{"id":515,"date":"2020-09-15T09:00:03","date_gmt":"2020-09-15T03:30:03","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=515"},"modified":"2021-07-02T15:01:51","modified_gmt":"2021-07-02T09:31:51","slug":"kseeb-sslc-class-10-maths-solutions-chapter-2-ex-2-5","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-maths-solutions-chapter-2-ex-2-5\/","title":{"rendered":"KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.5\u00a0"},"content":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.5<\/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 2.5.<\/p>\n

Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.5<\/h2>\n

Question 1.
\nSides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.
\ni) 7 cm, 24 cm, 25 cm,
\nii) 3 cm, 8 cm, 6 cm.
\niii) 50 cm, 80 cm, 100 cm.
\niv) 13 cm, 12 cm, 5 cm.
\nSolution:
\nIn \u22a5\u2206ABC, \u2220B = 90\u00b0.
\nLet AB = a, BC = b, Hypotenuse AC = c then
\nAC2<\/sup> = AB2<\/sup> + BC2<\/sup>
\nc2<\/sup> = a2<\/sup> + b2<\/sup>
\n\u2234 Here diagonal is the greatest side.
\ni) a, b,c
\n7 cm, 24 cm, 25 cm,
\nc2<\/sup> = a2<\/sup> + b2<\/sup>
\n252<\/sup> = (7)2<\/sup> + (24)2<\/sup>
\n625 = 49 + 576
\n625 = 625
\n625 = 49 + 576 625 = 625
\n\u2234 This is right angled triangle.
\nMeasurement of Hypotenuse, c = 25 cm.<\/p>\n

ii) a c b
\n3 cm, 8 cm, 6 cm.
\nc2<\/sup> = a2<\/sup> + b2<\/sup>
\n82<\/sup> = (3)2<\/sup> + (6)2<\/sup>
\n64 = 9 + 36
\n64 \u2260 45
\n\u2234 These are not sides of right angled triangle.<\/p>\n

iii) a b c
\n50 cm, 80 cm, 100 cm.
\nc2<\/sup> = a2<\/sup> + b2<\/sup>
\n1002<\/sup>= (50)2<\/sup> + (80)2<\/sup>
\n10000 = 2500 + 6400
\n10000 \u2260 8900
\n\u2234 These are not sides of right angled triangle.<\/p>\n

iv) a b c
\n12 cm, 5 cm, 13 cm,
\nc2<\/sup> = a2<\/sup> + b2<\/sup>
\n132<\/sup> = (12)2<\/sup> + (5)2<\/sup>
\n169 = 144 + 25
\n169 = 169
\n\u2234 These are sides of right angled triangle.
\nMeasurement of Hypotenuse =13 cm.<\/p>\n

Question 2.
\nPQR is a triangle right angled at P and M is a point on QR such that PM \u22a5 QR. Show that PM2<\/sup> = QM.MR.
\nSolution:
\n\"KSEEB
\nData: PQR is a triangle 9 right angled at P and M is a point on QR such that PM \u22a5 QR.
\nTo Prove: PM2<\/sup> = QM.MR
\nProof: In \u22a5D QPR, \u2220P = 90\u00b0,
\nQR2<\/sup> = QP2<\/sup> + PR2<\/sup> ……………. (i)
\nIn \u22a5D PMQ, \u2220M = 90\u00b0
\nQP2<\/sup> = PM2<\/sup> + QM2<\/sup> ………. (ii)
\nIn \u22a5D PMR, \u2220M = 90\u00b0
\nPR2<\/sup> = PM2<\/sup> + MR2<\/sup> ……………. (iii)
\nBy Adding eqn. (ii) and (iii)
\nQP2<\/sup> + PR2<\/sup> = PM2<\/sup> + QM2<\/sup> + PM2<\/sup> + MR2<\/sup>
\nQR2<\/sup> = 2PM2<\/sup> + QM2<\/sup> + MR2<\/sup> (. Eqn. (i))
\n(QM + MR)2<\/sup> = 2PM2<\/sup> + QM2<\/sup> + MR2<\/sup>
\nQM2<\/sup> + MR2<\/sup> + 2QM.MR = 2PM2<\/sup> + QM2<\/sup> + MR2<\/sup>
\nQM2<\/sup> – QM2<\/sup> + MR2<\/sup> – MR2<\/sup> + 2QM.MR = 2PM2<\/sup>
\n2QM.MR = 2PM2<\/sup>
\n2PM2<\/sup> = 2QM.MR
\n\u2234 PM2<\/sup> = QM.MR<\/p>\n

