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Triangles, Exercise (7.1), Q1. In quadrilateral ACBD, AC = AD and AB bisects  A (See the given figure)., Show that ABC  ABD . What can you say about BC and BD?, Difficulty Level:, Easy, Known/given:, AC = AD and AB bisects  A, To prove:, ABC  ABD and, what can be said about BC and BD., Reasoning:, We can show two sides and included angle of ABC are equals to corresponding sides, and included angle of ABD, by using SAS congruency criterion both triangles will be, congruent and by CPCT, BC and BD will be equal., , Solution:, In ABC and ABD,, AC = AD (Given), CAB = DAB(AB bisects A), AB = AB (Common), ABC  ABD (By SAS congruence rule),  BC = BD (By CPCT), Therefore, BC and BD are of equal lengths.

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Q3. AD and BC are equal perpendiculars to a line segment AB (See the given, figure). Show that CD bisects AB., Difficulty Level:, Easy, Known/given:, , AD ⊥ AB, BC ⊥ AB and AD = BC, , To prove:, CD bisects AB or OA = OB, Reasoning:, We can show two triangles OBC and OAD congruent by using AAS congruency rule and, then we can say corresponding parts of congruent triangles will be equal., , Solution:, In BOC and AOD,, BOC = AOD (Vertically opposite angles), CBO = DAO ( Each 90º ), BC = AD (Given), BOC AOD ( AAS congruence rule ),  BO = AO ( By CPCT ),  CD bisects AB., , Q4. l and m are two parallel lines intersected by another pair of parallel lines p and, q (see the given figure). Show that ABC  CDA., Difficulty Level:, Easy, Known/given:, l m and p q

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To prove:, ABC  CDA., , Reasoning:, We can show both the triangles congruent by using ASA congruency criterion, , Solution:, In ABC and CDA,, , BAC and DCA ( Alternate interior angles, as p || q ), AC = CA (Common), BCA and DAC ( Alternate interior angles, as l || m ), ABC  CDA (By ASA congruence rule), , 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 the given figure). Show that:, (i) APB  AQB, (ii) BP = BQ or B is equidistant from the arms of A, Difficulty Level:, Easy, What is known/given?, l is the bisector of an angle A and BP ⊥ AP and BQ ⊥ AQ, To prove:, APB  AQB and BP = BQ or B is equidistant from the arms of A, Reasoning:, We can show two triangles APB and AQB congruent by using AAS congruency rule and, then we can say corresponding parts of congruent triangles will be equal.

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Solution:, , In APB and AQB,, BAP = BAQ ( l is the angle bisector of A ), APB = AQB ( Each 90 ), AB = AB (Common), APB  AQB (By AAS congruence rule),  BP = BQ (By CPCT), , Or, it can be said that B is equidistant from the arms of A., , Q6. In the given figure, AC = AE, AB = AD and BAD = EAC. Show that, BC = DE., Difficulty Level:, Medium, Known/given:, AC = AE, AB = AD and BAD = EAC., To prove:, BC = DE., Reasoning:, We can show two triangles BAC and DAE congruent by using SAS congruency rule and, then we can say corresponding parts of congruent triangles will be equal. To show both, triangles congruent two pair of equal sides are given and add angle DAC on both sides in, given pair of angles BAD and angle EAC to find the included angle BAC and DAE.

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Solution:, It is given that BAD = EAC, BAD + DAC = EAC + DAC, BAC = DAE, In BAC and DAE,, AB = AD (Given), BAC = DAE (Prove above), AC = AE (Given), BAC  DAE (By SAS congruence rule),  BC = DE (By CPCT), , 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 the given figure)., Show that, (i) DAP  EBP, (ii) AD = BE, Difficulty Level:, Medium, Known/given, P is its mid-point of AB, BAD = ABE and EPA = DPB, To prove:, (i) DAP  EBP and (ii) AD = BE ., Reasoning:, We can show two triangles DAP and EBP congruent by using ASA congruency rule and, then we can say corresponding parts of congruent triangles will be equal. To show both, triangles congruent one pair of equal sides and one pair of equal angles are given and add, angle EPD on both sides in given pair of angles EPA and angle DPB to find the other pair, of angles APD and BPE

