Q.1 Complete the following statements:
(a) Two line segments are congruent if _______ .
(b) Among the congruent angles, one has a measure of 70°, the measure of the other angle is _______ .
(c) When we write \(\angle A = \angle B \), we actually mean
NCERT Solutions for Class 7 Maths Chapter 7 Congruence Of Triangles
Answer :
(a) Two line segments are congruent if _they have the same length_ .
(b) Among the congruent angles, one has a measure of 70°, the measure of the other angle is _70°_ .
(c) \(m \angle A = m \angle B \)
Q.3 If \(\triangle ABC \cong \triangle FED \) under the correspondence ABC ? FED, write all the corresponding congruent parts of the triangles.
NCERT Solutions for Class 7 Maths Chapter 7 Congruence Of Triangles
Answer :
Given: \(\triangle ABC \cong \triangle FED \)
and ABC? FED
\(\therefore \) \(\overline{AB}?\overline{FE},
\overline{BC}?\overline{ED},
\overline{AC}?\overline{FD} \)
\(\angle A?\angle F, \angle B?\angle E,\angle C?\angle D \)
Q.4 If \(\triangle DEF \cong \triangle BCA.\) Write the part of ABCA that correspond to
(i) \(\angle E\)
(ii) \(\overline EF \)
(iii) \(\angle F \)
(iv) \(\overline DF \)
NCERT Solutions for Class 7 Maths Chapter 7 Congruence Of Triangles
Answer :
Given: \(\triangle DEF \cong \triangle BCA.\)
(i) \(\angle E ? \angle C \)
(ii) \(\angle F ? \angle A \)
(iii) EF ? ZA
(iv) DF ? BA
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, \(\triangle ABE \cong \triangle CDB \)
NCERT Solutions for Class 7 Maths Chapter 7 Congruence Of Triangles
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) ?
NCERT Solutions for Class 7 Maths Chapter 7 Congruence Of Triangles
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.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?
NCERT Solutions for Class 7 Maths Chapter 7 Congruence Of Triangles
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 \(\triangle RAT \cong \) ?
NCERT Solutions for Class 7 Maths Chapter 7 Congruence Of Triangles
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)
NCERT Solutions for Class 7 Maths Chapter 7 Congruence Of Triangles
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?
NCERT Solutions for Class 7 Maths Chapter 7 Congruence Of Triangles
Answer :
(i)
i.e. perimeter of \(\triangle ART \)= Perimeter of \(\triangle PEN \)
(ii)
Perimeter of \(\triangle ABC \neq \) Perimeter of \(\triangle PQR \)
Q.9 If \(\triangle ABC and \triangle PQR \) are to be congruent, name one additional pair of corresponding parts. What criterion did you use?
NCERT Solutions for Class 7 Maths Chapter 7 Congruence Of Triangles
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
\(\therefore \triangle ABC \cong \triangle PQR \)
NCERT Solutions for Class 7 Maths Chapter 7 Congruence Of Triangles
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)