NCERT Solutions for Class 9 Maths Chapter 7 Triangles

NCERT
SOLUTIONS
CLASS - 9th

aglase .co

Book : Mathematics Ncert Solutions | Chapter-7 Maths

Class : 9th
Subject : Maths
Chapter : 7
Chapter Name : Triangles

Exercise 7.1

Q1 In quadrilateral ACBD, AC = AD and AB bisects ∠ A (see Fig). Show that ∆ ABC ≅ ∆ ABD. What
can you say about BC and BD?

Answer. In ∆ ABC and ∆ ABD,
AC = AD (Given)
∠CAB = ∠ DAB (AB bisects ∠A)
AB = AB (Common)
Therefore, ∆ ABC ≅ ∆ ABD( By SAS congruence rule)
BC = BD (By CPCT)
Therefore, BC and BD are Of equal lengths.

Page : 118 , Block Name : Exercise 7.1

Q2 ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see Fig). Prove that
(i) ∆ ABD ≅ ∆ BAC
(ii) BD = AC
(iii) ∠ ABD = ∠ BAC.

Answer. In ∠ABD and ∠BAC,
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Book : Mathematics Ncert Solutions | Chapter-7 Maths

AD = BC (Given)
∠DAB = ∠CBA(Given)
AB = BA (Common)
∴ △ABD = ΔBAC(By SAS congruence rule)

∴ BD = AC(ByCPCT)

And, ∠ABD = ∠BAC (By CPCT)

Page : 119 , Block Name : Exercise 7.1

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

Answer. In ∆ BOC and ∆ AOD,
∠BOC = ∠AOD( Vertically opposite angles )

∘
∠CBO = ∠DAO ( Each 90 )

BC = AD( Given )

∴ ΔBOC = △AOD (AAS congruence rule)

∴ BO = AO(ByCPCT)

⇒ CD bisects AB

Page : 119 , Block Name : Exercise 7.1

Q4 l and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig). Show that
∆ ABC ≅ ∆ CDA.

∠BAC = ∠DCA( Alternate interior angles, as p∥q)
Answer. In ∆ ABC and ∆ CDA,
AC = CA( Common )

∠BCA = ∠DAC( Alternate interior angles, as l∥m)

△ABC = ΔCDA(By ASA congruence rule)

Page : 119 , Block Name : Exercise 7.1

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 Figure). Show that:
(i) ∆ APB ≅ ∆ AQB

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

In △AP B and △AQB .
Answer. ∘
∠AP B = ∠AQB (Each90 )

∠P AB = ∠QAB(1 is the angle bisector of ∠A)

AB = AB( Common )

∴ △APB ≅ △AQB(By AAS congruence rule)

∴ BP = BQ(ByCPCT)

Or, it can be said that B is equidistant from the arms of ∠ A.

Page : 119 , Block Name : Exercise 7.1

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

Answer. It is given that ∠ BAD = ∠ EAC
∠BAD + ∠DAC = ∠ EAC + ∠DAC
∠BAC = ∠DAE

In ΔBAC and ΔDAE,

AB = AD(G iven )

∠BAC = ∠DAE (Proved above)
AC = AE (Given)
Therefore, ∆BAC ≅ ∆DAE (By SAS congruence rule)
Therefore, BC = DE (By CPCT)

Page : 120 , Block Name : Exercise 7.1

Q7 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) ∆ DAP ≅ ∆ EBP
(ii) AD = BE

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Answer. It is given that ∠EPA = ∠DPB
∠EPA + ∠DPE = ∠DPB + ∠DPE
Therefore, ∠DPA = ∠EPB
In ∠DAP and ∠EBP,
∠DAP = ∠EPB(Given)
AP = BP (P is mid-point of AB)
∠DPA = ∠EPB (From above)
Therefore, ∆ DAP ≅ ∆ EBP (ASA congruence rule)
Therefore, AD - BE (By CPCT)

Page : 120 , Block Name : Exercise 7.1

Q8 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 = 1/2 AB

Answer. (i) In ∆ AMC and ∆ BMD,
AM = BM (M is the mid-point Of AB)
∠AMC = ∠BMD (Vertically opposite angles)
CM = DM (Given)
Therefore, ∆ AMC ≅ ∆ BMD (By SAS congruence rule)
Therefore AC = BD (By CPCT)
And, ∠ACM = ∠BDM (By CPCT)
(ii) ∠ACM = ∠BDM
However, ∠ACM and ∠BDM are alternate interior angles.
Since alternate angles are equal,
It can be said that DB II AC
∠DBC + ∠ACB = 180° (Co-interior angles)
∠DBC + 90° = 180°
∠DBC = 90°
(iii) In ∆ DBC and ∆ ACB,
DB = AC (Already proved)
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∠DBC = ∠ACB(Each 90°)
BC = CB (Common)
Therefore, ∆ DBC ≅ ∆ ACB (SAS congruence rule)
(iv) ∆DBC ≅ ∆ACB
AB = DC (By CPCT)
AB = 2 CM
1
∴ CM = AB
2

