Free Congruence of Triangles Subjective Test 02 Practice Test - 7th grade

Question 1

In the figure, the two triangles are congruent. The corresponding parts are marked. Complete the congruent statement $Δ R A T ≅$ [1 MARK]

SOLUTION

Solution : Solution: 1 Mark

In the figure, the two triangles are congruent.

So, the corresponding congruent parts are:

$∠ A = ∠ O, ∠ R = ∠ W, ∠ T = ∠ N$

Side AT = Side ON, Side AR = Side OW

$∴$ We can write, $Δ R A T ≅ Δ W O N$

Question 2

If $Δ A B C$ and $Δ P Q R$ are to be congruent, name one additional pair of corresponding parts. Which criterion did you use? [2 MARKS]

SOLUTION

Solution : Naming: 1 Mark
Criterion: 1 Mark

Given, $Δ A B C$ and $Δ P Q R$ are congruent with,

$∠ B = ∠ Q = 90^{\circ}$

$∠ C = ∠ R$

For $Δ A B C$ and $Δ P Q R$ to be congruent, the side in between the equal angles needs to be equal.
$¯ ¯¯¯¯¯¯ ¯ B C = ¯ ¯¯¯¯¯¯¯ ¯ Q R$

$\Rightarrow Δ A B C$ and $Δ P Q R$ are congruent by ASA congruence rule.

Then one additional pair is BC = QR.

Question 3

Prove that in the following kite, $Δ A D C$ is congruent to $Δ A B C$ . Given that AD = AB and $∠ A D C$ = $∠ A B C$ = $90^{\circ}$
[2 MARKS]

SOLUTION

Solution :
Steps: 1 Mark
Proof: 1 Mark

In $Δ A D C a n d Δ A B C$

$A D = A B$ [Given]

$∠ A D C = ∠ A B C = 90^{o}$ [Given]

$A C = C A$ [common]

Hence, $Δ A D C ≅ Δ A B C$ [By RHS congruence condition]

Question 4

In the given figure, prove that: [2 MARKS]

$Δ A B D ≅ Δ A C D$

SOLUTION

Solution : Steps: 1 Mark
Proof: 1 Mark

In $Δ A B D$ and $Δ A C D$

$A B = D C$ [Given]

$B D = C A$ [Given]

$A D = A D$ [Common]

$\Rightarrow Δ A B D ≅ Δ A C D$ [SSS congruency criteria]

Question 5

Which congruence criterion is used in the following?
Given: $∠ M L N = ∠ F G H, ∠ N M L = ∠ H F G, M L = F G$
So $Δ L M N ≅ Δ G F H$ [2 MARKS]

SOLUTION

Solution : Steps: 1 Mark
Proof: 1 Mark

It is given that,

$∠ M L N = ∠ F G H$ ,

$∠ N M L = ∠ H F G$ ,

$M L = F G .$

$\Rightarrow$ The two angles and an included side of one triangle are equal to the corresponding angles and an included side of other triangles.

$∴ Δ L M N ≅ Δ G F H$ [By ASA congruence criterion]

Question 6

In the given figure, $A B ∥ C D$ and AB = CD. [3 MARKS]

Prove that: (i) $Δ A O B ≅ Δ D O C$

                 (ii) AO = DO

                 (iii) BO = CO

SOLUTION

Solution : Each proof: 1 Mark

In $Δ A O B$ and $Δ C O D$

$A B = C D$ [Given]

$∠ B A O = ∠ C D O$ [Alternate angles; as $A B ∥ C D$ ]

$∠ A B O = ∠ D C O$ [Alternate angles; as $A B ∥ C D$ ]

(i) $∴ Δ A O B ≅ Δ D O C$ [A.S.A congruency criteria]

(ii) $A O = D O$ [Corresponding sides of congruent triangles]

also, (iii) $B O = C O$ [Corresponding sides of congruent triangles]

Question 7

Prove that: [3 MARKS]

(i) $Δ A B C ≅ Δ A D C$

(ii) $∠ B = ∠ D$

SOLUTION

Solution : Each Proof: 1 Mark
Steps: 1 Marks

In $Δ A B C$ and $Δ A D C$

$A B = D C$ [Given]

$B C = A D$ [Given]

$A C = A C$ [Common side]

