Chapter 7: Congruence of Triangles

# Question: 1

Complete the following statements:

a.    Two line segments are congruent if __________.

b.   Among two congruent angles, one has a measure of $70°$; the measure of the other angle is _________.

c.    When we write $\angle \text{A}=\angle \text{B}$, we actually mean ___________.

## Solution

a.    They have the same length

b.   $70°$

c.    $m\angle \text{A}=m\angle \text{B}$

# Question: 2

Give any two real-life examples for congruent shapes.

## Solution

(i)             Two footballs

(ii)          Pages of same book

# Question: 3

If $\Delta \text{ABC}\cong \Delta \text{FED}$ under the correspondence $\text{ABC}↔\text{FED}$ write all the corresponding congruent parts of the triangles.

## Solution

Given: $\Delta \text{ABC}\cong \Delta \text{FED}$

The corresponding congruent parts of the triangles are: (i)             $\angle \text{A}↔\angle \text{F}$

(ii)          $\angle \text{B}↔\angle \text{E}$

(iii)      $\angle \text{C}↔\angle \text{D}$

(iv)       $\overline{\text{AB}}↔\overline{\text{FE}}$

(v)          $\overline{\text{BC}}↔\overline{\text{ED}}$

(vi)       $\overline{\text{AC}}↔\overline{\text{FD}}$

# Question: 4

## If $\Delta \text{DEF}\cong \Delta \text{BCA}$, write the part(s) of $\Delta \text{BCA}$ that correspond to

(i)             $\angle \text{E}$

(ii)          $\overline{\text{EF}}$

(iii)      $\angle \text{F}$

(iv)       $\overline{\text{DF}}$

## Solution

Given: $\Delta \text{DEF}\cong \Delta \text{BCA}$.

(i)             $\angle \text{E}↔\angle \text{C}$

(ii)          $\overline{\text{EF}}↔\overline{\text{CA}}$

(iii)      $\angle \text{F}↔\angle \text{A}$

(iv)       $\overline{\text{DF}}↔\overline{\text{BA}}$ # Question: 1

Which congruence criterion do you use in the following?

a.    Given : $\text{AC}=\text{DF}$

$\text{AB}=\text{DE}$

$\text{BC}=\text{EF}$

So, $\Delta \text{ABC}\cong \Delta \text{DEF}$ b.   Given: $\text{ZX}=\text{RP}$

$\text{RQ}=\text{ZY}$

$\angle \text{PRQ}=\angle \text{XZY}$

So, $\Delta \text{PQR}\cong \Delta \text{XYZ}$ c.    Given: $\angle \text{MLN}=\angle \text{FGH}$

$\angle \text{NML}=\angle \text{GFH}$

$\text{ML}=\text{FG}$

So, $\Delta \text{LMN}\cong \Delta \text{GFH}$ d.   Given: $\text{EB}=\text{DB}$

$\text{AE}=\text{BC}$

$\angle \text{A}=\angle \text{C}=\text{90}°$

So, $\Delta \text{ABE}\cong \Delta \text{CDB}$ ## Solution

a.    By $\text{SSS}$ congruence criterion,

since it is given that  $\text{BC}=\text{EF}$

The three sides of one triangle are equal to the three corresponding sides of another triangle.

Therefore, $\Delta \text{ABC}\cong \Delta \text{DEF}$

b.   By $\text{SAS}$ congruence criterion,
since it is given that  and $\angle \text{PRQ}=\angle \text{XZY}$

The two sides and one angle in one of the triangle are equal to the corresponding sides and the angle of other triangle.

Therefore, $\Delta \text{PQR}\cong \Delta \text{XYZ}$

c.    By $\text{ASA}$ congruence criterion,

since it is given that $\angle \text{MLN}=\angle \text{FGH,}$

The two angles and one side in one of the triangle are equal to the corresponding angles and side of other triangle.

Therefore, $\Delta \text{LMN}\cong \Delta \text{GFH}$

d.   By $\text{RHS}$ congruence criterion,

since it is given that $\text{EB}=\text{BD,}$  $\angle \text{A}=\angle \text{C}=\text{90}°$

Hypotenuse and one side of a right angled triangle are respectively equal to the hypotenuse and one side of another right angled triangle.

Therefore, $\Delta \text{ABE}\cong \Delta \text{CDB}$

# Question: 2

You want to show that $\Delta \text{ART}\cong \Delta \text{PEN}$

a.    If you have to use $\text{SSS}$ criterion, then you need to show

(i) $\text{AR}=$

(ii) $\text{RT}=$

(iii) $\text{AT}=$

b.   If it is given that $\angle \text{T}=\angle \text{N}$ and you are to use SAS criterion, you need to have

(i) $\text{RT}=$

and

(ii) $\text{PN}=$

c.    If it is given that $\text{AT}=\text{PN}$ and you are to use ASA criterion, you need to have