Page : 120 , Block Name : Exercise 7.1

Exercise 7.2

Q1. In an isosceles triangle ABC, with AB = AC, the bisectors of ∠ B and ∠ C intersect each other at
O. Join A to O. Show that :
(i) OB = OC
(ii) AO bisects ∠ A

Answer.

(i) It is given that in triangle ABC, AB = AC
∠ACB = ∠ABC (Angles opposite to equal sides Of a triangle are equal)
1 1
∠ACB = ∠ABC
2 2

∠OCB =∠OBC
Therefore, OB = OC (Sides opposite to equal angles of a triangle are also equal)
(ii) In ∆OAB and ∆OAC,
AO = AO(Common)
AB = AC (Given)
OB = OC (Proved above)
Therefore, ∆OAB ≅ ∆OAC (By SSS congruence rule)
∠BAO = ∠CAO (CPCT)
Therefore, AO bisects ∠A.

Page : 123 , Block Name : Exercise 7.2

Q2 In ∆ ABC, AD is the perpendicular bisector of BC (see Fig). Show that ∆ ABC is an isosceles
triangle in which AB = AC.

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Answer. In ∆ADC and ∆ADB,
AD = AD (Common)
∠ADC = ∠ADB (Each 90°)
CD = BD (AD is the perpendicular bisector of BC)
Therefore, ∆ADC ≅ ∆ADB (By SAS congruence rule)
AB = AC (By CPCT)
Therefore, ABC is an isosceles triangle in which AB = AC

Page : 123 , Block Name : Exercise 7.2

Q3 ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB
respectively (see Fig). Show that these altitudes are equal.

Answer. In ∆AEB and ∆AFC,
∠AEB and ∠AFC (Each 90°)
∠A = ∠A (Common angle)
AB = AC (Given)
Therefore, ∆AEB ≅ ∆AFC (By AAS congruence rule)
BE = CF (By CPCT)

Page : 124 , Block Name : Exercise 7.2

Q4 ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig). Show that
(i) ∆ ABE ≅ ∆ ACF
(ii) AB = AC, i.e., ABC is an isosceles triangle.

Answer. (i) ∆ ABE and ∆ ACF,

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∘
∠ABE and ∠ACF (Each90 )

∠A = ∠A( Common angle )

BE = CF (G iven )

Therefore, ∆ ABE ≅ ∆ ACF(By AAS congruence rule)
(ii) It has already been proved that
∆ ABE ≅ ∆ ACF
Therefore, AB = AC(By CPCT)

Page : 124 , Block Name : Exercise 7.2

Q5 ABC and DBC are two isosceles triangles on the same base BC (see Fig. 7.33). Show that ∠ ABD =
∠ ACD

Answer.

Let us join AD.
In ∆ABD and ∆ACD,
AB = AC (Given)
BD = CD (Given)
AD = AD (Common side)
Therefore, ∆ABD ≅ ∆ACD (By SSS congruence rule)
∠ABD = ∠ACD (By CPCT)

Page : 124 , Block Name : Exercise 7.2

Q6 ∆ABC is an isosceles triangle in which AB = AC.Side BA is produced to D such that AD = AB (see
Fig). Show that ∠ BCD is a right angle.

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Answer. In ∆ABC,
AB = AC( Given)

∴ ∠ACB = ∠ABC (Angles opposite to equal sides of a triangle are also equal)

In ∆ACD,
AC = AD( Given)

∴ ∠ADC = ∠ACD (Angles opposite to equal sides of a triangle are also equal)

In ∆BCD,
∠ABC + ∠BCD + ∠ADC = 1800 (Angle sum property of a triangle)
∠ACB + ∠ACB +∠ACD + ∠ACD = 180°
2(∠ACB + ∠ACD) = 180°
2(∠BCD) = 180°
Therefore, ∠BCD = 90°

Page : 124 , Block Name : Exercise 7.2

Q7 ABC is a right angled triangle in which ∠ A = 90° and AB = AC. Find ∠ B and ∠ C.

