**NCERT Solutions Class 9 Maths Chapter 7 Triangles** – Here are all the NCERT solutions for Class 9 Maths Chapter 7. This solution contains questions, answers, images, explanations of the complete Chapter 7 titled The Fun They Had of Maths taught in class 9. If you are a student of class 9 who is using NCERT Textbook to study Maths, then you must come across Chapter 7 Triangles. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 9 Maths Chapter 7 Triangles in one place.

## NCERT Solutions Class 9 Maths Chapter 7 Triangles

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For a better understanding of this chapter, you should also see summary of Chapter 7 Triangles , Maths, Class 9.

Class | 9 |

Subject | Maths |

Book | Mathematics |

Chapter Number | 7 |

Chapter Name |
Triangles |

### NCERT Solutions Class 9 Maths chapter 7 Triangles

Class 9, Maths chapter 7, Triangles solutions are given below in PDF format. You can view them online or download PDF file for future use.

### Triangles

Q.1:In quadrilateral ACBD, AC = AD and AB bisects ∠ A (see Fig). Show that ∆ ABC ≅ ∆ ABD. What can you say about BC and BD?

Ans :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.

Q.2:ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see Fig). Prove that (i) ∆ ABD ≅ ∆ BAC (ii) BD = AC (iii) ∠ ABD = ∠ BAC.

Ans :In ∠ABD and ∠BAC, AD = BC (Given) ∠DAB = ∠CBA(Given) AB = BA (Common) \(\begin{array}{l}{\therefore \triangle \mathrm{ABD}=\Delta \mathrm{BAC}(\mathrm{By} \text { SAS congruence rule) }} \\ {\therefore \mathrm{BD}=\mathrm{AC}(\mathrm{By} \mathrm{CPCT})}\end{array}\) And, ∠ABD = ∠BAC (By CPCT)

Q.3:AD and BC are equal perpendiculars to a line segment AB (see Fig.). Show that CD bisects AB.

Ans :In ∆ BOC and ∆ AOD, \(\begin{array}{l}{\angle \mathrm{BOC}=\angle \mathrm{AOD}(\text { Vertically opposite angles })} \\ {\angle \mathrm{CBO}=\angle \mathrm{DAO}\left(\text { Each } 90^{\circ}\right)}\end{array}\) \(\begin{array}{l}{\mathrm{BC}=\mathrm{AD}(\text { Given })} \\ {\therefore \Delta \mathrm{BOC}=\triangle \mathrm{AOD} \text { (AAS congruence rule) }} \\ {\therefore \mathrm{BO}=\mathrm{AO}(\mathrm{By} \mathrm{CPCT})} \\ {\Rightarrow \mathrm{CD} \text { bisects } \mathrm{AB}}\end{array}\)

Q.4:l and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig). Show that ∆ ABC ≅ ∆ CDA.

Ans :In ∆ ABC and ∆ CDA, \(\begin{array}{l}{\angle B A C=\angle D C A(\text { Alternate interior angles, as } p \| q)} \\ {A C=C A(\text { Common })}\end{array}\) \(\begin{array}{l}{\angle B C A=\angle D A C(\text { Alternate interior angles, as } l \| m)} \\ { \triangle A B C = \Delta C D A(B y \text { ASA congruence rule) }}\end{array}\)

Q.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) ∆ APB ≅ ∆ AQB (ii) BP = BQ or B is equidistant from the arms of ∠ A.

Ans :\(\begin{array}{l}{\text { In } \triangle A P B \text { and } \triangle A Q B \text { . }} \\ {\angle A P B=\angle A Q B\left(E a c h 90^{\circ}\right)}\end{array}\) \(\begin{array}{l}{\angle P A B=\angle Q A B(1 \text { is the angle bisector of } \angle A)} \\ {A B=A B(\text { Common })}\end{array}\) \(\begin{array}{l}{\therefore \triangle \mathrm{APB} \cong \triangle \mathrm{AQB}(\mathrm{By} \text { AAS congruence rule) }} \\ {\therefore \mathrm{BP}=\mathrm{BQ}(\mathrm{By} \mathrm{CPCT})}\end{array}\) Or, it can be said that B is equidistant from the arms of ∠ A.