$\Rightarrow Δ A B C ≅ Δ A D C$ [SSS congruency criteria]

$∴ ∠ B = ∠ D$ [Corresponding parts of congruent triangles]

Question 8

In the given figure, prove that: [3 MARKS]

(i) $Δ A C B ≅ Δ E C D$

(ii) $A B = E D$

SOLUTION

Solution : Concept : 1 Mark
Proof : 2 Marks

In $Δ A C B$ and $Δ E C D$

$A C = E C$ [Given]

$B C = D C$ [Given]

$∠ A C B = ∠ E C D$ [Vertically opposite angles]

$\Rightarrow Δ A C B ≅ Δ E C D$ (By SAS condition)

$∴ A B = E D$ [Corresponding parts of congruent triangles]

Question 9

(a) Show with an example that two triangles can't be congruent using AAA criterion.

(b) Which congruence criterion will you use in the following?

Given: AC = DF

AB = DE

BC = EF

So, ΔABC ≅ ΔDEF

[3 MARKS]

SOLUTION

Solution :
(a) Proof: 2 Marks
(b) Answer: 1 Mark

(a)Consider the two triangles :

In the two triangles,

$∠ A B C$ = $∠ P Q R$

$∠ B A C$ = $∠ Q P R$

$∠ A C B$ = $∠ P R Q$

But clearly, $Δ A B C$ is not congruent to $Δ P Q R$ .
As the sides of triangle are not equal.
Thus, AAA cannot be a congruence condition.

It actually tells that the two triangles are similar, but not congruent.

(b) Since, the three sides of the first triangle is equal to the corresponding three sides of the second triangle, by SSS congruence criterion ΔABC is congruent to ΔDEF.

Question 10

(a) In the given figure, prove that: BD = BC.

(b) $In the given figure, Δ A B C and Δ O B C are both isosceles.$
Prove that $Δ A O C$ is congruent to $Δ A O B$ .

[4 MARKS]

SOLUTION

Solution : Each Proof: 2 Marks

(a) In $Δ A B C$ and $Δ A B D$

$A B = A B$ [Common side]

$∠ A B C = ∠ A B D = 90^{\circ}$ [Given]

$A C = A D$ [Given]

$\Rightarrow Δ A B C ≅ Δ A B D$ [RHS congruency criteria]

$∴ B D = B C$ [Corresponding parts of congruent triangles]

(b)

$In Δ A O B and Δ A O C, A B = A C (Sides of isosceles Δ A B C) . O B = O C (Sides of isosceles Δ O B C) . A O = A O (Common side) ∴ Δ A O B ≅ Δ A O C (by SSS congruence)$

Question 11

In the isosceles $Δ A B C$ with AB = AC. E and O are the midpoints of AB and AC respectively. 2OD = DE and $∠ O C D$ = $70^{o}$ . Prove using congruency that $∠ E A O$ = $70^{o}$ . [4 MARKS]

SOLUTION

Solution :
Steps: 2 Marks
Proof: 2 Marks

In $Δ A B C$ ,

OD + OE = DE

Multiplying both sides with 2:

2OD + 2OE = 2DE

DE + 2OE = 2DE (2OD = DE; given in question)

2OE = DE

So, OD = OE -------------- (1)

In $Δ A O E a n d Δ D O C$ ,

OD = OE [From (1)]

$∠ A O E$ = $∠ D O C$ [Vertically Opposite Angles]

AO = OC [O is the mid-point of AC]

$Δ A O E ≅ Δ D O C$ [By SAS condition]

Hence, $∠ E A O$ = $∠ O C D$ = $70^{o}$ [Corresponding parts of corresponding triangles]

Question 12

In the given figure; $∠ 1 = ∠ 2$ and AB = AC. Prove that: [4 MARKS]

(i) $∠ B = ∠ C$

(ii) BD = DC

(iii) AD is perpendicular to BC.

