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NCERT exemplar Class 9 Maths solutions chapter 7 discusses a triangle and its properties which are essential in mathematics. The NCERT exemplar Class 9 Maths chapter 7 solutions are curated to get a detailed overview of steps required to solve the problems. These NCERT exemplar Class 9 Maths chapter 7 solutions are prepared by highly qualified and seasoned subject matter experts to provide supporting material while attempting NCERT Class 9 Maths Questions. The NCERT exemplar Class 9 Maths solutions chapter 7 develops a progressive grasp of triangles’ concepts and follows the CBSE Class 9 Syllabus
Question:1
Which of the following is not a criterion for congruence of triangles?
(A) SAS
(B) ASA
(C) SSA
(D) SSS
Answer:
[C] SSA is correct optionQuestion:2
If AB = QR, BC = PR and CA = PQ, then
Answer:
[B]Question:3
In , AB = AC and . Then is equal to
Answer:
[B] Solution.Question:4
In ABC, BC = AB and B = . Then A is equal to
Answer:
[C]Question:5
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
Answer:
[A]Question:6
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
Answer:
[A]Question:7
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°
Answer:
[B]Question:8
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
Answer:
[D]Question:5
In PQR, if R > Q, then
(A) QR > PR
(B) PQ > PR
(C) PQ < PR
(D) QR < PR
Answer:
[B]Question:10
In triangles ABC and PQR, AB = AC, C = P and B =Q. The two triangles are
(A) isosceles but not congruent
(B) isosceles and congruent
(C) congruent but not isosceles
(D) neither congruent nor isosceles
Answer:
[A]Question:1
Answer:
Given : In triangles ABC and PQR, A = Q andB =RQuestion:2
Answer:
Side BC should be equal to side PR.Question:3
Answer:
If two sides and the included angle of one triangle are equal to the two side and the included angle of the other triangle then only the triangles can be congruent by SAS criterion of congruence otherwise not.Question:4
Answer:
[True]Question:5
Answer:
Not possible.Question:7
If PQR EDF, then is it true to say that PR = EF? Give reason for your answer.
Answer:
TrueQuestion:8
Answer:
[PR]Question:9
AD is a median of the triangle ABC. Is it true that AB + BC + CA > 2AD?
Give reason for your answer.
Answer:
TrueQuestion:10
Answer:
TrueQuestion:11
Answer:
[Not possible]Question:12
Answer:
YesQuestion:4
In Figure, BA AC, DE DF such that BA = DE and BF = EC. Show that ABC DEF.
Answer:
Given, BA ACQuestion:5
Q is a point on the side SR of a DPSR such that PQ = PR. Prove that PS > PQ.
Answer:
Given: Q is point on side SR in DPSR and PQ = PRQuestion:6
S is any point on side QR of a PQR. Show that:PQ + QR + RP > 2 PS.
Answer:
Given, PQR, S is any point on QR.Question:7
D is any point on side AC of a ABC with AB = AC. Show that CD < BD.
Answer:
Given: ABC is a triangle and D is any point on AC.Question:8
Answer:
Given :Question:9
Answer:
Given: ABC is a isosceles triangle AB = ACQuestion:10
Answer:
Given, ABC is an isosceles triangleQuestion:11
In given figure, AD is the bisector of BAC. Prove that AB > BD.
Answer:
Given: In ABC, AD is bisector of BAC.Question:1
Find all the angles of an equilateral triangle.
Answer:
Question:2
Answer:
Given : ABC is an triangle and let O is a point on AB.Question:3
ABC is an isosceles triangle with AB = AC and D is a point on BC such that AD BC (Figure). 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?
Answer:
[ABD = ACD is defect.]Question:4
Answer:
Given : P is a point on the bisector at ÐABC.Question:5
Answer:
Given: ABCD is a quadrilateral in which AB = BC and AD = CDQuestion:6
ABC is a right triangle with AB = AC. Bisector of A meets BC at D. Prove that BC = 2 AD.
Answer:
Given : ABC is a right triangle AB = ACQuestion:7
Answer:
ABCD is square. O is any interior point of the square and OAB is an equilateral triangle.Question:8
Answer:
Question:9
Answer:
Given: ABC is isosceles triangle AB = ACQuestion:10
Answer:
Question:11
Show that in a quadrilateral ABCD, AB + BC + CD + DA < 2 (BD + AC)
Answer:
Given : ABCD is a quadrilateralQuestion:12
Show that in a quadrilateral ABCD, AB + BC + CD + DA > AC + BD
Answer:
Given, ABCD is a quadrilateralQuestion:13
Answer:
ABC is triangle and D is mid-point at ACQuestion:14
Answer:
Given : ABC is right angle triangleQuestion:15
Answer:
Given: l and m intersect at point O and P is a point on a line n passing through O andQuestion:16
Answer:
Given : ABCD is trapeziumQuestion:17
Answer:
Given: ABCD is quadrilateralQuestion:18
Answer:
Given, ABC is right angle triangleQuestion:19
Answer:
ABCD is quadrilateralQuestion:20
Answer:
Given: ABC is triangle and AC is longest sideSalient topics covered in NCERT exemplar Class 9 Maths solutions chapter 7 are mentioned below:
◊ Learning of trivial properties and theorems about triangles.
◊ Type of triangles.
◊ Fundamental inequalities of a triangle.
◊ Congruency of triangles and methods to prove it.
◊ NCERT exemplar Cass 9 Maths chapter 7 solutions discuss examples such as “the sum of two sides of a triangle will be more than me third side’s length,” which is very important and useful inequality.
These Class 9 Maths NCERT exemplar chapter 7 solutions provide a student with basic concepts of Triangles so that the student can solve challenging geometry problems and coordinate geometry. Fundamental inequalities of a triangle will help the student solve many vectors and complex number-related problems later in higher classes. These Class 9 Maths NCERT exemplar solutions chapter 7 Triangles consist of a wide range of practice problems available in the exemplar which provides a student with adequate opportunity to learn and practice the concepts of triangles. The Class 9 Maths NCERT exemplar solutions chapter 7 Triangles builds a good practice regime along with clarity on the concepts such that students can easily attempt other books as RD Sharma Class 9 Maths, NCERT Class 9 Maths, RS Aggarwal Class 9 Maths, et cetera.
A distinct feature of these solutions can be accessed through NCERT exemplar Class 9 Maths solutions chapter 7 pdf download, where students are made available the pdf versions of these solutions which comes in handy while studying NCERT exemplar Class 9 Maths chapter 7 in an offline mode.
Chapter No. | Chapter Name |
Chapter 1 | |
Chapter 2 | |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | |
Chapter 7 | |
Chapter 8 | |
Chapter 9 | |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | |
Chapter 13 | |
Chapter 14 | |
Chapter 15 |
If two triangles are similar, their corresponding interior angles will be the same; however, the side length will not be the same. If two triangles are congruent, then their corresponding angles and side-lengths will be the same.
No, it is not possible to make a triangle with these three side lengths. Sum of two sides must be more than the third side length.
As, 5 + 8 <14
Therefore, 5,8, and 14 cannot be the sides of a triangle.
If two triangles are congruent, they will have the same interior angles and the same sides. Therefore, they must be similar, and however, if two triangles are similar, they need not be congruent.
It is highly recommended to understand and practice all the congruencies mentioned in the chapter along with attempting practice questions on the same. The NCERT exemplar Class 9 Maths solutions chapter 7 will help you understand the problems in a detailed manner.
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