Subject: Compulsory Maths

Two triangles are congruent if they have exactly the same three sides and exactly the same three angles. Similar triangle has the same shape, but the size may be different. Two triangles are similar if and only if the corresponding sides are in proportion and the corresponding angles are congruent.

The triangles having same size and shape are called congruent triangles. Two triangles are congruent when the three sides and three angles of one triangle have the measurements as three sides and three angles of another triangle. The symbol for congruent is ≅.

In the following figure, ΔABC and ΔPQR are congruent. We denote this as ΔABC ≅ ΔPQR.

**Postulate (SAS)**

If two sides and the angle between them in one triangle are congruent to the corresponding parts in another triangle, then the triangles are congruent.

In the given figure,

AB ≅ PQ Sides (S)

∠B ≅ ∠Q Angle (A)

BC ≅ QR Side (S)

Therefore, ΔABC ≅ ΔPQR

**Theorem (ASA)**

A unique triangle is formed by two angles and the included side.

Therefore, if two angles and the included side of one triangle are congruent to two angles and the included side of the another triangle, then the triangles are congruent.

In the figure,

∠B ≅ ∠E Angle (A)

BC ≅ EF Side (S)

∠C ≅ ∠F Angle (A)

Therefore, ΔABC ≅ ΔDEF

**Theorem ( AAS)**

A unique triangle is formed by two angles and non-included side. Therefore, if two angles and the side opposite to one of them in a triangle are congruent to the corresponding parts in another triangle, then the triangles are congruent.

In the figure,

∠A ≅ ∠X Angle (A)

∠C ≅ ∠Z Angle (A)

BC ≅ YZ Side (S)

Therefore, ΔABC ≅ ΔXYZ

**Theorem (SSS)**

A unique triangle is formed by specifying three sides of a triangle, where the longest side (if there is one) is less than the sum of the two shorter sides.

Therefore, if their sides of a triangle are congruent to three sides of another triangle, then the triangles are congruent.

In the figure

AB ≅ PQ Sides (S)

BC ≅ QR Sides (S)

CA ≅ RP Sides (S)

Therefore, ΔABC ≅ ΔPQR

- If the corresponding sides of a triangle are proportion to another triangle then the triangles are similar.
**Example**

If∠A ≅ ∠D and ∠B ≅ ∠E, Then ΔABC ∼ ΔDEF - If the corresponding angle of a triangle is congruent to another triangle then, the triangles are similar.
**Example**

If \(\frac{AB}{DE}\) = \(\frac{BC}{EF}\) = \(\frac{AC}{DF}\), then ΔABC ∼ ΔDEF - Conversing first and the second method we can prove triangle similar as their sides being proportional and angles congruent.
**Example**

∠A ≅ ∠D and \(\frac{AB}{DE}\) = \(\frac{AC}{DF}\) then ΔABC ∼ ΔDEF

When the lines are parallel in a triangle, then they intersect each other which divides the sides of a triangle proportionally.

**Verification:**

In ΔPQR and ΔSPT

Statements | Reasons |

ST⁄⁄QR | Given |

\(\angle\)PST \(\cong\) \(\angle\)QSR | Corresponding angles |

ΔPQR \(\cong\) ΔSPT | Common Angle P |

\(\frac{PS}{SQ}\) = \(\frac{PT}{TR}\) | ST⁄⁄QR |

**Example**

Given the following triangles, find the length of x.

Solution:

The triangles are similar by AA rule.So, the ratio of lengths are equal.

\(\frac{6}{3}\) = \(\frac{10}{x}\)

or, 6x = 30

or, x = \(\frac{30}{6}\)

\(\therefore\) x = 5 cm

- There is three easy way to prove similarity. If two pairs of corresponding angles in a pair of triangles are congruent, then the triangles are similar.
- When the three angle pairs are all equal, the three pairs of the side must be proportion.
- When triangles are congruent and one triangle is placed on the top of other sides and angles that are in the same position are called corresponding parts.
- Congruent and similar shapes can make calculations and design work easier.

- It includes every relationship which established among the people.
- There can be more than one community in a society. Community smaller than society.
- It is a network of social relationships which cannot see or touched.
- common interests and common objectives are not necessary for society.

Find angles, arms of the given triangles & show if they are congurentor not?

Solution:

Here, AB=1.9 cm DE=1.7 cm

BC=1.7 cm EF=1.3 cm

CA=2.1 cm DF=2.1 cm

\(\angle\)A=107^{o}\(\angle\)D=90^{o}

\(\angle\)B=57^{o}\(\angle\)E=35^{o}

\(\angle\)C=16^{o}\(\angle\)F=55^{o}

The angles arms of the triangle is not equal so it is not congurent.

Find angles, arms of the given triangles & show if they are congurentor not?

Solution:

Here, AB=1.6 cm DE=1.9 cm

BC=1.3 cm EF=1.4 cm

CA=1.9 cm DF=1.9 cm

\(\angle\)A=45^{o}\(\angle\)D=40^{o}

\(\angle\)B=100^{o}\(\angle\)E=70^{o}

\(\angle\)C=35^{o}\(\angle\)F=70^{o}

The angles arms of the triangle is not equal so it is not congurent.

