{"id":381,"date":"2020-09-15T05:09:22","date_gmt":"2020-09-14T23:39:22","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=381"},"modified":"2021-07-02T15:01:52","modified_gmt":"2021-07-02T09:31:52","slug":"kseeb-sslc-class-10-maths-solutions-chapter-2-ex-2-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-maths-solutions-chapter-2-ex-2-3\/","title":{"rendered":"KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.3"},"content":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.3<\/strong>\u00a0are part of KSEEB SSLC Class 10 Maths Solutions<\/a>. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.3.<\/p>\n

Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.3<\/h2>\n

Question 1.
\nState which pairs of triangles In the following figures are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:
\n\"KSEEB
\nSolution:
\n(i)
\n\"KSEEB
\nIn \u2206ABC and \u2206PQR,
\n\u2220A = \u2220P = 60\u00b0
\n\u2220B=\u2220Q = 80\u00b0
\n\u2220C =\u2220R = 40\u00b0
\n\u2234 SimilarIty criterion of two triangles is A.A.A.
\n\u2234 \u2206ABC ~ \u2206PQR.<\/p>\n

(ii)
\n\"KSEEB
\nIn \u2206ABC and PQR, we have
\n\\(\\frac{A B}{Q R}=\\frac{B C}{P R}=\\frac{A C}{P Q}=\\frac{1}{2}\\)
\ni.e., sides of one triangle are proportional to the sides of the other triangle.
\n\u2234SImilarity criterion is S.S.S.
\n\u2234 \u2206ABC ~ \u2206PQR.<\/p>\n

(iii)
\n\"KSEEB
\n\u2206LMP and \u2206DEF are not similar triangles because one pair of sides are not proportional to each other.
\n\\(\\quad \\frac{\\mathrm{MP}}{\\mathrm{DE}}=\\frac{\\mathrm{LP}}{\\mathrm{DF}}=\\frac{2}{4}=\\frac{3}{6}=\\frac{1}{2}\\)
\nBut, \\(\\frac{\\mathrm{LM}}{\\mathrm{EF}}=\\frac{2.7}{5} \\neq \\frac{1}{2}\\)
\n\u2234 These are not similar triangles<\/p>\n

(iv)
\n\"KSEEB
\nIn \u2206MNL and \u2206PQR, we have
\n\u2220M = \u2220Q = 70\u00b0
\nBut, \\(\\frac{\\mathrm{MN}}{\\mathrm{PQ}} \\neq \\frac{\\mathrm{ML}}{\\mathrm{QR}}\\)
\n\u2234 These are not similar triangles<\/p>\n

(v)
\n\"KSEEB
\nIn \u2206ABC and \u2206DEF, we have
\n\\(\\frac{A B}{D F}=\\frac{B C}{E F}=\\frac{1}{2}\\)
\n\u2220A = \u2220F = 80\u00b0
\n\u2234 Similarly criterion for these triangles is S.A.S
\n\u2234 \u2206ABC ~ \u2206DEF<\/p>\n

(vi)
\n\"KSEEB
\nSum of three angles of is 180\u00b0, but in \u2206DEF. sum of two angles. \u2220F= 30\u00b0. In \u2206PQR, \u2220P = 70\u00b0.
\nIn \u2206DEF and \u2206PQR. we have
\n\u2220D = \u2220P = 70\u00b0
\n\u2220E =\u2220R = 80\u00b0
\n\u2220F = \u2220R = 30\u00b0
\n\u2234 Three angles of these triangles are equal to each other.
\n\u2234 SimIlarity criterion here is A.A.A.
\n\u2234 \u2206DEF ~ \u2206PQR.<\/p>\n

Question 2.
\nIn the following figure. \u2206OBA ~ \u2206ODC, \u2220BOC = 125\u00b0 and \u2220CDO = 70\u00b0. Find \u2220DOC, \u2220DCO and \u2220OAB.
\n\"KSEEB
\nSolution:
\n\u2206OBA ~ \u2206ODC (data)
\nIn \u2206OBA and \u2206ODC, we have
\n\u2220ODC = \u2220OBA = 70\u00b0 (. Alternate angle)
\n\u2220AOB = 180\u00b0- 125\u00b0 = 55\u00b0 ( Adjacent angle)
\n\u2234 \u2220AOB = \u2220DOC = 55\u00b0 (: VertIcally opposite angles)
\n\u2220OAB = \u2220OCD = 55\u00b0 (. Alternate angles)
\n\u2234 \u2220DOC=55\u00b0
\n\u2220DCO = 55%
\n\u2220OAB= 55%.<\/p>\n

