Mathematics NCERT Grade 9, Chapter 7: Triangles- Students have already studied about triangles and its various properties. A closed figure formed by three intersecting lines is called a triangle. In this chapter, students will study in detail about the congruence of trianglesrules of congruence, some more properties of triangles and inequalities in a triangle.
Through first section one can recall congruence.
• Two figures are congruent if they are of the same shape and of the same size.
• Two circles of the same radii are congruent.
• Two squares of the same sides are congruent.
• Two triangles are congruent if their corresponding parts are congruent.
After that, students will learn about the criteria for the congruence of triangles. Some congruence rules are discussed in this section.

SAS(Side-Angle-Side) Congruence Rule: Two triangles are congruent if two sides and the included angle of one triangle are equal to the sides and the included angle of the other triangle.

ASA(Angle-Side-Side) Congruence Rule: Two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of other triangle.

AAS Congruence Rule: Two triangles are congruent if any two pairs of angles and one pair of corresponding sides are equal.

Later on, Some properties of a triangle are discussed involving isosceles triangles. Theorems related to isosceles triangles involve angles opposite to equal sides and vice-versa.
Other than the given 3 congruence criteria, 2 more congruence rules are given:
• SSS(Side-Side-side) congruence rule: If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.
• RHS(Right angle-Hypotenuse-Side) congruence rule: If in two right triangles the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle, then the two triangles are congruent.​
The next section discusses inequalities in a triangle. Theorems discussed in the following section:
• If two sides of a triangle are unequal, the angle opposite to the longer side is larger (or greater).
• In any triangle, the side opposite to the larger (greater) angle is longer.
• The sum of any two sides of a triangle is greater than the third side.
Ample questions are given in the form of solved examples and exercise questions.
For the final conclusion, summary of the chapter is discussed.

#### Question 1:

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

∠CAB = ∠DAB (AB bisects ∠A)

AB = AB (Common)

∴ ΔABC ≅ ΔABD (By SAS congruence rule)

∴ BC = BD (By CPCT)

Therefore, BC and BD are of equal lengths.

##### Video Solution for triangles (Page: 118 , Q.No.: 1)

NCERT Solution for Class 9 maths - triangles 118 , Question 1

#### Question 2:

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. In ΔABD and ΔBAC,

∠DAB = ∠CBA (Given)

AB = BA (Common)

∴ ΔABD ≅ ΔBAC (By SAS congruence rule)

∴ BD = AC (By CPCT)

And, ∠ABD = ∠BAC (By CPCT)

##### Video Solution for triangles (Page: 119 , Q.No.: 2)

NCERT Solution for Class 9 maths - triangles 119 , Question 2

#### Question 3:

AD and BC are equal perpendiculars to a line segment AB (See the given figure). Show that CD bisects AB. In ΔBOC and ΔAOD,

∠BOC = ∠AOD (Vertically opposite angles)

∠CBO = ∠DAO (Each 90º)

∴ ΔBOC ≅ ΔAOD (AAS congruence rule)

∴ BO = AO (By CPCT)

⇒ CD bisects AB.

##### Video Solution for triangles (Page: 119 , Q.No.: 3)

NCERT Solution for Class 9 maths - triangles 119 , Question 3

#### Question 4:

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

∠BAC = ∠DCA (Alternate interior angles, as p || q)

AC = CA (Common)

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

∴ ΔABC ≅ ΔCDA (By ASA congruence rule)

##### Video Solution for triangles (Page: 119 , Q.No.: 4)

NCERT Solution for Class 9 maths - triangles 119 , Question 4

#### 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 the given figure). Show that:

(i) ΔAPB ≅ ΔAQB

(ii) BP = BQ or B is equidistant from the arms of ∠A. In ΔAPB and ΔAQB,

∠APB = ∠AQB (Each 90º)

∠PAB = ∠QAB (l is the angle bisector of ∠A)

AB = AB (Common)

∴ ΔAPB ≅ ΔAQB (By AAS congruence rule)

∴ BP = BQ (By CPCT)

