NCERT Solutions for Class 9 Maths Chapter 7 Triangles

NCERT Solutions for Class 9 Maths Chapter 7 Triangles

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1

Ex 7.1 Class 9 Maths Question 1.
In quadrilateral ACBD, AC = AD and AB bisects ∠ A (see figure). Show that ∆ABC ≅ ∆ABD. What can you say about BC and BD?

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1 Q1
Solution:
In quadrilateral ACBD, we have AC = AD and AB being the bisector of ∠A.
Now, In ∆ABC and ∆ABD,
AC = AD (Given)
∠ CAB = ∠ DAB ( AB bisects ∠ CAB)
and AB = AB (Common)
∴ ∆ ABC ≅ ∆ABD (By SAS congruence axiom)
∴ BC = BD (By CPCT)

Ex 7.1 Class 9 Maths Question 2.
ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see figure). Prove that
NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1 Q2
(i) ∆ABD ≅ ∆BAC
(ii) BD = AC
(iii) ∠ABD = ∠ BAC
Solution:
In quadrilateral ACBD, we have AD = BC and ∠ DAB = ∠ CBA

(i) In ∆ ABC and ∆ BAC,
AD = BC (Given)
∠DAB = ∠CBA (Given)
AB = AB (Common)
∴ ∆ ABD ≅ ∆BAC (By SAS congruence)

(ii) Since ∆ABD ≅ ∆BAC
⇒ BD = AC [By C.P.C.T.]

(iii) Since ∆ABD ≅ ∆BAC
⇒ ∠ABD = ∠BAC [By C.P.C.T.]

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.
NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1 Q3
Solution:
In ∆BOC and ∆AOD, we have
∠BOC = ∠AOD
BC = AD [Given]
∠BOC = ∠AOD [Vertically opposite angles]
∴ ∆OBC ≅ ∆OAD [By AAS congruency]
⇒ OB = OA [By C.P.C.T.]
i.e., O is the mid-point of AB.
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 ∆ABC = ∆CDA.
NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1 Q4
Solution:
∵ p || q and AC is a transversal,
∴ ∠BAC = ∠DCA …(1) [Alternate interior angles]
Also l || m and AC is a transversal,
∴ ∠BCA = ∠DAC …(2)
[Alternate interior angles]
Now, in ∆ABC and ∆CDA, we have
∠BAC = ∠DCA [From (1)]
CA = AC [Common]
∠BCA = ∠DAC [From (2)]
∴ ∆ABC ≅ ∆CDA [By ASA congruency]

Ex 7.1 Class 9 Maths Question 5.
Line l is the bisector of an ∠ A and ∠ B is any point on l. BP and BQ are perpendiculars from B to the arms of LA (see figure). Show that
(i) ∆APB ≅ ∆AQB
(ii) BP = BQ or B is equidistant from the arms ot ∠A.
NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1 Q5
Solution:
We have, l is the bisector of ∠QAP.
∴ ∠QAB = ∠PAB
∠Q = ∠P [Each 90°]
∠ABQ = ∠ABP
[By angle sum property of A]
Now, in ∆APB and ∆AQB, we have
∠ABP = ∠ABQ [Proved above]
AB = BA [Common]
∠PAB = ∠QAB [Given]
∴ ∆APB ≅ ∆AQB [By ASA congruency]
Since ∆APB ≅ ∆AQB
⇒ BP = BQ [By C.P.C.T.]
i. e., [Perpendicular distance of B from AP]
= [Perpendicular distance of B from AQ]
Thus, the point 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.
NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1 Q6
Solution:
We have, ∠BAD = ∠EAC
Adding ∠DAC on both sides, we have
∠BAD + ∠DAC = ∠EAC + ∠DAC
⇒ ∠BAC = ∠DAE
Now, in ∆ABC and ∆ADE. we have
∠BAC = ∠DAE [Proved above]
AB = AD [Given]
AC = AE [Given]
∴ ∆ABC ≅ ∆ADE [By SAS congruency]
⇒ BC = DE [By C.P.C.T.]