Question 3.
\nIn the following figure, ABD is a triangle right angled at A and AC \u22a5BD. Show that
\ni) AB2<\/sup>= BC.BD
\nii) AC2<\/sup> = BC.DC
\niii) AD2<\/sup> = BD.CD
\n\"KSEEB
\nSolution:
\nData: In \u2206ABD, \u2220A = 90\u00b0,
\nAC \u22a5 BD.
\nTo Proved: AB2<\/sup> = BC.BD
\nii) AC2<\/sup> = BC.DC
\niii) AD2<\/sup> = BD.CD<\/p>\n

i) AB2<\/sup> = BC.BD
\n\u2206ACB ~ \u2206BAD (. Theorem7)
\n\\(\\frac{\\mathrm{AB}}{\\mathrm{BD}}=\\frac{\\mathrm{BC}}{\\mathrm{AB}}\\)
\n\u2234 AB2<\/sup>= BC \u00d7 BD.<\/p>\n

ii) AC2<\/sup> = BC.DC
\n\u2206BCA ~ \u2206ACD
\n\\(\\frac{\\mathrm{AB}}{\\mathrm{AD}}=\\frac{\\mathrm{AC}}{\\mathrm{CD}}=\\frac{\\mathrm{BC}}{\\mathrm{AC}}\\)
\n\u2234 AC \u00d7 AC = BC \u00d7 CD
\n\u2234 AC2<\/sup> = BC \u00d7 CD<\/p>\n

iii) AD2<\/sup> = BD.CD
\n\u2206ACD ~ \u2206BAD
\n\\(\\frac{\\mathrm{AD}}{\\mathrm{BD}}=\\frac{\\mathrm{CD}}{\\mathrm{AD}}=\\frac{\\mathrm{AC}}{\\mathrm{AB}}\\)
\n\u2234 AD \u00d7 AD = BD \u00d7 DC
\n\u2234 AD2<\/sup> = BD \u00d7 DC<\/p>\n

Question 4.
\nABC is an isosceles triangle right angled at C. Prove that AB2<\/sup> = 2AC2<\/sup>.
\nSolution:
\nData: ABC is an isosceles triangle right angled at C.
\n\"KSEEB
\nTo Prove: AB2<\/sup> = 2AC2<\/sup>.
\nIn \u22a5\u2206ACB, \u2220C = 90\u00b0
\n\u2234 AB2<\/sup>= AC2<\/sup> + BC2<\/sup>
\n(\u2235 Pythagoras Theorem)
\nAB2<\/sup> = AC2<\/sup> + AC2<\/sup> (\u2235 AC = BC)
\nAB2<\/sup> = 2 AC2<\/sup>.<\/p>\n

Question 5.
\nABC is an isosceles triangle with AC = BC. If AB2<\/sup> = 2AC2<\/sup>, prove that ABC is a right-angled triangle.
\nSolution:
\nData: ABC is an isosceles triangle with AC = BC.
\n\"KSEEB
\nAB2<\/sup> = 2 AC2<\/sup>.
\nTo Prove: \u2206ABC is a right angled triangle
\nAB2<\/sup> = 2AC2<\/sup> (Data)
\nAB2<\/sup> = AC2<\/sup> + AC2<\/sup>
\nAB2<\/sup> = AC2<\/sup> + BC2<\/sup> (\u2235 AC = BC)
\nNow, in \u2206ABC, square of one side is equal to squares of other two sides.
\n\u2206ABC is a right angled triangle, Opposite angle to AB, i.e., \u2220C is 90\u00b0.<\/p>\n

