Ex. 1Multiple choice
If ∆ ABC ≅ ∆ PQR and ∆ ABC is not congruent to ∆ RPQ, then which of the following is not true:
- (A)BC = PQ
- (B)AC = PR
- (C)QR = BC
- (D)AB = PQ
Show solution
Solution : Answer (A)
Chapter 7 – Triangles
Class 9 Mathematics · 54 questions · 24 with answers
Solved examples
Ex. 1Short answerExercise 7.1
In the two triangles ABC and DEF, AB = DE and AC = EF. Name two angles from the two triangles that must be equal so that the two triangles are congruent. Give reason for your answer.
Show solution
Solution: The required two angles are ∠A and ∠E. When ∠A = ∠E, ∆ ABC ≅ ∆ EDF by SAS criterion.
Ex. 2Short answerExercise 7.1
In triangles ABC and DEF, ∠A = ∠D, ∠B = ∠E and AB = EF. Will the two triangles be congruent? Give reasons for your answer.
Show solution
Solution: Two triangles need not be congruent, because AB and EF are not corresponding sides in the two triangles.
Ex. 1Short answerExercise 7.2
In Fig 7.1, PQ = PR and ∠Q = ∠R. Prove that ∆ PQS ≅ ∆ PRT.
Show solution
Solution : In ∆ PQS and ∆ PRT, PQ = PR (Given) ∠Q = ∠R (Given) and ∠QPS = ∠RPT (Same angle) Therefore, ∆ PQS ≅ ∆ PRT (ASA)
Ex. 2Short answerExercise 7.2
In Fig.7.2, two lines AB Fig. 7.1 and CD intersect each other at the point O such that BC || DA and BC = DA. Show that O is the mid- point of both the line-segments AB and CD.
Show solution
Solution : BC || AD (Given) Therefore, ∠CBO = ∠DAO (Alternate interior angles) and ∠BCO = ∠ADO (Alternate interior angles) Also, BC = DA (Given) So, ∆ BOC ≅ ∆ AOD (ASA) Therefore, OB = OA and OC = OD, i.e., O is Fig. 7.2 the mid-point of both AB and CD.
Ex. 3Short answerExercise 7.2
In Fig.7.3, PQ > PR and QS and RS are the bisectors of ∠Q and ∠R, respectively. Show that SQ > SR.
Show solution
Solution : PQ > PR (Given) Therefore, ∠R > ∠Q(Angles opposite the longer side is greater) So, ∠SRQ > ∠SQR(Half of each angle) Therefore, SQ > SR(Side opposite the greater angle will be longer) Fig. 7.3 TRIANGLES 67
Ex. 1Short answerExercise 7.3
In Fig. 7.9, ABC is a right triangle and right Fig. 7.8 angled at B such that ∠BCA = 2 ∠BAC. Show that hypotenuse AC = 2 BC.
Show solution
Solution: Produce CB to a point D such that BC = BD and join AD. In ∆ ABC and ∆ ABD, we have BC = BD (By construction) AB = AB (Same side) Fig. 7.9 ∠ABC = ∠ABD (Each of 90°) Therefore, ∆ ABC ≅ ∆ ABD (SAS) So, ∠CAB = ∠DAB (1) (CPCT) and AC = AD (2) Thus, ∠CAD = ∠CAB + ∠BAD = x + x = 2x [From (1)] (3) and ∠ACD = ∠ADB = 2x [From (2), AC = AD] (4) That is, ∆ ACD is an equilateral triangle. [From (3) and (4)] or AC = CD, i.e., AC = 2 BC (Since BC = BD)
Ex. 2Short answerExercise 7.3
Prove that if in two triangles two angles and the included side of one triangle are equal to two angles and the included side of the other triangle, then the two triangles are congruent.
Show solution
Solution: See proof of Theorem 7.1 of Class IX Mathematics Textbook.
Ex. 3Short answerExercise 7.3
If the bisector of an angle of a triangle also bisects the opposite side, prove that the triangle is isosceles.
