NCERT Solutions for Class 7 Maths Chapter 7 Congruence Of Triangles

Q.1 Which congruence criterion do you use in the following?
(a) Given:
AC = DF
AB = DE
BC = EF
So, \(\triangle ABC \cong \triangle DEF \)


(b) Given:
ZX = RP
RQ = ZY
\(\angle PRQ = \angle XZY \)
So, \(\triangle PQR \cong \triangle XYZ \)


(c) Given: \(\angle MLN = \angle FGH \)
\(\angle NML = \angle GFH \)
ML = FG
So, \(\triangle LMN \cong \triangle GFH \)

(d) Given: EB = DB
AE = BC
\(\angle A = \angle C = 90° \)
So, \(\triangle ABE \cong \triangle CDB \)

Answer :

(a) \(\triangle ABC \cong \triangle DEF \) (BY SSS rule)
(b) \(\triangle PQR \cong \triangle XYZ \) (BY SAS rule)
(c) \(\triangle LMN \cong \triangle GFH \) (BY ASA rule)
(d) \(\triangle ABE \cong \triangle CDB \) (BY RHS rule)

Q.2 You want to show that \(\triangle ART = \triangle PEN, \)


(a) If you have to use SSS criterion, then you need to show
(i) AR = (ii) RT = (iii) AT =
(b) If it is given that ?T = ?N and you are to use SAS criterion, you need to have
(i) RT = and (ii) PN =
(c) If it is given that AT = PN and you are to use ASA criterion, you need to have
(i) ? (ii) ?

Answer :

(a) For SSS criterion, we need
(i) AR = PE
(ii) RT = EN
(iii) AT = PN
(b) For SAS criterion, we need
(i) RT = EN and
(ii) PN = AT
(c) For ASA criterion, we need
(i) \(\angle A = \angle P \) (ii) \(\angle T = \angle N \)

Q.3 You have to show that \(\triangle AMP \cong \triangle AMQ \). In the fallowing proof, supply the missing reasons.

Answer :

Q.4 In ?ABC, \(\angle A = 30° \angle B = 40° and \angle C = 110° \)
In \(\triangle PQR, \angle P = 30°, \angle Q = 40° and \angle R = 110°.\)
A student says that \(\triangle ABC = \triangle PQR \) by AAA congruence criterion.
Is he justified? Why or why not?

Answer :

The student is not justified because there is not criterion for AAA congruence rule.

Q.5 In the figure, the two triangles are congruent. The corresponding parts are marked. We can write \(\triangle RAT \cong \) ?

Answer :

In \(\triangle RAT and \triangle WON \)
AT=ON (Given)
AR=OW (Given)
\(\angle A = \angle O \) (Given)
\(\therefore \triangle RAT \cong \triangle WON \)(By SAS rule)

Q.6 Complete the congruence statement:

\(\triangle BCA \cong ?
\triangle QRS \cong ? \)

Answer :

(i) In \(\triangle BCA and \triangle BTA \)
BC = BT (Given)
\(\angle CBA= \angle TBA (Given) \)
BA= BA (Common)
\(\therefore \triangle BCA = \triangle BTA \) (by SAS rule)
(ii) In \(\triangle QRS and \triangle TPQ \)
RS = PQ (Given)
QS =TQ (Given)
PT= QR (Given)
\(\triangle QRS \cong \triangle TPQ \) (by AAA rule)

Q.7 In a squared sheet, draw two triangles of equal areas such that:
(i) the triangles are congruent.
(ii) the triangle are not congruent.
What can you say about their perimeters?

Answer :

(i)
i.e. perimeter of \(\triangle ART \)= Perimeter of \(\triangle PEN \)
(ii)
Perimeter of \(\triangle ABC \neq \) Perimeter of \(\triangle PQR \)

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

Answer :

Q.9 If \(\triangle ABC and \triangle PQR \) are to be congruent, name one additional pair of corresponding parts. What criterion did you use?

Answer :

In \(\triangle ABC and \triangle PQR \)
\(\angle ABC = \angle PQR \)
\(\angle BCA = \angle PRQ \)
The other additional pair of corresponding part is BC = QR
\(\therefore \triangle ABC \cong \triangle PQR \)

Q.10 Explain, why \(\triangle ABC \cong \triangle FED \)

Answer :

In \(\triangle ABC and \triangle FED \)
\(\angle B = \angle E = 90° \) (Given)
\(\angle A = \angle F \) (Given)
\(\therefore \angle A + \angle B = \angle E + \angle F \)
\( 180° – \angle C = 180° – \angle D \)
[Angle sum property of triangles]
\(\therefore \angle C =\angle D \)
BC = ED (Given)
\(\therefore \triangle ABC = \triangle FED \) (By ASA rule)