Question 6.
\nABC is an equilateral triangle of side 2a. Find each of its altitudes.
\nSolution:
\nData: ABC is an equilateral triangle of side 2a.
\n\"KSEEB
\nTo Prove: Altitude of \u2206ABC,
\nAD =?
\nIn equilateral triangle
\nD bisect base.
\n\u2234 AB = BC = CA = 2a. 3 D a
\nIf BC = 2a,
\n\\(\\frac{1}{2} \\mathrm{BC}=\\mathrm{a}\\) unit
\n\u2234 BD = DC = a.
\nNow, in \u22a5\u2206ADB, \u2220D = 90\u00b0
\n\u2234 AD2<\/sup> + BD2<\/sup> = AB2<\/sup>
\nAD2<\/sup> + a2<\/sup> = (2a)2<\/sup>
\nAD2<\/sup> + a2<\/sup> = 4a2<\/sup>
\n\u2234 AD2<\/sup> = 4a2<\/sup> – a2<\/sup>
\nAD2<\/sup> = 3a2<\/sup>.
\n\\(\\sqrt{A D^{2}}=\\sqrt{3 a^{2}}\\)
\n\u2234Altitude,\\(\\mathrm{AD}=\\sqrt{3} \\mathrm{a}\\) unit.<\/p>\n

Question 7.
\nProve that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
\n\"KSEEB
\nSolution:
\nData: ABCD is a rhombus.
\nHere, AB = BC = CD = DA.
\nDiagonals AC and BD intersects at \u2019O’.
\nTo Prove: AB2<\/sup> + BC2<\/sup> + CD2<\/sup> + DA2<\/sup> = AC2<\/sup> + BD2<\/sup>.
\nIn rhombus diagonals bisects perpendicularly.
\n\u2234 \u2220AOB = \u2220AOD = 90\u00b0.
\nIn \u22a5\u2206AOB,
\nAB2<\/sup> = AO2<\/sup> + BO2<\/sup> ……..(i)
\nIn \u22a5\u2206BOC,
\nAC2<\/sup> = BO2<\/sup> + CO2<\/sup> ……..(ii)
\nIn \u22a5\u2206COD,
\nCD2<\/sup> = OC2<\/sup> + OD2<\/sup> ……….(iii)
\nIn \u22a5\u2206AOD,
\nAD2<\/sup> = AO2<\/sup> + OD2<\/sup>…………(iv)
\nBy Adding equations (i) + (ii) + (iii) + (iv)
\nAB2<\/sup> + BC2<\/sup> + CD2<\/sup> + DA2<\/sup> =
\n= AO2<\/sup> + BO2<\/sup> + BO2<\/sup> + CO2<\/sup> + CO2<\/sup> + DO2<\/sup> + AO2<\/sup> + OD2<\/sup>
\n= 2AO2<\/sup> + 2BO2<\/sup> + 2CO2<\/sup> + 2DO2<\/sup>
\n= 2AO2<\/sup> + 2CO2<\/sup> + 2BO2<\/sup> + 2DO2<\/sup>
\nNow, RHS = 2AO2<\/sup> + 2CO2<\/sup> + 2BO2<\/sup> + 2DO2<\/sup>
\n\\(=2 \\times\\left(\\frac{1}{2} \\mathrm{AC}\\right)^{2}+2 \\times\\left(\\frac{1}{2} \\mathrm{AC}\\right)^{2}+2 \\times\\left(\\frac{1}{2} \\mathrm{BD}\\right)^{2}+2 \\times\\left(\\frac{1}{2} \\mathrm{BD}\\right)^{2}\\)
\n\\(=2 \\times \\frac{1}{4} \\mathrm{AC}^{2}+2 \\times \\frac{1}{4} \\mathrm{AC}^{2}+2 \\times \\frac{1}{4} \\mathrm{BD}^{2}+2 \\times \\frac{1}{4} \\mathrm{BD}^{2}\\)
\n\\(=\\frac{1}{2} A C^{2}+\\frac{1}{2} A C^{2}+\\frac{1}{2} B D^{2}+\\frac{1}{2} B D^{2}\\)
\nRHS = AC2<\/sup> + BD2<\/sup>
\n\u2234 LHS = RHS
\n\u2234 AB2<\/sup> + BC2<\/sup> + CD2<\/sup> + DA2<\/sup> = AC2<\/sup> + BD2<\/sup>.<\/p>\n

