• Skip to main content
  • Skip to secondary menu
  • Skip to primary sidebar
  • Skip to footer

CBSE Tuts

CBSE Maths notes, CBSE physics notes, CBSE chemistry notes

  • NCERT Solutions
    • NCERT Solutions for Class 12 English Flamingo and Vistas
    • NCERT Solutions for Class 11 English
    • NCERT Solutions for Class 11 Hindi
    • NCERT Solutions for Class 12 Hindi
    • NCERT Books Free Download
  • TS Grewal
    • TS Grewal Class 12 Accountancy Solutions
    • TS Grewal Class 11 Accountancy Solutions
  • CBSE Sample Papers
  • NCERT Exemplar Problems
  • English Grammar
    • Wordfeud Cheat
  • MCQ Questions

NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles

NCERT Exemplar Class 9 Maths Solutions Triangles

Exercise 7.1: Multiple Choice Questions (MCQs)

Question 1:
Which of the following is not a criterion for congruence of triangles?
(a) SAS               (b) ASA                     (c) SSA                        (d) SSS
Thinking Process
For triangle to be congruent it is very important that the equal angles are included between the pairs of equal sides. So, SAS congruence rule holds but not ASS or SSA rule.
Solution:
(c) We know that, two triangles are congruent, if the sides and angles of one triangle are equal to the corresponding sides and angles of the other triangle.
Also, criterion for congruence of triangles are SAS (Side-Angle-Side), ASA (Angle-Side- Angle), SSS (Side-Side-Side) and RHS (right angle-hypotenuse-side).
So, SSA is not a criterion for congruence of triangles.

Question 2:
If AB = QR, BC = PR and CA = PQ, then
(a) ΔABC ≅ ΔQRP             (b) ΔCBA ≅ ΔPRQ
(c) ΔBAC ≅ ΔRQP            (d) ΔPQR ≅ ΔBCA
Solution:
(b) We know that, if ΔRST is congruent to ΔUVW i.e., ΔRST = ΔUVW, then sides of ΔRST fall on corresponding equal sides of ΔUVW and angles of ΔRST fall on corresponding equal angles of ΔUVW.
Here, given AB = QR, BC = PR and CA = PQ, which shows that AB covers QR, BC covers PR and CA covers PQ i.e., A correspond to Q, B correspond to R and C correspond to P.
or A↔Q,B↔R,C↔P
Under this correspondence,
ΔABC ≅ ΔQRP, so option (a) is incorrect,
or ΔBAC ≅ ΔRQP, so option (c) is incorrect,
or ΔCBA ≅ ΔPRQ, so option (b) is correct,
or ΔBCA ≅ ΔRPQ, so option (d) is incorrect.

Question 3:
In ΔABC,if AB = AC and ∠B = 50°, then ∠C is equal to
(a) 40°                  (b) 50°                    (c) 80°                     (d)130°
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-1

Question 4:
In ΔABC,if BC = AB and ∠B = 80°, then ∠A is equal to
(a) 80°                 (b) 40°                      (c) 50°                         (d) 100°
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-2

Question 5:
In ΔPQR, if ∠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
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-3

Question 6:
If D is a point on the side BC of a ΔABC such that AD bisects ∠BAC. Then,
(a) BD = CD             (b) BA > BD                  (c) BD > BA                  (d)CD > CA
Thinking Process
(i) Firstly, use the property, exterior angle of a triangle is greater than interior opposite
angle.
(ii) Secondly, use the property that in a triangle, the side opposite to the greater angle is longer.
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-4

Question 7:
It is given that ΔABC = ΔFDE 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°
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-5

Question 8:
If two sides of a triangle are of lengths 5 cm and 1.5 cm, then the length of third side of the triangle cannot be
(a) 3.6 cm                (b) 4.1 cm                    (c) 3.8 cm                         (d) 3.4 cm
Thinking Process
Use the condition that, sum of any two sides of a triangle is greater than third side and difference of any two sides is less than the third side.
Solution:
(d) Given, the length of two sides of a triangle are 5 cm and 1.5 cm, respectively.
Let sides AB = 5 cm and CA = 1.5 cm
We know that, a closed figure formed by three intersecting lines (or sides) is called a triangle, if difference of two sides < third side and sum of two sides > third side
∴ 5-1.5 < BC and 5+1.5 > BC
=> 3.5 < BC and 6.5 > BC
Here, we see that options (a), (b) and (c) satisfy the above inequality but option (d) does not satisfy the above inequality.