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

##### Video Solution for triangles (Page: 119 , Q.No.: 5)

NCERT Solution for Class 9 maths - triangles 119 , Question 5

#### Question 6:

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

∠BAD + ∠DAC = ∠EAC + ∠DAC

∠BAC = ∠DAE

In ΔBAC and ΔDAE,

∠BAC = ∠DAE (Proved above)

AC = AE (Given)

∴ ΔBAC ≅ ΔDAE (By SAS congruence rule)

∴ BC = DE (By CPCT)

##### Video Solution for triangles (Page: 120 , Q.No.: 6)

NCERT Solution for Class 9 maths - triangles 120 , Question 6

#### 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 the given figure). Show that

(i) ΔDAP ≅ ΔEBP It is given that ∠EPA = ∠DPB

⇒ ∠EPA + ∠DPE = ∠DPB + ∠DPE

⇒ ∠DPA = ∠EPB

In DAP and EBP,

∠DAP = ∠EBP (Given)

AP = BP (P is mid-point of AB)

∠DPA = ∠EPB (From above)

∴ ΔDAP ≅ ΔEBP (ASA congruence rule)

∴ AD = BE (By CPCT)

##### Video Solution for triangles (Page: 120 , Q.No.: 7)

NCERT Solution for Class 9 maths - triangles 120 , Question 7

#### 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 the given figure). Show that:

(i) ΔAMC ≅ ΔBMD

(ii) ∠DBC is a right angle.

(iii) ΔDBC ≅ ΔACB

(iv) CM = AB (i) In ΔAMC and ΔBMD,

AM = BM (M is the mid-point of AB)

∠AMC = ∠BMD (Vertically opposite angles)

CM = DM (Given)

∴ ΔAMC ≅ ΔBMD (By SAS congruence rule)

∴ 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 || AC

⇒ ∠DBC + ∠ACB = 180º (Co-interior angles)

⇒ ∠DBC + 90º = 180º

⇒ ∠DBC = 90º

(iii) In ΔDBC and ΔACB,

∠DBC = ∠ACB (Each 90 )

BC = CB (Common)

∴ ΔDBC ≅ ΔACB (SAS congruence rule)

(iv) ΔDBC ≅ ΔACB

∴ AB = DC (By CPCT)

⇒ AB = 2 CM

∴ CM = AB

##### Video Solution for triangles (Page: 120 , Q.No.: 8)

NCERT Solution for Class 9 maths - triangles 120 , Question 8

#### Question 1:

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 (i) It is given that in triangle ABC, AB = AC

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

⇒ ∠OCB = ∠OBC

⇒ OB = OC (Sides opposite to equal angles of a triangle are also equal)

(ii) In ΔOAB and ΔOAC,

AO =AO (Common)

AB = AC (Given)

OB = OC (Proved above)

Therefore, ΔOAB ≅ ΔOAC (By SSS congruence rule)

⇒ ∠BAO = ∠CAO (CPCT)

⇒ AO bisects ∠A.

##### Video Solution for triangles (Page: 123 , Q.No.: 1)

NCERT Solution for Class 9 maths - triangles 123 , Question 1

#### Question 2:

In ΔABC, AD is the perpendicular bisector of BC (see the given figure). Show that ΔABC is an isosceles triangle in which AB = AC. CD = BD (AD is the perpendicular bisector of BC)

∴AB = AC (By CPCT)

Therefore, ABC is an isosceles triangle in which AB = AC.

##### Video Solution for triangles (Page: 123 , Q.No.: 2)

NCERT Solution for Class 9 maths - triangles 123 , Question 2

#### Question 3:

ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see the given figure). Show that these altitudes are equal. In ΔAEB and ΔAFC,

∠AEB and ∠AFC (Each 90º)

∠A = ∠A (Common angle)

AB = AC (Given)

∴ ΔAEB ≅ ΔAFC (By AAS congruence rule)

⇒ BE = CF (By CPCT)

##### Video Solution for triangles (Page: 124 , Q.No.: 3)

NCERT Solution for Class 9 maths - triangles 124 , Question 3

#### Question 4:

ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see the given figure). Show that