Question 8.
\nIn the following figure, O is a point in the interior of a triangle ABC, OD \u22a5 BC, OE \u22a5 AC, OF \u22a5 AB. Show that
\ni) OA2<\/sup> + OB2<\/sup> + OC2<\/sup> – OD2<\/sup> – OE2<\/sup> – OF2<\/sup> = AF2<\/sup> + BD2<\/sup> + CE2<\/sup>
\nii) AF2<\/sup> + BD2<\/sup> + CE2<\/sup> = AE2<\/sup> + CD2<\/sup> + BF2<\/sup>.
\n\"KSEEB
\nSolution:
\nData: O is a point in the interior of a triangle ABC, OD \u22a5 BC, OE \u22a5 AC, OF\u22a5 AB.
\nTo Proved: i) OA2<\/sup> + OB2<\/sup> + OC2<\/sup> – OD2<\/sup> – OE2<\/sup> – OF2<\/sup> = AF2<\/sup> + BD2<\/sup> + CE2<\/sup>
\nii) AF2<\/sup> + BD2<\/sup> + CE2<\/sup> = AE2<\/sup> + CD2<\/sup> + BF2<\/sup><\/p>\n

(i) OA2<\/sup> + OB2<\/sup> + OC2<\/sup> – OD2<\/sup> – OE2<\/sup> – OF2<\/sup> = AF2<\/sup> + BD2<\/sup> + CE2<\/sup>
\n\"KSEEB
\nOA2<\/sup> = AF2<\/sup> + OF2<\/sup> \u2192 (i)
\nOB2<\/sup> = BD2<\/sup> + OD2<\/sup> \u2192 (ii)
\nOC2<\/sup> = OE2<\/sup> + EC2<\/sup> \u2192 (iii)
\nBy Adding equations (i) + (ii) + (iii),
\nOA2<\/sup> + OB2<\/sup>+ OC2<\/sup> = AF2<\/sup> + OF2<\/sup> + BD2<\/sup> + OD2<\/sup> + OE2<\/sup> +EC2<\/sup>
\n\u2234 OA2<\/sup> + OB2<\/sup> + OC2<\/sup> – OE2<\/sup> – OF2<\/sup> – OD2<\/sup> = AF2<\/sup> + BD2<\/sup> + CE2<\/sup><\/p>\n

(ii) AF2<\/sup> + BD2<\/sup> + CE2<\/sup> = AE2<\/sup> + CD2<\/sup> + BF2<\/sup>.
\nOA2<\/sup> = AF2<\/sup> + OF2<\/sup>
\n\u2234 AF2<\/sup> = OA2<\/sup> – OF2<\/sup> \u2192 (i)
\nOB2<\/sup> = BD2<\/sup> + OD2<\/sup>
\n\u2234BD2<\/sup> = OB2<\/sup> – OD2<\/sup> \u2192 (ii)
\nOC2<\/sup> = OE2<\/sup> + EC2<\/sup>
\n\u2234 CE2<\/sup> = OC2<\/sup> – OE2<\/sup> \u2192 (iii)
\nFrom adding equations (i) + (ii) + (iii),
\nAF2<\/sup> + BD2<\/sup> + CE2<\/sup> = OA2<\/sup> – OF2<\/sup> + OB2<\/sup> – OD2<\/sup> + OC2<\/sup> – OE2<\/sup>
\nAF2 <\/sup>+ BD2<\/sup> + CE2<\/sup> = OA2<\/sup> – OE2<\/sup> + OB2<\/sup> – OF2<\/sup> + OC2<\/sup> – OE2<\/sup>
\n\u2234 AF2<\/sup> + BD2<\/sup> + CE2<\/sup> = AE2<\/sup> + FB2<\/sup> + CD2<\/sup> .<\/p>\n