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Solution:, 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 ), DAP  EBP ( ASA congruence rule ),  AD = BE ( By CPCT ), , 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 the given figure). Show that:, , (i ), ( ii ), ( iii ), ( iv ), , AMC  BMD, DBC is a right angle., DBC  ACB, 1, CM = AB, 2, , Difficulty Level:, Medium, Known/given:, M is the mid-point of hypotenuse AB, C = 90 and DM = CM, To prove:, ( i ) AMC  BMD, , ( ii ), ( iii ), ( iv ), , DBC is a right angle., DBC  ACB, 1, CM = AB, 2, , Reasoning:, We can show two triangles AMC and BMD congruent by using SAS congruency rule and, then we can say corresponding parts of congruent triangles will be equal means angle, ACM will equal to angle BDM, which are alternate interiors angle and can conclude DB, is parallel to AC. Now it will help to find angle DBC by co-interior angles. Similarly,, triangles DBC and ACB will be congruent by using SAS criterion and CM will be half of, AB by using M mid-point.

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Triangles, 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, , Difficulty Level:, Easy, Known/given:, Triangle ABC is isosceles in which AB=AC also OB and OC are bisectors of angle B and, angle C, To prove:, (i) OB = OC (ii) AO bisects ∠A, Reasoning:, As OB and OC are bisectors of angle B and angle C means half of angle B will be equal, to half of angle c which will help us to conclude OB is equal to OC. Now We can show, two triangles OAB and OAC congruent by using SSS congruency rule and then we can, say corresponding parts of congruent triangles will be equal., , Solution:, ( 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, 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 )

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Therefore, OAB  OAC (By SSS congruence rule), BAO = CAO ( CPCT ),  AO bisects A., , Q2. In ABC, AD is the perpendicular bisector of BC (see the given figure)., Show that ABC is an isosceles triangle in which AB = AC., Difficulty Level:, Easy, Known/given:, AD is perpendicular bisector of BC means ADB = ADC = 90 and BD=DC, To prove:, ABC is an isosceles triangle in which AB = AC., Reasoning:, We can show two triangles ADB and ADC congruent by using SAS congruency rule and, then we can say corresponding parts of congruent triangles will be equal., , Solution:, , In ADC and ADB,, AD = AD ( Common ), ADC = ADB (Each 900 ), CD = BD ( AD is the perpendicular bisector of BC ), ADC  ADB ( By SAS congruence rule ),  AB = AC ( By CPCT ), Therefore, ABC is an isosceles triangle in which AB = AC.

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Q3. 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., Difficulty Level:, Easy, known/given:, Sides AB=AC, BE ⊥ AC and CF ⊥ AB, To prove:, Altitudes BE and CF are equal or BE = CF, Reasoning:, We can show two triangles ABE and ACF congruent by using AAS congruency rule and, then we can say corresponding parts of congruent triangles will be equal., , Solution:, , In AEB and AFC,, AEB and AFC ( Each 90º ), A = A ( Common angle ), AB = AC ( Given ), AEB  AFC ( By AAS congruence rule ),  BE = CF ( By CPCT ), , Q4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal, (see the given figure). Show that, , (i ), ( ii ), , ABE  ACF, AB = AC, i.e., ABC is an isosceles triangle., , Difficulty Level:, Easy, Known/given:, Altitudes, BE = CF, BE ⊥ AC and CF ⊥ AB

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To prove:, , (i ), ( ii ), , ABE  ACF, AB = AC, i.e., ABC is an isosceles triangle., , Reasoning:, We can show two triangles ABE and ACF congruent by using AAS congruency rule and, then we can say corresponding parts of congruent triangles will be equal., , Solution:, ( i ) In ABE and ACF,, AEB andAFC ( Each 90º ), A = A ( Common angle ), BE = CF ( Given ), ABE  ACF ( By AAS congruence rule ), , ( ii ), , It has already been proved that, ABE  ACF,  AB = AC ( By CPCT ), , Q5. ABC and DBC are two isosceles triangles on the same base BC (see the given, figure). Show that ∠ABD = ∠ACD., Difficulty Level:, Easy, Known/given:, ABC and DBC are two isosceles triangles., To prove:, ABD = ACD, Reasoning:, First of all, we can join point A and D then we can show two triangles ADB and ADC, congruent by using SSS congruency rule after that we can say corresponding parts of, congruent triangles will be equal.