(i) ABE ≅ ACF

(ii) AB = AC, i.e., ABC is an isosceles triangle. (i) In ΔABE and ΔACF,

∠AEB = ∠AFC (Each 90º)

∠A = ∠A (Common angle)

BE = CF (Given)

∴ ΔABE ≅ ΔACF (By AAS congruence rule)

(ii) It has already been proved that

ΔABE ≅ ΔACF

⇒ AB = AC (By CPCT)

##### Video Solution for triangles (Page: 124 , Q.No.: 4)

NCERT Solution for Class 9 maths - triangles 124 , Question 4

#### Question 5:

ABC and DBC are two isosceles triangles on the same base BC (see the given figure). Show that ∠ABD = ∠ACD.  In ΔABD and ΔACD,

AB = AC (Given)

BD = CD (Given)

∴ ΔABD ΔACD (By SSS congruence rule)

⇒ ∠ABD = ∠ACD (By CPCT)

##### Video Solution for triangles (Page: 124 , Q.No.: 5)

NCERT Solution for Class 9 maths - triangles 124 , Question 5

#### Question 6:

ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see the given figure). Show that ∠BCD is a right angle. In ΔABC,

AB = AC (Given)

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

In ΔACD,

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

In ΔBCD,

∠ABC + ∠BCD + ∠ADC = 180º (Angle sum property of a triangle)

⇒ ∠ACB + ∠ACB +∠ACD + ∠ACD = 180º

⇒ 2(∠ACB + ∠ACD) = 180º

2(∠BCD) = 180º

⇒ ∠BCD = 90º

##### Video Solution for triangles (Page: 124 , Q.No.: 6)

NCERT Solution for Class 9 maths - triangles 124 , Question 6

#### Question 7:

ABC is a right angled triangle in which ∠A = 90º and AB = AC. Find ∠B and ∠C. It is given that

AB = AC

⇒ ∠C = ∠B (Angles opposite to equal sides are also equal)

In ΔABC,

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

⇒ 90º + ∠B + ∠C = 180º

⇒ 90º + ∠B + ∠B = 180º

⇒ 2 ∠B = 90º

⇒ ∠B = 45º

∴ ∠B = ∠C = 45º

##### Video Solution for triangles (Page: 124 , Q.No.: 7)

NCERT Solution for Class 9 maths - triangles 124 , Question 7

#### Question 8:

Show that the angles of an equilateral triangle are 60º each. Let us consider that ABC is an equilateral triangle.

Therefore, AB = BC = AC

AB = AC

⇒ ∠C = ∠B (Angles opposite to equal sides of a triangle are equal)

Also,

AC = BC

⇒ ∠B = ∠A (Angles opposite to equal sides of a triangle are equal)

Therefore, we obtain

∠A = ∠B = ∠C

In ΔABC,

∠A + ∠B + ∠C = 180°

⇒ ∠A + ∠A + ∠A = 180°

⇒ 3∠A = 180°

⇒ ∠A = 60°

⇒ ∠A = ∠B = ∠C = 60°

Hence, in an equilateral triangle, all interior angles are of measure 60º.

##### Video Solution for triangles (Page: 124 , Q.No.: 8)

NCERT Solution for Class 9 maths - triangles 124 , Question 8

#### Question 1:

ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see the given figure). If AD is extended to intersect BC at P, show that

(i) ΔABD ≅ ΔACD

(ii) ΔABP ≅ ΔACP

(iii) AP bisects ∠A as well as ∠D.

(iv) AP is the perpendicular bisector of BC. (i) In ΔABD and ΔACD,

AB = AC (Given)

BD = CD (Given)

∴ ΔABD ≅ ΔACD (By SSS congruence rule)

⇒ ∠BAP = ∠CAP …. (1)

(ii) In ΔABP and ΔACP,

AB = AC (Given)

∠BAP = ∠CAP [From equation (1)]

AP = AP (Common)

∴ ΔABP ≅ ΔACP (By SAS congruence rule)

⇒ BP = CP (By CPCT) … (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)]

∴ ΔBDP ≅ ΔCDP (By S.S.S. Congruence rule)

⇒ ∠BDP = ∠CDP (By CPCT) … (3)

Hence, AP bisects ∠D.