(i) $?$

(ii) $?$ ## Solution

a.    Using SSS criterion, $\Delta \text{ART}\cong \Delta \text{PEN}$

(i) $\text{AR}=\text{PE}$

(ii) $\text{RT}=\text{EN}$

(iii) $\text{AT}=\text{PN}$

b.   Given: $\angle \text{T}=\angle \text{N}$

Using SAS criterion, $\Delta \text{ART}\cong \Delta \text{PEN}$

(i) $\text{RT}=\text{EN}$

(ii) $\text{PN}=\text{AT}$

c.    Given: $\text{AT}=\text{PN}$

Using ASA criterion, $\Delta \text{ART}\cong \Delta \text{PEN}$

(i) $\angle \text{RAT}=\angle \text{EPN}$

(ii) $\angle \text{RTA}=\angle \text{ENP}$

# Question: 3

You have to show that $\Delta \text{AMP}\cong \Delta \text{AMQ}$. In the following proof, supply the missing reasons. Steps Reasons (i) $\text{PM}=\text{QM}$ (i) …………. (ii) $\angle \text{PMA}=\angle \text{QMA}$ (ii) …………. (iii) $\text{AM}=\text{AM}$ (iii) …………. (iv) $\Delta \text{AMP}\cong \Delta \text{AMQ}$ (iv) ………….

## Solution

 Steps Reasons (i) $\text{PM}=\text{QM}$ (i) Given (ii) $\angle \text{PMA}=\angle \text{QMA}$ (ii) Given (iii) $\text{AM}=\text{AM}$ (iii) Common (iv) $\Delta \text{AMP}\cong \Delta \text{AMQ}$ (iv) $\text{SAS}$ congruence rule

# Question: 4

In , $\angle \text{B}=\text{40}°$ and $\angle \text{C}=\text{110}°$.

In , $\angle \text{Q}=\text{40}°$ and $\angle \text{R}=\text{110}°$.

A student says that $\Delta \text{ABC}\cong \Delta \text{PQR}$  by $\text{AAA}$ congruence criterion. Is he justified? Why or why not?

## Solution

No, because the 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.

# Question: 5

In the figure, the two triangles are congruent. The corresponding parts are marked. We can write $\Delta \text{RAT}\cong ?$  ## Solution

In the figure, given two triangles are congruent. So, the corresponding parts are:

$\text{A}↔\text{O}$,             $\text{R}↔\text{W}$,                   $\text{T}↔\text{N}$,

We can write,

$\Delta \text{RAT}\cong \Delta \text{WON}$            [By $\text{SAS}$ congruence rule]

# Question: 6

Complete the congruence statement:  $\Delta \text{BCA}\cong ?$                                  $\Delta \text{QRS}\cong ?$

## Solution

In $\Delta \text{BAT}$ and $\Delta \text{BAC}$, given triangles are congruent so the corresponding parts are:

$\text{B}↔\text{B}$,              $\text{A}↔\text{A}$,             $\text{T}↔\text{C}$,

Thus,

$\Delta \text{BCA}\cong \Delta \text{BTA}$       [By $\text{SSS}$ congruence rule]

In $\Delta \text{QRS}$ and $\Delta \text{TPQ}$, given triangles are congruent so the corresponding parts are:

$\text{P}↔\text{R}$,              $\text{T}↔\text{Q}$,             $\text{Q}↔\text{S}$,

Thus,

$\Delta \text{QRS}\cong \Delta \text{TPQ}$       [By $\text{SSS}$ congruence rule]

# Question: 7

In a squared sheet, draw two triangles of equal areas such that

(i)             the triangles are congruent.

(ii)          the triangles are not congruent.

What can you say about their perimeters?

## Solution

In a squared sheet, draw $\Delta \text{ABC}$ and $\Delta \text{PQR}$.

When two triangles have equal areas and

(i)             these triangles are congruent, i.e., $\Delta \text{ABC}\cong \Delta \text{PQR}$ [By $\text{SSS}$ congruence rule]

Then, their perimeters are same because length of sides of first triangle are equal to the length of sides of another triangle by $\text{SSS}$ congruence rule.

(ii)          But, if the triangles are not congruent, then their perimeters are not same because lengths of sides of first triangle are not equal to the length of corresponding sides of another triangle.

# Question: 8

## Solution

Let us draw two triangles $\text{PQR}$ and $\text{ABC}$.  All angles are equal, two sides are equal except one side. Hence, $\Delta \text{PQR}$ are not congruent to $\Delta \text{ABC}$.

# Question: 9

If $\Delta \text{ABC}$ and $\Delta \text{PQR}$ are to be congruent, name one additional pair of corresponding parts. What criterion did you use?  ## Solution

$\Delta \text{ABC}$ and $\Delta \text{PQR}$ are congruent. Then one additional pair is $\overline{\text{BC}}=\overline{\text{QR}}$

Given:

$\angle \text{C}=\angle \text{R}$

$\overline{\text{BC}}=\overline{\text{QR}}$

Therefore,

$\Delta \text{ABC}\cong \Delta \text{PQR}$       [By $\text{ASA}$ congruence rule]

# Question: 10

Explain, why $\Delta \text{ABC}\cong \Delta \text{FED}$  ## Solution

Given:

In $\Delta \text{ABC}$ and $\Delta \text{FED}$

$\angle \text{A}=\angle \text{F}$  [Given]

Using angle sum property in both triangles and we find that $\angle \text{C}=\angle \text{D}$

Also, $\text{BC}=\text{ED}$  [Given]

Therefore,

$\Delta \text{ABC}\cong \Delta \text{FED}$              [By $\text{ASA}$ congruence rule]