ICSE Class 7 Mathematics · ML Aggarwal Solutions Understanding ICSE Mathematics Class 7 · 49 solved questions
If △ABC and △DEF are congruent under the correspondence ABC ↔ FED, write all the corresponding congruent parts of the triangles.
Under the correspondence ABC ↔ FED, the vertices correspond as A ↔ F, B ↔ E and C ↔ D.
The corresponding congruent parts are:
∠A = ∠F, ∠B = ∠E, ∠C = ∠D
AB = FE, BC = ED, AC = FD
Hence, the corresponding congruent parts are ∠A = ∠F, ∠B = ∠E, ∠C = ∠D, AB = FE, BC = ED and AC = FD.
If △DEF ≅ △BCA, then write the part(s) of △BCA that correspond to
(i) ∠E
(ii) $\overline{EF}$
(iii) ∠F
(iv) $\overline{DF}$
Under the congruence △DEF ≅ △BCA, the vertices correspond as D ↔ B, E ↔ C and F ↔ A.
(i) ∠E corresponds to ∠C.
(ii) $\overline{EF}$ corresponds to $\bm{\overline{CA}}$.
(iii) ∠F corresponds to ∠A.
(iv) $\overline{DF}$ corresponds to $\bm{\overline{BA}}$.
In the figures given below, the lengths of the sides of the triangles are indicated. By using SSS congruency rule, state which pairs of triangles are congruent. In case of congruent triangles, write the result in symbolic form:
(i)
(ii)
(i) Here in △ABC, AB = 3.3 cm, BC = 3 cm and CA = 2.5 cm, while in △PQR, QP = 3.3 cm, PR = 3 cm and RQ = 2.5 cm.
Comparing the sides,
⇒ AB = QP = 3.3 cm, BC = PR = 3 cm and CA = QR = 2.5 cm.
Since the three sides of one triangle are equal to the three sides of the other triangle, by the SSS congruence rule the triangles are congruent (A ↔ Q, B ↔ P, C ↔ R).
Hence, the triangles are congruent and △ABC ≅ △QPR.
(ii) Here in △ABC, AB = 2 cm, BC = 4 cm and CA = 5 cm, while in △PQR, PQ = 4 cm, PR = 5 cm and QR = 2.5 cm.
The sides of △ABC are 2 cm, 4 cm and 5 cm, whereas the sides of △PQR are 2.5 cm, 4 cm and 5 cm. The side 2 cm of △ABC has no equal side in △PQR.
Since the three sides of one triangle are not equal to the three sides of the other triangle, the SSS rule cannot be applied.
Hence, the two triangles are not congruent.
In the adjoining figure, AB = 5 cm, AC = 5 cm, BD = 2.5 cm and CD = 2.5 cm.
(i) State the three pairs of equal parts in △ADB and △ADC.
(ii) Is △ADB ≅ △ADC? Give reasons.
(iii) Is ∠B = ∠C? Why?
(i) The three pairs of equal parts in △ADB and △ADC are:
AB = AC (5 cm each)
BD = CD (2.5 cm each)
AD = AD (common)
(ii) From part (i), the three sides of △ADB are equal to the three sides of △ADC.
Hence, △ADB ≅ △ADC by the SSS congruence rule.
(iii) Since △ADB ≅ △ADC, the corresponding angles ∠B and ∠C are equal.
Hence, ∠B = ∠C as they are corresponding parts of congruent triangles.
In the adjoining figure, AB = AC and D is the mid-point of $\overline{BC}$.
(i) State the three pairs of equal parts in △ADB and △ADC.
(ii) Is △ADB ≅ △ADC? Give reasons.
(iii) Is ∠B = ∠C? Why?
(i) The three pairs of equal parts in △ADB and △ADC are:
AB = AC (given)
BD = CD (D is the mid-point of BC)
AD = AD (common)
(ii) From part (i), the three sides of △ADB are equal to the three sides of △ADC.
Hence, △ADB ≅ △ADC by the SSS rule of congruency.
(iii) Since △ADB ≅ △ADC, the corresponding angles ∠B and ∠C are equal.
Hence, ∠B = ∠C as they are corresponding parts of congruent triangles.
In the figure given below, the measures of some parts of the triangles are indicated. By using SAS rule of congruency, state whether the pair of triangles is congruent. In case of congruent triangles, write the result in symbolic form.
In △ABC, AB = 2.5 cm, AC = 2.8 cm and the included ∠A = 80°. In △DEF, DE = 2.5 cm, DF = 2.8 cm and the included ∠D = 70°.