Question 9:
In ΔPQR, if ∠R > ∠Q, then
(a) QR > PR                  (b) PQ > PR                   (c) PQ < PR                  (d) QR < PR
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-6

Question 10:
In ΔABC and ΔPQR, if AB = AC, ∠C =∠P and ∠B = ∠Q, then the two triangles are
(a) isosceles but not congruent       (b) isosceles and congruent
(c) congruent but not isosceles       (d) Neither congruent nor isosceles
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-7

Question 11:
In ΔABC and ΔDEF, 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
Solution:
(b) Given, in ΔABC and ΔDEF, AB = DF and ∠A = ∠D
We know that, two triangles will be congruent by ASA rule, if two angles and the included side of one triangle are equal to the two angles and the included side of other triangle.
∴ AC = DE

Exercise 7.2: Very Short Answer Type Questions

Question 1:
In ΔABC and ΔPQR, ∠A = ∠Q and ∠B =∠R. Which side of ΔPQR should be equal to side AB of ΔABC, so that the two triangles are congruent? Give reason for your answer.
Solution:
We have given, in ΔABC and ΔPQR, ∠A = ∠Q and ∠B = ∠R
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-8
Since, AB and QR are included between equal angles. Hence, the side of ΔPQR is QR which should be equal to side AB of ΔABC, so that the triangles are congruent by the rule ASA.

Question 2:
In ΔABC and ΔPQR, ∠A = ∠Q and ∠B = ∠R. Which side of  ΔPQR should be equal to side BC of ΔABC, so that the two triangles are congruent? Give reason for your answer.
Solution:
We have given, in ΔABC and ΔPQR,
∠A = ∠Q and ∠B = ∠R
Since, two pairs of angles are equal in two triangles.
We know that, two triangles will be congruent by AAS rule, if two angles and the side of one triangle are equal to the two angles and the side of other triangle.
∴ BC = RP
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-9

Question 3:
‘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?
Solution:
No, because in the congruent rule, the two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle i.e., SAS rule.

Question 4:
‘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?
Solution:
No, because sides must be corresponding sides.

Question 5:
Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm? Give reason for your answer.
Solution:
No, it is not possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm because here we see that sum of the lengths of two sides is equal to third side i.e., 4+3 = 7.
As we know that, the sum of any two sides of a triangle is greater than its third side, so given statement is not correct.

Question 6:
It is given that ΔABC ≅ ΔRPQ. Is it true to say that BC = QR? Why?
Solution:
No, we know that two triangles are congruent, if the sides and angles of one triangle are equal to the corresponding side and angles of other triangle.
Here  ΔABC ≅ ΔRPQ
AB = RP, BC = PQ and AC = RQ Hence, it is not true to say that BC = QR.

Question 7:
If ΔPQR ≅ ΔEOF, then is it true to say that PR = EF? Give reason for your answer.
Solution:
Yes, if ΔPQR ≅ ΔEDF, then it means that corresponding angles and their sides are equal because we know that, two triangles are congruent, if the sides and angles of one triangle are equal to the corrosponding sides and angles of other triangle.
Here, ΔPQR ≅ ΔEDF
∴ PQ = ED, QR = DF and PR = EF
Hence, it is true to say that PR = EF.

Question 8:
In ΔPQR, ∠P = 70° and ∠R = 30°. Which side of this triangle is the longest? Give reason for your answer.
Solution:
Given, in ΔPQR, ∠P = 70° and ∠R = 30°.
We know that, sum of all the angles of a triangle is 180°.
∠P + ∠Q + ∠R = 180°
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-10
We know that here ∠Q is longest, so side PR is longest.
[∴ since in a triangle, the side opposite to the largest angle is the longest]

Question 9:
AD is a median of the ΔABC. Is it true that AB + BC +CA > 2AD? Give reason for your answer.
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-11

Question 10:
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 2AM? Give reason for your answer?
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-12

Question 11:
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.
Solution:
No. Here, we see that  9 + 7 =16 < 17
i.e., the sum of two sides of a triangle is less than the third side.
Hence, it contradicts the property that the sum of two sides of a triangle is greater than the third side. Therefore, it is not possible to construct a triangle with given sides.

Question 12:
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.
Solution:
Yes, because in each case the sum of two sides is greater than the third side. i.e., 7 + 4>8,
8+ 4 >7,  7 + 8 >4
Hence, it is possible to construct a triangle with given sides.

