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

### Question 1

What is the criteria for any two plane figures to be congruent? [1 MARK]

#### SOLUTION

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

### Question 2

If ΔPNE≅ΔCAR, If PN = CR then name all the other corresponding parts of ΔPEN and ΔCAR. [2 MARKS]

#### SOLUTION

Solution :All parts: 2 Marks

Given that,

ΔPEN≅ΔCAR and

PN = CR

Corresponding parts of congruent triangle are congruent.

Therefore, the corresponding sides of congruent triangle are equal.

⇒PE=CA, EN=AR, PN=CR.

⇒ Also all the corresponding angles of congruent triangles are equal.

⇒∠P=∠C,∠E=∠A,∠N=∠R.

### Question 3

What are congruent angles? Justify with an example that in all the right-angled triangles, at least one pair of angles will be congruent. [2 MARKS]

#### SOLUTION

Solution :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∘.

Example: Consider two right-angled triangles ΔAJU and ΔNIV

In ΔAJU,∠AJU=90∘

and in ΔNIV,∠NIV=90∘

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.

### Question 4

(a) If all the sides of a triangle are equal to the sides of another triangle, will both the triangles be congruent to each other?

(b) If AB is parallel to CD then △ABO should be congruent to △CDO always. Is this right? [2 MARKS]

#### SOLUTION

Solution :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.

⇒ Both triangles look like the mirror image of each other.

⇒ 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.

### Question 5

For the given figures, complete the congruence statements: [2 MARKS]

ΔBCA≅ ? ΔQRS≅ ?

#### SOLUTION

Solution :Each part: 1 Mark

In the given figure,

In ΔBCA and ΔBTA,

BC = BT (Given)

CA = TA (Given)

BA = BA (Common side)

Thus, ΔBCA≅ΔBTA [By SSS congruence rule]

In ΔQRS and ΔTPQ,

QT = QS (Given)

PQ = RS (Given)

PT = QR (Given)

Thus, ΔQRS≅ΔTPQ [By SSS congruence rule]

### Question 6

In the figure given below, CT = TR and AT = A'T. Is CA∥A′R ? If yes, give a proof for the same. [3 MARKS]

#### SOLUTION

Solution :Application of theorem: 1 Mark

Steps: 2 Marks

In ΔCAT and ΔRA′T

CT=RT [Given]

∠CTA=∠RTA′ [Vertically Opposite Angles]

AT=A′T [Given]

∴ΔCAT≅ΔRA′T [By SAS congruence rule]

⇒∠CAT=∠RA′T [Corresponding parts of congruent triangles]

But ∠CAT and ∠RA′T are alternate interior angles.

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

⇒CA∥A′R.

### Question 7

You went to eat pizza with 3 of your friends. You ordered a small pizza which was equally divided into 4 slices. Prove that all these slices are congruent to each other. [3 MARKS]

#### SOLUTION

Solution :Steps: 1 Mark

Proof: 2 Marks

In Δ1 and Δ2:

∠AOD=∠COD=90∘ (Diagonals of square intersect at right angles)

AD=CD (Sides of a square; hypotenuse)

OD=DO (Common)

Hence, Δ1≅Δ2 (By RHS congruence rule) ---------------------1

Similarly, Δ4≅Δ3 (By RHS congruence rule) ---------------------2

In Δ1 and Δ4

∠AOD=∠AOB=90∘ (Diagonals of square intersect at right angles)

AD=AB (Sides of a square; hypotenuse)

OA=AO (Common)

Hence, Δ1≅Δ4 (By RHS congruence rule) -------------------3

Similarly, Δ2≅Δ3 (By RHS congruence rule) ----------------4

From 1, 2, 3 and 4 we can say that all the triangles i.e. Δ1, Δ2, Δ3 and Δ4 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.

### Question 8

In two triangles, two angles and one side of the first triangle are equal to the two angles and one side of the second triangle. Will these two triangles always be congruent?[3 MARKS]

#### SOLUTION

Solution :Proof: 1 Mark

Steps: 2 Marks

Consider two triangles, ΔABC and ΔPQR in which,

∠ABC=∠PQR

∠ACB=∠PRQ

AB=PQ

We know that,

∠ABC+∠ACB+∠BAC=1800

∠BAC=1800−(∠ABC+∠ACB)....(i)

Similarly,

∠QPR=1800−(∠PQR+∠PRQ)......(ii)

From (i) and (ii),

∠BAC=∠QPR

Now, In ΔABC and ΔPQR

∠ABC=∠PQR [Given]

∠BAC=∠QPR [ Proved above]

AB=PQ [Given]

⇒ΔABC≅ΔPQR [ ASA congruency rule]

⇒ These triangles are always congruent.