Here two sides are equal (2.5 cm and 2.8 cm), but the included angles are not equal (80° ≠ 70°).
Hence, the two triangles are not congruent.
In the figure given below, the measures of some parts of the triangles are indicated. By using SAS rule of congruency, state whether the pair of triangles is congruent. In case of congruent triangles, write the result in symbolic form.
In △ABC, CA = 2.5 cm, CB = 3 cm and the included ∠C = 35°. In △PQR, PR = 2.5 cm, PQ = 3 cm and the included ∠P = 35°.
Comparing,
⇒ CA = PR = 2.5 cm, CB = PQ = 3 cm and included ∠C = ∠P = 35°.
By the SAS rule of congruency the triangles are congruent (C ↔ P, A ↔ R, B ↔ Q).
Hence, the triangles are congruent and △ABC ≅ △RQP.
In the figure given below, the measures of some parts of the triangles are indicated. By using SAS rule of congruency, state whether the pair of triangles is congruent. In case of congruent triangles, write the result in symbolic form.
In △DEF, FD = 3.5 cm, FE = 3 cm and the included ∠F = 40°. In △PQR, QP = 3.5 cm, QR = 3 cm and the included ∠Q = 40°.
Comparing,
⇒ FD = QP = 3.5 cm, FE = QR = 3 cm and included ∠F = ∠Q = 40°.
By the SAS rule of congruency the triangles are congruent (F ↔ Q, D ↔ P, E ↔ R).
Hence, the triangles are congruent and △DEF ≅ △PRQ.
In the figure given below, the measures of some parts of the triangles are indicated. By using SAS rule of congruency, state whether the pair of triangles is congruent. In case of congruent triangles, write the result in symbolic form.
In △ABC, AB = 4 cm, BC = 3 cm and the included ∠B = 80°. In △PQR, PQ = 4 cm, QR = 3 cm and ∠R = 80°.
Comparing,
⇒ AB = PQ = 4 cm and BC = QR = 3 cm.
The angle included between the equal sides PQ and QR is ∠Q, but the given 80° angle in △PQR is ∠R, which lies opposite the 4 cm side PQ.
Since the 80° angle of △PQR is not included between the equal sides, the SAS rule cannot be applied.
Hence, the two triangles are not congruent.
By applying SAS congruence rule, you want to establish that △PQR ≅ △FED. It is given that PQ = FE and RP = DF. What additional information is needed to establish the congruence?
For the SAS rule, the angle included between the two given pairs of sides must be equal.
In △PQR, the sides PQ and RP include the angle ∠P, while in △FED the sides FE and DF include the angle ∠F.
Hence, the additional information needed is ∠P = ∠F.
You want to show that △ART ≅ △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 =
For △ART ≅ △PEN, the vertices correspond as A ↔ P, R ↔ E and T ↔ N.
(a) Using the SSS criterion, the three sides must match:
(i) AR = PE
(ii) RT = EN
(iii) AT = PN
(b) Using the SAS criterion with ∠T = ∠N, the two sides forming these angles must be equal:
(i) RT = EN
(ii) PN = AT
You have to show that △AMP ≅ △AMQ. In the following proof, supply the missing reasons.
| Steps | Reasons |
|---|---|
| (i) PM = QM | (i) ... |
| (ii) ∠PMA = ∠QMA | (ii) ... |
| (iii) AM = AM | (iii) ... |
| (iv) △AMP ≅ △AMQ | (iv) ... |
| Steps | Reasons |
|---|---|
| (i) PM = QM | (i) Given |
| (ii) ∠PMA = ∠QMA | (ii) Given |
| (iii) AM = AM | (iii) Common |
| (iv) △AMP ≅ △AMQ | (iv) SAS rule of congruency |
In the adjoining figure:
(i) State three pairs of equal parts in △PSR and △RQP.
(ii) Is △PSR ≅ △RQP? Give reasons.
(iii) Is PS = RQ? Why?
(iv) Is ∠S = ∠Q? Why?
(i) The three pairs of equal parts in △PSR and △RQP are:
SR = QP (given)
∠SRP = ∠QPR (given)
PR = RP (common)
(ii) From part (i), two sides and the included angle of △PSR are equal to two sides and the included angle of △RQP.
Hence, △PSR ≅ △RQP by the SAS rule of congruency.
(iii) Since △PSR ≅ △RQP, the corresponding sides PS and RQ are equal.
Hence, PS = RQ as they are corresponding parts of congruent triangles.
(iv) Since △PSR ≅ △RQP, the corresponding angles ∠S and ∠Q are equal.