Answer.

It is given that
∴ ∠C = ∠B( Angles opposite to equal sides are also equal)

In ΔABC .

AB = AC ∠A + ∠B + ∠C = 180 (Angle sum property of a triangle)
∘

∘ ∘
90 + ∠B + ∠C = 180
∘ ∘
90 + ∠B + ∠B = 180
∘
2∠B = 90
∘
∠B = 45
∘
∴ ∠B = ∠C = 45

Page : 124 , Block Name : Exercise 7.2

Q8 Show that the angles of an equilateral triangle are 60° each.

Answer.

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Book : Mathematics Ncert Solutions | Chapter-7 Maths

Let us consider that ABC is an equilateral triangle.
Therefore, AB = BC + AC
AB = AC
Therefore, ∠C = ∠B (Angles opposite to equal sides of a triangle are equal)
Also,
AB = AC
∠B = ∠A (Angles opposite to equal sides of a triangle are equal)
Therefore, we obtain
∠A = ∠B = ∠C
In △ABC,
∘
∠A + ∠B + ∠C = 180
∘
∠A + ∠A + ∠A = 180
∘
∠A = 180
∘
∠A = 60
∘
∴ ∠A = ∠B = ∠C = 60

Hence, in a equilateral triangle, all interior angles are of measure 60°.

Page : 124 , Block Name : Exercise 7.2

Exercise 7.3

Q1 ∆ ABC and ∆ DBC are two isosceles triangles on the same base BC and vertices A and D are on the
same side of BC (see Figure). If AD is extended to intersect BC at P, show that
(i) ∆ ABD ≅ ∆ ACD
(ii) ∆ ABP≅ ∆ ACP
(iii) AP bisects ∠ A as well as ∠ D.
(iv) AP is the perpendicular bisector of BC

Answer. (i) In △ABD and △ACD,

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Book : Mathematics Ncert Solutions | Chapter-7 Maths

AB = AC(G wen )

BD = CD( Given )

AD = AD (Common)

△ABD ≅ △ACD(By SSS congruence rule)

∠BAD = ∠CAD(ByCP CT )

∠BAP = ∠CAP ….(1)
(ii) In △ABP and △ACP,
AB = AC( Given )

∠BAP = ∠CAP [ From equation (1)]

AP = AP ( Common )

∴ △ABP ≅ △ACP ( By SAS congruence rule)
∴ BP = CP (ByCP CT ) ….(2)

(iii) From equation (1),
∠BAP = ∠CAP

Hence, AP bisects ∠A,
In ΔBDP and ΔCDP
BD = CD( Given )

DP = DP( Common )

BP = CP[ From equation (2)]

∴ △BDP ≅ ΔCDP (By SSS congruence rule)
△∠BDP = ∠CDP (ByCP CT ) …(3)
Hence, AP bisects ∠D
(iv) △BDP ≅ ΔCDP
∴ ∠BP D = ∠CP D(ByCP CT ) ….(4)
∘
∠BP D + ∠CP D = 180 ( Linear pair angles)
∘
∠BP D + ∠BP D = 180
∘
2∠8P D = 180 [ From Equation (4)]
∘
∠BP D = 90 … (5)

FRom equations (2) and (5), it can be said that AP is the perpendicular bisector of BC.

Page : 128 , Block Name : Exercise 7.3

Q2 AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC
(ii) AD bisects ∠ A.

Answer.

(i) In ΔBAD and △CAD,
∘
∠ADB = ∠ADC ( Each 90 as AD is an altitude)

AB = AC(G iven )

AD = AD( Common )

∴ △BAD ≅ △CAD (By RHS Congruence rule)
△BD = CD (By CPCT)

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Hence, AD bisects BC.
(ii) Also, by CPCT,
∠BAD = ∠CAD

Hence, AD bisects ∠A

Page : 128 , Block Name : Exercise 7.3

Q3 Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and
QR and median PN of ∆ PQR (see Figure). Show that:
(i) ∆ ABM ≅ ∆ PQN
(ii) ∆ ABC ≅ ∆ PQR

Answer. (i) In △ABC, AM is the median to BC.
1
∴ BM = BC
2

In ΔPQR, PN is the median to QR.
1
∴ QN = QR
2

However, BC = QR
1 1
∴ BC = QR
2 2

∴ BM = QN … (1)

In △ABM and ΔP QN ,
AB = PQ(Given)
BM = QN[ From Equation (1)]

AM = PN( Given )

ABM ≅ ΔPQN (By SSS congruence rule …(2)
∠ABM = ∠PQN(ByCPCT)

∠ABC = ∠PQR

(iii) In △ABC and ΔP QR,
AB = PQ( Given )

∠ABC = ∠PQR[From Equation (2)]

BC = QR( Given )

∴ △ABC ≅ ΔP QR (By SAS congruence rule)

Page : 128 , Block Name : Exercise 7.3

Q4 BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the
triangle ABC is isosceles.

