Free Congruence of Triangles Subjective Test 01 Practice Test - 7th grade 

In geometry, two figures or objects are congruent if they have the same shape and size.

All parts: 2 Marks



Given that,

$\Delta PEN\cong \Delta CAR$ and

PN = CR

Corresponding parts of congruent triangle are congruent.

Therefore,  the corresponding sides of congruent triangle are equal.

$\Rightarrow PE=CA,EN=AR,PN=CR$.

$\Rightarrow$ Also all the corresponding angles of congruent triangles  are equal.

$\Rightarrow \angle P=\angle C,\angle E=\angle A,\angle N=\angle R$.

Definition: 1 Mark
Proof: 1 Mark

If two angles have the same measurement, they are congruent. Also, if two angles are congruent, their measurements are same.

We know that in every right-angled triangle, one angle is ${90}^{\circ }$.

Example: Consider two right-angled triangles $\Delta AJU$ and $\Delta NIV$

In $\Delta AJU,\angle AJU={90}^{\circ }$

and in  $\Delta NIV,\angle NIV={90}^{\circ }$

Since one angle is same in both, we can say that they have one pair of congruent angles.

So, if you take any two right-angled triangles, at least one pair of angles will be congruent, i.e. equal.

 Reason: 1 Mark each

(a)If three sides of one triangle are equal to the  three sides of the other triangle, then the two  triangles are congruent to each other by SSS congruence criterion.
$\Rightarrow$ Both triangles look like the mirror image of each other.
$\Rightarrow$ Both the  triangles  superimpose on each other.

So, if the sides of a triangle are congruent to the sides of another triangle, the two triangles will be congruent.

(b) Nothing is given or can be said about any of the corresponding sides in this case, As, AAA is not a rule for congruency, the triangles formed may or may not be congruent, depending on if the corresponding parts are equal or not.

Application of theorem: 1 Mark
Steps: 2 Marks

In $\Delta CATand\Delta R{A}^{\prime }T$

$CT=RT$         [Given]

$\angle CTA=\angle RT{A}^{\prime }$   [Vertically Opposite Angles]

$AT=A^{\prime }T$   [Given]

$\therefore \Delta CAT\cong \Delta R{A}^{\prime }T$   [By SAS congruence rule]

$\Rightarrow \angle CAT=\angle R{A}^{\prime }T$   [Corresponding parts of congruent triangles]

But $\angle CATand\angle R{A}^{\prime }T$ are alternate interior angles.

If the pair of alternate interior angles is equal then the lines are parallel.

$\Rightarrow CA\parallel A^{\prime }R$.  

Steps: 1 Mark
Proof: 2 Marks

In $\Delta 1and\Delta 2$:

$\angle AOD=\angle COD={90}^{\circ }$ (Diagonals of square intersect at right angles)

$AD=CD$ (Sides of a square; hypotenuse)

$OD=DO$ (Common)

Hence,  $\Delta 1\cong \Delta 2$ (By RHS congruence rule) ---------------------1

Similarly, $\Delta 4\cong \Delta 3$ (By RHS congruence rule) ---------------------2

In $\Delta 1and\Delta 4$

$\angle AOD=\angle AOB={90}^{\circ }$ (Diagonals of square intersect at right angles)

$AD=AB$ (Sides of a square; hypotenuse)

$OA=AO$ (Common)

Hence, $\Delta 1\cong \Delta 4$ (By RHS congruence rule) -------------------3

Similarly, $\Delta 2\cong \Delta 3$ (By RHS congruence rule) ----------------4

From 1, 2, 3 and 4 we can say that all the triangles i.e. $\Delta 1$, $\Delta 2$, $\Delta 3$ and $\Delta 4$ are congruent to each other.

5, 6, 7 and 8 have relatively small area. Since they have same shape and size, they are also congruent. So, we can say that all the slices of the pizza are congruent to each other.

Proof: 1 Mark
Steps: 2 Marks

Consider two triangles, $\Delta ABC$ and $\Delta PQR$ in which,



$\angle ABC=\angle PQR$

$\angle ACB=\angle PRQ$

$AB=PQ$

We know that, 

$\angle ABC+\angle ACB+\angle BAC={180}^0$

$\angle BAC={180}^0-(\angle ABC+\angle ACB)....\left(i\right)$

Similarly,

$\angle QPR={180}^0-(\angle PQR+\angle PRQ)......\left(ii\right)$

From (i) and (ii), 

$\angle BAC=\angle QPR$

Now, In  $\Delta ABC$ and $\Delta PQR$

$\angle ABC=\angle PQR$  [Given]

$\angle BAC=\angle QPR$   [ Proved above]

$AB=PQ$  [Given]

$\Rightarrow \Delta ABC\cong \Delta PQR$  [ ASA congruency rule]

$\Rightarrow$ These triangles are always congruent.

Properties: 1 Mark
Proof: 1 Mark
Steps: 2 Marks

In a rectangle opposite sides are equal and parallel.

In $\Delta OADand\Delta OCB$,

$\angle ODA=\angle OBC$

[Alternate interior angles; $AD\parallel BC$ and BD as transversal]

AD = BC  [Opposite sides of a rectangle are equal]

$\angle OAD=\angle OCB$  

[Alternate interior angles; $AD\parallel BC$ and AC as transversal]

Hence $\Delta OAD\cong \Delta OCB$   [By ASA congruence rule]

Equating the corresponding parts of congruent triangles, we get:

AO = CO

BO = DO

$\Rightarrow$  Diagonals of a rectangle bisect each other.

Properties: 1 Mark
Each proof: 1 Mark

In $\Delta ADBand\Delta ADC$

$AB=AC$    [Given]

$\angle ADB=\angle ADC={90}^{\circ }$   [Given]

$AD=AD$   [common]

Hence, $\Delta ADB\cong \Delta ADC$ [By RHS congruence rule…….(1)]

From (1), $\angle BAD=\angle CAD$   [Corresponding parts of congruent triangles]

From (1), $BD=DC$   [Corresponding parts of congruent triangles]

Each Part: 2 Marks

(a)

In $\Delta BDAand\Delta CDA$

$\angle B=\angle C$    [Given]

$\angle BAD=\angle CAD$   [Given, DA is an angle bisector ]

$AD=AD$   [Common side]

$\Rightarrow \Delta BDA\cong \Delta CDA$  [  AAS criteria] 


(b) we observe that in the given figures, there are no pairs of congruent sides. Since all of the congruency theorems call for at least one pair of congruent sides, there isn't enough information to prove that the triangles are congruent. Two triangles cannot be proved congruent just by AAA because triangles with same angles can have different sizes.

 $\angle$A + $\angle$B + $\angle$C = 180${}^{\circ }$

 $\angle$A = 180${}^{\circ }$– (50${}^{\circ }$ + 60${}^{\circ }$) = 180${}^{\circ }$ – 110${}^{\circ }$ = 70${}^{\circ }$

Proof: 2 Marks
Steps: 2 Marks

In $\Delta ABCand\Delta BAD$,

AD = BC [Given] 

$\angle DAB=\angle CBA$   [Given]

$AB=BA$   [Common]

$\therefore \Delta ABC\cong \Delta BAD$ (By SAS congruence rule)

$\Rightarrow DB=AC$   [Corresponding parts of corresponding triangles] ...... (1)

In $\Delta ADCand\Delta BCD$

$AD=BC$    [Given]

$DB=AC$    [From (1)]

$DC=CD\left(Common\right)$

$\Delta ADC\cong \Delta BCD$  [By SSS congruence rule]

$\Rightarrow \angle ACD=\angle BDC$  [Corresponding parts of corresponding triangles]