NCERT Solutions Class 9 Maths Chapter 7 Triangles – Here are all the NCERT solutions for Class 9 Maths Chapter 7. This solution contains questions, answers, images, explanations of the complete Chapter 7 titled Triangles of Maths taught in class 9. If you are a student of class 9 who is using NCERT Textbook to study Maths, then you must come across Chapter 7 Triangles. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 9 Maths Chapter 7 Triangles in one place.
NCERT Solutions Class 9 Maths Chapter 7 Triangles
Here on AglaSem Schools, you can access to NCERT Book Solutions in free pdf for Maths for Class 9 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.
For a better understanding of this chapter, you should also see summary of Chapter 7 Triangles , Maths, Class 9.
Class | 9 |
Subject | Maths |
Book | Mathematics |
Chapter Number | 7 |
Chapter Name |
Triangles |
NCERT Solutions Class 9 Maths chapter 7 Triangles
Class 9, Maths chapter 7, Triangles solutions are given below in PDF format. You can view them online or download PDF file for future use.
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Question & Answer
Q.1: In quadrilateral ACBD, AC = AD and AB bisects ∠ A (see Fig). Show that ∆ ABC ≅ ∆ ABD. What can you say about BC and BD?

Ans : In ∆ ABC and ∆ ABD, AC = AD (Given) ∠CAB = ∠ DAB (AB bisects ∠A) AB = AB (Common) Therefore, ∆ ABC ≅ ∆ ABD( By SAS congruence rule) BC = BD (By CPCT) Therefore, BC and BD are Of equal lengths.
Q.2: ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see Fig). Prove that
(i) ∆ ABD ≅ ∆ BAC
(ii) BD = AC
(iii) ∠ ABD = ∠ BAC.

Ans : In ∠ABD and ∠BAC, AD = BC (Given) ∠DAB = ∠CBA(Given) AB = BA (Common) \(\begin{array}{l}{\therefore \triangle \mathrm{ABD}=\Delta \mathrm{BAC}(\mathrm{By} \text { SAS congruence rule) }} \\ {\therefore \mathrm{BD}=\mathrm{AC}(\mathrm{By} \mathrm{CPCT})}\end{array}\) And, ∠ABD = ∠BAC (By CPCT)
Q.3: AD and BC are equal perpendiculars to a line segment AB (see Fig.). Show that CD bisects AB.

Ans : In ∆ BOC and ∆ AOD, \(\begin{array}{l}{\angle \mathrm{BOC}=\angle \mathrm{AOD}(\text { Vertically opposite angles })} \\ {\angle \mathrm{CBO}=\angle \mathrm{DAO}\left(\text { Each } 90^{\circ}\right)}\end{array}\) \(\begin{array}{l}{\mathrm{BC}=\mathrm{AD}(\text { Given })} \\ {\therefore \Delta \mathrm{BOC}=\triangle \mathrm{AOD} \text { (AAS congruence rule) }} \\ {\therefore \mathrm{BO}=\mathrm{AO}(\mathrm{By} \mathrm{CPCT})} \\ {\Rightarrow \mathrm{CD} \text { bisects } \mathrm{AB}}\end{array}\)
Q.4: l and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig). Show that ∆ ABC ≅ ∆ CDA.

Ans : In ∆ ABC and ∆ CDA, \(\begin{array}{l}{\angle B A C=\angle D C A(\text { Alternate interior angles, as } p \| q)} \\ {A C=C A(\text { Common })}\end{array}\) \(\begin{array}{l}{\angle B C A=\angle D A C(\text { Alternate interior angles, as } l \| m)} \\ { \triangle A B C = \Delta C D A(B y \text { ASA congruence rule) }}\end{array}\)
Q.5: Line l is the bisector of an angle ∠ A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠ A (see Figure). Show that:
(i) ∆ APB ≅ ∆ AQB
(ii) BP = BQ or B is equidistant from the arms of ∠ A.

Ans : \(\begin{array}{l}{\text { In } \triangle A P B \text { and } \triangle A Q B \text { . }} \\ {\angle A P B=\angle A Q B\left(E a c h 90^{\circ}\right)}\end{array}\) \(\begin{array}{l}{\angle P A B=\angle Q A B(1 \text { is the angle bisector of } \angle A)} \\ {A B=A B(\text { Common })}\end{array}\) \(\begin{array}{l}{\therefore \triangle \mathrm{APB} \cong \triangle \mathrm{AQB}(\mathrm{By} \text { AAS congruence rule) }} \\ {\therefore \mathrm{BP}=\mathrm{BQ}(\mathrm{By} \mathrm{CPCT})}\end{array}\) Or, it can be said that B is equidistant from the arms of ∠ A.
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