{"id":46958,"date":"2023-02-07T10:00:11","date_gmt":"2023-02-07T04:30:11","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=46958"},"modified":"2023-02-08T09:13:19","modified_gmt":"2023-02-08T03:43:19","slug":"ml-aggarwal-class-9-solutions-for-icse-maths-chapter-10-chapter-test","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/ml-aggarwal-class-9-solutions-for-icse-maths-chapter-10-chapter-test\/","title":{"rendered":"ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 10 Triangle Chapter Test"},"content":{"rendered":"

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 10 Triangle Chapter Test<\/h2>\n

Question 1.
\nIn triangles ABC and DEF, \u2220A = \u2220D, \u2220B = \u2220E and AB = EF. Will the two triangles be congruent? Give reasons for your answer.
\nSolution:
\nIn \u2206ABC and \u2206DEF
\n\u2220A = \u2220D
\n\u2220B = \u2220E
\nAB = EF
\nIn \u2206ABC, two angles and included side is given but in \u2206DEF, corresponding angles are equal but side is not included of there angle.<\/p>\n

Question 2.
\nIn the given figure, ABCD is a square. P, Q and R are points on the sides AB, BC and CD respectively such that AP= BQ = CR and \u2220PQR = 90\u00b0. Prove that
\n(a) \u2206PBQ = \u2206QCR
\n(b) PQ = QR
\n(c) \u2220PRQ = 45\u00b0
\n\"ML
\nSolution:
\nGiven: In the given figure, ABCD is a square P, Q and R are the points on the sides AB, BC and CD respectively such that AP = BQ = CR, \u2220PQR = 90\u00b0
\nTo prove : (a) \u2206PBQ \u2245 \u2206QCR
\n(b) PQ = QR
\n(c) \u2220PRQ = 45\u00b0
\nProof: AB = BC = CD (Sides of square)
\nand AP = BQ = CR (Given)
\nSubtracting, we get
\nAB – AP = BC – BQ = CD – CR
\n\u21d2 PB = QC = RD
\nNow in APBQ and \u2220QCR
\nPB = QC (Proved)
\nBQ = CR (Given)
\n\u2220B = \u2220C (Each 90\u00b0)
\n\u2234 \u2206PBQ = \u2206QCR (SAS axiom)
\n\u2234 PQ = QR (c.p.c.t.)
\nBut \u2220PQR = 90\u00b0 (Given)
\n\u2220RPQ = \u2220PRQ (Angles opposite to equal angles)
\nBut \u2220RPQ + \u2220PRQ = 90\u00b0
\n\u2220RPQ = \u2220PRQ = \\(\\frac{90^{\\circ}}{2}\\) = 45\u00b0<\/p>\n

Question 3.
\nIn the given figure, AD = BC and BD = AC. Prove that \u2220ADB = \u2220BCA.
\n\"ML
\nSolution:
\nGiven : In the figure,
\nAD = BC, BD = AC
\nTo prove : \u2220ADB = \u2220BCA
\nProof: In \u2206ADB and \u2206ACB
\nAB = AB (Common)
\nAD = BC (Given)
\nBD = AC (Given)
\n\u2206ADB = \u2206ACB (SSS axiom)
\n\u2234 \u2220ADB = \u2220BCA (c.p.c.t.)<\/p>\n

Question 4.
\nIn the given figure, OA \u22a5 OD, OC \u22a5 OB, OD = OA and OB = OC. Prove that AB = CD.
\n\"ML
\nSolution:
\nGiven : In the figure, OA \u22a5 OD, OC \u22a5 OB. OD = OA, OB = OC
\n\"ML
\nTo prove : AB = CD
\nProof: \u2220AOD = \u2220COB (each 90\u00b0)
\nAdding \u2220AOC (both sides)
\n\u2220AOD + \u2220AOC = \u2220AOC + \u2220COB
\n\u21d2 \u2220COD = \u2220AOB
\nNow, in \u2206AOB and \u2206DOC
\nOA = OD (given)
\nOB = OC (given)
\n\u2220AOB = \u2220COD (proved)
\n\u2234 \u2206AOB = \u2206DOC (SAS axiom)
\n\u2234 AB = CD (c.p.c.t)<\/p>\n