From the given congurent triangle.Write congurent sides and angles.

Solution:

Here given, congurent triangle is PQ and LM, QR and MN, PR and LN. So that, congurent angles is \(\angle\)P and(\angle\)L,(\angle\)R and(\angle\)N,(\angle\)Q and(\angle\)M.

From the given congurent triangle.Write congurent sides and angles.

Solution:

Here given, congurent triangle is XY and AB, YZ and BC, XZ and AC. So that, congurent angles is \(\angle\)X and(\angle\)A,(\angle\)Y and(\angle\)B,(\angle\)Z and(\angle\)C.

In the triangle shown below, A'C is parellel to AC.Find the measure of x and y.

Solution:

In \(\triangle\) ABC and \(\triangle\)A'BC'

\(\angle\)BAC≈\(\angle\)BA'C',\(\angle\)ACB≈\(\angle\)BC'A' being corresponding angles.

so, \(\triangle\)ABC ~\(\triangle\)A'BC'

\(\frac{AB}{A'B}\)=\(\frac{BC}{BC'}\)=\(\frac{AC}{A'C'}\)

\(\frac{30+x}{30}\)=\(\frac{15+y}{y}\)=\(\frac{22}{14'}\)

Taking 1st and 3rd \(\frac{30+x}{30}\)=\(\frac{22}{14}\)

14(30+x)=22\(\times\)30

420+14x=660

14x=660-420

14x=240

x=\(\frac{240}{14}\)=17.1 cm

Taking 2nd and 3rd

\(\frac{15+y}{y}\)=\(\frac{22}{14}\)=\(\frac{11}{7}\)

11 y=105+7 y

11y-7y=105

4y=105

y=\(\frac{105}{4}\)=26.25 cm

Find the values of unknown sides and angle of given congurent triangled calculating the values of x and y.

Solution,

Here, 5y-3^{o}=62^{o}[ \(\therefore\)

or, 5y-3^{o}=62^{o}

or, 5y=62^{o}+3^{o}=65^{o}

or, y=\(\frac{65^o}{5}\)=13^{o}

Likewise, 4x-4^{o}=48^{o}

or, 4x-4^{o}=48^{o}

or, 4x=48^{o}+4^{o}=52^{o}

or, x=\(\frac{52^o}{4}\)=13^{o}

\(\therefore\)x=13^{o} and y=13^{o} Ans.

Solution:

We know that,

\(\angle\)A+\(\angle\)B+\(\angle\)C=180^{o}

or, 122^{o}+38^{o}+\(\angle\)C=180^{o}

or,\(\angle\)C=180^{o}-160^{o}=20^{o}

Again,\(\angle\)R=\(\angle\)C [\(\therefore\) ]

or,\(\angle\)R=\(\angle\)A

or, x=20^{o}

Likewise,\(\angle\)P=\(\angle\)A

or, y+10^{o}=122^{o}

or, y=112^{o}

and\(\angle\)Q=\(\angle\)B

or,\(\angle\)Q=38^{o}

\(\therefore\) x=20^{o} and y = 112^{o} Ans.

Solution:

Here, \(\angle\)A+ \(\angle\)B + \(\angle\)C[\(\therefore\) sum of interior angles of a triangle is 180^{o}]

or, 56^{o}+\(\angle\)B+64^{o}=180^{o}

or,\(\angle\)B=180^{o}-120^{o}=60^{o}

Again, AC=PR

or, 3x-0.6=x+3

or, 3x-x=3+0.6

or, 2x=3.6

or, x=\(\frac{3.6}{2}\)=1.8

\(\therefore\)PR=AC=(3x-0.6)cm

=(3\(\times\)1.8-0.6)cm

=(5.4 - 0.6) cm

=95.4 - 0.6) cm

=4.8 cm

AB=PQ=3.1 cm [ \(\therefore\) corresponding sides of congurent triangles]

so,\(\angle\)B=60^{o}, AC=PR=4.8 cm, AB=PQ=3.1 cm

Solution:

Here, \(\angle\)L+ \(\angle\)M+ \(\angle\)N [\(\therefore\) sum of interior angles of a triangle is 180^{o}]

or, 45^{o} + 70^{o} + \(\angle\)N=180^{o}

or, \(\angle\)N=180^{o}-115^{o}=65^{o}

Again, \(\angle\)X=45^{o} and \(\angle\)Z=65^{o} [\(\therefore\) Corresponding angles of congurent triangle]

Likewise, MN=YZ

or, 2 y=2.5

or, y=1.25

2x+1=3

or, 2x=2

or, x=1

so that , \(\angle\)L= \(\angle\)X=45^{o}, \(\angle\)N= \(\angle\)Z=65^{o}, MN=YZ=2y=2.5 cm and LM=XY=3 cm, x=1, y= 1.25 Ans.