Show solution
Solution : We are given a point D on side BC of a ∆ ABC such that ∠BAD = ∠CAD and BD = CD (see Fig. 7.10). We are to prove that AB = AC. Produce AD to a point E such that AD = DE and then join CE. Now, in ∆ ABD and ∆ ECD, we have Fig. 7.10 TRIANGLES 69 BD = CD (Given) AD = ED (By construction) and ∠ADB = ∠EDC (Vertically opposite angles) Therefore, ∆ ABD ≅ ∆ ECD (SAS) So, AB = EC (1) (CPCT) and ∠BAD = ∠CED (2) Also, ∠BAD = ∠CAD (Given) Therefore, ∠CAD = ∠CED [From (2)] So, AC = EC [Sides opposite the equal angles] (3) Therefore, AB = AC [From (1) and (3)]
Ex. 4Short answerExercise 7.3
S is any point in the interior of ∆ PQR. Show that SQ + SR < PQ + PR.
Show solution
Solution : Produce QS to intersect PR at T (See Fig. 7.11). From ∆ PQT, we have PQ + PT > QT(Sum of any two sides is greater than the third side) i.e., PQ + PT > SQ + ST (1) From ∆ TSR, we have ST + TR > SR (2) Fig. 7.11 Adding (1) and (2), we get PQ + PT + ST + TR > SQ + ST + SR i.e., PQ + PT + TR > SQ + SR i.e., PQ + PR > SQ + SR or SQ + SR < PQ + PR
Questions
Which of the following is not a criterion for congruence of triangles?
- (A)SAS
- (B)ASA
- (C)SSA
- (D)SSS
Show answer
(C) SSA
If AB = QR, BC = PR and CA = PQ, then
- (A)∆ ABC ≅ ∆ PQR
- (B)∆ CBA ≅ ∆ PRQ
- (C)∆ BAC ≅ ∆ RPQ
- (D)∆ PQR ≅ ∆ BCA
Show answer
(B) ∆ CBA ≅ ∆ PRQ
In ∆ ABC, AB = AC and ∠B = 50°. Then ∠C is equal to
- (A)40°
- (B)50°
- (C)80°
- (D)130°
Show answer
(B) 50°
In ∆ ABC, BC = AB and ∠B = 80°. Then ∠A is equal to
- (A)80°
- (B)40°
- (C)50°
- (D)100°
Show answer
(C) 50°
In ∆ PQR, ∠R = ∠P and QR = 4 cm and PR = 5 cm. Then the length of PQ is
- (A)4 cm
- (B)5 cm
- (C)2 cm
- (D)2.5 cm
Show answer
(A) 4 cm
D is a point on the side BC of a ∆ ABC such that AD bisects ∠BAC. Then
- (A)BD = CD
- (B)BA > BD
- (C)BD > BA
- (D)CD > CA
Show answer
(B) BA > BD
It is given that ∆ ABC ≅ ∆ FDE and AB = 5 cm, ∠B = 40° and ∠A = 80°. Then which of the following is true?
- (A)DF = 5 cm, ∠F = 60°
- (B)DF = 5 cm, ∠E = 60°
- (C)DE = 5 cm, ∠E = 60°
- (D)DE = 5 cm, ∠D = 40°
Show answer
(B) DF = 5 cm, ∠E = 60°
Two sides of a triangle are of lengths 5 cm and 1.5 cm. The length of the third side of the triangle cannot be
- (A)3.6 cm
- (B)4.1 cm
- (C)3.8 cm
- (D)3.4 cm
Show answer
(D) 3.4 cm
In ∆ PQR, if ∠R > ∠Q, then
- (A)QR > PR
- (B)PQ > PR
- (C)PQ < PR
- (D)QR < PR
Show answer
(B) PQ > PR
In triangles ABC and PQR, AB = AC, ∠C = ∠P and ∠B = ∠Q. The two triangles
- (A)isosceles but not congruent
- (B)isosceles and congruent
- (C)congruent but not isosceles
- (D)neither congruent nor isosceles
Show answer
(A) isosceles but not congruent
In triangles ABC and DEF, AB = FD and ∠A = ∠D. The two triangles will be congruent by SAS axiom if
- (A)BC = EF
- (B)AC = DE
- (C)AC = EF
- (D)BC = DE TRIANGLES 65
Show answer
(B) AC = DE
In triangles ABC and PQR, ∠A = ∠Q and ∠B = ∠R. Which side of ∆ PQR should be equal to side AB of ∆ ABC so that the two triangles are congruent? Give reason for your answer.