Question 5.
\nn the given figure, PQ || BA and RS CA. If BP = RC, prove that:
\n(i) \u2206BSR = \u2206PQC
\n(ii) BS = PQ
\n(iii) RS = CQ.
\n\"ML
\nSolution:
\nGiven : In the given figure,
\nPQ || BA, RS || CA
\nBP = RC
\n\"ML
\nTo prove :
\n(i) \u2206BSR = \u2206PQC
\n(ii) BS = PQ
\n(iii) RS = CQ
\nProof: BP = RC
\n\u2235 BC – RC = BC – BP
\n\u2234 BR = PC
\nNow, in \u2206BSR and \u2206PQC
\n\u2220B = \u2220P (corresponding angles)
\n\u2220R = \u2220C (corresponding angles)
\nBR = PC (proved)
\n\u2234 \u2206BSR \u2245 \u2206PQC (ASA axiom)
\n\u2234 BS = PQ (c.p.c.t.)
\nRS = CQ (c.p.c.t.)<\/p>\n

Question 6.
\nIn the given figure, AB = AC, D is a point in the interior of \u2206ABC such that \u2220DBC = \u2220DCB. Prove that AD bisects \u2220BAC of \u2206ABC.
\nSolution:
\nGiven : In the figure given, AB = AC
\nD is a point in the interior of \u2206ABC
\nSuch that \u2220DBC = \u2220DCB
\nTo prove : AD bisects \u2220BAC
\nConstruction : Join AD and produced it to BC in E
\nProof: In \u2206ABC,
\nAB = AC
\n\u2234 \u2220B = \u2220C (Angles opposite to equal sides) and \u2220DBC = \u2220DCB (Given)
\nSubtracting, we get
\n\u2220B – \u2220DBC = \u2220C – \u2220DCB
\n\"ML
\n\u21d2 \u2220ABD = \u2220ACD
\nNow in \u2206ABD and \u2206ACD
\nAD = AD (Common)
\n\u2220ABD = \u2220ACD (Proved)
\nAB=AC (Given)
\n\u2234 \u2206ABD \u2245 \u2206ACD (SAS axiom)
\n\u2234 \u2220BAD = \u2220CAD (c.p.c.t.)
\n\u2234 AD is bisector of \u2220BAC<\/p>\n

Question 7.
\nIn the adjoining figure, AB || DC. CE and DE bisects \u2220BCD and \u2220ADC respectively. Prove that AB = AD + BC.
\nSolution:
\nGiven :
\nIn the given figure, AB || DC
\nCE and DE bisects \u2220BCD and \u2220ADC respectively
\n\"ML
\nTo prove : AB = AD + BC
\nProof: \u2235 AD || DC and ED is the transversal
\n\u2234 \u2220AED = \u2220EDC (Alternate angles)
\n= \u2220ADC (\u2235 ED is bisector of \u2220ADC)
\n\u2234 AD = AE …(i) (Sides opposite to equal angles)
\nSimilarly,
\n\u2220BEC = \u2220ECD = \u2220ECB
\n\u2234 BC = EB …(ii)
\nAdding (i) and (ii),
\nAD + BC = AE + EB = AB
\n\u2234 AB = AD + BC<\/p>\n

Question 8.
\nIn \u2206ABC, D is a point on BC such that AD is the bisector of \u2220BAC. CE is drawn parallel to DA to meet BD produced at E. Prove that \u2206CAE is isosceles.
\nSolution:
\nGiven : In \u2206ABC,
\nD is a point on BC such that AD is the bisector of \u2206BAC
\nCE || DA to meet BD produced at E
\nTo prove : \u2206CAE is an isosceles
\nProof: \u2235 AD || EC and AC is its transversal
\n\u2234 \u2220DAC = \u2220ACE (Alternate angles)
\nand \u2220BAD = \u2220CEA (Corresponding angles)
\n\"ML
\nBut \u2220BAD = \u2220DAC (\u2235 AD is bisector of \u2220BAC)
\n\u2234 \u2220ACE = \u2220CAE
\nAE = AC (Sides opposite to equal angles)
\n\u2234 \u2206ACE is an isosceles triangle.<\/p>\n