Q.6:In Figure, AC = AE, AB = AD and ∠ BAD = ∠ EAC. Show that BC = DE.

Ans :It is given that ∠ BAD = ∠ EAC ∠BAD + ∠DAC = ∠ EAC + ∠DAC \(\begin{array}{l}{\angle B A C=\angle D A E} \\ {\text { In } \Delta B A C \text { and } \Delta D A E,} \\ {A B=A D(G \text { iven })}\end{array}\) ∠BAC = ∠DAE (Proved above) AC = AE (Given) Therefore, ∆BAC ≅ ∆DAE (By SAS congruence rule) Therefore, BC = DE (By CPCT)

Q.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) ∆ DAP ≅ ∆ EBP (ii) AD = BE

Ans :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)

Q.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 = 1/2 AB

Ans :(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) ∠DBC = ∠ACB(Each 90°) BC = CB (Common) Therefore, ∆ DBC ≅ ∆ ACB (SAS congruence rule) (iv) ∆DBC ≅ ∆ACB AB = DC (By CPCT) AB = 2 CM \(\therefore C M=\frac{1}{2} A B\)

Q.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

Ans :(i) It is given that in triangle ABC, AB = AC ∠ACB = ∠ABC (Angles opposite to equal sides Of a triangle are equal) \(\frac{1}{2} \angle A C B=\frac{1}{2} \angle A B C\) ∠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.

Q.2:In ∆ ABC, AD is the perpendicular bisector of BC (see Fig). Show that ∆ ABC is an isosceles triangle in which AB = AC.

Ans :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

Q.3: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.

Ans :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)

Q.4: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.

Ans :(i) ∆ ABE and ∆ ACF, \(\begin{array}{l}{\angle A B E \text { and } \angle A C F\left(E a c h 90^{\circ}\right)} \\ {\angle A=\angle A(\text { Common angle })} \\ {B E=C F(G \text { iven })}\end{array}\) Therefore, ∆ ABE ≅ ∆ ACF(By AAS congruence rule) (ii) It has already been proved that ∆ ABE ≅ ∆ ACF Therefore, AB = AC(By CPCT)

Q.5:ABC and DBC are two isosceles triangles on the same base BC (see Fig). Show that ∠ ABD = ∠ ACD

Ans :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)

Q.6:∆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.

Ans :In ∆ABC, \(\begin{array}{l}{\mathrm{AB}=\mathrm{AC}(\text { Given) }} \\ {\therefore \angle \mathrm{ACB}=\angle \mathrm{ABC} \text { (Angles opposite to equal sides of a triangle are also equal) }}\end{array}\) In ∆ACD, \(\begin{array}{l}{\mathrm{AC}=\mathrm{AD}(\text { Given) }} \\ {\therefore \angle \mathrm{ADC}=\angle \mathrm{ACD} \text { (Angles opposite to equal sides of a triangle are also equal) }}\end{array}\) 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°

Q.7:ABC is a right angled triangle in which ∠ A = 90° and AB = AC. Find ∠ B and ∠ C.

Ans :It is given that AB = AC\(\begin{array}{l}{\therefore \angle C=\angle B(\text { Angles opposite to equal sides are also equal) }} \\ {\text { In } \Delta A B C \text { . }} \\ {\angle A+\angle B+\angle C=180^{\circ} \text { (Angle sum property of a triangle) }} \\ {90^{\circ}+\angle B+\angle C=180^{\circ}} \\ {90^{\circ}+\angle B+\angle B=180^{\circ}}\end{array}\) \(\begin{array}{l}{2 \angle B=90^{\circ}} \\ {\angle B=45^{\circ}} \\ {\therefore \angle B=\angle C=45^{\circ}}\end{array}\)

Q.8:Show that the angles of an equilateral triangle are 60° each.

Ans :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 \(\begin{array}{l}{\text { In } \triangle A B C,} \\ {\angle A+\angle B+\angle C=180^{\circ}} \\ {\angle A+\angle A+\angle A=180^{\circ}} \\ {\angle A=180^{\circ}} \\ {\angle A=60^{\circ}} \\ {\therefore \angle A=\angle B=\angle C=60^{\circ}}\end{array}\) Hence, in a equilateral triangle, all interior angles are of measure 60°.