Find the angles & length of side of the given pairs of triangles and check if they are similar or not.

Solution:

Here, \(\angle\)A=52^{o} \(\angle\)=88^{o} \(\angle\)C=40^{o}

\(\angle\)P=52^{o} \(\angle\Q=88^{o} \(\angle\)R=40^{o}

AB=1 cm BC=1.2 cm CA=2 cm

PQ=1.3 cm QR=1.6 cm PR=2.6 cm

\(\angle\)A= \(\angle\)P, \(\angle\)B= \(\angle\)Q and \(\angle\)C= \(\angle\)R

\(\frac{AB}{PQ}\)=\(\frac{1.0}{1.3}\),\(\frac{BC}{QR}\),\(\frac{1.0}{1.3}\),=\(\frac{CA}{RP}\)=\(\frac{1.0}{1.3}\)

Hence, given triangle is similar triangle.

Solution:

Here, \(\angle\) A=40^{o}\(\angle\) B=70^{o}\(\angle\) C=70^{o}

\(\angle\) P=41^{o}\(\angle\) Q=62^{o}\(\angle\) R=77^{o}

AB=1.5 cm BC=1 cm CA=1.5 cm

PQ=1.1 cm QR=1 cm RP=1.6 cm

Here,\(\angle\)A \(\neq \)\(\angle\)P,\(\angle\)B\(\neq \)\(\angle\)Q and \(\angle\)C\(\neq \) \(\angle\)R

Hence, given triangle is not similar triangle.

Solution:

Here, \(\angle\)A=38^{o} \(\angle\)B=90^{o} \(\angle\)C=52^{o}

\(\angle\)P=38^{o} \(\angle\)Q=90^{o} \(\angle\)r=52^{o}

AB=1.6 cm BC=1.2 cm

CA=2 cm PQ=0.8 cm

QR=0.6 cm PR=1 cm

\(\angle\)A= \(\angle\)P, \(\angle\)B= \(\angle\)Q and \(\angle\)C= \(\angle\)R

and \(\frac{AB}{PQ}\)=\(\frac{1.6}{0.8}\)=2, \(\frac{BC}{QR}\)=\(\frac{1.2}{0.6}\)=2, \(\frac{CA}{RP}\)=\(\frac{2}{1}\)=2

Hence, given triangle is similar triangle.

In the given figure if BC||DE and \(\angle\)CED=30^{o}

a) \(\triangle\)ADE ~ \(\triangle\)ABC

b) Find DE and \(\angle\)ACB

Solution:

Here,In \(\triangle\)ADE and \(\triangle\)ABC

\(\angle\)DAE=\(\angle\)ABC

\(\therefore\)\(\triangle\)ADE~\(\triangle\)ABC

so, \(\frac{DE}{BC}\)=\(\frac{AE}{AC}\)

or,\(\frac{DE}{7.2}\)=\(\frac{6+4}{6}\)

or, \(\frac{DE}{7.2}\)=\(\frac{10}{6}\)

or, DE\(\times\)6=10\(\times\)7.2

or, DE=\(\frac{10\times7.2}{6}\)=12 cm

Again, \(\angle\)ACB=\(\angle\)CDE=30^{o}

Hence, DE=12 cm and \(\angle\)ACD=30^{o}Ans.

Are these triangles congruent or not?

From the given figure,

∠STU ≅ ∠SVW and TU ≅ VW

Here, ∠TSU and ∠VSW are vertical angles. Since vertical angles are congruent,

∠TSU ≅ ∠VSW.

Finally, put the three congruency statements in order. ∠STU is between ∠TSU and TU, and ∠SVW is between ∠VSW and VW in the diagram.

∠TSU ≅ ∠VSW (**A**ngle)

∠STU ≅ ∠SVW (**A**ngle)

TU ≅ VW (**S**ide)

Hence, the given triangles are congurent as it forms AAS theorem.

Why these triangles are congruent?

From the given figure,

BC ≅ BH and ∠BCF≅∠BHG.

Here, ∠CBF and ∠GBH are vertical angles. Since vertical angles are congruent,

∠CBF ≅ ∠GBH.

Finally, put the three congruency statements in order. BC is between ∠BCF and ∠CBF, and BH is between ∠BHG and ∠GBH in the diagram.

∠BCF ≅ ∠BHG Angle

BC ≅ BH Side

∠CBF ≅ ∠GBH Angle

Hence, the congruent sides and angles form ASA. The triangles are congruent by the ASA Theorem.

Prove these triangles are congruent.

From the figure,

∠XWY≅∠YWZ and ∠WXY≅∠WZY.

Here, the triangles share WY. By the reflexive property of congruence, WY ≅ WY.

Finally, put the three congruency statements in order. ∠WXY is between ∠XWY and WY, and ∠WZY is between ∠YWZ and WY in the diagram.

∠XWY ≅ ∠YWZ Angle

∠WXY ≅ ∠WZY Angle

WY ≅ WY Side

Hence, the congruent sides and angles form AAS. The triangles are congruent by the AAS Theorem.

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