Question 9.
\nIn the figure (it) given below, ABC is a right angled triangle at B, ADEC and BCFG are squares. Prove that AF = BE.
\n\"ML
\nAns.
\nGiven. In right \u2206ABC, \u2220B = 90\u00b0
\nADEC and BCFG are squares on the sides
\nAC and BC of \u2206ABC respectively AF and BE are joined.
\nTo prove. AE = BE
\nProof. \u2220ACE = \u2220BCF
\nAdding \u2220ACB both sides
\n\u2220ACB + \u2220ACE = \u2220ACB + \u2220BCF
\n\u21d2 \u2220\u2220BCE = \u2220ACF
\nNow in \u2206BCE and \u2206ACF,
\nCF = AC (sides of a square)
\nBC = CF (sides of a square)
\n\u2220BCE = \u2220ACF (proved)
\n\u2234 \u2206BCE = \u2206ACF (SAS postulate)
\n\u2234 BE = AF (c.p.c.t.)
\nHence proved.<\/p>\n

Question 10.
\nIn the given figure, BD = AD = AC. If \u2220ABD = 36\u00b0, find the value of x.
\n\"ML
\nSolution:
\nGiven : In the figure, BD = AD = AC
\n\u2220ABD = 36\u00b0
\nTo find : Measure of x.
\nProof: In \u2206ABD,
\nAD = BD (given)
\n\u2234 \u2220ABD = \u2220BAD = 36\u00b0 (\u2235 \u2220ABD = 36\u00b0)
\n\u2234 Ext. \u2220ADC = \u2220ABD + \u2220BAD (sum of interior opposite angles)
\n= 36\u00b0 + 36\u00b0 = 72\u00b0
\nBut in \u2206ADC
\nAD = AC
\n\u2234 \u2220ADC = \u2220ACD = 72\u00b0
\nand Ext. \u2220PBC = \u2220ABC + \u2220ACD
\n= 36\u00b0 + 72\u00b0 = 108\u00b0
\n\u2234 x= 108\u00b0<\/p>\n

Question 11.
\nIn the adjoining figure, TR = TS, \u22201 = 2\u22202 and \u22204 = 2\u22203.
\nProve that RB = SA.
\nSolution:
\n\"ML
\nGiven : In the figure, RST is a triangle
\nTR = TS,
\n\u22201 = 2\u22202 and \u22204 = 2\u22203
\nTo prove : RB = SA
\nProof: \u22201 = \u22204 (Vertically opposite angles)
\nBut 2\u22202 = \u22201 and 2\u22203 = 4
\n\u2234 2\u22202 = 2\u22203
\n\u2234 \u22202 = \u22203
\n\u2234 But \u2220TRS = \u2220TSR (\u2235 TR = TS given)
\n\u2234 \u2220TRS – \u2220BRS = \u2220TSR – \u2220ASR
\n\u21d2 \u2220ARB = \u2220BSA
\nNow in \u2206RBT and \u2206SAT
\n\u2220T = \u2220T (Common)
\nTR = TS (Given)
\nand \u2220TRB = \u2220TSA (proved)
\n\u2234 \u2206RBT \u2245 \u2206SAT
\n\u2234 RB = SA<\/p>\n

