NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1.
- Triangles Class 9 Ex 7.1
- Triangles Class 9 Ex 7.2
- Triangles Class 9 Ex 7.3
- Triangles Class 9 Ex 7.4
- Triangles Class 9 Ex 7.5
| Board | CBSE |
| Textbook | NCERT |
| Class | Class 9 |
| Subject | Maths |
| Chapter | Chapter 7 |
| Chapter Name | Triangles |
| Exercise | Ex 7.1 |
| Number of Questions Solved | 8 |
| Category | NCERT Solutions |
NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1
Ex 7.1 Class 9 Maths Question 1.
In a quadrilateral ACBD, AC = AD, and AB bisects ∠A (see figure). Show that Δ ABC ≅ ΔABD. What can you say about and BD?
Solution:
Given : In quadrilateral ACBD.
AC = AD and ∠BAC = ∠DAB
To prove : ΔABC = ΔABD
Proof: In ΔABC and ΔABD, AC = AD [given]
∠BAC = ∠DAB [given]
and B = AB [common side]
Hence, ΔABC = ΔABD [by SAS congrunce rule]
Then, BC = BD
Thus, BC and BD are equal.
Ex 7.1 Class 9 Maths Question 2.
ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see figure). Prove that
(i) Δ ABD ≅ Δ BAC
(ii) BD = AC
(iii) Δ ABD = Δ BAC
Solution:
Given : In quadrilateral ABCD.
AD = BC and ∠DAB = ∠CBA
To prove :
(i) ΔABD ≅ ΔBAC
(ii) BD= AC
(iii) ∠ABD = ∠BAC
Proof : (i) In ΔABD and ΔBAC,
AD-BC [given]
ΔDAB = ΔCBA [given]
and AB = AB [common side]
∴ Δ ABD ≅ Δ BAC [by SAS congruence rule]
(ii) From part (i), Δ ABD ≅ ΔBAC
Then, BD = AC [by CPCT]
(iii) From part (i), Δ ABD ≅ Δ BAC
Then, ∠ABD = ∠BAC [by CPCT]
Hence proved.
Ex 7.1 Class 9 Maths Question 3.
AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB.
Solution:
Given: AD ⊥ AB and BC ⊥ AB
To prove: CD bisects AB.
Proof: In Δ AOD and Δ BOC.
AD = BC [given]
∠OAD = ∠OBC [each 90°]
and ∠AOD =∠BOC [vertically opposite angles]
∠AOD = ∠BOC [by AAS congruence rule]
Then,OA = OB [by CPCT]
Thus, CD bisects AB.
Ex 7.1 Class 9 Maths Question 4.
l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that
ΔABC = ΔCDA.
Solution:
Given : l || m and p || q
To prove : ΔABC = ΔCDA
Proof : In ΔABC and ΔADC,
∠BAC = ∠ACD [alternative interior angles as p || q]
∠ACB = ∠DAC [alternative interior angles as l || m]
AC = AC [common side]
ΔABC ≅ ΔCDA [by AAS congruence rule]
Ex 7.1 Class 9 Maths Question 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 fig.). Show that
(i) ΔAPB = ΔAQB
(ii) BP = BQ or B is equidistant from the arms of ∠A
Solution:
Consider ΔAPB and ΔAQB, we have
∠APB = ∠AQB =90°
∠PAB = ∠QAB [∵ AB bisects ∠PAQ]
AB = AB [common]
∴ By AAS congruence axiom, we have
ΔAPB ≅ ΔAQB, which proves (i)
=> BP = BQ [by CPCT]
i.e., B is equidistant from the arms of ∠A, which proves (ii).
Ex 7.1 Class 9 Maths Question 6.
In figure, AC = AE, AB – AD and ∠BAD = ∠EAC. Show that BC = DE.
Solution:
In Δ ABC and Δ ADE,
We have AB = AD (Given)
∠BAD = ∠EAC (Given) …(i)
On adding ∠DAC on both sides in Eq. (i)
∠BAD + ∠DAC = ∠EAC + ∠DAC
∠BAC = ∠DAE
and AC – AE [given]
Δ ABC ≅ Δ ADE [by AAS congruence rule]
BC = DE [by CPCT]
Ex 7.1 Class 9 Maths Question 7.
AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA ∠DPB (see figure). Show that
(i) Δ DAP ≅ ΔEBP
(ii) AD = BE
Solution:
We have AP = BP [∵ P is the mid-point of AB (Given) ]…(i)
∠EPA = ∠DPB (Given) …. (ii)
∠BAD = ∠ABE (Given) …(iii)
On adding ∠EPD on both sides in Eq. (ii),
we have => ∠EPA + ∠EPD = ∠DPB + ∠EPD
=> ∠DPA = ∠EPB …(iv)
Now, in Δ DAP and Δ EBP, we have …..(iv)
∠DPA = ∠EPB [from eq.(iv)]
∠DAP = ∠EBP (Given)
and AP = BP [from eq.(i)]
ΔDAP ≅ ΔEBP [by ASA congruence axiom]
Hence,AD = BE [by CPCT]
Ex 7.1 Class 9 Maths Question 8.
In right triangle ABC, right-angled at C, M is the 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 fig.). Show that:
(i) Δ AMC = Δ BMD
(ii) ∠DBC is a right angle
(iii) Δ DBC = Δ ACB
(iv) CM=\(\cfrac { 1 }{ 2 }\)AB
Solution:
Given: Δ ACB in which ZC = 90° and M is the mid-point of AB.
To Prove :
(i) Δ AMC ≅ Δ BMD
(ii) ∠DBC is a right angle
(iii) Δ DBC ≅ ΔACB
(iv) CM = \(\cfrac { 1 }{ 2 }\) AB
Proof : Consider Δ AMC and Δ BMD,
we have AM – BM
CM = DM
∠AMC = ∠BMD
∴ Δ AMC = Δ BMD
AC = DB
and ∠1-∠2
But ∠1 and ∠2 are alternate angles.
=> BD || CA
Now, BD || CA and BC is transversal.
∴ ∠ACB + ∠CBD = 180° =>90°+ ∠CBD = 180° => ∠CBD =90°
In ΔDBC and ΔACB, we have
CB = BC [common]
DB = AC [using eq.(i)]
∠CBD = ∠BCA [each= 90°]
ΔDBC ≅ = ΔACB [by ASA congruence axiom]
=> DB = AB => \(\frac { 1 }{ 2 } AB = \frac{ 1 }{ 2 } \) DC => \(\frac { 1 }{ 2 } \) AB=CM
=> CM = \(\frac { 1 }{ 2 } \) [∵ CM = \(\frac { 1 }{ 2 } \) DC]
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