squares of the other two sides then the angle opposite to the first side is a right angle.

Given: A ΔPQR in which PR^2 = PQ^2 + QR^2
To prove: /_Q = 90°
Construction: Draw a ΔXYZ such that XY = PQ, YZ = QR and /_Y = 90° So, by Pythagora's theorem we get,
XZ2 = XY2 + YZ2
⇒ XZ2 = PQ2 + QR2 ........ (i), [since XY = PQ and YZ = QR]
But, PR2 = PQ2 + QR2 ....... (ii), [given]

From (i) and (ii) we get,
PR2= XZ2 ⇒ PR = XZ
Now, in ΔPQR and ΔXYZ, we get
PQ = XY,
QR = YZ and
PR = XZ

Therefore ΔPQR ≅ ΔXYZ
Hence /_Q = /_Y = 90°

#### Word problems using the Converse of Pythagorean Theorem:

1. The side of a triangle are of length 4.5 cm, 7.5 cm and 6 cm. Is this triangle a right triangle? If so, which side is the hypotenuse?

Solution:
We know that hypotenuse is the longest side. If 4.5 cm, 7.5 cm and 6 cm are the lengths of angled triangle, then 7.5 cm will be the hypotenuse.
Using the converse of Pythagoras theorem, we get
(7.5)2 = (6)2 + (4.5)2
⇒ 56.25 = 36 + 20.25
⇒ 56.25 = 56.25
Since, both the sides are equal therefore, 4.5 cm, 7.5 cm and 6 cm are the side of the right angled triangle having hypotenuse 7.5 cm.

2. The side of a triangle are of length 8 cm, 15 cm and 17 cm. Is this triangle a right triangle? If so, which side is the hypotenuse?
Solution:
We know that hypotenuse is the longest side. If 8 cm, 15 cm and 17 cm are the lengths of angled triangle, then 17 cm will be the hypotenuse.
Using the converse of Pythagoras theorem, we get
(17)2 = (15)2 + (8)2
⇒ 289 = 225 + 64
⇒ 289 = 289
Since, both the sides are equal therefore, 8 cm, 15 cm and 17 cm are the side of the right angled triangle having hypotenuse 17 cm.

3. The side of a triangle are of length 9 cm, 11 cm and 6 cm. Is this triangle a right triangle? If so, which side is the hypotenuse?
Solution:
We know that hypotenuse is the longest side. If 9 cm, 11 cm and 6 cm are the lengths of angled triangle, then 11 cm will be the hypotenuse.
Using the converse of Pythagoras theorem, we get
(11)2 = (9)2 + (6)2
⇒ 121 = 81 + 36
⇒ 121 ≠ 117
Since, both the sides are not equal therefore 9 cm, 11 cm and 6 cm are not the side of the right angled triangle.
The above examples of the converse of Pythagorean Theorem will help us to determine the right triangle when the sides of the triangles will be given in the questions.

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