Question 12.
\n(a) In the figure (1) given below, find the value of x.
\n(b)In the figure (2) given below, AB = AC and DE || BC. Calculate
\n(i) x
\n(ii) y
\n(iii) \u2220BAC
\n(c) In the figure (1) given below, calculate the size of each lettered angle.
\n\"ML
\nSolution:
\n(a) We have to calculate the value of x.
\n\"ML
\nNow, in \u2206 ABC
\n\u22205 = 36\u00b0 (1)
\nAlso, 36\u00b0 + \u2220 1 + \u2220 5 = 180\u00b0 [ \u2235AC = BC] [sum of all angles in a triangle is 180\u00b0]
\n\u21d2 36\u00b0+ \u22201 + 36\u00b0 = 180\u00b0 [from (1)]
\n\u21d2 72\u00b0 + \u2220 1 = 180\u00b0 \u21d2 \u2220 1 = 180\u00b0 – 72\u00b0
\n\u21d2 \u2220 1 = 108\u00b0 ….. (2)
\nAlso, \u22201 + \u22202 = 180\u00b0 (Linear pair)
\n\u21d2 108\u00b0 + \u22202 = 180\u00b0 [From (2)]
\n\u21d2 \u22202 = 180\u00b0 – 108\u00b0 \u21d2 \u22202 = 72\u00b0 …..(3)
\nAlso, \u22202 = \u22203 (AC = AD)
\n\u21d2 \u2220 3 = 72\u00b0 [From (3)] (4)
\nNow, in \u2206 ACD
\n\u22202 + \u22203 + \u22204 = 180\u00b0 [sum of all angles in a triangle is 180\u00b0]
\n\u21d2 72\u00b0 + 72\u00b0 + \u22204 = 180\u00b0 [From (3) and (4)]
\n\u21d2 144\u00b0 + \u22204 = 180\u00b0 \u21d2 \u22204 = 180\u00b0 – 144\u00b0
\n\u21d2 \u22204 = 36\u00b0 (5)
\n\u2234 ABP is a St. line
\n\u2234 \u22205 + \u22204 + x=180\u00b0
\n36\u00b0 + 36\u00b0\u2019+ x = 180\u00b0 [From (1) and (5)]
\n72\u00b0 + x = 180 \u21d2 x = 108\u00b0
\nHence, value of x = 108\u00b0<\/p>\n

(b) Given. AB = AC, and DE || BC
\n\u2220ADE = (x + y -36)\u00b0
\n\u2220 ABC = 2x\u00b0and \u2220 ACB = (y – 2)\u00b0
\n\"ML
\nTo Calculate. (i) x (ii) y (iii) \u2220 BAC
\nNow, in \u2206ABC
\n\u2234 AB = AC
\n2x = y – 2
\n[In a triangle equal sides here equal angle opposite to them]
\n2x – y = -2 …..(1)
\n\u2234 DE || BC,
\nx + y – 36 = 2x [corresponding angles]
\n\u21d2 x + y – 2x = 36 \u21d2 -x + y = 36 …..(2)
\nFrom equation (1) and (2),
\n\"ML
\nSubstituting the value of x in equation (1), we get
\n2 \u00d7 34 -y = – 2 \u21d2 68 – y = – 2
\n\u21d2 68 + 2 = y \u21d2 70 = y \u21d2 y = 70
\nHence, value of x = 34\u00b0 and value of y = 70\u00b0<\/p>\n

(iii) In \u2206 ABC
\n\u2220BAC + 2 x\u00b0 + (y – 2)\u00b0 = 180\u00b0 [sum of all angles in a triangle is 180\u00b0]
\n\u21d2 \u2220 BAC + 2 \u00d7 34\u00b0+ (70 – 2)\u00b0= 180\u00b0 (Substituting the value of x and y)
\n\u21d2 \u2220BAC + 68\u00b0+ 68\u00b0 = 180\u00b0
\n\u21d2 \u2220BAC = 180\u00b0- 136\u00b0 \u21d2 \u2220BAC = 44\u00b0
\nHence, value of \u2220 BAC = 44\u00b0<\/p>\n

