NCERT Grade 7-Congruence of Triangles-Answers – MySchoolPage

NCERT Grade 7-Congruence of Triangles-Answers

NCERT Solutions for Class 7 Maths

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1.    (a) They have the same length

(b) 70°

(c) m ∠ A = m ∠ B

2.    (i) Two football             (ii) Two teacher’s tables 3.    Given: ΔABC ≅ ΔFED.

The corresponding congruent parts of the triangles are:

4.    ΔDEF ≅ ΔBCA.

5.    (a) By SSS congruence criterion, since it is given that AC = DF, AB = DE, BC = EF

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

Therefore, ΔABC ≅ ΔDEF

(b) By SAS congruence criterion, since it is given that RP = ZX, RQ = ZY and ∠ PRQ = ∠ 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, ΔPQR ≅ ΔXYZ

(c) By ASA congruence criterion, since it is given that ∠ MLN = ∠ FGH, ∠ NML = ∠ HFG, ML = FG.

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

Therefore, ΔLMN ≅ ΔGFH

(d) By RHS congruence criterion, since it is given that EB = BD, AE = CB, ∠ A = ∠ C = 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, ΔABE ≅ ΔCDB

6.   (a) Using SSS criterion,    ΔART ≅ ΔPEN

(i) AR = PE              (ii) RT = EN                       (iii) AT = PN

(b) Given: ∠ T = ∠ N

Using SAS criterion,                ΔART ≅ ΔPEN

(i) RT = EN              (ii) PN = AT

(c) Given: AT = PN

Using ASA criterion,                ΔART ≅ ΔPEN

(i) ∠ RAT = ∠ EPN     (ii) ∠ RTA = ∠ ENP

7.    Sol.

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

A ↔ O, R ↔ W, T ↔ N.

We can write,  ΔRAT ≅ ΔWON           [By SAS congruence rule]

10.  In ΔBAT and ΔBAC, given triangles are congruent so the corresponding parts are:

B ↔ B, A ↔ A, T ↔ C.

Thus, ΔBAC ≅ ΔBAT                             [By SSS congruence rule]

In ΔQRS and ΔTPQ, given triangles are congruent so the corresponding parts are:

Thus, ΔQRS ≅ ΔTPQ                             [By SSS congruence rule]

11.  In a squared sheet, draw ΔABC and ΔPQR.

When two triangles have equal areas and

(i) These triangles are congruent, i.e., Δ ABC ≅ Δ PQR     [By 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 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.

12.  Let us draw two triangles PQR and ABC.

All angles are equal, two sides are equal except one side. Hence, Δ PQR are not congruent to Δ ABC.

13.  Δ ABC and Δ PQR are congruent. Then one additional pair is BC = QR.

Given: ∠ B = ∠ Q = 90° ∠ C = ∠ R

(iii) Therefore,       Δ ABC ≅ Δ PQR        [By ASA congruence rule]

14.  Given: ∠ A = ∠ F, BC = ED, ∠ B = ∠ E In Δ ABC and Δ FED,

∠ B = ∠ E = 90°

∠ A = ∠ F

BC = ED

Therefore,        Δ ABC ≅ Δ FED               [By RHS congruence rule]

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