Class 9 Maths Triangles Chapter 7 Exercise 7.1

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Class 9 Maths Chapter 7.1

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Solution of Ex- 7.1

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Answers to NCERT  class 9 maths chapter 7 exercise 7.1 

Class 9 Maths Chapter 7 Exercise 7.1 Question 1

In quadrilateral ACBD, AC=AD and AB bisect ∠A(see figure) show that ∆ABC=∆ABD. What can you say about BC and BD?

Exercise 7.1 Class 9 Question 1 Solution

Now, in ∆s ABC and ABD, we have:

AC=AD                 (given)

∠CAB=∠BAD    (∵AB bisects ∠A)

AB=AB               (common)

∴ By SAS congruence criterion, we have:

∆ABC≅∆ABD 

⟹ BC=BD they are equal.

(∵ Corresponding parts of congruent triangles are equal)

Class 9 Maths Chapter 7 Exercise 7.1 Question 2

ABCD is a quadrilateral in which AD=BC and ∠DAB=∠CBA(see figure). Prove that 

(i)∆ABD=∆BAC

(ii)BD=AC

(iii)∠ABD=∠BAC

Exercise 7.1 Class 9 Question 2 Solution

In ∆s ABD and BAC, we have :

AD=BC             (given)

∠DAB=∠CBA   (given)

AB=BA             (common)

∴ By SAS criterion of congruence, we have:

∆ABD=∆BAC , which proves (i)

From(i), (ii) BD=AC (C.P.CT.)

And, (iii) )∠ABD=∠BAC (∵ Corresponding parts of

 congruent triangles (CPCT)are equal)

Class 9 Maths Chapter 7 Exercise 7.1 Question 3

AD and BC are equal perpendiculars to a line segment AB(see figure). Show that CD bisects AB.

Exercise 7.1 Class 9 Question 3 Solution

Since AB and CD intersect at O, therefore

∠AOD=∠BOC ……(i) (vertically opposite angles)

 In ∆s AOD and BOC, we have :

∠AOD=∠BOC             (from (i))

∠DAO=∠CBO             (each =90°) 

And, AD=BC                (given)

∴ By ASS criterion of congruence, we have:

∆AOD=∆BOC 

⟹ OA=OB (∵ Corresponding parts of congruent triangles(CPCT) are equal)

i.e. O is the mid point AB

Hence,  CD bisects AB.

Class 9 Maths Chapter 7 Exercise 7.1 Question 4

l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure), show that ∆ABC≅∆CDA.

Exercise 7.1 Class 9 Question 4 Solution

Since l and m are two parallel lines intersected by another pair of parallel lines p and q, therefore AD||BC and AB||CD.

∠BAC=∠DCA            (alternate angles)……….(i)

Again, BC||AD and AC is a transversal

∠BCA=∠DCA           (alternat angles)……….(ii)

Now, in ∆s ABC and CDA,we have:

∠BAC=∠DCA           [proved above in (i)]

∠BCA=∠DAC          [proved above in (ii)]

And AC=CA              (common)

∴ By ASA criterion of congruence,

∆ABC≅∆CDA 

Class 9 Maths Chapter 7 Exercise 7.1 Question 5

Line l is the bisector of an angle ∠A and ∠B is any point on l. BP and BQ are perpendicular from B to the arms of  ∠A (see figure). Show that:

(i)      ΔAPB≅ ΔAQB

(ii) BP=BQ or B is equidistant  form the arms of ∠A

Exercise 7.1 Class 9 Question 5 Solution

In ∆s APB and AQB, we have:

∠APB=∠AQB             (∵each 90°)

∠PAB=∠QAB              (∵AB bisects∠PAQ)

AB=AB                        (common)

By AAS congruence criterion, we have:

∆APB≅∆AQB , which proves (i)

⟹ BP=PQ (∵ corresponding parts of congruent triangles are equal)

i.e. B  is equidistant from the arms of ∠A which proves (ii)

Class 9 Maths Chapter 7 Exercise 7.1 Question 6

In the given figure, AC=AE, AB=AD, and ∠EAC. Show that BC=DE.

Exercise 7.1 Class 9 Question 6 Solution

In ∆s ABC and ADE, we have:

AB=AD                       (given)

∠BAC=∠DAE 

∵∠BAD=∠EAC 

⇒∠BAD+∠DAC=∠EAC+∠DAC 

⇒∠BAC=∠DAE 

AB=AE                        (given)

∴ By SAS criterion of congruence, we have:

∆ABC≅∆ADE

⟹BC=DE

(∵  Corresponding parts of congruent triangles are equal)

Class 9 Maths Chapter 7 Exercise 7.1 Question 7

AB is a line segment and P is its midpoint. D and E are points on the same side of AB such that ∠BAD=∠ABE and ∠EPA=∠DPB(see figure). Show that:

(i). ∆DAP≅∆EBP

(ii). AD=BE

Exercise 7.1 Class 9 Question 7 Solution

We have :∠EPA=∠DPB

∴ ∠EPA+ ∠DPE=∠DPB+ ∠DPE 

Or   ∠DPA=∠EPB  ………….(i)

Now, in ∆s DAP and EBP,we have

∠DPA=∠EPB            [from (i)]

AP=BP                       (given)

And, ∠DAP=∠EBP   (given)

So, AD=BE, which proves (ii)(∵Corresponding parts of congruent triangles are equal)

Class 9 Maths Chapter 7 Exercise 7.1 Question 8

In the right triangle ABC, right-angled at C, M is the midpoint of hypotenuse AB. C is joined to M and produced to a point D such that DM= CM. point D is joined to point B(see figure). Show that:

(i).∆AMC≅∆BMD

(ii).∠DBC is a right angle.

(iii).∆DBC≅∆ACB

(iv).CM= `frac{1}{2}` AB

Exercise 7.1 Class 9 Question 8 Solution

Given:

∠C=90°,AM=BM,DM=CM 

To prove 

(i)∆AMC≅∆BMD

(ii). ∆DBC=90°

(iii). ∆DBC≅∆ACB

(iv). CM=`frac{1}{2}` AB

Proof:

(i). In ∆AMC and ∆BMD

AM=BM               (given)

MC=MD                (given)

∠1=∠2                  (Vertically opposite angles)

∴∆AMC≅∆BMD (by SAS)

And so,  AC=DB and ∠4=∠3            (by CPCT)

(ii).∠4=∠3          (proved above)

∠4=∠3                 (But these are alternate interior angles)

∠C+∠B=180°       (Cointerior angles)

⟹ ∠B=180°-90°=90° (∵∠C=90°)

(iii). In ∆DBC and ∆ACB

DB=AC                (proved)

∠B=∠C                (each 90°)

BC=CB                (common)

∴∆DBC≅∆ACB   (By SAS)

And so, DC=AB (By CPCT)

(iv)DC=AB

`frac{1}{2}=DC=frac{1}{2}` AB 

CM=`frac{1}{2}` AB 

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Related Topics 

1. Exercise 7.3

2. Exercise 7.2 

3. All Theorems