 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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 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,
∠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,
∠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 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,
∠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,
CD = BD (AD is the perpendicular bisector of BC)
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 : In ∆ABD and ∆ACD,
AB = AC (Given)
BD = CD (Given)
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
(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)
(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)
∴ ∠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 
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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