Hence, ∠S = ∠Q as they are corresponding parts of congruent triangles.
In the adjoining figure, AB = DC and ∠ABC = ∠DCB.
(i) State three pairs of equal parts in △ABC and △DCB.
(ii) Is △ABC ≅ △DCB? Give reasons.
(iii) Is AC = DB? Why?
(i) The three pairs of equal parts in △ABC and △DCB are:
AB = DC (given)
∠ABC = ∠DCB (given)
BC = CB (common)
(ii) From part (i), two sides and the included angle of △ABC are equal to two sides and the included angle of △DCB.
Hence, △ABC ≅ △DCB by the SAS rule of congruency.
(iii) Since △ABC ≅ △DCB, the corresponding sides AC and DB are equal.
Hence, AC = DB as they are corresponding parts of congruent triangles.
In the quadrilateral, AC = AD and AB bisects ∠CAD.
(i) State three pairs of equal parts in △ABC and △ABD.
(ii) Is △ABC ≅ △ABD? Give reasons.
(iii) Is BC = BD? Why?
(iv) Is ∠C = ∠D? Why?
(i) The three pairs of equal parts in △ABC and △ABD are:
AC = AD (given)
∠CAB = ∠DAB (AB bisects ∠CAD)
AB = AB (common)
(ii) From part (i), two sides and the included angle of △ABC are equal to two sides and the included angle of △ABD.
Hence, △ABC ≅ △ABD by the SAS rule of congruency.
(iii) Since △ABC ≅ △ABD, the corresponding sides BC and BD are equal.
Hence, BC = BD as they are corresponding parts of congruent triangles.
(iv) Since △ABC ≅ △ABD, the corresponding angles ∠C and ∠D are equal.
Hence, ∠C = ∠D as they are corresponding parts of congruent triangles.
You want to establish △DEF ≅ △MNP, using ASA rule of congruence. You are given that ∠D = ∠M and ∠F = ∠P. What additional information is needed to establish the congruence?
For the ASA rule, the side included between the two given angles must be equal.
In △DEF, the side included between ∠D and ∠F is DF, while in △MNP the side included between ∠M and ∠P is MP.
Hence, the additional information needed is DF = MP.
In the adjoining figure, two triangles are congruent. The corresponding parts are marked. We can write △RAT ≅ ?
Matching the marked corresponding parts of the two triangles, the vertices correspond as R ↔ W, A ↔ O and T ↔ N.
Hence, △RAT ≅ △WON.
If △ABC and △PQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?
In the figure, ∠B = ∠Q = 90° and ∠C = ∠R. To make the triangles congruent, we take the side included between these two angles to be equal.
The additional pair of corresponding parts is BC = QR.
Hence, the criterion used is the ASA congruence rule.
Given below are measurements of some parts of two triangles. Examine whether the two triangles are congruent or not, by ASA congruence rule. In case of congruence, write it in symbolic form.
| △DEF | △PQR |
|---|---|
| (i) ∠D = 60°, ∠F = 80°, DF = 5 cm | ∠Q = 60°, ∠R = 80°, QR = 5 cm |
| (ii) ∠D = 60°, ∠F = 80°, DF = 6 cm | ∠Q = 60°, ∠R = 80°, QP = 6 cm |
| (iii) ∠E = 80°, ∠F = 30°, EF = 5 cm | ∠P = 80°, PQ = 5 cm, ∠R = 30° |
(i) In △DEF, DF = 5 cm is the side included between ∠D = 60° and ∠F = 80°. In △PQR, QR = 5 cm is the side included between ∠Q = 60° and ∠R = 80°.
Comparing,
⇒ ∠D = ∠Q = 60°, ∠F = ∠R = 80° and the included sides DF = QR = 5 cm.
By the ASA rule, the triangles are congruent (D ↔ Q, F ↔ R, E ↔ P).
Hence, the triangles are congruent and △DEF ≅ △QPR.
(ii) In △DEF, ∠D = 60° and ∠F = 80°. By the angle sum property,
⇒ ∠E = 180° − (60° + 80°) = 40°
The given side DF = 6 cm is included between ∠D = 60° and ∠F = 80°.
In △PQR, ∠Q = 60° and ∠R = 80°. By the angle sum property,
⇒ ∠P = 180° − (60° + 80°) = 40°
The given side QP = 6 cm is included between ∠Q = 60° and ∠P = 40°.
Thus, the 6 cm side lies between the 60° and 80° angles in △DEF, but between the 60° and 40° angles in △PQR. So DF and QP are not corresponding sides, and the ASA rule cannot be applied.
Hence, the two triangles are not congruent.