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Solution:, , Let us join AD., In ABD and ACD,, AB = AC ( Given ), BD = CD ( Given ), AD = AD ( Common side ), ABD  ACD ( By SSS congruence rule ), ABD = ACD ( By CPCT ), , Q6. ABC 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., Difficulty Level:, Easy, Known/given:, Sides AB=AC and AD=AB, To prove:, ∠BCD is a right angle.

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Reasoning:, We can use the property angles opposite to equal sides are equal and then by the angle, sum property in triangle BCD we can show the required result., , Solution:, In ABC,, , AB = AC ( Given ), ACB = ABC ( Angles opposite to equal sides of a triangle are also equal ), , In ACD,, AC = AD, 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 = 1800, 2 ( ACB + ACD ) = 1800, 2 ( BCD ) = 1800, BCD = 900, , Q7. ABC is a right-angled triangle in which A = 900 and AB = AC., Find ∠B and ∠C., Difficulty Level:, Easy, Known/given:, ABC is right-angled triangle and sides AB=AC, To Find:, , Value of ∠B and ∠C.

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Reasoning:, , We can use the property angles opposite to equal sides are equal and then by the, angle sum property in triangle ABC we can find the value of ∠B and ∠C., Solution:, , It is given that, AB = AC, C = B ( Angles opposite to equal sides are also equal ), In ABC,, A + B + C = 1800 ( Angle sum property of a triangle ), 900 + B + C = 1800, 900 + B + B = 1800, 2B = 900, B = 450, B = C = 450, , Q8. Show that the angles of an equilateral triangle are 60º each., Difficulty Level:, Easy, Known/given:, Triangle ABC is an equilateral triangle., To prove:, The angles of an equilateral triangle are 60º each., Reasoning:, We can use the property angles opposite to equal sides are equal and then by the angle, sum property in triangle ABC we can show the value of each angle is 60 degree.

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Solution:, , 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 ABC,, , A + B + C = 1800, A + A + A = 1800, 3A = 1800, A = 600, A = B = C = 600, , Hence, in an equilateral triangle, all interior angles are of measure 600.

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Triangles, 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 the given 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., Difficulty Level:, Medium, Known/given:, ABC and DBC are two isosceles triangles., To prove:, i) ABD  ACD, ii) ABP  ACP, iii) AP bisects  A as well as  D., iv) AP is the perpendicular bisector of BC., Reasoning:, We can use Isosceles triangle property to show triangles ABD and ACD by, SSS congruency and then we can say corresponding parts of congruent triangles are, equal, Similarly ABP and ACP by SAS then with the help of these we can prove other, required results., , Solution:, i) In ABD and ACD , AB = AC (Given), BD = CD (Given) AD = AD (Common), ABD  ACD (By SSS congruence rule),  BAD =  CAD (By CPCT),  BAP =  CAP …. (1)

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ii) In ABP and ACP , AB = AC (Given),  BAP =  CAP [From equation (1)], AP = AP (Common),  ABP  ACP (By SAS congruence rule), BP = CP (By CPCT), … (2), iii) From Equation (1),,  BAP =  CAP, Hence, AP bisects  A., In BDP and CDP ,, BD = CD (Given), DP = DP (Common), BP = CP [From equation (2)],  BDP  CDP (By SSS Congruence rule),   BDP =  CDP (By CPCT) … (3), Hence, AP bisects  D., iv) BDP  CDP,   BPD =  CPD (By CPCT) …. (4),  BPD +  CPD = 1800 (Linear pair angles),  BPD +  BPD = 1800, 2  BPD = 1800 [From Equation (4)], … (5),  BPD = 900, From Equations (2) and (5), it can be said that AP is the perpendicular bisector of BC., , Q2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that, i) AD bisects BC, ii) AD bisects  A., Difficulty Level:, Easy, Known/given:, AD is an altitude of an isosceles triangle ABC in which AB = AC., To prove:, i) AD bisects BC, ii) AD bisects  A., Reasoning:, We can show triangle BAD and CAD congruent by using RHS congruency criterion and, then we can say corresponding parts of congruent triangle are equal.