Show answer
QR; They will be congruent by ASA.
In triangles ABC and PQR, ∠A = ∠Q and ∠B = ∠R. Which side of ∆ PQR should be equal to side BC of ∆ ABC so that the two triangles are congruent? Give reason for your answer.
Show answer
RP; They will be congruent by AAS.
“If two sides and an angle of one triangle are equal to two sides and an angle of another triangle, then the two triangles must be congruent.” Is the statement true? Why?
Show answer
No; Angles must be included angles.
“If two angles and a side of one triangle are equal to two angles and a side of another triangle, then the two triangles must be congruent.” Is the statement true? Why?
Show answer
No; Sides must be corresponding sides.
Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and
Show answer
No; Sum of the two sides = the third side.
cm? Give reason for your answer. 6. It is given that ∆ ABC ≅ ∆ RPQ. Is it true to say that BC = QR? Why? 7. If ∆ PQR ≅ ∆ EDF, then is it true to say that PR = EF? Give reason for your answer.
Show answer
Yes; They are corresponding sides.
In ∆ PQR, ∠P = 70° and ∠R = 30°. Which side of this triangle is the longest? Give reason for your answer.
Show answer
PR; Side opposite the greater angle is longer.
AD is a median of the triangle ABC. Is it true that AB + BC + CA > 2 AD? Give reason for your answer.
Show answer
Yes; AB + BD > AD and AC + CD > AD.
M is a point on side BC of a triangle ABC such that AM is the bisector of ∠BAC. Is it true to say that perimeter of the triangle is greater than 2 AM? Give reason for your answer.
Show answer
Yes; AB + BM > AM and AC + CM > AM.
Is it possible to construct a triangle with lengths of its sides as 9 cm, 7 cm and 17 cm? Give reason for your answer.
Show answer
No; Sum of two sides is less than the third side.
Is it possible to construct a triangle with lengths of its sides as 8 cm, 7 cm and 4 cm? Give reason for your answer.
Show answer
Yes, because in each case the sum of two sides is greater than the third side.
ABC is an isosceles triangle with AB = AC and BD and CE are its two medians. Show that BD = CE.
In Fig.7.4, D and E are points on side BC of a ∆ ABC such that BD = CE and AD = AE. Show that ∆ ABD ≅ ∆ ACE.
This question refers to a figure in the original PDF.
CDE is an equilateral triangle formed on a side CD of a square ABCD (Fig.7.5). Show that Fig. 7.4 ∆ ADE ≅ ∆ BCE. Fig. 7.5
This question refers to a figure in the original PDF.
In Fig.7.6, BA ⊥ AC, DE ⊥ DF such that BA = DE and BF = EC. Show that ∆ ABC ≅ ∆ DEF.
This question refers to a figure in the original PDF.
Q is a point on the side SR of a ∆ PSR such that PQ = PR. Prove that PS > PQ.
S is any point on side QR of a ∆ PQR. Show that: PQ + QR + RP > 2 PS.
D is any point on side AC of a ∆ ABC with AB = AC. Fig. 7.6 Show that CD < BD.
This question refers to a figure in the original PDF.
In Fig. 7.7, l || m and M is the mid-point of a line segment AB. Show that M is also the mid-point of any line segment CD, having its end points on l and m, respectively.