Answer.

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In △BEC and ΔCF B ,
∘
∠BEC = ∠CF B (Each90 )

BC = CB(Common)

BE = CF(Given)

∴ △BEC ≅ ΔCF B (By RHS congruency)
∴ ∠BCE = ∠CBF (ByCP CT )

∴ AB = AC (Sides opposite to equal angles of a triangle are equal)
Hence, △ABC is isosceles.

Page : 128 , Block Name : Exercise 7.3

Q5 ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠ B = ∠ C.

Answer.

In △AP B and △AP C ,
∘
∠AP B = ∠AP C (Each90 )

AB = AC(Given)

AP = AP (Common)

∴ △AP B ≅ △AP C (Using RHS congruence rule)
∴ ∠B = ∠C (By using CPCT)

Page : 128 , Block Name : Exercise 7.3

Exercise 7.4

Q1 Show that in a right angled triangle, the hypotenuse is the longest side.

Answer.

Let us consider a right-angled triangle ABC, right-angled at B.
In △ABC ,

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∘
∠A + ∠B + ∠C = 180 (Angle sum property of a triangle)

∘ ∘
∠A + 90 + ∠C = 180
∘
∠A + ∠C = 90

Hence, the other two angles have to be acute(i.e., less than 90 ).
∘

∠B is the largest angle in △ABC.

∠B > ∠A and ∠B > ∠C

AC > BC and AC > AB

[In any triangle, the side opposite to the larger (greater) angle is longer.]
Therefore, AC is the largest side in ΔABC
However, AC is the hypotenuse of △ABC. Therefore, hypotenuse is the longest side in a right-angled
triangle.

Page : 132 , Block Name : Exercise 7.4

Q2 In Figure, sides AB and AC of ∆ ABC are extended to points P and Q respectively. Also, ∠ PBC <
∠ QCB. Show that AC > AB.

Answer. In the given figure,
∘
∠ABC + ∠P BC = 180 ( Linear pair)
∘
∠ABC = 180 − ∠P BC … (1)

Also,
∘
∠ACB + ∠QCB = 180

∠ACB = 180
∘
− ∠QCB …(2)
As ∠P BC < ∠QCB
∘ ∘
180 − ∠P BC > 180 − ∠QCB

∠ABC > ∠ACB[ From Equations (1) and (2)]

AC > AB( Side opposite to the larger angle is larger.)

Hence proved AC > AB

Page : 132 , Block Name : Exercise 7.4

Q3 In Figure, ∠ B < ∠ A and ∠ C < ∠ D. Show that AD < BC.

Answer. In △AOB

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∠B < ∠A
…(1)
AO < BO( Side opposite to smaller angle is smaller)

In ΔCOD ,

∠C < ∠D

OD < OC( Side opposite to smaller angle is smaller) …(2)
On adding Equations (1) and (2), we obtain
AO + OD < BO + OC

AD < BC , proved

Page : 132 , Block Name : Exercise 7.4

Q4 AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Figure).
Show that ∠ A > ∠ C and ∠ B > ∠ D.

Answer.

In △ABC
AB < BC (AB is the smallest side of quadrilateral ABCD)
∠2 < ∠1 ( Angle opposite to the smaller side is smaller)...(1)

In △ADC
AD < CD (CD is the largest side of quadrilateral ABCD)

∠4 < ∠3(Angle opposite to the smaller side i smaller)...(2)

On adding equations(1) and (2),we obtain

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∠2 + ∠4 < ∠1 + ∠3

∠C < ∠A

∠A > ∠C

Let us join BD.

In △ABD
AB < AD(AB is the smallest side of quadriateral ABCD)

∠8 < ∠5(Angle opposite to the smaller side is smaller)...(3)
In ΔBDC
BC < CD(CD is the largest side of quadriateral ABCD)

∠7 < ∠6 (Angle opposite to the smaller side is smaller)...(4)
On adding equations (3) and (4),we obtain
∠8 + ∠7 < ∠5 + ∠6

∠D < ∠B

∠B > ∠D( Hence, proved )

Page : 132 , Block Name : Exercise 7.4

Q5 In Figure, PR > PQ and PS bisects ∠ QPR. Prove that ∠ PSR > ∠ PSQ.