Q.1:∆ 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

Ans :(i) In \( \triangle A B D \text { and } \triangle A C D, \) \( \begin{array}{l}{A B=A C(G \text { wen })} \\ {B D=C D(\text { Given })} \\ {A D=A D \text { (Common) }}\end{array} \) \( \triangle A B D \cong \triangle A C D(B y \text { SSS congruence rule) } \) \( \angle B A D=\angle C A D(B y C P C T) \) \( \angle B A P=\angle C A P \) ….(1) (ii) In \( \triangle \mathrm{ABP} \text { and } \triangle \mathrm{ACP} , \) \( A B=A C(\text { Given }) \) \( \begin{array}{l}{\angle B A P=\angle C A P[\text { From equation }(1)]} \\ {A P=A P(\text { Common })}\end{array} \) \( \therefore \triangle \mathrm{ABP} \cong \triangle \mathrm{ACP} \) ( By SAS congruence rule) \( \therefore B P=C P(B y C P C T) \) ….(2) (iii) From equation (1), \( \angle B A P=\angle C A P \) Hence, AP bisects \( \angle A \), In \( \Delta B D P \text { and } \Delta C D P \) \( \begin{array}{l}{\mathrm{BD}=\mathrm{CD}(\text { Given })} \\ {\mathrm{DP}=\mathrm{DP}(\text { Common })} \\ {\mathrm{BP}=\mathrm{CP}[\text { From equation }(2)]}\end{array} \) \( \therefore \triangle B D P \cong \Delta C D P \) (By SSS congruence rule) \( \triangle \angle B D P=\angle C D P(B y C P C T) \) …(3) Hence, AP bisects \( \angle \mathrm{D} \) (iv) \( \triangle B D P \cong \Delta C D P \) \( \therefore \angle B P D=\angle C P D(B y C P C T) \) ….(4) \( \angle B P D+\angle C P D=180^{\circ}(\text { Linear pair angles) } \) \( \angle B P D+\angle B P D=180^{\circ} \) \( 2 \angle 8 P D=180^{\circ}[\text { From Equation }(4)] \) \( \angle B P D=90^{\circ} \quad \dots(5) \) FRom equations (2) and (5), it can be said that AP is the perpendicular bisector of BC.

Q.2:AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that (i) AD bisects BC (ii) AD bisects ∠ A.

Ans :(i) In \( \Delta B A D \text { and } \triangle C A D ,\) \( \angle \mathrm{ADB}=\angle \mathrm{ADC}\left(\text { Each } 90^{\circ} \text { as } \mathrm{AD} \text { is an altitude) }\right. \) \( \begin{array}{l}{A B=A C(G \text { iven })} \\ {A D=A D(\text { Common })}\end{array} \) \( \therefore \triangle B A D \cong \triangle C A D \) (By RHS Congruence rule) \( \triangle B D=C D \) (By CPCT) Hence, AD bisects BC. (ii) Also, by CPCT, \( \angle B A D=\angle C A D \) Hence, AD bisects \( \angle \mathrm{A} \)

Q.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 ∆ PQR (see Figure). Show that: (i) ∆ ABM ≅ ∆ PQN (ii) ∆ ABC ≅ ∆ PQR

Ans :(i) In \( \triangle A B C, A M \) is the median to BC. \( \therefore \) \( \mathrm{BM}=\frac{1}{2} \mathrm{BC} \) In \( \Delta \mathrm{PQR} \), PN is the median to QR. \( \therefore Q N=\frac{1}{2} Q R \) However, BC = QR \( \therefore \frac{1}{2} B C=\frac{1}{2} Q R \) \( \therefore \mathrm{BM}=\mathrm{QN} \quad \ldots(1) \) In \( \triangle A B M \text { and } \Delta P Q N, \) AB = PQ(Given) \( \begin{array}{l}{\mathrm{BM}=\mathrm{QN}[\text { From Equation }(1)]} \\ {\mathrm{AM}=\mathrm{PN}(\text { Given })} \\ {\text { } \mathrm{ABM} \cong \Delta \mathrm{PQN} \text { (By SSS congruence rule }} \\ {\angle \mathrm{ABM}=\angle \mathrm{PQN}(\mathrm{By} \mathrm{CPCT})} \\ {\angle \mathrm{ABC}=\angle \mathrm{PQR}}\end{array} \) …(2) (iii) In \( \triangle A B C \text { and } \Delta P Q R \), \( \begin{array}{l}{\mathrm{AB}=\mathrm{PQ}(\text { Given })} \\ {\angle \mathrm{ABC}=\angle \mathrm{PQR}[\mathrm{From} \text { Equation }(2)]} \\ {\mathrm{BC}=\mathrm{QR}(\text { Given })}\end{array} \) \( \therefore \triangle A B C \cong \Delta P Q R \) (By SAS congruence rule)