Question 3.
\nDiagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that \\(\\frac{\\mathrm{OA}}{\\mathrm{OC}}=\\frac{\\mathrm{OB}}{\\mathrm{OD}}\\)
\n\"KSEEB
\nSolution:
\nData: Diagonals AC and BD of a trapezium ABCD with AB || DC. intersect each other at the point O.
\nTo Prove: \\(\\frac{\\mathrm{OA}}{\\mathrm{OC}}=\\frac{\\mathrm{OB}}{\\mathrm{OD}}\\)
\nIn the trapezium ABCD AB || DC.
\n\u2234 \u2220OCD = \u2220OAB (Alternate angles)
\n\u2220ODC = \u2220OBA (Alternate angles)
\n\u2220DOC = \u2220AOB (Vertically opposite angles)
\n\u2234 These are equlangular triangles.
\n\u2234 These triangles are similar.
\n\u2234 \u2206ODC ~ \u2206OAB
\n\\(\\quad \\frac{\\mathrm{OA}}{\\mathrm{OC}}=\\frac{\\mathrm{OB}}{\\mathrm{OD}}\\)
\n\u2235 Corresponding sides are in proportional.<\/p>\n

Question 4.
\nIn the following figure. \\(\\frac{\\mathrm{QR}}{\\mathrm{QS}}=\\frac{\\mathrm{QT}}{\\mathrm{PR}}\\) and \u22201 = \u22202. Show that \u2206PQS ~ \u2206TQR.
\n\"KSEEB
\nSolution:
\nData: Here \\(\\frac{\\mathrm{QR}}{\\mathrm{QS}}=\\frac{\\mathrm{QT}}{\\mathrm{PR}}\\)and \u22201 = \u22202
\nTo prove: \u2206PQS ~ \u2206TQR.
\nIn \u2206PQS and \u2206TQR, we have
\n\\(\\frac{\\mathrm{Q} \\mathrm{R}}{\\mathrm{QS}}=\\frac{\\mathrm{QT}}{\\mathrm{PR}}\\)
\n\u2220PQR = \u2220PRQ (\u2235 \u22201 = \u22202)
\nHere, Similarity criterion used here is side, angle, side (SAS).
\n\u2234 \u2206PQS ~ \u2206TQR.<\/p>\n

Question 5.
\nS and T are points on sides PR and QR of \u2206PQR such that \u2220P = \u2220RTS. Show that \u2206RPQ ~ \u2206RTS.
\n\"KSEEB
\nSolution:
\nData: S and T are points on sides PR and QR of \u2206PQR such that \u2220P = \u2220RTS.
\nTo Prove: \u2206RPQ ~ \u2206RTS.
\nIn \u2206RPQ and \u2206RTS,
\n\u2220P = \u2220RTS (Data)
\n\u2220PRQ = \u2220SRT (Common)
\n\u2234 3rd angle \u2220PRQ = \u2220SRT
\n\u2234 These are equiangularangular triangles.
\n\u2234 Here A.A.A. similarity criterion.
\n\u2234 \u2206RPQ ~ \u2206RTS.<\/p>\n

Question 6.
\nIn the given fIgure. \u2206ABE \u2245 \u2206ACD. show that \u2206ADE ~ \u2206ABC.
\n\"KSEEB
\nSolution:
\nData: \u2206ABE \u2245 \u2206ACD.
\nTo Prove: \u2206ADE ~ \u2206ABC.
\n\u2206ABE \u2245 \u2206ACD
\nAB = AC
\nAD = AE.
\nThen DC = BE
\nAB = AC
\nAD + DB = AE + EC
\n\u2234 DB = EC
\n(\u2235 DA = AE)
\nIn \u2206ADE and \u2206ABC,
\n\\(\\frac{\\mathrm{AD}}{\\mathrm{AB}}=\\frac{\\mathrm{AE}}{\\mathrm{AC}}=\\frac{\\mathrm{DE}}{\\mathrm{BC}}\\)
\nHere, corresponding sides are in proportion.
\n\u2234 Similarity criterion for \u2206 is side, side, side
\n\u2234 \u2206ADE ~ \u2206ABC<\/p>\n

