- NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1
- NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2
- NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3
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 ΔABC ≅ ΔABD. What can you say about BC and BD?
Solution:
To prove: ΔABC ≅ ΔABD
Proof: In
ΔABC and ΔABD, AC = AD [given]
∠BAC = ∠DAB
[given]
and AB = AB
[common side]
Hence, ΔABC ≅ ΔABD [by SAS congruence
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) ΔABD ≅ ΔBAC
(ii) BD = AC
(iii) ∠ABD = ∠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]
∠DAB = ∠CBA [given]
and AB = AB
[common side]
∴ ΔABD ≅ ΔBAC [by SAS congruence rule]
(ii) From
part (i), ΔABD ≅ ΔBAC
Then, BD = AC [by
CPCT]
(iii) From
part (i), ΔABD ≅ ΔBAC
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
= BC, AD
⊥ AB and BC ⊥ AB
To
prove: CD bisects AB.
Proof: In ΔAOD and ΔBOC.
AD = BC [given]
∠OAD = ∠OBC [each 90°]
and ∠AOD = ∠BOC [vertically opposite angles]
ΔAOD ≅ ΔBOC [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 [alternate 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 figure). Show that
(i) ΔAPB ≅ ΔAQB
(ii) BP = BQ or B is equidistant from the arms
of ∠A
Solution:
(i) Consider ΔAPB and ΔAQB, we have
∠APB = ∠AQB = 90°
∠PAB = ∠QAB [∵ AB bisects ∠PAQ]
AB = AB [common]
∴ ΔAPB ≅ ΔAQB [by AAS congruence rule]
(ii) From part (i), ΔAPB ≅ ΔAQB.
Then,
BP = BQ [by
CPCT]
i.e.,
B is equidistant from the arms of ∠A.
Ex 7.1 Class 9 Maths Question 6.
In figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC =
DE.
Solution:
In ΔABC and ΔADE,
AB = AD [Given]
∠BAD = ∠EAC [Given] …(i)
On adding ∠DAC on both sides in Eq. (i), we get
∠BAD + ∠DAC = ∠EAC + ∠DAC
∠BAC = ∠DAE
and AC = AE
[given]
ΔABC ≅ ΔADE
[by SAS 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) ΔDAP ≅ ΔEBP
(ii) AD = BE
Solution:
(i)
∠EPA = ∠DPB [Given] …. (i)
∠BAD = ∠ABE [Given] .…(ii)
On adding ∠EPD on both sides in Eq. (i), we
get
∠EPA + ∠EPD = ∠DPB + ∠EPD
∠DPA = ∠EPB …(iii)
Now, in ΔDAP and ΔEBP, we have
∠DPA = ∠EPB [from eq.(iii)]
∠DAP = ∠EBP [from eq.(ii)]
AP
= BP [∵ P is the mid-point of AB (Given)]
ΔDAP ≅ ΔEBP [by ASA congruence rule]
(ii) Since ΔDAP ≅ ΔEBP
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 figure). Show that
(i) ∆AMC ≅ ∆BMD
(ii) ∠DBC is a right angle
(iii) ∆DBC ≅ ∆ACB
(iv) CM = ½ AB
Solution:
(i) In ∆AMC and ∆BMD,
CM = DM [Given]
∠AMC = ∠BMD [Vertically opposite angles]
AM = BM [Since M is the mid-point of AB]
∴ ∆AMC ≅ ∆BMD [By SAS congruence rule]
(ii) Since ∆AMC ≅ ∆BMD
⇒ ∠MAC = ∠MBD [By
CPCT]
But they form a pair of alternate interior angles.
∴ AC || DB
Now, BC is a transversal which intersects parallel lines AC and DB,
∴ ∠BCA + ∠DBC = 180° [Co-interior angles]
But ∠BCA = 90° [∆ABC is right angled at
C]
∴ 90° + ∠DBC = 180°
⇒ ∠DBC = 90°
(iii) Again, since ∆AMC ≅ ∆BMD
∴ AC = BD [By CPCT]
Now, in ∆DBC and ∆ACB, we have
BD = CA [Proved
above]
∠DBC = ∠ACB [Each 90°]
BC = CB [Common]
∴ ∆DBC ≅ ∆ACB [By SAS congruence rule]
(iv)
As ∆DBC ≅ ∆ACB
DC = AB [By CPCT]
But DM = CM [Given]
∴ CM = ½ DC = ½ AB
⇒ CM = ½ AB
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