SOLUTION

Solution : Steps: 1 Mark
Each proof: 1 Mark

In $Δ A D B$ and $Δ A D C$

$∠ 1 = ∠ 2$ [Given]

$\Rightarrow ∠ B A D = ∠ C A D$

$A D = A D$ [Common side]

$A B = A C$ [Given]

$\Rightarrow Δ A D B ≅ Δ A D C$ [SAS congruency criteria]

(i) $∴ ∠ B = ∠ C$ [Corresponding parts of congruent triangles]

(ii) $B D = D C$ [Corresponding parts of congruent triangles]

(iii) $Δ A D B ≅ Δ A D C$ [proved above]

$∠ A D B + ∠ A D C = 180 °$ [Linear pair]

$\Rightarrow ∠ A D B = ∠ A D C$ [c.p.c.t]

$∴ ∠ A D B + ∠ A D B = 180 °$

$\Rightarrow 2 ∠ A D B = 180 °$

$\Rightarrow ∠ A D B = \frac{180 °}{2}$

$= 90 °$

$\Rightarrow A D ⊥ B C$

Question 13

(a) In the given figure, prove that:

(i) PQ = RS

(ii) PS = QR

(b) Which congruence criterion will you use in the following?

(i) Given: ∠ MLN = ∠ FGH

∠ NML = ∠ GFH

ML = FG

So, ΔLMN ≅ ΔGFH

(ii) Given: EB = DB

AE = BC

∠ A = ∠ C = 90°

So, ΔABE ≅ ΔCDB

[4 MARKS]

SOLUTION

Solution : Each part: 2 Marks

(a) In $Δ P S R$ and $Δ R Q P$

$∠ P S R = ∠ R Q P$ [Given]

$∠ S P R = ∠ Q R P$ [Given]

$P R = P R$ [Common]

$\Rightarrow Δ P S R ≅ Δ R Q P$ [AAS congruency criteria]

(i) $∴ P Q = R S$ [Corresponding parts of congruent triangles]

(ii) Also, $P S = Q R$ [Corresponding parts of congruent triangles]

(b) (i) ASA, as two angles and the side included between these angles of ΔLMN, are equal to two angles and the side included between these angles of ΔGFH.

(ii) RHS, as in the given two right-angled triangles, one side and the hypotenuse are respectively equal.

Question 14

In the given figure, prove that: [4 MARKS]

(i) $Δ X Y Z ≅ Δ X P Z$

(ii) YZ = PZ

(iii) $∠ Y X Z = ∠ P X Z$

SOLUTION

Solution : Steps: 1 Mark
Each Proof: 1 Mark

(i) In $Δ X Y Z$ and $Δ X P Z$

$X Y = X P$ [Given]

$∠ X Y Z = ∠ X P Z = 90^{\circ}$ [Given]

$X Z = X Z$ [Common]

$\Rightarrow Δ X Y Z ≅ Δ X P Z$ [RHS congruency criteria]

(ii) $Y Z = P Z$ [Corresponding parts of congruent triangles]

(iii) $∠ Y X Z = ∠ P X Z$ [Corresponding parts of congruent triangles]

Question 15

(a) In the given figure, prove that:

(i) $Δ A B C ≅ Δ D C B$

(ii) AC = DB

(b) In the figure, it is given that $∠$ A = $90^{\circ}$ , AB = AC, and D is the mid point of BC. Find $∠$ ADC.

[4 MARKS]

SOLUTION

Solution : Each Part: 2 Marks

(a)

(i) In $Δ A B C$ and $Δ D C B$

$A B = D C$ [Given]

$∠ A B C = ∠ D C B = 90^{\circ}$ [Given]

$B C = B C$ [Common side]

$\Rightarrow Δ A B C ≅ Δ D C B$ [SAS congruency criteria]

(ii) $∴ A C = D B$ [Corresponding parts of congruent triangles]

(b)

AB = AC (Given)
BD = DC (D is mid point of BC)
AD = AD (Common side)

Therefore, $△$ ADB $≅$ $△$ ADC [SSS congruency criteria]

thus, $∠$ ADB = $∠$ ADC (c.p.c.t.) ...(1)

but, $∠$ ADB + $∠$ ADC = $180^{\circ}$ [linear pair]

$∴ ∠$ ADC + $∠$ ADC = $180^{\circ}$

$\Rightarrow 2 ∠$ ADC = $180^{\circ}$

$\Rightarrow ∠ A D C = \frac{180 °}{2}$

$\Rightarrow ∠ A D C = 90 °$

Free Congruence of Triangles Subjective Test 02 Practice Test - 7th grade

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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