Question 7.
\nIn following figure. altitudes AD and CE of \u2206ABC Intersect each other at the point P. Show that:
\n\"KSEEB
\ni) \u2206AEP ~ \u2206CDP
\nii) \u2206ABD ~ \u2206CBE
\niii) \u2206AEP ~ \u2206A\u00d0B
\niv) \u2206PDC ~ \u2206BEC.
\nSolution:
\nData: altitudes AD and CE of ABC
\nIntersect each other at the point P.
\nTo Prove: i) \u2206AEP ~ \u2206CDP
\nii) \u2206ABD ~ \u2206CBE
\niii) \u2206AEP ~ \u2206ADB
\niv) \u2206PDC ~ \u2206BEC<\/p>\n

i) In \u2206AEP and \u2206CDP.
\n\u2220AEP = \u2220CDP = 90\u00b0 (data)
\n\u2220APE = \u2220CPD (Vertically opposite angle)
\n\u2234 \u2220PAE = \u2220PCD
\nThese are equiangular triangles.
\nSimilarity criterion for \u2206 is A.A.A
\n\u2234 \u2206AEP ~ \u2206CDP<\/p>\n

(ii) In \u2206ABD and \u2206CBE.
\n\u2220ADB = \u2220CEB = 90\u00b0 (data)
\n\u2220ABD = \u2220CBE (common)
\n\u2234 \u2220DAB = \u2220BCE
\nThese are equiangular triangles.
\n\u2234 Similarity criterion for is A.A.A.
\n\u2234 \u2206ABD ~ \u2206CBE<\/p>\n

(iii) In \u2206AEP and \u2206ADB.
\n\u2220AEB = \u2220ADB = 90\u00b0 (data)
\n\u2220PAE = \u2220DAB (common)
\n\u2234 \u2220APE = \u2220ABD.
\n\u2234 These are equiangular triangles.
\n\u2234 Similarity criterion for is A.A.A.
\n\u2234 \u2206AEP ~ \u2206ADB<\/p>\n

(iv) In \u2206PDC and \u2206BEC.
\n\u2220PDC = \u2220BEC = 90\u00b0 (data)
\n\u2220PCD = \u2220BCE (common)
\n\u2234 \u2220CPD = \u2220CBE
\n\u2234 These are equiangularangular triangles.
\nSimilarity criterion for \u2206 is A.A.A.
\n\u2234 \u2206PDC ~ \u2206BEC.<\/p>\n

Question 8.
\nE is a point on the side AD produced of a parallelogram ABCD and BE Intersects CD at F. Show that \u2206ABE ~ \u2206CFB.
\n\"KSEEB
\nSolution:
\nData: E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F.
\nTo Prove: \u2206ABE ~ \u2206CFB
\nIn \\(\\) ABCD Adjacent angles are equal.
\nLet \u2220DAB = \u2220BCD = 70\u00b0
\n\u2220DAB = \u2220EAF = 70\u00b0 (\u2235 Corresponding angle)
\nIn \u2206EDF, \u2220DEF = 30\u00b0 then,
\n\u2220EFD = 80\u00b0.
\n\u2220EFD = \u2220BFC = 80\u00b0 (vertIcally opposite angles)
\nIn \u2206FBC, \u2220FBC = 30\u00b0. ,
\nNow in \u2206ABE and \u2206CFB,
\n\u2206EAB = \u2206BCF = 70\u00b0
\n\u2206AEB = \u2206FBC = 30\u00b0
\n\u2206ABE = \u2206BFC = 80\u00b0
\n\u2234 Similarity criterion for \u2206 is A.A.A.
\n\u2234 \u2206ABE ~ \u2206CFB<\/p>\n

