NCERT Solutions of class 9 Maths exercise 7.1 of chapter 7- Triangles - Future Study Point

# NCERT Solutions of class 9 Maths exercise 7.1 – Triangles

NCERT Solutions of class 9 Maths exercise 7.1 of chapter 7- Triangles are the solutions of unsolved questions of class 9 NCERT maths textbook, these NCERT solutions are solved by an expert of maths as per the CBSE norms. The NCERT solutions of exercise 7.1 of chapter 7-Triangles are based on the congruency of the triangle. Every year one or two questions of 3 to 4 marks are always asked from this exercise 7.1, so it is very important for class 9 students to study NCERT solutions of exercise 7.1.

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Exercise 7.2-Triangle

Exercise 7.3 – Triangle

Exercise 7.4-Triangle

NCERT Solutions of Class 9 Science : Chapter 1 to Chapter 15

Q1.In quadrilateral ACBD, AC = AD, and AB bisect ∠A (see the given figure). Show that ΔABC≅ΔABD. What can you say about BC and BD?

Ans.

AB is the bisector of ∠A

So, ∠BAC = ∠DAB

TO PROVE: ΔABC≅ΔABD

PROOF: In ΔABC and ΔABD

We have

AB = AB (common)

∠BAC = ∠DAB (given)

ΔABC≅ΔABD (SAS rule)

BC = BD (by CPCT)

Q2. ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see the given figure ).Prove that

(i) ΔABD ≅ΔBAC

(ii) BD = AC

(iii) ∠ABD = ∠BAC

Ans.

(i) GIVEN: ABCD is a quadrilateral

AD = BC and ∠DAB = ∠CBA

TO PROVE: ΔABD ≅ΔBAC

PROOF: In ΔABD and ΔBAC

AB = AB (common)

∠DAB = ∠CBA

ΔABD ≅ΔBAC (SAS)

Hence proved

(ii) BD = AC (By CPCT)

(iii) ∠ABD = ∠BAC(By CPCT)

Q3. AD and BC are equal perpendiculars to a line segment AB (see the given figure).Show that CD bisects AB.

Ans.

AB ⊥ BC and AB ⊥ BC

TO PROVE: CD bisects AB i.e AO = BO

PROOF: In ΔAOD and ΔBOD

Since BOA and COD is a line

∠DOA = ∠BOC (vertically opposite angle)

ΔBOC ≅ ΔAOD(AAS)

AO = BO (By CPCT)

Hence Proved

See the video for the solutions of Q1 TO Q3

Q4. l and m are two parallel lines intersected by another pair of parallel lines p and q(see the given figure). Show that ΔABC ≅ CDA.

Ans.

GIVEN: p ∥ q and l∥ m

TO PROVE: ΔABC ≅ ΔCDA

PROOF:In the figure

l∥ m and AC is the transversal(given)

So, ∠DAC = ∠ACB (alternate angles)

p ∥ q and AC is the transversal(given)

∠BAC = ∠ACD(alternate angles)

AC = AC(given)

ΔABC ≅ Δ CDA(ASA rule)

Hence proved

Q5.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 the given figure). Show that

(i) ΔAPB ≅ΔAQB

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

Ans.

(i) GIVEN: ,BP and BQ are perpendiculars from B to the arms of ∠A

∴∠AQB = ∠APB = 90°

Line l is the bisector of an angle ∠A and

∴∠QAB = ∠PAB

TO PROVE: ΔAPB ≅ΔAQB

PROOF:∠AQB = ∠APB = 90°(given)

AB = AB (common)

∠QAB = ∠PAB (given)

ΔAPB ≅ΔAQB (AAS rule)

(ii) BP = BQ(by CPCT)

Hence proved

Q6. In the given figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC =DE.

Ans.

GIVEN: AC = AE

TO PROVE: BC =DE

PROOF: We have to prove BC =DE, which are sides of the triangle ΔABC and ΔADE

AC = AE(given)

∠BAD + ∠DAC = ∠EAC + ∠DAC

∠BAC = ∠DAE(proved)

BC = DE (by CPCT )

Hence proved

See the video for the solutions of Q4 TO Q6

Q7. AB is a line segment and p is its midpoint. D and E are points on the same sides of AB such that ∠BAD = ∠ABE  and ∠EPA=∠DPB(see the given figure). Show that

(i) ΔDAP≅ ΔEBP

Ans.(i)

GIVEN: ∠DAP = ∠EBP

∠EPA = ∠DPB

P is the mid point of AB

∴AP = BP

TO PROVE:

(i) ΔDAP≅ ΔEBP

PROOF: In ΔDAP and ΔEBP

∠EPA=∠DPB(given)

∠EPA +∠EPD =∠DPB+ ∠EPD

∠APD = ∠EPB

AP = BP(given)

ΔDAP≅ ΔEBP(ASA rule)

Q8.In right triangle ABC ,right angled at C , M is th mid point 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 the given figure). Show that

(i) ΔAMC ≅ BMD

(ii) ∠DBC is a right angle

(iii) ΔDBC ≅ ΔACB

Ans.

(i) GIVEN:ΔABC is a right triagle ,∠C = 90°

M is the midpoint of AB

∴BM = AM

AM = AB/2

DM = CM

TO PROVE: ΔAMC ≅ BMD

PROOF:DM = CM (given)

BM = AM (given)

∠BMD = ∠AMC (Vertically opposite angle)

ΔAMC ≅ BMD(SAS rule)

Hence proved

(ii) Since ΔAMC ≅ BMD(proved)

∠DBM = ∠CAM (By CPCT)

∠DBM  and ∠CAM are alternate angles

∴DB ∥ AC

∠DBC + ∠ACB = 180° (co-interior angles)

∠ACB = 90°

∠DBC + 90° = 180°

∠DBC = 90°

Hence proved

(iii) Since ΔAMC ≅ BMD(proved)

BD = AC (By CPCT)

BC = BC(common)

∠DBC = ∠ACB =90°(Proved above)

ΔDBC ≅ ΔACB (SAS rule)

proved above

(iv) CM = DM(given)

CM + DM = DC

CM + CM = DC

2CM = DC

CM = 1/2 DC

Since we have proved

ΔDBC ≅ ΔACB

So, DC = AB (by CPCT)

Therefore

CM = 1/2 AB

Hence proved

See the video of the solutions for question 7 and question 8

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