Question 9.
\nA ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.
\n\"KSEEB
\nSolution:
\nIn \u22a5\u2206ACB, \u2220C = 90\u00b0, BC = ?
\nAC2<\/sup> + CB2<\/sup> = AB2<\/sup>
\n(8)2<\/sup> + CB2<\/sup> = (10)2<\/sup>
\n64 + CB2<\/sup> = 100
\nCB2<\/sup> = 100 – 64
\nCB2<\/sup> = 36
\n\u2234 CB = 6
\n\u2234 Ladder is at a distance of 6m from the base of the wall.<\/p>\n

Question 10.
\nA guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
\n\"KSEEB
\nSolution:
\nIn \u22a5\u2206PQR, \u2220Q = 90\u00b0, QR = ?
\n\u2234 PQ2<\/sup> + QR2<\/sup> = PR2<\/sup>
\n(18)2<\/sup> + QR2<\/sup> = (24)2<\/sup>
\n324 + QR2<\/sup> = 576
\nQR2<\/sup> = 576 – 324
\nQR2<\/sup> = 252
\n\u2234 QR = 15.8 m.<\/p>\n

Question 11.
\nAn aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after \\(1 \\frac{1}{2}\\) hours?
\n\"KSEEB
\nSolution:
\nDistance travelled by aeroplace trowards North is \\(1 \\frac{1}{2}\\) hours :
\n\\(=1000 \\times 1 \\frac{1}{2}\\)
\n\\(=1000 \\times \\frac{3}{2}\\)
\n= 1500 km.
\nDiatance travelled by aeroplane towards West in \\(1 \\frac{1}{2}\\) Hours :
\n\\(=1200 \\times 1 \\frac{1}{2}\\)
\n\\(=1200 \\times \\frac{3}{2}\\)
\n= 1800 km.
\nIn \u22a5\u2206AOB,
\nAB2<\/sup> = OA2<\/sup> + OB2<\/sup>
\n= (1500)2<\/sup> + (1800)2<\/sup>
\n= 2250000 + 3240000 = 5490000
\n\\(\\mathrm{AB}=\\sqrt{5490000}\\)
\n\\(A B=\\sqrt{90000 \\times 61}\\)
\n\\(A B=300 \\sqrt{61} \\mathrm{km}\\) km
\n\u2234 Two planes are 300V6T km. apart after 14 hours<\/p>\n

Question 12.
\nTwo poles of heights 6 m and 11m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
\n\"KSEEB
\nSolution:
\nPole AB = 6 m.
\nPole CD = 11 m.
\nDistance between poles BD = 12 m.
\nDistance between feet of the poles, AC = ?
\nABDM is a rectangle, AB = MD = 6 m.
\nBD = AM = 12 m.
\nIn \u22a5\u2206AMC, \u2220AMC = 90\u00b0
\nAM = 12 m, CM = 5 m. AC = ?
\nAC2<\/sup> = AM2<\/sup> + CM2<\/sup>
\n= (12)2<\/sup> + (5)2<\/sup>
\n= 144 + 25
\nAC2<\/sup> = 169
\n\u2234 AC = 13 m.<\/p>\n