Question 9.
\nIn the given figure. \u2206ABC and \u2206AMP are two right triangles, right angled at B and M respectively. Prove that:
\ni) \u2206ABC ~ \u2206AMP
\nii) \\(\\frac{\\mathrm{CA}}{\\mathrm{PA}}=\\frac{\\mathrm{BC}}{\\mathrm{MP}}\\)
\n\"KSEEB
\nSolution:
\nData: \u2206ABC and \u2206AMP are two right triangles, right angled at B and M respectvely.
\nTo Prove: i) \u2206ABC ~ \u2206AMP
\nii) \\(\\frac{\\mathrm{CA}}{\\mathrm{PA}}=\\frac{\\mathrm{BC}}{\\mathrm{MP}}\\)
\n(i) In \u2206ABC and \u2206AMP,
\n\u2220ABC = \u2220AMP = 90\u00b0 (data)
\n\u2220CAB = \u2220MAP (common)
\n\u2234 \u2220ACB = \u2220MPA
\n\u2234 These are equiangular triangles.
\nSimilarity criterion for \u2206 is A.A.A.
\n\u2234 \u2206ABC ~ \u2206AMP<\/p>\n

(ii) \u2206ABC ~ \u2206AMP(Proved)
\n\u2234 Corresponding sides are in proportion.
\nLC and LP are corresponding angles.
\n\u2234 Adjacent sides are CA, PA.
\nSimilarly BC and MP are adjacent sides.
\n\\(\\quad \\frac{\\mathrm{CA}}{\\mathrm{PA}}=\\frac{\\mathrm{BC}}{\\mathrm{MP}}\\)<\/p>\n

Question 10.
\nCD and OH are respectively the bisectors of \u2220ACB and \u2220EGF such that D and H lie on sides AB and FE of \u2206ABC and \u2206EFG respectively.
\n\"KSEEB
\nIf \u2206ABC ~ \u2206EFG, show that
\ni) \\(\\frac{C D}{G H}=\\frac{A C}{F G}\\)
\nii) \u2206DCB ~ \u2206HGE
\niii) \u2206DCA ~ \u2206HGF
\nSolution:
\nData : CD and OH are respectively the bisectors of \u2220ACB and \u2220EGF such that D and H 11e on sides AB and FE of \u2206ABC and \u2206EFG respectively.
\n\u2206ABC ~ \u2206EFG
\nTo Prove: i) \\(\\frac{C D}{G H}=\\frac{A C}{F G}\\)
\nii) \u2206DCB ~ \u2206HGE
\niii) \u2206DCA ~ \u2206HGF
\nProof: \u2206ABC ~ \u2206EFG (data given)
\n\u2234 Their corresponding sides are in proportion.
\n\\(\\quad \\frac{\\mathrm{AB}}{\\mathrm{EF}}=\\frac{\\mathrm{BC}}{\\mathrm{FG}}=\\frac{\\mathrm{AC}}{\\mathrm{EG}}\\)
\n\u2220B = \u2220F, \u2220A = \u2220E, \u2220C = \u2220G.<\/p>\n

(i) In \u2206ADC and \u2206EHG,
\n\u2220A = \u2220E .
\n\u2220ACD = \u2220EGH
\n\u2234 Their sides are in proportion.
\n\\(\\frac{\\mathrm{CD}}{\\mathrm{GH}}=\\frac{\\mathrm{AC}}{\\mathrm{FG}}\\)<\/p>\n

(ii) In \u2206DCB and \u2206HGE.
\n\\(\\frac{\\mathrm{CD}}{\\mathrm{GH}}=\\frac{\\mathrm{BC}}{\\mathrm{GF}}\\)
\n\u2234 \u2206DCB ~ \u2206HGE<\/p>\n

(iii) In \u2206DCA and \u2206HGE,
\n\\(\\frac{\\mathrm{DC}}{\\mathrm{GH}}=\\frac{\\mathrm{AD}}{\\mathrm{EH}}=\\frac{\\mathrm{AC}}{\\mathrm{EG}}\\)
\n\u2234 \u2206DCA ~ \u2206HGE<\/p>\n