(iv) ΔBDP ≅ ΔCDP

∴ ∠BPD = ∠CPD (By CPCT) …. (4)

∠BPD + ∠CPD = 180 (Linear pair angles)

∠BPD + ∠BPD = 180 2∠BPD = 180 [From equation (4)]

∠BPD = 90 … (5)

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

##### Video Solution for triangles (Page: 128 , Q.No.: 1)

NCERT Solution for Class 9 maths - triangles 128 , Question 1

#### Question 2:

AD is an altitude of an isosceles triangles ABC in which AB = AC. Show that AB = AC (Given)

⇒ BD = CD (By CPCT)

(ii) Also, by CPCT,

##### Video Solution for triangles (Page: 128 , Q.No.: 2)

NCERT Solution for Class 9 maths - triangles 128 , Question 2

#### Question 3:

Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ΔPQR (see the given figure). Show that:

(i) ΔABM ≅ ΔPQN

(ii) ΔABC ≅ ΔPQR (i) In ΔABC, AM is the median to BC.

∴ BM = BC

In ΔPQR, PN is the median to QR.

∴ QN = QR

However, BC = QR BC = QR

⇒ BM = QN … (1)

In ΔABM and ΔPQN,

AB = PQ (Given)

BM = QN [From equation (1)]

AM = PN (Given)

∴ ΔABM ≅ ΔPQN (SSS congruence rule)

∠ABM = ∠PQN (By CPCT)

∠ABC = ∠PQR … (2)

(ii) In ΔABC and ΔPQR,

AB = PQ (Given)

∠ABC = ∠PQR [From equation (2)]

BC = QR (Given)

⇒ ΔABC ≅ ΔPQR (By SAS congruence rule)

##### Video Solution for triangles (Page: 128 , Q.No.: 3)

NCERT Solution for Class 9 maths - triangles 128 , Question 3

#### Question 4:

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

∠BEC = ∠CFB (Each 90°)

BC = CB (Common)

BE = CF (Given)

∴ ΔBEC ≅ ΔCFB (By RHS congruency)

⇒ ∠BCE = ∠CBF (By CPCT)

∴ AB = AC (Sides opposite to equal angles of a triangle are equal)

Hence, ΔABC is isosceles.

##### Video Solution for triangles (Page: 128 , Q.No.: 4)

NCERT Solution for Class 9 maths - triangles 128 , Question 4

#### Question 5:

ABC is an isosceles triangle with AB = AC. Drawn AP ⊥ BC to show that ∠B = ∠C. In ΔAPB and ΔAPC,

∠APB = ∠APC (Each 90º)

AB =AC (Given)

AP = AP (Common)

∴ ΔAPB ≅ ΔAPC (Using RHS congruence rule)

⇒ ∠B = ∠C (By using CPCT)

##### Video Solution for triangles (Page: 128 , Q.No.: 5)

NCERT Solution for Class 9 maths - triangles 128 , Question 5

#### Question 1:

Show that in a right angled triangle, the hypotenuse is the longest side. Let us consider a right-angled triangle ABC, right-angled at B.

In ΔABC,

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

∠A + 90º + ∠C = 180°

∠A + ∠C = 90°

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

∴ ∠B is the largest angle in ΔABC.

⇒ ∠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 ΔABC. Therefore, hypotenuse is the longest side in a right-angled triangle.

##### Video Solution for triangles (Page: 132 , Q.No.: 1)

NCERT Solution for Class 9 maths - triangles 132 , Question 1

#### Question 2:

In the given figure sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB. In the given figure,

∠ABC + ∠PBC = 180° (Linear pair)

⇒ ∠ABC = 180° − ∠PBC ... (1)

Also,

∠ACB + ∠QCB = 180°

∠ACB = 180° − ∠QCB … (2)

As ∠PBC < ∠QCB,

⇒ 180º − ∠PBC > 180º − ∠QCB

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

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

##### Video Solution for triangles (Page: 132 , Q.No.: 2)

NCERT Solution for Class 9 maths - triangles 132 , Question 2

#### Question 3:

In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC. In ΔAOB,

∠B < ∠A

⇒ AO < BO (Side opposite to smaller angle is smaller) ... (1)

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

##### Video Solution for triangles (Page: 132 , Q.No.: 3)

NCERT Solution for Class 9 maths - triangles 132 , Question 3

#### Question 4:

AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see the given figure). Show that ∠A > ∠C and ∠B > ∠D.  Let us join AC.