Answer. As P R > P Q
∠P QR > ∠P RQ( Angle opposite to larger side is larger) ...(1)
PS is the bisector of ∠QP R.
∠QP S = ∠RP S … (2)

∠P SR is the exterior angle of ΔP QS

∠P SR = ∠P QR + ∠QP S − (3)

∠P SQ is the exterior angle of ΔP RS .
…(4)
∠P SQ = ∠P RQ + ∠RP S

Adding Equations (1) and (2), we obtain

∠P QR + ∠QP S > ∠P RQ + ∠RP S

∠PSR > ∠PSQ[ Using the values of Equations (3) and (4)]

Page : 132 , Block Name : Exercise 7.4
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Book : Mathematics Ncert Solutions | Chapter-7 Maths

Q6 Show that of all line segments drawn from a given point not on it, the perpendicular line segment is
the shortest.

Answer.

Let us take a line l and from point P (i.e., not on line l), draw two line segments PN and PM. Let PN be
perpendicular to line L and PM is drawn at some other angle.
In △PNM
∘
∠N = 90
∘
∠P + ∠N + ∠M = 180 ( Angle sum property of a triangle)

∘
∠P + ∠M = 90

Clearly, ∠M is an acute angle
∠M < AN

P N < P M ( side opposite to the smaller angle is smaller)

Similarly, by drawing different line segments from P to l, it can be proved that PN is smaller in
comparison to them.
Therefore, it can be observed that of all line segments drawn from a given point not on it, the
perpendicular line segment is the shortest.

Page : 133 , Block Name : Exercise 7.4

Exercise 7.5

Q1 ABC is a triangle. Locate a point in the interior of ∆ ABC which is equidistant from all the vertices
of ∆ ABC.

Answer. Circumference of a triangle is always equidistant from all the vertices of that triangle.
Circumference is the point where perpendicular bisectors of all the sides of the triangle meet together.

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In △ABC, we can find the circumference by drawing the perpendicular bisectors of sides AB, BC,
AND CA of this triangle. O is the point where these bisectors are meeting together. Therefore, O is the
point which is equidistant from all the vertices of △ABC.

Page : 133 , Block Name : Exercise 7.5

Q2 In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.

Answer. The point which is equidistant from all the sides of a triangle is called the incentre of the
triangle. Incentre of a triangle is the intersection point of the angle bisectors of the interior angles of that
triangle.

Here, in △ABC, we can find the incentre of this triangle by drawing the angle bisectors of the interior
angles of this triangle. I is the point where these angle bisectors are intersecting each other. Therefore, I
is the point equidistant from all the sides of △ABC

Page : 133 , Block Name : Exercise 7.5

Q3 In a huge park, people are concentrated at three points (see Figure):
A : where there 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 exit. Where should an ice cream parlour be set up so that
maximum number of persons can approach it?
(Hint : The parlour should be equidistant from A, B and C)

Answer. Maximum number of persons can approach the ice-cream parlour if it is equidistant from A, B
and C from a triangle. In a triangle, the circumcentre is the only point that is equidistant from its

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vertices. So, the ice-cream parlour should be set up at the circumcentre O of △ABC

In this situation, maximum number of persons can approach it. We can find circumcentre O of this
triangle by drawing perpendicular bisectors of the ised of this triangle.

Page : 133 , Block Name : Exercise 7.5

Q4 . Complete the hexagonal and star shaped Rangolies [see Figure (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?

Answer. It can be observed that hexagonal-shaped rangoli has 6 equilateral triangles in it.

√3 √3
2 2
Area of ΔOAB = ( side ) = (5)
4 4

√3 25√3
2
= (25) = cm
4 4

25√3 75√3
Area of hexagonal-shaped rangoli = 6 × 4
=
2
cm
2

√3 √3
Area of equilateral triangle having its sides as 1 cm = (1) = cm
4
2

4
2

Number of equilateral triangles of 1 cm side that can be filled In this hexagonal-shaped rangoli

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75√3

2
= = 150
√3

4

Star-shaped rangoli has 12 equilateral triangles of side 5 cm in it.

√3
Area of star-shaped rangoli = 12 × 4
× (5)
2
= 75√3

75√3
Number of equilateral triangles of 1 cm side that can be filled in this star-shaped rangoli = √3
= 300

4

Therefore, star-shaped rangoli has more equilateral triangles in it.

Page : 133 , Block Name : Exercise 7.5

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