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Solution:, , i) In BAD and CAD ,,  ADB =  ADC (Each 900 as AD is an altitude), AB = AC (Given), AD = AD (Common),  BAD  CAD (By RHS Congruence rule),  BD = CD (By CPCT), Hence, AD bisects BC., ii) Also, by CPCT,,  BAD =  CAD, Hence, AD bisects  A., , 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 the given figure)., Show that:, i) ABM  PQN, ii) ABC  PQR, Difficulty Level:, Medium, Known/given:, Sides AB=PQ, BC=QR and AM = PN also AM and PN are medians., To prove:, i) ABM  PQN, ii) ABC  PQR, Reasoning:, Using the median property, we can show Triangles ABM and PQN are congruent by, SSS congruency then we can say corresponding parts of congruent triangles are equal, and then Triangles ABC and PQR by SAS congruency.

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Solution:, i) In ABC , AM is the median to BC., BM = 1 BC, 2, , In PQR , PN is the median to QR., QN = 1 QR, 2, However, BC = QR,  1 BC = 1 QR, 2, 2, BM = QN … (1), , In ABM and PQN ,, AB = PQ (Given), BM = QN [From Equation (1)], AM = PN (Given), ABM  PQN (By SSS congruence rule),  ABM =  PQN (By CPCT),  ABC =  PQR … (2), ii) In ABC and PQR , AB = PQ (Given),  ABC =  PQR [From Equation (2)], BC = QR (Given),  ABC  PQR (By SAS congruence rule), , Q4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence, rule, prove that the triangle ABC is isosceles., Difficulty Level:, Easy, Known/given:, BE and CF are two equal altitudes of a triangle ABC., To prove:, Triangle ABC is isosceles by using RHS congruence rule.

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Reasoning:, We can show triangles BEC and CFB congruent by using RHS congruency and then we, can say corresponding parts of congruent triangles are equal to prove the required result., Solution:, , In BEC and CFB ,,  BEC =  CFB (Each 900 ), BC = CB (Common), BE = CF (Given),  BEC  CFB (By RHS congruency),   BCE =  CBF (By CPCT), AB = AC (Sides opposite to equal angles of a triangle are equal), Hence, ABC is isosceles., , Q5. ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that, B = C ., Difficulty Level:, Easy, Known/given:, ABC is an isosceles triangle with AB = AC and AP ⊥ BC, To prove:, B = C, Reasoning:, , We can show triangles APB and APC congruent by using RHS congruency and, then we can say corresponding parts of congruent triangles are equal to show, the required result.

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Solution:, , In APB and APC ,, APB = APC, AB = AC, AP = AP, APB = APC, B = C, , (Each 900 ), (Given), (Common), (Using RHS congruence rule), (CPCT)

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Triangles, Exercise (7.4), Q1. Show that in a right-angled triangle, the hypotenuse is the longest side., Difficulty Level:, Easy, Known/given:, A right-angled triangle., To prove:, Hypotenuse is the longest side., Reasoning:, In a triangle if one angle is of 90 degree then the other two angles have to be acute and, we can use the fact that in any triangle, the side opposite to the larger (greater) angle, is longer., Solution:, , Let us consider a right-angled triangle ABC, right-angled at B., In ∆ABC,, , 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 ABC., B  A and B  C, , AC  BC and AC  AB