(iii) In △DEF, ∠E = 80° and ∠F = 30°. By the angle sum property,
⇒ ∠D = 180° − (80° + 30°) = 70°
The given side EF = 5 cm is included between ∠E = 80° and ∠F = 30°.
In △PQR, ∠P = 80° and ∠R = 30°. By the angle sum property,
⇒ ∠Q = 180° − (80° + 30°) = 70°
The given side PQ = 5 cm is included between ∠P = 80° and ∠Q = 70°.
Thus, the 5 cm side lies between the 80° and 30° angles in △DEF, but between the 80° and 70° angles in △PQR. So EF and PQ are not corresponding sides, and the ASA rule cannot be applied.
Hence, the two triangles are not congruent.
In the adjoining figure, measures of some parts are indicated.
(i) State three pairs of equal parts in triangles ABC and ABD.
(ii) Is △ABC ≅ △BAD? Give reasons.
(iii) Is BC = AD? Why?
(i) The three pairs of equal parts in △ABC and △BAD are:
∠BAC = ∠ABD (30° each)
∠ABC = ∠BAD (30° + 45° = 75° each)
AB = BA (common)
(ii) From part (i), two angles and the included side AB of △ABC are equal to two angles and the included side BA of △BAD.
Hence, △ABC ≅ △BAD by the ASA rule of congruence.
(iii) Since △ABC ≅ △BAD, the corresponding sides BC and AD are equal.
Hence, BC = AD as they are corresponding parts of congruent triangles.
In the adjoining figure, ray AZ bisects ∠DAB as well as ∠DCB.
(i) State the three pairs of equal parts in triangles BAC and DAC.
(ii) Is △BAC ≅ △DAC? Give reasons.
(iii) Is AB = AD? Justify your answer.
(iv) Is CD = CB? Give reasons.
(i) The three pairs of equal parts in △BAC and △DAC are:
∠BAC = ∠DAC (AZ bisects ∠DAB)
∠BCA = ∠DCA (AZ bisects ∠DCB)
AC = AC (common)
(ii) From part (i), two angles and the included side AC of △BAC are equal to two angles and the included side AC of △DAC.
Hence, △BAC ≅ △DAC by the ASA rule of congruency.
(iii) Since △BAC ≅ △DAC, the corresponding sides AB and AD are equal.
Hence, AB = AD as they are corresponding parts of congruent triangles.
(iv) Since △BAC ≅ △DAC, the corresponding sides CB and CD are equal.
Hence, CD = CB as they are corresponding parts of congruent triangles.
Explain why △ABC ≅ △FED.
From the figure, ∠B = ∠E = 90°, ∠A = ∠F and BC = ED.
In △ABC, by the angle sum property,
⇒ ∠C = 180° − ∠A − ∠B = 180° − ∠A − 90° = 90° − ∠A
In △FED, by the angle sum property,
⇒ ∠D = 180° − ∠E − ∠F = 180° − 90° − ∠F = 90° − ∠F
Since ∠A = ∠F,
⇒ ∠D = 90° − ∠A = ∠C
Now in △ABC and △FED:
∠B = ∠E (90° each)
BC = ED (given)
∠C = ∠D (proved)
Here BC is the side included between ∠B and ∠C, and ED is the side included between ∠E and ∠D.
Hence, △ABC ≅ △FED by the ASA rule of congruency.
Given below are the measurements of some parts of triangles. Examine whether the two triangles are congruent or not, using RHS congruence rule. In case of congruent triangles, write the result in symbolic form:
| △ABC | △PQR |
|---|---|
| (i) ∠B = 90°, AC = 8 cm, AB = 3 cm | ∠P = 90°, PR = 3 cm, QR = 8 cm |
| (ii) ∠A = 90°, AC = 5 cm, BC = 9 cm | ∠Q = 90°, PR = 8 cm, PQ = 5 cm |
(i) In △ABC, ∠B = 90°, so AC = 8 cm is the hypotenuse and AB = 3 cm is one side. In △PQR, ∠P = 90°, so QR = 8 cm is the hypotenuse and PR = 3 cm is one side.
Comparing,
⇒ ∠B = ∠P = 90°, hypotenuse AC = QR = 8 cm and side AB = PR = 3 cm.
By the RHS rule, the triangles are congruent (B ↔ P, A ↔ R, C ↔ Q).
Hence, the triangles are congruent and △ABC ≅ △RPQ.
(ii) In △ABC, ∠A = 90°, so BC = 9 cm is the hypotenuse. In △PQR, ∠Q = 90°, so PR = 8 cm is the hypotenuse.