### Question 9

Given: EB = BD, AE = CB, ∠A=∠C=90∘

Which congruence criterion do you use to prove ΔABE≅ΔCDB? [3 MARKS]

#### SOLUTION

Solution :Answer: 1 Mark

Explanation: 2 Marks

In ΔAEB and ΔCBD,

EB=BD [Given]

AE=CB [Given]

∠A=∠C=90∘ [Given]

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

∴ΔABE≅ΔCDB [RHS congruence criterion]

### Question 10

Prove that the diagonals of a rectangle bisect each other. [4 MARKS]

#### SOLUTION

Solution :Properties: 1 Mark

Proof: 1 Mark

Steps: 2 Marks

In a rectangle opposite sides are equal and parallel.

In ΔOAD and ΔOCB,

∠ODA=∠OBC

[Alternate interior angles; AD∥BC and BD as transversal]

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

∠OAD=∠OCB

[Alternate interior angles; AD∥BC and AC as transversal]

Hence ΔOAD≅ΔOCB [By ASA congruence rule]

Equating the corresponding parts of congruent triangles, we get:

AO = CO

BO = DO

⇒ Diagonals of a rectangle bisect each other.

### Question 11

ABC is an isosceles triangle with AB=AC. Prove: [4 MARKS]

(i) ΔADB≅ΔADC

(ii) ∠BAD=∠CAD

(iii) BD=CD

#### SOLUTION

Solution :Properties: 1 Mark

Each proof: 1 Mark

In ΔADB and ΔADC

AB=AC [Given]

∠ADB=∠ADC=90∘ [Given]

AD=AD [common]

Hence, ΔADB≅ΔADC [By RHS congruence rule…….(1)]

From (1), ∠BAD=∠CAD [Corresponding parts of congruent triangles]

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

### Question 12

(a) DA bisects ∠BAC and ∠B=∠C. Prove that ΔBDA≅ΔCDA.

(b) If these triangles are congruent, choose the property by which they are congruent.

[4 MARKS]

#### SOLUTION

Solution :Each Part: 2 Marks

(a)

In ΔBDA and ΔCDA

∠B=∠C [Given]

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

AD=AD [Common side]

⇒ΔBDA≅ΔCDA [ 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.

### Question 13

(a) In the given figure, show that ΔAMP≅ΔAMQ.

(b)

In the given figure, AC = CE and AB ∥ ED. The value of x is ___ units.

[4 MARKS]

#### SOLUTION

Solution :Each part: 2 Marks

(a)

In ΔAMP and ΔAMQ,

PM=QM [Given]

∠AMP=∠AMQ [Given]

AM=AM [Common side]

⇒ΔAMP≅ΔAMQ [SAS congruency criteria]

(b)

In ΔABC and ΔEDC,AC=CE (given)∠BAC=∠DEC (since AB||DE and AE is a transversal, so they are alternate angles)∠ACB=∠ECD (vertically opposite angles)∴ΔABC≅ΔEDC (A.S.A. congruence criteria)∴AB=DE(sides of congruent triangles)∴x+10=2x−5⇒x−2x=−5−10⇒−x=−15⇒x=15 units

### Question 14

(a) Observe the given triangles and explain, why is ΔABC≅ΔFED?

(b) In a ΔABC, ∠B = 50∘ and ∠C is 60∘. Find ∠A.

[4 MARKS]

#### SOLUTION

Solution :(a) Proof: 2 Marks

(b) Steps: 1 Mark

Final answer: 1 Mark

(a) In ΔABC and ΔFED,

∠B=∠E=90∘ [Given]

∠A=∠F [Given]

BC=ED [Given]

⇒ Two angles and one side of ΔABC are equal to two angles and one side of ΔFED.

Therefore, ΔABC≅ΔFED [AAS congruence rule]

(b) Sum of the angles of a triangle = 180∘∠A + ∠B + ∠C = 180∘

∠A = 180∘– (50∘ + 60∘) = 180∘ – 110∘ = 70∘

### Question 15

In the given figure, ∠DAB=∠CBA and AD=BC. Prove that ∠ACD=∠BDC. [4 MARKS]

#### SOLUTION

Solution :Proof: 2 Marks

Steps: 2 Marks

In ΔABC and ΔBAD,

AD = BC [Given]

∠DAB=∠CBA [Given]

AB=BA [Common]

∴ΔABC≅ΔBAD (By SAS congruence rule)

⇒DB=AC [Corresponding parts of corresponding triangles] ...... (1)

In ΔADC and ΔBCD

AD=BC [Given]

DB=AC [From (1)]

DC=CD(Common)

ΔADC≅ΔBCD [By SSS congruence rule]

⇒∠ACD=∠BDC [Corresponding parts of corresponding triangles]