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[In any triangle, the side opposite to the larger (greater) angle is longer.], Therefore, AC is the largest side in ABC., However, AC is the hypotenuse of ABC. Therefore, hypotenuse is the longest side in a, right-angled triangle., , Q2. In the given figure sides AB and AC of ABC are extended to points P and Q, respectively. Also, PBC  QCB. Show that AC  AB., Difficulty Level:, Easy, Known/given:, PBC  QCB ., To prove:, AC  AB ., Reasoning:, By using linear pair, we can find inequality of interior angles and then we can use the fact, that in any triangle, the side opposite to the larger (greater) angle is longer., , Solution:, In the given figure,, ABC, ABC, Also,, ACB, ACB, , + PBC = 180 ( Linear pair ), = 180 − PBC, ... (1), , + QCB = 180, = 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.), Hence proved, AC  AB

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Q3. In the given figure, B  A and C  D. Show that AD  BC., Difficulty Level:, Easy, Known/given:, B  A and C  D ., To prove:, AD  BC, Reasoning:, We can use the fact that In any triangle, the side opposite to the larger (greater) angle is, longer We can add both the triangles result to get the required result., , Solution:, In AOB,, , B  A, AO  BO (Side opposite to smaller angle is smaller ), , ... (1), , In COD,, , 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, , Q4. 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., Difficulty Level:, Easy

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Known/give:, AB and CD are respectively the smallest and longest sides., To prove:, A  C and B  D, Reasoning:, First of all we can join vertex A to C then triangles ABC and ADC will be formed now, we can use the fact that In any triangle, the side opposite to the larger (greater) angle is, longer We can add both the triangles result to get the first required result, Similarly we, can join vertex B and D and use the fact to get the other required result., , Solution:, Let us join AC., , 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 is smaller ) ... ( 2 )

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On adding Equations (1) and (2), we obtain, 2 + 4  1 + 3, , C  A, A  C, Let us join BD., , In ABD,, , In BDC,, , AB < AD ( AB is the smallest side of quadrilateral ABCD ), 8  5 ( Angle opposite to the smaller side is smaller ) ...( 3), , 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 ( Hence, proved ), , Q5. In the given figure, PR > PQ and PS bisects QPR. Prove that, PSR  PSQ., , Difficulty Level:, Easy, Known/given:, PR > PQ and PS bisects QPR, To prove:, PSR  PSQ.

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Reasoning:, We can use exterior angle sum property to find the required inequality., , Solution:, 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 PQS., PSR = PQR + QPS ... ( 3), PSQ is the exterior angle of PRS., 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)], , Q6. Show that of all line segments drawn from a given point not on it, the, perpendicular line segment is the shortest., Difficulty Level:, Easy, Known/given:, Line segments drawn from a given point is not on line., To prove:, The perpendicular line segment is the shortest., Reasoning:, We know that in a triangle if one angle is 90 degree then other have to be acute.

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Solution:, , 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 = 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.

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Triangles, 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., Solution:, 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 ABC, 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 ABC., , Q2. In a triangle locate a point in its interior which is equidistant from all the sides, of the triangle., Solution:, 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.

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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., , Q3. 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), Solution:, 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 ABC.

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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., , Q4. 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?, , Solution:, It can be observed that hexagonal-shaped rangoli has 6 equilateral triangles of side 5cm, in it.

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3, 2, ( side ), 4, 3, 25 3 2, 2, =, cm, ( 5cm ) =, 4, 4, 25 3 2 75 3 2, cm =, cm, Area of hexagonal-shaped rangoli = 6 , 4, 2, Area of OAB =, , 3, 3 2, (1cm) 2 =, cm, 4, 4, Number of equilateral triangles of 1 cm side that can be filled in this hexagonal-shaped, 75 3 2, cm, 75 3 4, Rangoli = 2, =, , = 150, 2, 3 2, 3, cm, 4, Area of equilateral triangle having its side as 1cm =, , Star-shaped rangoli has 12 equilateral triangles of side 5 cm in it., , Area of star-shaped rangoli = 12 , , 3, 2,  ( 5cm ) = 75 3cm 2, 4, , Number of equilateral triangles of 1 cm side that can be filled in this star-shaped, 75 3cm2, 4, rangoli =, = 75 3 , = 300, 3 2, 3, cm, 4, Therefore, star-shaped rangoli has more equilateral triangles of side 1 cm in it.