Q.4:BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

Ans :In \( \triangle B E C \text { and } \Delta C F B \) , \( \angle B E C=\angle C F B\left(E a c h 90^{\circ}\right) \) \( \begin{array}{l}{\mathrm{BC}=\mathrm{CB}(\mathrm{Common})} \\ {\mathrm{BE}=\mathrm{CF}(\mathrm{Given})}\end{array} \) \( \therefore \triangle B E C \cong \Delta C F B \) (By RHS congruency) \( \therefore \angle B C E=\angle C B F(B y C P C T) \) \( \therefore A B=A C \) (Sides opposite to equal angles of a triangle are equal) Hence, \( \triangle \mathrm{ABC} \) is isosceles.

Q.5:ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠ B = ∠ C.

Ans :In \( \triangle A P B \text { and } \triangle A P C \) , \( \angle A P B=\angle A P C\left(E a c h 90^{\circ}\right) \) \( \begin{aligned} A B &=A C(G i v e n) \\ A P &=A P(C o m m o n) \end{aligned} \) \( \therefore \triangle A P B \cong \triangle A P C \) (Using RHS congruence rule) \( \therefore \angle B=\angle C \) (By using CPCT)

Q.1:Show that in a right angled triangle, the hypotenuse is the longest side.

Ans :Let us consider a right-angled triangle ABC, right-angled at B. In \( \triangle A B C \), \( \begin{array}{l}{\angle A+\angle B+\angle C=180^{\circ} \text { (Angle sum property of a triangle) }} \\ {\angle A+90^{\circ}+\angle C=180^{\circ}} \\ {\angle A+\angle C=90^{\circ}}\end{array} \) Hence, the other two angles have to be acute(i.e., less than \( 90^{\circ} \) ). \( \angle B \text { is the largest angle in } \triangle A B C . \) \( \begin{array}{l}{\angle B>\angle A \text { and } \angle B>\angle C} \\ {A C>B C \text { and } A C>A B}\end{array} \) [In any triangle, the side opposite to the larger (greater) angle is longer.] Therefore, AC is the largest side in \( \Delta A B C \) However, AC is the hypotenuse of \( \triangle \mathrm{ABC} \). Therefore, hypotenuse is the longest side in a right-angled triangle.

Q.2:In Figure, sides AB and AC of ∆ ABC are extended to points P and Q respectively. Also, ∠ PBC < ∠ QCB. Show that AC > AB.

Ans :In the given figure, \( \angle A B C+\angle P B C=180^{\circ}(\text { Linear pair) } \) \( \angle A B C=180^{\circ}-\angle P B C \quad \ldots(1) \) Also, \( \angle A C B+\angle Q C B=180^{\circ} \) \( \angle A C B=180^{\circ}-\angle Q C B \) …(2) As \( \angle P B C<\angle Q C B \) \( 180^{\circ}-\angle P B C>180^{\circ}-\angle Q C B \) \( \begin{array}{l}{\angle A B C>\angle A C B[\text { From Equations }(1) \text { and }(2)]} \\ {A C>A B(\text { Side opposite to the larger angle is larger.) }}\end{array} \) Hence proved AC > AB

Q.3:In Figure, ∠ B < ∠ A and ∠ C < ∠ D. Show that AD < BC.

Ans :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 AD < BC

Q.4:AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Figure). Show that ∠ A > ∠ C and ∠ B > ∠ D.