(c) Given. \u2220BAE = 54\u00b0, \u2220DEC = 80\u00b0 and AB = BC.
\n\"ML
\nTo calculate. The value of x, y and z.
\nNow \u2220 2 = 80\u00b0 …………(1) (vertically opposite angles
\n\u2234 AC and BD cut at point E)
\nIn \u2206 ABE,
\n54\u00b0 +x + \u22202 = 180\u00b0 (sum of all angles in triangle is 180\u00b0)
\n\u21d2 54\u00b0+ x + 80\u00b0 = 180\u00b0 (\u2235 \u22202 = 80\u00b0)
\n\u21d2 134\u00b0 + x = 180\u00b0 \u21d2 x = 180\u00b0 – 134\u00b0
\n\u21d2 x = 46\u00b0
\nNow, \u22201 + 80\u00b0 = 180\u00b0 (Linear pair)
\n\u22201 = 180\u00b0-80\u00b0 \u21d2 \u2220l = 100\u00b0 …….(2)
\nAlso, AB = BC (given)
\n\u22203 = 54\u00b0
\n(In a triangle equal sides have equal angles)
\nNow, in \u2206 ABC
\n54\u00b0 + (x + y) + \u22203 = 180\u00b0 (substituting the value of x and \u2220 3)
\n\u21d2 154\u00b0 + y = 180\u00b0 \u21d2 y = 180\u00b0 – 154\u00b0
\n\u21d2 y = 26\u00b0 ….(3)
\n\u2234 AB || CD, \u2234 x + y = z [corresponding angles]
\n\u21d2 46\u00b0 + 26\u00b0 = z [From (2) and (3)]
\n\u21d2 z = 46\u00b0 + 26\u00b0 \u21d2 z = 72\u00b0
\nHence, value of x = 46\u00b0, y = 26\u00b0 and z = 72\u00b0 Ans.<\/p>\n

Question 13.
\n(a) In the figure (1) given below, AD = BD = DC and \u2220ACD = 35\u00b0. Show that
\n(i) AC > DC
\n(ii) AB > AD.
\n(b) In the figure (2) given below, prove that
\n(i) x + y = 90\u00b0
\n(ii) z = 90\u00b0
\n(iii) AB = BC
\n\"ML
\nSolution:
\n(a) Given: In the figure given,
\nAD = BD = DC
\n\u2220ACD = 35\u00b0
\nTo prove : (i) AC > DC, (ii) AB > AD
\nProof: In \u2206 ADC, AD = DC
\n\u2234 \u2220DAC = \u2220DCA = 35\u00b0
\n\u21d2 \u2220ADC = 180\u00b0 – (\u2220DAC + \u2220DCA)
\n\u2234 \u2220ADC = 180\u00b0 – (35\u00b0 + 35\u00b0)
\n= 180\u00b0 – 70\u00b0= 110\u00b0
\nand Ext. \u2220ADB = \u2220DAC + \u2220DCA = 35\u00b0 + 35\u00b0 = 70\u00b0
\n\"ML
\n\u2235 AD = BD
\n\u2220BAD = \u2220ABD
\nBut \u2220BAD + \u2220ABD = 180\u00b0 – \u2220ADB
\n\u21d2 \u2220ABD + \u2220ABD = 180\u00b0 – 70\u00b0 = 110\u00b0
\n\u21d2 2\u2220ABD = 110\u00b0 \u21d2 \u2220ABD = \\(\\frac{110^{\\circ}}{2}\\) = 55\u00b0
\n(i) Now v \u2220ADC > \u2220DAC
\n\u2234 AC > DC
\nand \u2220ADB > \u2220ABD
\n\u2234 AB > AD<\/p>\n

(b) Given. \u2220 EAC = \u2220 BAC = x
\n\u2220 ABD = \u2220 DBC = y
\n\u2220BDC = z
\nTo prove. (i) x + y = 90\u00b0
\n(ii) z = 90\u00b0
\n(iii) AB = BC
\nProof. (i) \u2234 AE || BC
\n\u2234 \u2220 ACB = x [Alternate angles] …….(1)
\nIn \u2206 ABC
\nx + (y + y) + \u2220 ACB = 180\u00b0 [sum of all angles in a triangle is 180\u00b0]
\n\u21d2 x + 2y + x = 180\u00b0 [From (1)]
\n\u21d2 2x + 2y =180\u00b0
\n\u21d2 2 (x + y) = 180\u00b0 (proved) ……..(2)
\n\u21d2 x + y = 90\u00b0
\n(ii) Now, in \u2206 BCD,
\ny + z + \u2220BCD = 180\u00b0 [sum of all angles in a triangle is 180\u00b0]
\n\u21d2 y + z + x = 180\u00b0
\n\u21d2 90\u00b0 + z = 180\u00b0 [From (2) , x + y = 90\u00b0]
\n\u21d2 z = 90\u00b0 (proved) …..(3)
\n(iii) In \u2206 ABC
\n\u2220 BAC \u2220 B AC = x (each same value)
\n\u2234 AB = CB (In a triangle equal angles has equal sides) (proved)<\/p>\n