Question 11.
\nIn the following figure. E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD \u22a5 BC and EF \u22a5 AC, prove that \u2206ABD ~ \u2206ECF
\n\"KSEEB
\nSolution:
\nData: E Is a point on side CB produced of an isosceles \u2206ABC with AB = AC. AD \u22a5 BC and EF \u22a5 AC,
\nTo Prove: \u2206ABD ~ \u2206ECF
\nIn \u2206ABD and \u2206ECF.
\n\u2220ADB = \u2220EFC = 90\u00b0 (data)
\n\u2220ABD = \u2220FCE (\u2235 \u2220B = \u2220C)
\n\u2220BAD = \u2220FEC
\n\u2234 Equiangular triangles.
\n\u2234 Similarity criterion for triangles is A.A.A.
\n\u2234 \u2206ABD ~ \u2206ECF.<\/p>\n

Question 12.
\nSides AB and BC and median AD of a \u00bfABC are respectively proportional to sides PQ and QR and median PM of \u2206PQR. (Sec figure gIven below) Show that \u2206ABC ~ \u2206PQR
\n\"KSEEB
\nSolution:
\nData: Sides AB and BC and median AD of a \u2206ABC are respectively proportional to sides P9 and QR and medIan PM of \u2206PQR.
\nTo Prove: \u2206ABC ~ \u2206PQR
\nIn \u2206ABC and \u2206PQR,
\nSide \\(\\frac{\\mathrm{AB}}{\\mathrm{PG}}=\\frac{\\mathrm{BC}}{\\mathrm{QR}}=\\frac{\\mathrm{AD}}{\\mathrm{PM}}\\)
\nD is the mid-point of BC
\n\u2234 BD = DC
\nM is the mid-point of QR.
\n\u2234 QM = MR.
\nIn \u2206ADC and \u2206PMR,
\n\\(\\frac{\\mathrm{AD}}{\\mathrm{PM}}=\\frac{\\mathrm{DC}}{\\mathrm{MR}}=\\frac{\\mathrm{AC}}{\\mathrm{PR}} \\quad \\left(\\frac{\\mathrm{BC}}{\\mathrm{QR}}=\\frac{\\frac{1}{2} \\mathrm{BC}}{\\frac{1}{2} \\mathrm{QR}}\\right)\\)
\nSimilarity criterion is S.S.S.
\n\u2234 \u2206A\u00d0C ~ \u2206PMR
\nNow, In \u2206ABC and \u2206PQR.
\n\\(\\frac{A B}{P Q}=\\frac{B C}{Q R}=\\frac{A C}{P R}\\)
\n\u2234 SimilarIty criterion is S.S.S.
\n\u2234 \u2206ABC ~ \u2206PQR.<\/p>\n

Question 13.
\nD is a point on the side BC of a triangle ABC such that \u2220ADC = \u2220BAC. Show that CA2<\/sup> = CB.CD.
\n\"KSEEB
\nSolution:
\nData: D is a point on the side BC of a triangle \u2206ABC such that \u2220ADC = \u2220BAC.
\nTo Prove: CA2<\/sup> = CB \u00d7 CD
\nLet \u2220ADC = \u2220BAC = 100\u00b0
\nIn \u2206ABC, If \u2220B = 50\u00b0,then \u2220C = 30\u00b0
\nIn \u2206ADC, If \u2220C = 30\u00b0. then \u2220DAC = 50\u00b0
\nIn \u2206BCP, \u2220A= 100\u00b0, \u2220B= 50\u00b0, \u2220C= 30\u00b0
\nIn \u2206ADC, \u2220ADC = 100. \u2220DAC = 50\u00b0.
\n\u2220ACD = 30\u00b0
\nSimilarity criterion of \u2206 is A.A.A.
\n\u2234 In \u2206ABC and \u2206ADC,
\n\\(\\frac{\\mathrm{CA}}{\\mathrm{BC}}=\\frac{\\mathrm{DC}}{\\mathrm{CA}}\\)
\n\u2234 CA \u00d7 CA = BC \u00d7 DC
\n\u2234 CA2<\/sup> = BC \u00d7 DC.<\/p>\n