The hypotenuse of △ABC is 9 cm while the hypotenuse of △PQR is 8 cm, so the hypotenuses are not equal.
Hence, the two triangles are not congruent.
In the adjoining figure, measurements of some parts are given.
(i) State the three pairs of equal parts in △PQS and △PRS.
(ii) Is △PQS ≅ △PRS? Give reasons.
(iii) Is S mid-point of $\overline{QR}$? Why?
(i) The three pairs of equal parts in △PQS and △PRS are:
∠PSQ = ∠PSR (90° each)
PQ = PR (3 cm each, hypotenuses)
PS = PS (common)
(ii) From part (i), the right angle, hypotenuse and one side of △PQS are equal to the right angle, hypotenuse and one side of △PRS.
Hence, △PQS ≅ △PRS by the RHS rule of congruency.
(iii) Since △PQS ≅ △PRS, the corresponding sides QS and RS are equal, so S divides QR into two equal parts.
Hence, S is the mid-point of QR because QS = RS as corresponding parts of congruent triangles.
In the adjoining figure, O is mid-point of $\overline{AB}$ and ∠A = ∠B. Show that △AOC ≅ △BOD.
In △AOC and △BOD:
∠OAC = ∠OBD (given, ∠A = ∠B)
AO = BO (O is the mid-point of AB)
∠AOC = ∠BOD (vertically opposite angles)
Here AO is the side included between ∠OAC and ∠AOC, and BO is the side included between ∠OBD and ∠BOD.
Hence, △AOC ≅ △BOD by the ASA rule of congruency.
Fill in the blanks:
(i) Two line segments are congruent if ....
(ii) Among two congruent angles, one has a measure of 63°; the measure of the other angle is ....
(iii) When we write ∠A = ∠B, we actually mean ....
(iv) The side included between ∠M and ∠N of △MNP is ....
(v) The side QR of △PQR is included between angles ....
(vi) If two triangles ABC and PQR are congruent under the correspondence A ↔ R, B ↔ P and C ↔ Q, then in symbolic form it can be written as △ABC ≅ ....
(vii) If △DEF ≅ △SRT, then the correspondence between vertices is ....
(i) Two line segments are congruent if they have the same length.
(ii) The measure of the other angle is 63°.
(iii) When we write ∠A = ∠B, we actually mean m∠A = m∠B.
(iv) The side included between ∠M and ∠N of △MNP is MN.
(v) The side QR of △PQR is included between angles ∠Q and ∠R.
(vi) In symbolic form it can be written as △ABC ≅ △RPQ.
(vii) The correspondence between vertices is D ↔ S, E ↔ R, F ↔ T.
State whether the following statements are true (T) or false (F):
(i) All circles are congruent.
(ii) Circles having equal radii are congruent.
(iii) Two congruent triangles have equal areas and equal perimeters.
(iv) Two triangles having equal areas are congruent.
(v) Two squares having equal areas are congruent.
(vi) Two rectangles having equal areas are congruent.
(vii) All acute angles are congruent.
(viii) All right angles are congruent.
(ix) Two figures are congruent if they have same shape.
(x) A two rupee coin is congruent to a five rupee coin.
(xi) All equilateral triangles are congruent.
(xii) Two equilateral triangles having equal perimeters are congruent.
(xiii) If two legs of one right angle triangle are equal to two legs of another right angle triangle, then the two triangles are congruent by SAS rule.
(xiv) If three angles of two triangles are equal, then triangles are congruent.
(xv) If two sides and one angle of one triangle are equal to two sides and one angle of another triangle, then the triangle are congruent.
(i) False — circles can have different radii.
(ii) True — circles with equal radii have the same shape and size.
(iii) True — congruent triangles have equal areas and equal perimeters.
(iv) False — equal areas do not guarantee equal shape and size.
(v) True — equal areas mean equal sides, so the squares are congruent.
(vi) False — rectangles of equal area can have different length and breadth.
(vii) False — acute angles can have different measures.
(viii) True — all right angles measure 90°, so they are congruent.
(ix) False — figures must have the same shape and the same size.
(x) False — the two coins differ in size.
(xi) False — equilateral triangles can have different side lengths.
(xii) True — equal perimeters mean equal sides, so the triangles are congruent.
(xiii) True — the two legs include the right angle, so SAS applies.
(xiv) False — equal angles give similar, not necessarily congruent, triangles.
(xv) False — congruence holds only when the equal angle is the included angle (SAS).
Which one of the following is not a standard criterion of congruency of two triangle?