Exercise 7.3: Short Answer Type Questions

Question 1:
ABC is an isosceles triangle with AB = AC and BD, CE are its two medians. Show that BD = CE.
Solution:
Given ΔABC is an isosceles triangle in which AB = AC and BD, CE are its two medians.
To show BD = CE.
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-13

Question 2:
In figure, D and E are points on side BC of a ΔABC such that BD = CE and AD = AE. Show that ΔABD ≅ ΔACE.
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-14
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-15

Question 3:
In the given figure, ΔCDE is an equilateral triangle formed on a side CD of a square ABCD. Show that ΔADE ≅ ΔBCE.
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-16
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-17

Question 4:
In figure, BA ⊥ AC, DE ⊥ DF such that BA = DE and BF = EC. Show that ΔABC ≅ ΔDEF.
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-18
Thinking Process
Use the RHS congruence rule to show the given result
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-19

Question 5:
If Q is a point oh the side SR of a ΔPSR such that PQ = PR, then prove that PS > PQ.
Thinking Process
Use the property of a triangle that if two sides are equal then their opposite angles are also equal. Also use the property that side opposite to greater angle is longer.
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-20

Question 6:
S is any point on side QR of a ΔPQR. Show that PQ + QR + RP > 2 PS.
Thinking Process
Use the inequality of a triangle i.e., sum of two sides of a triangle is greater than the third side. Further, show the required result.
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-21

Question 7:
D is any point on side AC of a ΔABC with AB = AC. Show that CD < BD.
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-22

Question 8:
In given figure 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.
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-23
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-24

Question 9:
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 = ∠ABC.
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-25

Question 10:
Bisectors of the angles B and C of an isosceles ΔABC with AB = AC intersect each other at O. Show that external angle adjacent to ∠ABC is equal to ∠BOC.
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-26
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-27

Question 11:
In following figure if AD is the bisector of ∠BAC, then prove that AB > BD.
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-28
Solution:
Given ABC is a triangle such that AD is the bisector of ∠BAC. To prove AB > BD.
Proof Since, AD is the bisector of ∠BAC.
But ∠BAD = CAD …(i)
∴  ∠ADB > ∠CAD
[exterior angle of a triangle is greater than each of the opposite interior angle]
∴ ∠ADB > ∠BAD [from Eq. (i)]
AB > BD [side opposite to greater angle is longer]
Hence proved.

Exercise 7.4: Long Answer Type Questions

Question 1:
Find all the angles of an equilateral triangle.
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-29

Question 2:
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 figure. Prove that the image is as far behind the mirror as the object is in front of the mirror.
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-30
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-31

Question 3:
ABC is an isosceles triangle with AB = AC and D is a point on BC such that AD ⊥BC (see figure). To prove that ∠BAD = ∠CAD, a student proceeded as follows
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-32
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-33

Question 4:
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.
Solution:
Given we have P is a point on the bisector of ∠ABC and draw the line through P parallel to BA and meet BC at Q.
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-34
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-35

Question 5:
ABCD is a quadrilateral in which AB = BC and AD = CD. Show that BD bisects both the angles ABC and ADC.
Thinking Process
Firstly, use the property that if two sides of a triangle are equal, then their opposite angles are equal. Further, show that  ΔBAD and ΔBCD are congruent by SAS rule.
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-36

Question 6:
ABC is a right triangle with AB = AC. If bisector of ∠A meets BC at D,then prove that BC = 2AD.
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-37
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-38

Question 7:
O is a point in the interior of a square ABCD such that OAB is an equilateral triangle. Show that ΔOCD is an isosceles triangle.
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-39

Question 8:
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.
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-40

Question 9:
If ABC is an isosceles triangle in which AC = BC, AD and BE are respectively two altitudes to sides BC and AC, then prove that AE = BD.
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-41
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-42

Question 10:
Prove that sum of any two sides of a triangle is greater than twice the median with respect to the third side.
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-43

Question 11:
Show that in a quadrilateral ABCD, AB + BC + CD + DA< 2 (BD + AC)
Thinking Process
Firstly, draw a quadrilateral ABCD. Further use the property of a triangle that sum of two sides of a triangle is greater than third side and show the required result.
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-44
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-45

Question 12:
Show that in a quadrilateral ABCD, AB + BC + CD + DA > AC + BD.
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-46

Question 13:
In a A ABC, D is the mid-point of side AC such that BD = ½ AC. Show that ∠ABC is a right angle.
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-47
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-48

Question 14:
In a right triangle, prove that the line-segment joining the mid-point of the hypotenuse to the opposite vertex is half the hypotenuse.
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-49
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-50

