# NCERT Exemplar Class 9 Maths Solutions Chapter 7 Triangles

In this chapter, we provide NCERT Exemplar Problems Solutions for Class 9 Maths Solutions Chapter 7 Triangles for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Exemplar Problems Solutions for Class 9 Maths Solutions Chapter 7 Triangles pdf, free NCERT Exemplar Problems Solutions for Class 9 Maths Solutions Chapter 7 Triangles book pdf download. Now you will get step by step solution to each question.

## 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: Question 4:
In ΔABC,if BC = AB and ∠B = 80°, then ∠A is equal to
(a) 80°                 (b) 40°                      (c) 50°                         (d) 100°
Solution: 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: 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: 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: 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: 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: 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

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

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° 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:
Solution: 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: 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. 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.

Solution: Question 3:
In the given figure, ΔCDE is an equilateral triangle formed on a side CD of a square ABCD. Show that ΔADE ≅ ΔBCE.

Solution: Question 4:
In figure, BA ⊥ AC, DE ⊥ DF such that BA = DE and BF = EC. Show that ΔABC ≅ ΔDEF.

Thinking Process
Use the RHS congruence rule to show the given result
Solution: 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: 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: Question 7:
D is any point on side AC of a ΔABC with AB = AC. Show that CD < BD.
Solution: 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.

Solution: 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: 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:  Question 11:
In following figure if AD is the bisector of ∠BAC, then prove that AB > BD.

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.
[exterior angle of a triangle is greater than each of the opposite interior angle]
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: 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.

Solution: 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 Solution: 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. 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: Question 6:
ABC is a right triangle with AB = AC. If bisector of ∠A meets BC at D,then prove that BC = 2AD.
Solution:  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: 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: 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:  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: 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:  Question 12:
Show that in a quadrilateral ABCD, AB + BC + CD + DA > AC + BD.
Solution: 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:  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:  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: 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:  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: 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:  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:  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: Question 21:
If ABCD is quadrilateral such that AB = AD and CB = CD, then prove that AC is the perpendicular bisector of BD.
Solution:
Construction Join AC and BD.
To prove AC is the perpendicular bisector of BD. All Chapter NCERT Exemplar Problems Solutions For Class 9 maths

—————————————————————————–

All Subject NCERT Exemplar Problems Solutions For Class 9

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share ncertexemplar.com to your friends.