In ΔABC,

AB < BC (AB is the smallest side of quadrilateral ABCD)

∴ ∠2 < ∠1 (Angle opposite to the smaller side is smaller) ... (1)

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

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

∠2 + ∠4 < ∠1 + ∠3

⇒ ∠C < ∠A

⇒ ∠A > ∠C

Let us join BD. In ΔABD,

∴ ∠8 < ∠5 (Angle opposite to the smaller side is smaller) ... (3)

In ΔBDC,

BC < CD (CD is the largest side of quadrilateral 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

##### Video Solution for triangles (Page: 132 , Q.No.: 4)

NCERT Solution for Class 9 maths - triangles 132 , Question 4

#### Question 5:

In the given figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR >∠PSQ. As PR > PQ,

∴ ∠PQR > ∠PRQ (Angle opposite to larger side is larger) ... (1)

PS is the bisector of ∠QPR.

∴∠QPS = ∠RPS ... (2)

∠PSR is the exterior angle of ΔPQS.

∴ ∠PSR = ∠PQR + ∠QPS ... (3)

∠PSQ is the exterior angle of ΔPRS.

∴ ∠PSQ = ∠PRQ + ∠RPS ... (4)

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

∠PQR + ∠QPS > ∠PRQ + ∠RPS

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

##### Video Solution for triangles (Page: 132 , Q.No.: 5)

NCERT Solution for Class 9 maths - triangles 132 , Question 5

#### Question 6:

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

∠N = 90º

∠P + ∠N + ∠M = 180º (Angle sum property of a triangle)

∠P + ∠M = 90º

Clearly, ∠M is an acute angle.

∴ ∠M < ∠N

⇒ PN < PM (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.

##### Video Solution for triangles (Page: 133 , Q.No.: 6)

NCERT Solution for Class 9 maths - triangles 133 , Question 6

#### Question 1:

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

Circumcentre of a triangle is always equidistant from all the vertices of that triangle. Circumcentre is the point where perpendicular bisectors of all the sides of the triangle meet together. In ΔABC, we can find the circumcentre 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 ΔABC.

##### Video Solution for triangles (Page: 133 , Q.No.: 1)

NCERT Solution for Class 9 maths - triangles 133 , Question 1

#### Question 2:

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

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 ΔABC, 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 ΔABC.

##### Video Solution for triangles (Page: 133 , Q.No.: 2)

NCERT Solution for Class 9 maths - triangles 133 , Question 2

#### Question 3:

In a huge park people are concentrated at three points (see the given 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 parlor should be equidistant from A, B and C)

Maximum number of persons can approach the ice-cream parlour if it is equidistant from A, B and C. Now, A, B and C form a triangle. In a triangle, the circumcentre is the only point that is equidistant from its vertices. So, the ice-cream parlour should be set up at the circumcentre O of ΔABC. In this situation, maximum number of persons can approach it. We can find circumcentre O of this triangle by drawing perpendicular bisectors of the sides of this triangle.

##### Video Solution for triangles (Page: 133 , Q.No.: 3)

NCERT Solution for Class 9 maths - triangles 133 , Question 3

#### Question 4:

Complete the hexagonal and star shaped rangolies (see the given figures) 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? It can be observed that hexagonal-shaped rangoli has 6 equilateral triangles in it. Area of ΔOAB  Area of hexagonal-shaped rangoli    Star-shaped rangoli has 12 equilateral triangles of side 5 cm in it. Area of star-shaped rangoli =   Therefore, star-shaped rangoli has more equilateral triangles in it.

##### Video Solution for triangles (Page: 133 , Q.No.: 4)

NCERT Solution for Class 9 maths - triangles 133 , Question 4

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