Question 15:
Two lines l and m intersect at the point 0 and P is a point on a line n passing through the point 0 such that P is equidistant from l and m. Prove that n is the bisector of the angle formed by l and m.
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-51

Question 16:
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.
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-52
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-53

Question 17:
If ABCD is a quadrilateral such that diagonal AC bisects the angles A and C,then prove that AB = AD and CB = CD.
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-54

Question 18:
If ABC is a right angled triangle such that AB = AC and bisector of angle C intersects the side AB at D, then prove that AC + AD = BC.
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-55
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-56

Question 19:
If AB and CD are the smallest and largest sides of a quadrilateral ABCD, out of ∠B and ∠D decide which is greater.
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-57
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-58

Question 20:
Prove that in a triangle, other than an equilateral triangle, angle opposite the longest side is greater than 2/3 of a right angle.
Solution:
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-59

Question 21:
If ABCD is quadrilateral such that AB = AD and CB = CD, then prove that AC is the perpendicular bisector of BD.
Solution:
Given In quadrilateral ABCD, AB= AD and CB = CD.
Construction Join AC and BD.
To prove AC is the perpendicular bisector of BD.
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-60
NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles img-61

NCERT Exemplar ProblemsNCERT Exemplar MathsNCERT Exemplar Science

Primary Sidebar

NCERT Exemplar problems With Solutions CBSE Previous Year Questions with Solutoins CBSE Sample Papers
  • The Summer Of The Beautiful White Horse Answers
  • Job Application Letter class 12 Samples
  • Science Lab Manual Class 9
  • Letter to The Editor Class 12 Samples
  • Unseen Passage For Class 6 Answers
  • NCERT Solutions for Class 12 Hindi Core
  • Invitation and Replies Class 12 Examples
  • Advertisement Writing Class 11 Examples
  • Lab Manual Class 10 Science

Recent Posts

  • Understanding Diversity Question Answer Class 6 Social Science Civics Chapter 1 NCERT Solutions
  • Our Changing Earth Question Answer Class 7 Social Science Geography Chapter 3 NCERT Solutions
  • Inside Our Earth Question Answer Class 7 Social Science Geography Chapter 2 NCERT Solutions
  • Rulers and Buildings Question Answer Class 7 Social Science History Chapter 5 NCERT Solutions
  • On Equality Question Answer Class 7 Social Science Civics Chapter 1 NCERT Solutions
  • Role of the Government in Health Question Answer Class 7 Social Science Civics Chapter 2 NCERT Solutions
  • Vital Villages, Thriving Towns Question Answer Class 6 Social Science History Chapter 9 NCERT Solutions
  • New Empires and Kingdoms Question Answer Class 6 Social Science History Chapter 11 NCERT Solutions
  • The Delhi Sultans Question Answer Class 7 Social Science History Chapter 3 NCERT Solutions
  • The Mughal Empire Question Answer Class 7 Social Science History Chapter 4 NCERT Solutions
  • India: Climate Vegetation and Wildlife Question Answer Class 6 Social Science Geography Chapter 8 NCERT Solutions
  • Traders, Kings and Pilgrims Question Answer Class 6 Social Science History Chapter 10 NCERT Solutions
  • Environment Question Answer Class 7 Social Science Geography Chapter 1 NCERT Solutions
  • Understanding Advertising Question Answer Class 7 Social Science Civics Chapter 7 NCERT Solutions
  • The Making of Regional Cultures Question Answer Class 7 Social Science History Chapter 9 NCERT Solutions

Footer

Maths NCERT Solutions

NCERT Solutions for Class 12 Maths
NCERT Solutions for Class 11 Maths
NCERT Solutions for Class 10 Maths
NCERT Solutions for Class 9 Maths
NCERT Solutions for Class 8 Maths
NCERT Solutions for Class 7 Maths
NCERT Solutions for Class 6 Maths

SCIENCE NCERT SOLUTIONS

NCERT Solutions for Class 12 Physics
NCERT Solutions for Class 12 Chemistry
NCERT Solutions for Class 11 Physics
NCERT Solutions for Class 11 Chemistry
NCERT Solutions for Class 10 Science
NCERT Solutions for Class 9 Science
NCERT Solutions for Class 7 Science
MCQ Questions NCERT Solutions
CBSE Sample Papers
NCERT Exemplar Solutions LCM and GCF Calculator
TS Grewal Accountancy Class 12 Solutions
TS Grewal Accountancy Class 11 Solutions