Question 13.
\nD and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.
\nProve that AE2<\/sup> + BD2<\/sup> = AB2<\/sup> + DE2<\/sup>.
\n\"KSEEB
\nSolution:
\nData: In \u2206ABC, \u2220C = 90\u00b0, D and E are points on the sides CA and CB respectively
\nTop Prove: AE2<\/sup> + BD2<\/sup> = AB2<\/sup> + DE2<\/sup>
\nIn \u22a5\u2206ACE, \u2220C = 90\u00b0
\n\u2234 AE2<\/sup> = AC2<\/sup> + CE2<\/sup> ………. (i)
\nIn \u22a5\u2206DEB, \u2220C = 90\u00b0
\n\u2234 BD2<\/sup> = DC2<\/sup> + CB2<\/sup> ………… (ii)
\nFrom adding equations (i) + (ii)
\nAE2<\/sup> + BD2<\/sup> = AC2<\/sup> + CE2<\/sup> + DC2<\/sup> + CB2<\/sup>
\n= AC2<\/sup> + CB2<\/sup> + DC2<\/sup> + CE2<\/sup>
\n\u2234 AE2<\/sup> + BD2<\/sup> = AB2<\/sup> + DE2<\/sup> ( . Theorem 8).<\/p>\n

Question 14.
\nThe perpendicular from A on side BC of a \u2206ABC intersects BC at D such that DB = 3CD (see the following figure) Prove that 2AB2<\/sup> = 2AC2<\/sup> + BC2<\/sup>.
\n\"KSEEB
\nSolution:
\nIn \u2206ACD,
\nAC2<\/sup> = AD2<\/sup> + DC2<\/sup>
\nAD2<\/sup> = AC2<\/sup> – DC2<\/sup>……….. (1)
\nIn \u2206ABD,
\nAB2<\/sup> = AD2<\/sup> + DB2<\/sup>
\nAD2<\/sup> = AB2<\/sup> – DB2<\/sup>………. (2)
\nFrom equations (1) and (2),
\nAC2<\/sup> – DC2<\/sup> = AB2<\/sup> – DB2<\/sup> ………… (3)
\n3DC = DB (data given)
\n\\(\\mathrm{DC}=\\frac{\\mathrm{BC}}{4}, \\text { and } \\mathrm{DB}=\\frac{3 \\mathrm{BC}}{4}\\) ……….. (3)
\nSubstituting eqn. (4) in eqn. (3),
\n\\(\\mathrm{AC}^{2}-\\left(\\frac{\\mathrm{BC}}{4}\\right)^{2}=\\mathrm{AB}^{2}-\\left(\\frac{3 \\mathrm{BC}}{4}\\right)^{2}\\)
\n\\(A C^{2}-\\frac{B C^{2}}{16}=A B^{2}-\\frac{9 B C^{2}}{16}\\)
\n\\(\\frac{16 \\mathrm{AC}^{2}-\\mathrm{BC}^{2}}{16}=\\frac{16 \\mathrm{AB}^{2}-9 \\mathrm{BC}^{2}}{16}\\)
\n16AC2<\/sup> – BC2<\/sup> = 16 AB2<\/sup> – 9BC2<\/sup>
\n16AB2<\/sup> – 16AC2<\/sup> = 9 BC2<\/sup> – BC2<\/sup>
\n16AB2<\/sup> – 16AC2<\/sup> = 8BC2<\/sup>
\n8(2AB2<\/sup> – 2AC2<\/sup> = BC2<\/sup>)
\n2AB2<\/sup> – 2AC2<\/sup> = BC2<\/sup>
\n2AB2<\/sup> = 2AC2<\/sup> \u2260 BC2<\/sup><\/p>\n