⇒ SSS, SAS and ASA are standard congruence criteria, but SSA is not a valid congruence criterion.
Hence, option 2 is the correct option.
If △ABC ≅ △PQR and ∠CAB = 65°, then ∠RPQ is
⇒ Since △ABC ≅ △PQR, ∠A = ∠P, i.e. ∠CAB = ∠RPQ = 65°.
Hence, option 1 is the correct option.
If △ABC ≅ △EFD, then the correct statement is
⇒ Under △ABC ≅ △EFD, the vertices correspond as A ↔ E, B ↔ F and C ↔ D, so ∠A = ∠E.
Hence, option 3 is the correct option.
If △ABC ≅ △PQR, then the correct statement is
⇒ Under △ABC ≅ △PQR, the corresponding sides are AB = PQ, BC = QR and AC = PR.
Hence, option 4 is the correct option.
If ∠D = ∠P, ∠E = ∠Q and DE = PQ, then △DEF ≅ △PQR, by the congruence rule:
⇒ Two angles (∠D, ∠E) and the included side DE are equal to two angles (∠P, ∠Q) and the included side PQ, which is the ASA rule.
Hence, option 2 is the correct option.
In △ABC and △PQR, BC = QR and ∠C = ∠R. To establish △ABC ≅ △PQR by SAS congruence rule, the additional information required is
⇒ For SAS, the sides forming ∠C and ∠R must be equal. ∠C lies between BC and CA, and ∠R lies between QR and RP, so we need CA = RP, i.e. AC = PR.
Hence, option 1 is the correct option.
In the adjoining figure, the lengths of the sides of two triangles are given. The correct statement is
⇒ In △ABC, AB = 2.5 cm, AC = 2 cm, BC = 3.5 cm and in △PQR, PQ = 2 cm, PR = 3.5 cm, QR = 2.5 cm. Matching the sides, A ↔ Q, B ↔ R, C ↔ P, so △ABC ≅ △QRP.
Hence, option 2 is the correct option.
In the adjoining figure, M is the mid-point of both AC and BD. Then
⇒ In △AMB and △CMD, AM = CM, BM = DM and ∠AMB = ∠CMD (vertically opposite angles), so △AMB ≅ △CMD by SAS. Hence the corresponding angles ∠1 = ∠4.
Hence, option 2 is the correct option.
In the adjoining figure, △PQR ≅ △STU. What is length of TU?
⇒ Under △PQR ≅ △STU, the side TU corresponds to QR, so TU = QR = 6 cm.
Hence, option 2 is the correct option.
In the adjoining figure, △ABC and △DBC are on the same base BC. If AB = DC and AC = DB, then which of the following statement is correct?
⇒ In △ABC and △DCB, AB = DC, AC = DB and BC = CB (common), so by SSS, △ABC ≅ △DCB.
Hence, option 3 is the correct option.
The two triangles shown in the adjoining figure are:
⇒ In the first triangle, the marked side is opposite the 75° angle, while in the second triangle, the marked side lies between the 75° and 55° angles. So the marked equal sides are not corresponding sides, and the triangles are not congruent.
Hence, option 4 is the correct option.
In the adjoining figure, △ABC ≅ △PQR. The values of x and y are:
⇒ Since △ABC ≅ △PQR, ∠A = ∠P and ∠B = ∠Q.
⇒ ∠A = ∠P gives 75° = (2y + 5)°
⇒ 2y = 70
⇒ y = 35.
⇒ ∠B = ∠Q gives (x − 7)° = 70°
⇒ x = 77.
Hence, option 2 is the correct option.
Statement I: Two equilateral triangles are always congruent.
Statement II: Each angle of an equilateral triangle is 60°
Two equilateral triangles can have different side lengths, so they are not always congruent.
∴ Statement I is false.
Each angle of an equilateral triangle is 60°.
∴ Statement II is true.
Hence, option 2 is the correct option.
Statement I: In the given figure, △PLQ ≅ △PMQ.
Statement II: When three sides of a triangle are equal to the three sides of another triangle, they are congruent, as per the SSS congruence criterion.
In △PLQ and △PMQ,
PL = PM (From figure)
LQ = MQ (From figure)
PQ = PQ (Common side)
∴ △PLQ ≅ △PMQ (By SSS axiom)
∴ Statement I is true.
When three sides of a triangle are equal to the three sides of another triangle, they are congruent, as per the SSS congruence criterion.
∴ Statement II is true.
Hence, option 3 is the correct option.
Statement I: If △ABC ≅ △PQR, it means that A ↔ P, B ↔ Q and C ↔ R.
Statement II: SAS is a valid criterion for congruence of triangles.