Ans :In ΔABC, AB < BC (AB is the smallest side of quadrilateral ABCD) ∴ ∠2 < ∠1 (Angle opposite to the smaller side is smaller) … (1) In ΔADC, AD < CD (CD is the largest side of quadrilateral ABCD) ∴ ∠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, AB < AD (AB is the smallest side of quadrilateral ABCD) ∴ ∠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

Q.5:In Figure, PR > PQ and PS bisects ∠ QPR. Prove that ∠ PSR > ∠ PSQ.

Ans :As \( P R>P Q \) \( \angle P Q R>\angle P R Q(\text { Angle opposite to larger side is larger) } \) ...(1) PS is the bisector of \( \angle Q P R \). \( \angle Q P S=\angle R P S \quad \ldots(2) \) \( \angle P S R \text { is the exterior angle of } \Delta P Q S \) \( \angle P S R=\angle P Q R+\angle Q P S \quad-(3) \) \( \begin{array}{l}{\angle P S Q \text { is the exterior angle of } \Delta P R S \text { . }} \\ {\angle P S Q=\angle P R Q+\angle R P S}\end{array} \) …(4) \( \begin{array}{l}{\text { Adding Equations }(1) \text { and }(2), \text { we obtain }} \\ {\angle P Q R+\angle Q P S>\angle P R Q+\angle R P S}\end{array} \) \( \angle \mathrm{PSR}>\angle \mathrm{PSQ}[\text { Using the values of Equations }(3) \text { and }(4)] \)

Q.6:Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Ans :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 \( \triangle \mathrm{PNM} \) \( \begin{array}{l}{\angle N=90^{\circ}} \\ {\angle P+\angle N+\angle M=180^{\circ}(\text { Angle sum property of a triangle) }} \\ {\angle P+\angle M=90^{\circ}}\end{array} \) Clearly, \( \angle M \) is an acute angle \( \begin{array}{l}{\angle M < A N} \\ {P N < P M (\text { side opposite to the smaller angle is smaller}) }\end{array} \) 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.

Q.1:ABC is a triangle. Locate a point in the interior of ∆ ABC which is equidistant from all the vertices of ∆ ABC.

Ans :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. In \( \triangle \mathrm{ABC} \), 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 \( \triangle \mathrm{ABC} \).

Q.2:In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.

Ans :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 \( \triangle \mathrm{ABC} \), 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 \( \triangle \mathrm{ABC} \)

Q.3: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)

Ans :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 vertices. So, the ice-cream parlour should be set up at the circumcentre O of \( \triangle \mathrm{ABC} \) 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.

Q.4:Complete the hexagonal and star shaped Rangolis [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?

Ans :It can be observed that hexagonal-shaped rangoli has 6 equilateral triangles in it. \( \begin{array}{l}{\text { Area of } \Delta \mathrm{OAB}=\frac{\sqrt{3}}{4}(\text { side })^{2}=\frac{\sqrt{3}}{4}(5)^{2}} \\ {=\frac{\sqrt{3}}{4}(25)=\frac{25 \sqrt{3}}{4} \mathrm{cm}^{2}}\end{array} \) Area of hexagonal-shaped rangoli \( =6 \times \frac{25 \sqrt{3}}{4}=\frac{75 \sqrt{3}}{2} \mathrm{cm}^{2} \) Area of equilateral triangle having its sides as 1 cm \( =\frac{\sqrt{3}}{4}(1)^{2}=\frac{\sqrt{3}}{4} \mathrm{cm}^{2} \) Number of equilateral triangles of 1 cm side that can be filled In this hexagonal-shaped rangoli \( =\frac{\frac{75 \sqrt{3}}{2}}{\frac{\sqrt{3}}{4}}=150 \) Star-shaped rangoli has 12 equilateral triangles of side 5 cm in it. Area of star-shaped rangoli = \( 12 \times \frac{\sqrt{3}}{4} \times(5)^{2}=75 \sqrt{3} \) Number of equilateral triangles of 1 cm side that can be filled in this star-shaped rangoli = \( \frac{75 \sqrt{3}}{\frac{\sqrt{3}}{4}}=300 \) Therefore, star-shaped rangoli has more equilateral triangles in it.

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