Ex 7.1 Class 9 Maths Question 7.
AS 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) ∆DAP ≅ ∆EBP
(ii) AD = BE
NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1 Q7
Solution:
We have, P is the mid-point of AB.
∴ AP = BP
∠EPA = ∠DPB [Given]
Adding ∠EPD on both sides, we get
∠EPA + ∠EPD = ∠DPB + ∠EPD
⇒ ∠APD = ∠BPE

(i) Now, in ∆DAP and ∆EBP, we have
∠PAD = ∠PBE [ ∵∠BAD = ∠ABE]
AP = BP [Proved above]
∠DPA = ∠EPB [Proved above]
∴ ∆DAP ≅ ∆EBP [By ASA congruency]

(ii) Since, ∆ DAP ≅ ∆ EBP
⇒ AD = BE [By C.P.C.T.]

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
NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1 Q8
(iv) CM = 1/2 AB
Solution:
Since M is the mid – point of AB.
∴ BM = AM

(i) In ∆AMC and ∆BMD, we have
CM = DM [Given]
∠AMC = ∠BMD [Vertically opposite angles]
AM = BM [Proved above]
∴ ∆AMC ≅ ∆BMD [By SAS congruency]

(ii) Since ∆AMC ≅ ∆BMD
⇒ ∠MAC = ∠MBD [By C.P.C.T.]
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, ∆AMC ≅ ∆BMD [Proved above]
∴ AC = BD [By C.P.C.T.]
Now, in ∆DBC and ∆ACB, we have
BD = CA [Proved above]
∠DBC = ∠ACB [Each 90°]
BC = CB [Common]
∴ ∆DBC ≅ ∆ACB [By SAS congruency]

(iv) As ∆DBC ≅ ∆ACB
DC = AB [By C.P.C.T.]
But DM = CM [Given]
∴ CM = 1/2DC = 12AB
⇒ CM = 1/2AB

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2 q1

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2 q1.1

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2 q2

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2 q3

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2 q4

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2 q5

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2 q6

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NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2 q7

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2 q8

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3 q1

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3 q1.1

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NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3 q2

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3 q3

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NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3 q4

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3 q5

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4 q1

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4 q2

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4 q3

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4 q4

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4 q4.1

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4 q5

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4 q5.1

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NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4 q7

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4 q8

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4 q9

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NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.5

Question 1.
ABC is a triangle. Locate a point in the interior of ∆ ABC which is equidistant from all the vertices of ∆ ABC.
Solution:
Suppose OM and ON be the perpendicular bisectors of sides BC and AC of ∆ ABC.
NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.5 q1
So, O is equidistant from two endpoints 0 and C of line segment BC as O lies on the perpendicular bisector of BC. Similarly, O is equidistant from C and A Hence, O be an orthocentre of ∆ABC.

Question 2.
In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.
Solution:
Suppose BN and CM be the bisectors of ∠ ABC and ∠ ACB, respectively intersect AC and AB at N and M, respectively.
NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.5 q2
Since, O lies on the bisector BN of ∠ ABC, so O will be equidistant from BA and BC. Again, O lies on the bisector CM of ∠ ACB.
So, O will be equidistant from CA and BC. Thus, O will be equidistant from AB, BC and CA Hence, O be a circumcentre of ∆ABC.

Question 3.
In a huge park, people are concentrated at three points (see figure)
A: where these are different slides and swings for children.
B: near which a man-made lake is situated.
C: which is near to a large parking and exist.
Where should an ice-cream parlor be set? up so that maximum number of persons can approach it?
[Hint The parlour should be equidistant from A, B and C.]
NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.5 q3
Solution:
The ice-cream parlor should be equidistant from A B and C for which the point of intersection of perpendicular bisectors of AB, BC, and CA should be situated.
So, O is the required point which is equidistant from A B and C.

Question 4.
Complete the hexagonal and star shaped Rangolies [see Fig. (i) and (ii)] by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles?
NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.5 q4
Solution:
We first divide the hexagon into six equilateral triangles of side 5cm as follow.
NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.5 q4.1
We take one triangle from six equilateral triangles as shown above and make as many equilateral triangles of one side 1 cm as shown in the figure.
NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.5 q4.2
The number of equilateral triangles of side 1 cm = 1 + 3 + 5 + 7 + 9 = 25
So, the total number of triangles in the hexagon = 6x 25 = 150
To find the number of triangles in the Fig. (ii), we adopt the same procedure.
So, the number of triangles in the Fig. (ii) = 12 x 25 = 30Q Hence, Fig. (ii) has more triangles.

Triangles Class 9 Extra Questions Maths Chapter 7

Extra Questions for Class 9 Maths Chapter 7 Triangles

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