The order of letters in △ABC ≅ △PQR gives the correspondence A ↔ P, B ↔ Q and C ↔ R.
∴ Statement I is true.
SAS is a valid congruence criterion.
∴ Statement II is true.
Hence, option 3 is the correct option.
State, giving reasons, whether the following pairs of triangles are congruent or not:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(i) The three sides of one triangle (3.5 cm, 3 cm, 2.7 cm) are equal to the three sides of the other triangle (3.5 cm, 3 cm, 2.7 cm).
Hence, the triangles are congruent by the SSS congruence rule.
(ii) Two angles 35°, 90° and the included side of one triangle are equal to two angles and the included side of the other triangle.
Hence, the triangles are congruent by the ASA congruence rule.
(iii) In the first triangle, the given angles are 36° and 54°. By the angle sum property,
⇒ Third angle = 180° − (36° + 54°) = 90°
The single-marked side is opposite this 90° angle, so it is the hypotenuse, and the double-marked side is one of the other two sides.
In the second triangle, the right angle is marked, the single-marked side is the hypotenuse and the double-marked side is one of the other two sides.
Thus, the right angles are equal (90° each), the hypotenuses are equal (single-marked) and one pair of sides is equal (double-marked).
Hence, the triangles are congruent by the RHS congruence rule.
(iv) The angles 70° and 50° give a third angle of 60°, and the other triangle has angles 50° and 60° giving a third angle of 70°. Two angles and the included side match.
Hence, the triangles are congruent by the ASA congruence rule.
(v) In the first triangle, the 75° angle is included between the 4 cm side and the 3 cm side.
In the second triangle, the 75° angle is formed by the 3 cm side and the unmarked side, while the 4 cm side lies opposite the 75° angle.
So the 4 cm side of the second triangle is not in the corresponding position with respect to the 75° angle.
Hence, the triangles are not congruent because the 4 cm side of the second triangle is opposite the 75° angle instead of forming it with the 3 cm side.
(vi) In the first triangle, the marked side is included between the 70° and 50° angles.
In the second triangle, the 70° angle is at the top vertex, so its third angle = 180° − (70° + 50°) = 60°. The marked side lies between the 60° and 50° angles, i.e. opposite the 70° angle.
So the marked side of the second triangle is not in the corresponding position with respect to the equal angles.
Hence, the triangles are not congruent because the marked side of the second triangle is opposite the 70° angle instead of being included between the 70° and 50° angles.
Given below are measurements of some parts of two triangles. Examine whether the two triangles are congruent or not. In case of congruence, give reasons and write it in symbolic form:
| △ABC | △PQR |
|---|---|
| (i) AB = 4 cm, BC = 5 cm, ∠B = 70° | QR = 4 cm, RP = 5 cm, ∠R = 70° |
| (ii) AB = 4 cm, BC = 5 cm, ∠B = 80° | PQ = 4 cm, RP = 5 cm, ∠R = 80° |
| (iii) BC = 6 cm, ∠A = 90°, ∠C = 50° | QR = 6 cm, ∠R = 50°, ∠Q = 40° |
| (iv) AB = 5 cm, ∠A = 90°, BC = 8 cm | PR = 5 cm, ∠P = 90°, QR = 8 cm |
(i) In △ABC, AB = 4 cm, BC = 5 cm and the included ∠B = 70°. In △PQR, QR = 4 cm, RP = 5 cm and the included ∠R = 70°.
Comparing, AB = QR = 4 cm, BC = RP = 5 cm and ∠B = ∠R = 70° (B ↔ R, A ↔ Q, C ↔ P).
Hence, the triangles are congruent by the SAS rule and △ABC ≅ △QRP.
(ii) Here AB = PQ = 4 cm and BC = RP = 5 cm.
In △ABC, the 80° angle ∠B is included between the equal sides AB and BC.
In △PQR, the angle included between the equal sides PQ and RP is ∠P, but the given 80° angle is ∠R, which is formed by RP and RQ and lies opposite the 4 cm side PQ.
Since the 80° angle of △PQR is not included between the equal sides, the equal parts are not in corresponding positions and the SAS rule cannot be applied.
Hence, the triangles are not congruent.
(iii) In △ABC, ∠A = 90° and ∠C = 50°, so ∠B = 40°, and BC = 6 cm lies between ∠B and ∠C. In △PQR, ∠R = 50°, ∠Q = 40°, so ∠P = 90°, and QR = 6 cm lies between ∠Q and ∠R.
