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Mathematics – Class 7 – Congruence Of Triangle – Exercise 7.2 – NCERT Exercise Solution

1. Which congruence criterion do you use in the following?

(a) Given: AC = DF

AB = DE

BC = EF

So, ΔABC ≅ ΔDEF

Solution:

Both the triangle is congruent by the SSS (side-side-side) rule.

According to the SSS congruence property, we know that two triangles are congruent if the three sides of one triangle are respectively equal to the three sides of the other triangle.

ΔABC ≅ ΔDEF

(b) Given: ZX = RP

RQ = ZY

∠PRQ = ∠XZY

So, ΔPQR ≅ ΔXYZ

Solution:

Given triangle is congruent by the rule of SAS (side-angle-side).

According to the SAS congruence property, we know that two triangles are congruent if the two sides and the included angle of one are respectively equal to the two sides and the included angle of the other.

ΔACB ≅ ΔDEF

(c) Given: ∠MLN = ∠FGH

∠NML = ∠GFH

∠ML = ∠FG

So, ΔLMN ≅ ΔGFH

Solution:

Given triangles are congruent by the rule of ASA (angle-side-angle)

According to the ASA congruence property, we know that two triangles are congruent if the two angles and the included side of one are respectively equal to the two angles and the included side of the other.

ΔLMN ≅ ΔGFH

(d) Given: EB = DB

AE = BC

∠A = ∠C = 90o

So, ΔABE ≅ ΔACD

Solution:

Given triangles are congruent by the rule of RHS (right-hand side)

According to the RHS congruence property, we know that two right triangles are congruent if the hypotenuse and one side of the first triangle are respectively equal to the hypotenuse and one side of the second.

ΔABE ≅ ΔACD

2. You want to show that ΔART ≅ ΔPEN,

(a) If you have to use SSS criterion, then you need to show

(i) AR = (ii) RT = (iii) AT =

Solution:

According to the SSS congruence property, we know that two triangles are congruent if the three sides of one triangle are respectively equal to the three sides of the other triangle.

For SSS (side-side-side) criterion, we need

(i) AR = PE

(ii) RT = EN

(iii) AT = PN

(b) If it is given that ∠T = ∠N and you are to use SAS criterion, you need to have

(i) RT = and (ii) PN =

Solution:

According to SAS congruence property, we know that two triangles are congruent if the two sides and the included angle of one are respectively equal to the two sides and the included angle of the other.

For SAS (side-angle-side) criterion, we need

(i) RT = EN

(ii) PN = AT

(c) If it is given that AT = PN and you are to use ASA criterion, you need to have

(i) ? (ii) ?

Solution:

According to the ASA congruence property, we know that two triangles are congruent if the two angles and the included side of one are respectively equal to the two angles and the included side of the other.

For ASA (angle-side-angle) criterion, we need

(i) ∠A = ∠P

(ii) ∠T = ∠N

3. You have to show that ΔAMP ≅ ΔAMQ.

In the following proof, supply the missing reasons.

Solution:

4. In ΔABC, ∠A = 30o, ∠B = 40o and ∠C = 110o

In ΔPQR, ∠P = 30o, ∠Q = 40o and ∠R = 110o

A student says that ΔABC ≅ ΔPQR by AAA congruence criterion. Is he justified? Why or Why not?

Solution:

No, the student is not justified because there is no criterion for the AAA congruence rule. 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.

Hence, in ΔABC, ∠A = 30o, ∠B = 40o and ∠C = 110o

In ΔPQR, ∠P = 30o, ∠Q = 40o and ∠R = 110o

Both is not congruent.

5. In the figure, the two triangles are congruent. The corresponding parts are marked. We can write ΔRAT ≅?

Solution:

According to the given figure,

In ΔRAT and ΔWON

AT (side) = ON (side)

∠TAR = ∠NOW

AR (side) = OW (side)

Hence, ΔRAT ≅ ΔWON by the property of SAS (side-angle-side)

6. Complete the congruence statement:

ΔBCA ≅ ?        ΔQRS ≅?

Solution:

Refer to fig (i)

First consider the ΔBCA and ΔBTA

From the given figure, it is given that,

BT = BC

∠C = ∠T

BA is common side for the ΔBCA and ΔBTA

Hence, ΔBCA ≅ ΔBTA

Similarly,

Refer fig. (ii)

Consider the ΔQRS and ΔTPQ

In the figure, it is given that

RS = PQ

∠RSQ = ∠PQT

QS = TQ

Hence, ΔQRS ≅ ΔTPQ (by SAS rule)

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:

(i)

In the above figure, ΔABC and ΔDEF :

(sides)

AB = DE

BC = EF

AC = DF

AB + BC + AC = DE + EF + DF

And also, ΔABC ≅ ΔDEF

So, we can say that perimeters of ΔABC and ΔDEF are equal.

(ii)

In the above figure, ΔABC AND ΔPQR

All sides are not equal

Hence, perimeter ΔABC ≠ ΔPQR .

So, the triangles are not congruent.

8. Draw a rough sketch of two triangles such that they have five pairs of congruent parts but still the triangles are not congruent.

Solution:

Let us draw ΔPQR and ΔTSU

In the above figure, all angles of two triangles are equal. But, out of three sides, only two sides are equal.

Hence, ΔPQR is not congruent to ΔTSU.

9. If ΔABC and ΔPQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?

Solution:

By observing the given figure, we can say that

∠ABC = ∠PQR

∠BCA = ∠PRQ

BC = QR

Hence, ΔABC ≅ ΔPQR (By ASA congruency rule)

10. Explain, why ΔABC ≅ ΔFED

Solution:

From the figure, it is given that,

∠ABC = ∠DEF = 90o

∠BAC = ∠DFE

BC = DE

ΔABC ≅ ΔFED (By ASA congruency rule)

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