Question 14.
\nIn the given figure, ABC and DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at P, show that
\n(i) \u2206ABD \u2245 \u2206ACD
\n(ii) \u2206ABP \u2245 \u2206ACP
\n(iii) AP bisects \u2220A as well as \u2220D
\n(iv) AP is the perpendicular bisector of BC.
\nSolution:
\nGiven : In the figure, two isosceles triangles ABC and DBC are on the same base BC. With vertices A and D on the same side of BC.
\nAD is joined and produced to meet BC at P.
\n\"ML
\nTo prove:
\n(i) \u2206ABD \u2245 \u2206ACD
\n(ii) \u2206ABP \u2245 \u2206ACP
\n(iii) AP bisects \u2220A as well as \u2220D
\n(iv) AP is the perpendicular bisector of BC
\nProof: \u2235 \u2206ABC and \u2206DBC are isosceles
\nAB = AC and DB = DC
\n(i) Now in \u2206ABD and \u2206ACP
\nAB = AC (Proved)
\nDB = DC (Proved)
\nAD = AD (Common)
\n\u2234 \u2206ABD \u2245 \u2206ACD (SSS axiom)
\n\u2234 \u2220BAD = \u2220CAD (c.p.c.t.)
\n\u2234ADP bisects \u2220A
\nand \u2220ADB = \u2220ADC (c.p.c.t.)
\nBut \u2220ADB + \u2220BDP = \u2220CAD + \u2220CDP = 180\u00b0
\n\u2234 \u2220BDP = \u2220CDP
\n\u2234 ADP bisects \u2220D also
\nNow in \u2206APB and \u2206ACD
\nAB = AC (Given)
\nAP = AP (Common)
\nand \u2220BAD = \u2220CAD (Proved)
\n\u2234 \u2220APB \u2245 \u2206ACP (ASA axiom)
\n\u2234 BP = CP (c.p.c.t)
\nand \u2220APB = \u2220APC
\nBut \u2220APB + \u2220APC = 180\u00b0
\n\u2234 \u2220APB = \u2220APC = 90\u00b0
\nand BP = PC
\n\u2234 AP is perpendicular bisector of BC<\/p>\n

Question 15.
\nIn the given figure, AP \u22a5 l and PR > PQ. Show that AR > AQ.
\nSolution:
\nGiven : In the given figure,
\nAP \u22a5 l and PR > PQ
\n\"ML
\nTo prove : AR > AQ
\nConstruction : Take a point S on l,
\nSuch that PS = PQ
\nJoin A and S
\nProof: In \u2206AQP and \u2206ASP
\nAP = AP (Common)
\nQP = SP (Given)
\n\u2220APQ = \u2220APS (Each 90\u00b0)
\n\u2234 \u2206APQ \u2245 \u2206APS
\n\u2234\u22201 = \u22202
\nAQ = AS (Sides opposite to equal angles)
\nIn \u2206ASR
\nExt. ASP > \u2220ARS
\n\u21d2 \u22202 > \u22203
\n\u21d2 \u22201 > \u22203 (\u2235 \u22201 = \u22202)
\n\u21d2 AR > AQ<\/p>\n