Question 15.
\nIn an equilateral triangle ABC, D is a point on side BC such that \\(B D=\\frac{1}{3} B C\\), Prove that 9AD2<\/sup> = 7AB2<\/sup>.
\nSolution:
\n\u2206ABC is an equilateral triangle.
\n\"KSEEB
\nAB = BC = AC = a
\nIn \u2206ABC, AE is perpendicular line.
\n\\(B E=E C=\\frac{B C}{2}\\)
\nAltitude, \\(\\mathrm{AE}=\\frac{\\mathrm{a} \\sqrt{3}}{2}\\)
\n\\(\\mathrm{BD}=\\frac{1}{3} \\mathrm{BC}(\\mathrm{Data})\\)
\n\\(B D=\\frac{a}{3}\\)
\nDE = BE – AD
\n\\(=\\quad \\frac{a}{2}-\\frac{a}{3}=\\frac{a}{6}\\)
\nIn \u2206ADE,
\nAD2<\/sup> = AE2<\/sup> + DE2<\/sup>
\n\\(=\\left(\\frac{\\mathrm{a} \\sqrt{3}}{2}\\right)^{2}+\\left(\\frac{\\mathrm{a}}{6}\\right)^{6}\\)
\n\\(=\\frac{3 a^{2}}{4}+\\frac{a^{2}}{36}=\\frac{27 a^{2}+a^{2}}{36}\\)
\n\\(=\\frac{28 \\mathrm{a}^{2}}{36}\\)
\n\\(=\\frac{7}{9} \\mathrm{a}^{2}\\)
\n\\(\\mathrm{AD}^{2}=\\frac{7}{9} \\mathrm{AB}^{2} \\quad 9 \\mathrm{AD}^{2}=7 \\mathrm{AB}^{2}\\)<\/p>\n

Question 16.
\nIn an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
\n\"KSEEB
\nSolution:
\nData: ABC is an equilateral triangle.
\nHere, AB = BC = CD.
\nAD \u22a5 BC.
\nTo Prove: 3AC2<\/sup> = 4AD2<\/sup>
\nIn \u22a5\u2206ADC, \u2220ADC = 90\u00b0
\n\u2234 AC2<\/sup> = AD2<\/sup> + DC2<\/sup>
\n\\(=A D^{2}+\\left(\\frac{1}{2} A C\\right)^{2}\\)
\n\\(\\left( \\begin{array}{l}{\\mathrm{DC}=\\frac{1}{2} \\mathrm{BC}} \\\\ {\\mathrm{DC}=\\frac{1}{2} \\mathrm{AC}}\\end{array}\\right)\\)
\n\\(\\mathrm{AC}^{2}=\\mathrm{AD}^{2}+\\frac{1}{4} \\mathrm{AC}^{2}\\)
\n\\(\\frac{\\mathrm{AC}^{2}}{1}-\\frac{1}{4} \\mathrm{AC}^{2}=\\mathrm{AD}^{2}\\)
\n\\(\\frac{4 \\mathrm{AC}^{2}-1 \\mathrm{AC}^{2}}{4}=\\mathrm{AD}^{2}\\)
\n\u2234 3AC2<\/sup> = 4AD2<\/sup>.<\/p>\n

Question 17.
\nTick the correct answer and justify:
\nIn \u2206ABC, \\(A B=6 \\sqrt{3} \\mathrm{cm}\\). AC = 12 cm. and BC = 6 cm. The angle B is
\nA) 120\u00b0
\nb) 60\u00b0
\nC) 90\u00b0
\nD) 45\u00b0
\nSolution:
\nC) 90\u00b0
\n\"KSEEB
\nJustification:
\na\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0b\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0c
\n\\(6 \\sqrt{3}\\)\u00a0 \u00a0 \u00a0 \u00a0 6\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a012
\nHere largest side is 12 cm.
\nIf square of the hypotenuse is equal to square of other two sides, then it is a right angled triangle.
\n\u2234 c2<\/sup> = a2<\/sup> + b2<\/sup>
\nAC2<\/sup> = AB2<\/sup> + BC2<\/sup>
\n\\((12)^{2}=(6 \\sqrt{3})^{2}+(6)^{2}\\)
\n144 = 36 \u00d7 3 + 36
\n144 = 108 + 36
\n144 = 144
\n\u2234 \u2206ABC is a right angled triangle and angle opposite to hypotenuse, \u2220B = 90\u00b0.<\/p>\n

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.5 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.5, 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 2.5\u00a0are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.5. Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.5 Question 1. Sides of triangles are given below. Determine which …<\/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\/515"}],"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=515"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/515\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=515"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=515"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=515"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}