Question 14.
\nSides AB and AC and median AD of a triangle \u2206ABC are respectively proportional to sides PQ and PR and mediam PM of another triangle \u2206PQR. Show that \u2206ABC ~ \u2206PQR.
\n\"KSEEB
\nSolution:
\nData: Sides AB and AC and median AD of a triangle \u2206ABC are respectively proportional to sides PQ and PR and medIan PM of another triangle \u2206PQR.
\nTo Prove: \u2206ABC ~ \u2206PQR
\nIn \u2206ABC and \u2206PQR
\n\\(\\frac{\\mathrm{AB}}{\\mathrm{PQ}}=\\frac{\\mathrm{AC}}{\\mathrm{PR}}=\\frac{\\mathrm{AD}}{\\mathrm{PM}}\\)
\nCorresponding sides are proportional.
\nSimilarity criterion for \u2206 is S.S.S
\n\u2234 \u2206ABC ~ \u2206PQR.<\/p>\n

Question 15.
\nA vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
\n\"KSEEB
\nSolution:
\nHeight of PQ =?
\nIn \u2206ABC and \u2206PQR,
\n\u2220B = \u2220Q = 90\u00b0
\n\u2234 Hypotenuse AC is proportional to PQ
\n\\(\\quad \\frac{\\mathrm{AB}}{\\mathrm{PQ}}=\\frac{\\mathrm{BC}}{\\mathrm{MR}}=\\frac{\\mathrm{AC}}{\\mathrm{PR}}\\)
\n\\(\\frac{6}{\\mathrm{PQ}}=\\frac{4}{28}\\)
\n\u2234 4 \u00d7 PQ = 6 \u00d7 28
\n\\(\\quad \\mathrm{PQ}=\\frac{6 \\times 28}{4} \\quad \\quad \\mathrm{PQ}=42 \\mathrm{m}\\)<\/p>\n

Question 16.
\nIf AD and PM are medians of triangles \u2206ABC and \u2206PQR respectively, where \u2206ABC ~ \u2206PQR. prove that \\(\\frac{\\mathrm{AB}}{\\mathrm{PQ}}=\\frac{\\mathrm{AD}}{\\mathrm{PM}}\\)
\n\"KSEEB
\nSolution:
\nData : AD and PM are medians of triangles \u2206ABC and \u2206PQR respectively. \u2206ABC ~ \u2206PQR.
\nTo Prove: \\(\\frac{\\mathrm{AB}}{\\mathrm{PQ}}=\\frac{\\mathrm{AD}}{\\mathrm{PM}}\\)
\n\u2206ABC ~ \u2206PQR.(data)
\n\u2234 we have \\(\\frac{\\mathrm{AB}}{\\mathrm{PQ}}=\\frac{\\mathrm{AC}}{\\mathrm{PR}}=\\frac{\\mathrm{BC}}{\\mathrm{QR}}\\)
\nD is the mid-point of BC. BD = DC.
\nM Is the mid-point of QR. QM = MR
\n\\(\\frac{\\mathrm{BC}}{\\mathrm{QR}}=\\frac{\\frac{1}{2} \\mathrm{BC}}{\\frac{1}{2} \\mathrm{QR}}=\\frac{\\mathrm{BD}}{\\mathrm{QM}}\\)
\n\\(\\quad \\frac{\\mathrm{BC}}{\\mathrm{QR}}=\\frac{\\mathrm{BD}}{\\mathrm{QM}}\\)
\nIn \u2206ABD and \u2206PQM, we have
\n\\(\\quad \\frac{\\mathrm{AB}}{\\mathrm{PQ}}=\\frac{\\mathrm{BD}}{\\mathrm{QM}}=\\frac{\\mathrm{AD}}{\\mathrm{PM}}\\)
\n\\(\\quad \\frac{\\mathrm{AB}}{\\mathrm{PQ}}=\\frac{\\mathrm{AD}}{\\mathrm{PM}}\\)<\/p>\n

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.3 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.3, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.3\u00a0are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.3. Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.3 Question 1. State which pairs of triangles In the following …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/381"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=381"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/381\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=381"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=381"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=381"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}