This question refers to a figure in the original PDF.
Bisectors of the angles B and C of an isosceles triangle with AB = AC intersect each other at O. BO is produced to a point M. Prove that ∠MOC = Fig. 7.7 ∠ABC.
This question refers to a figure in the original PDF.
Bisectors of the angles B and C of an isosceles triangle ABC with AB = AC intersect each other at O. Show that external angle adjacent to ∠ABC is equal to ∠BOC.
In Fig. 7.8, AD is the bisector of ∠BAC. Prove that AB > BD.
This question refers to a figure in the original PDF.
Find all the angles of an equilateral triangle.
Show answer
60°, 60°, 60°
The image of an object placed at a point A before a plane mirror LM is seen at the point B by an observer at D as shown in Fig. 7.12. Prove that the image is as far behind the mirror as the object is in front of the mirror. Fig. 7.12 [Hint: CN is normal to the mirror. Also, angle of incidence = angle of reflection].
This question refers to a figure in the original PDF.
ABC is an isosceles triangle with AB = AC and D is a point on BC such that AD ⊥ BC (Fig. 7.13). To prove that ∠BAD = ∠CAD, a student proceeded as follows: In ∆ ABD and ∆ ACD, AB = AC (Given) ∠B = ∠C (because AB = AC) and ∠ADB = ∠ADC Therefore, ∆ ABD ≅ ∆ ACD (AAS) So, ∠BAD = ∠CAD (CPCT) What is the defect in the above arguments? Fig. 7.13 [Hint: Recall how ∠B = ∠C is proved when AB = AC].
This question refers to a figure in the original PDF.
Show answer
It is defective to use ∠ABD = ∠ACD for proving this result. 19. ∠B will be greater.
P is a point on the bisector of ∠ABC. If the line through P, parallel to BA meet BC at Q, prove that BPQ is an isosceles triangle.
ABCD is a quadrilateral in which AB = BC and AD = CD. Show that BD bisects both the angles ABC and ADC.
ABC is a right triangle with AB = AC. Bisector of ∠A meets BC at D. Prove that BC = 2 AD.
O is a point in the interior of a square ABCD such that OAB is an equilateral triangle. Show that ∆ OCD is an isosceles triangle.
ABC and DBC are two triangles on the same base BC such that A and D lie on the opposite sides of BC, AB = AC and DB = DC. Show that AD is the perpendicular bisector of BC.
ABC is an isosceles triangle in which AC = BC. AD and BE are respectively two altitudes to sides BC and AC. Prove that AE = BD.
Prove that sum of any two sides of a triangle is greater than twice the median with respect to the third side.
Show that in a quadrilateral ABCD, AB + BC + CD + DA < 2 (BD + AC)
Show that in a quadrilateral ABCD, AB + BC + CD + DA > AC + BD
In a triangle ABC, D is the mid-point of side AC such that BD = AC. Show that ∠ABC is a right angle.
In a right triangle, prove that the line-segment joining the mid-point of the hypotenuse to the opposite vertex is half the hypotenuse. TRIANGLES 71
Two lines l and m intersect at the point O and P is a point on a line n passing through the point O such that P is equidistant from l and m. Prove that n is the bisector of the angle formed by l and m.
Line segment joining the mid-points M and N of parallel sides AB and DC, respectively of a trapezium ABCD is perpendicular to both the sides AB and DC. Prove that AD = BC.
ABCD is a quadrilateral such that diagonal AC bisects the angles A and C. Prove that AB = AD and CB = CD.
ABC is a right triangle such that AB = AC and bisector of angle C intersects the side AB at D. Prove that AC + AD = BC.
AB and CD are the smallest and largest sides of a quadrilateral ABCD. Out of ∠B and ∠D decide which is greater.
Prove that in a triangle, other than an equilateral triangle, angle opposite the longest side is greater than of a right angle.
ABCD is quadrilateral such that AB = AD and CB = CD. Prove that AC is the perpendicular bisector of BD.