Question 16.
\nIf O is any point in the interior of a triangle ABC, show that
\n\"ML
\nOA + OB + OC > \\(\\frac{1}{2}\\)
\n(AB + BC + CA).
\nSolution:
\nGiven : In the figure, O is any point in the interior of \u2206ABC.
\nTo prove : OA + OB + OC > \\(\\frac{1}{2}\\) (AB + BC + CA)
\nConstruct: Join B and C.
\nProof: In AOBC
\nOB + OC > BC …(i)
\n(Sum of two sides of a triangle is greater than its third side)
\nSimilarly OC + OA > CA
\nand OA + OB > AB
\nAdding are get,
\n(OB + OC + OC + OA + OA + OB) > BC + CA + AB
\n\u21d2 2(OA + OB + OC) > AB + BC +CA
\n\u21d2 OA + OB + OC > \\(\\frac{1}{2}\\) (AB + BC + CA)<\/p>\n

Question P.Q.
\nConstruct a triangle ABC given that base BC = 5.5 cm, \u2220 B = 75\u00b0 and height = 4.2 cm. Ans. Given. In a triangle ABC, Base BC = 5.5. cm, \u2220B = 750\u00b0 and height = 4.2 cm.
\n\"ML
\nSolution:
\nTo construct a triangle ABC.
\nSteps of Construction :
\n(1) Draw a line BC = 5.5 cm.
\n(2) Draw \u2220PBC = 75\u00b0.
\n(3) Draw the perpendicular bisector of BC and cut the BC at point D.
\n(4) Cut the DM at point E such that DE = 4.2 cm.
\n(5) Draw the line at point which is parallel to line BC.
\n(6) This parallel line cut the BP at point A.
\n(7) Join AC.
\n(8) ABC is the required triangle.<\/p>\n

Question P.Q.
\nConstruct a triangle ABC in which BC = 6.5 cm, \u2220 B = 75\u00b0 and \u2220 A = 45\u00b0. Also construct median of \u2206 ABC passing through B. Given. In \u2206 ABC , BC = 6.5 cm, \u2220B = 75\u00b0 and \u2220 A = 45\u00b0.
\nSolution:
\n(i) To construct a triangle ABC.
\n(ii) Construct median of \u2206 ABC passing through B.
\n\"ML
\nStep of Construction.
\n(1) Draw a line BC = 6.5 cm.
\n(2) Make \u2220PBC = 75\u00b0.
\n(3) Make \u2220BCQ = 60\u00b0.
\n(4) BP and CQ cut at point A.
\n(5) ABC is the required triangle.
\n(6) Draw the bisector of AC.
\n(7) The bisector line cut the line AC at point D.
\n(8) Join BD.
\n(9) BD is the required median of A ABC passing through B.<\/p>\n

Question P.Q.
\nConstruct triangle ABC given that AB – AC = 2.4 cm, BC = 6.5 cm. and \u2220 B = 45\u00b0. Ans. Given. A triangle ABC in which AB – AC = 2.4 cm, BC = 6.5 cm, \u2220B = 4.5\u00b0.
\nSolution:
\nTo construct a triangle ABC.
\n\"ML
\nSteps of Construction :
\n(1) Draw BC = 6.5 cm.
\n(2) Draw BP making angle 65\u00b0 with BC.
\n(3) From BP, cut BD = 2.4 cm.
\n(4) Join D and C.
\n(5) Draw perpendicular bisector of DC which cuts BP at A.
\n(6) Join A and C.
\n(7) ABC is the required triangle.<\/p>\n

ML Aggarwal Class 9 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 10 Triangle Chapter Test Question 1. In triangles ABC and DEF, \u2220A = \u2220D, \u2220B = \u2220E and AB = EF. Will the two triangles be congruent? Give reasons for your answer. Solution: In \u2206ABC and \u2206DEF \u2220A = \u2220D \u2220B = \u2220E AB = EF […]<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[3034],"tags":[],"yoast_head":"\nML Aggarwal Class 9 Solutions for ICSE Maths Chapter 10 Triangle Chapter Test - A Plus Topper<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.aplustopper.com\/ml-aggarwal-class-9-solutions-for-icse-maths-chapter-10-chapter-test\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 10 Triangle Chapter Test\" \/>\n<meta property=\"og:description\" content=\"ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 10 Triangle Chapter Test Question 1. 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