Comparing, ∠B = ∠Q = 40°, ∠C = ∠R = 50° and BC = QR = 6 cm (B ↔ Q, C ↔ R, A ↔ P).
Hence, the triangles are congruent by the ASA rule and △ABC ≅ △PQR.
(iv) In △ABC, ∠A = 90°, so BC = 8 cm is the hypotenuse and AB = 5 cm is one side. In △PQR, ∠P = 90°, so QR = 8 cm is the hypotenuse and PR = 5 cm is one side.
Comparing, ∠A = ∠P = 90°, hypotenuse BC = QR = 8 cm and side AB = PR = 5 cm (A ↔ P, B ↔ R, C ↔ Q).
Hence, the triangles are congruent by the RHS rule and △ABC ≅ △PRQ.
In the adjoining figure, ABC is an isosceles triangle with AB = AC and AD is one of its altitudes.
(i) State the three pairs of equal parts in △ADB and △ADC.
(ii) Is △ADB ≅ △ADC? Give reasons.
(iii) Is ∠B = ∠C? Why?
(iv) Is BD = DC? Why?
(i) The three pairs of equal parts in △ADB and △ADC are:
AB = AC (given)
∠ADB = ∠ADC (90° each, AD is an altitude)
AD = AD (common)
(ii) From part (i), the right angle, hypotenuse and one side of △ADB are equal to those of △ADC.
Hence, △ADB ≅ △ADC by the RHS rule of congruency.
(iii) Since △ADB ≅ △ADC, the corresponding angles ∠B and ∠C are equal.
Hence, ∠B = ∠C as they are corresponding parts of congruent triangles.
(iv) Since △ADB ≅ △ADC, the corresponding sides BD and DC are equal.
Hence, BD = DC as they are corresponding parts of congruent triangles.
In the adjoining figure, OA bisects ∠A and ∠ABO = ∠OCA. Prove that OB = OC.
In △OAB and △OAC:
∠OAB = ∠OAC (OA bisects ∠A)
∠ABO = ∠ACO (given)
Since two pairs of angles are equal, the third pair is also equal:
⇒ ∠AOB = ∠AOC
Also, OA = OA (common)
Now, ∠OAB = ∠OAC, OA = OA and ∠AOB = ∠AOC, so by the ASA rule:
⇒ △OAB ≅ △OAC
∴ OB = OC (corresponding parts of congruent triangles)
Hence, OB = OC.
In the figure given alongside, prove that
(i) AB = FC
(ii) AF = BC.
In the figure, DF ⊥ AC, EB ⊥ AC, DF = EB and AE = CD.
In △ABE and △CFD:
∠ABE = ∠CFD (90° each)
AE = CD (given, hypotenuses)
BE = FD (given)
By the RHS rule of congruency,
⇒ △ABE ≅ △CFD
(i) Since △ABE ≅ △CFD, the corresponding sides AB and CF are equal.
Hence, AB = FC.
(ii) From AB = FC and the order A, F, B, C on the line,
⇒ AF + FB = FB + BC
⇒ AF = BC
Hence, AF = BC.
If all the three altitudes of a triangle are equal, then prove that it is an equilateral triangle.
Let △ABC have altitudes AD, BE and CF drawn to the sides BC, CA and AB respectively, with AD = BE = CF = h (say).
The area of the triangle can be written in three ways using the three bases and their corresponding heights:
Area $= \frac{1}{2} × BC × AD = \frac{1}{2} × CA × BE = \frac{1}{2} × AB × CF$
Since AD = BE = CF = h,
⇒ $\frac{1}{2} × BC × h = \frac{1}{2} × CA × h = \frac{1}{2} × AB × h$
⇒ BC = CA = AB
Since all three sides are equal, the triangle is equilateral.
Hence, if all the three altitudes of a triangle are equal, then it is an equilateral triangle.
In the adjoining figure, if BA ∥ RP, QP ∥ BC and AQ = CR, then prove that △ABC ≅ △RPQ.
Given AQ = CR. Adding QC to both sides,
⇒ AQ + QC = CR + QC
⇒ AC = QR
Since BA ∥ RP, with AC as the transversal,
⇒ ∠BAC = ∠PRQ(alternate angles)
Since BC ∥ QP, with CQ as the transversal,
⇒ ∠BCA = ∠PQR (alternate angles)
Now in △ABC and △RPQ:
∠BAC = ∠PRQ
AC = RQ (proved)
∠BCA = ∠PQR
Here AC is the side included between ∠BAC and ∠BCA, and RQ is the side included between ∠PRQ and ∠PQR.
Hence, △ABC ≅ △RPQ by the ASA rule of congruency.
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