{"id":25288,"date":"2020-03-09T11:20:17","date_gmt":"2020-03-09T05:50:17","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=25288"},"modified":"2021-07-02T15:21:16","modified_gmt":"2021-07-02T09:51:16","slug":"kseeb-solutions-for-class-10-maths-chapter-2-additional-questions","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-10-maths-chapter-2-additional-questions\/","title":{"rendered":"KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Additional Questions"},"content":{"rendered":"

Students can Download Class 10 Maths Chapter 2 Triangles Additional Questions Questions, Notes Pdf, KSEEB Solutions for Class 10 Maths<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your in examinations<\/p>\n

Karnataka State Syllabus Class 10 Maths Chapter 2 Triangles Additional Questions<\/h2>\n

I Multiple choice questions:
\n<\/span><\/p>\n

Question 1.
\nTwo triangles are said to be similar if their corresponding sides are
\na) Proportional
\nb) Equal
\nc) Paralled
\nd) Perpendicular
\nAnswer:
\na) Proportional<\/p>\n

Question 2.
\nCongruent triangles are always,
\na) Equal
\nb) similar
\nc) Not similar
\nd) Not equal
\nAnswer:
\na) Equal<\/p>\n

Question 3.
\nWhich of the following is can form similar triangles i) 3, 6, 9 ii) 9,18, 27 iii) 2,4, 6 iv) 5, 6, 7
\na) i and ii
\nb) ii and iii
\nc) i and iii
\nd) i and iv
\nAnswer:
\na) i and ii<\/p>\n

Question 4.
\nBasic proportional theorem is proposed by
\na) Pythagoras
\nb) Thales
\nc) aryabhata
\nd) newton
\nAnswer:
\nb) Thales<\/p>\n

Question 5.
\nIf straight line is drawn parallel to one side of a triangle then it divides the other two sides proportionally is called.
\na) thales theorem
\nb) converse of thales theorem
\nc) Corollary of thales
\nd) Pythagoras theorem
\nAnswer:
\na) thales theorem<\/p>\n

Question 6.
\nIf straight-line divides two sides of a triangle proportionally then the straight line is parallel to the third side is called.
\na) converse of thales theorem
\nb) pythogoral theorem
\nc) thales theorem
\nd) converse of Pythagoras theorem
\nAnswer:
\na) converse of thales theorem<\/p>\n

Question 7.
\nIn the given fig. which of following is<\/p>\n

\"KSEEB
\nAnswer:
\n\"3\"<\/p>\n

\"KSEEB<\/p>\n

Question 8.
\nIn the given fig. which of the following Is true
\n\"KSEEB
\n\"KSEEB
\nAnswer:
\n\"KSEEB<\/p>\n

Question 9.
\nIn the following figure DE || BC. AD = 4 cm, DB = 6 cm and AE =5 cm, then the value of EC is
\n\"KSEEB
\na) 6.5 cm
\nc) 7.5 cm
\nb) 7 cm
\nd) 8.0 cm
\nAnswer:
\nc) 7.5 cm<\/p>\n

Question 10.
\nIn the following figure DE || BC then the value of \u2018x\u2019 is<\/p>\n

\"KSEEB
\nAnswer:
\n\"KSEEB<\/p>\n

Question 11.
\nIn a trapezium the line joining the midpoints of non-parallel sides is.
\na) Parallel to the parallel sides.
\nb) Half of the sum of the parallel side
\nc) both a and b
\nd) perpendicular to each other
\nAnswer:
\nc) both a and b<\/p>\n

Question 12.
\nIf two triangles are equiangular then their corresponding sides are
\na) proportional
\nb) equal
\ne) non-parallel
\nd) perpendicular
\nAnswer:
\na) proportional<\/p>\n

Question 13.
\nIf the three sides of a triangle are proportional to the corresponding three sides of another triangle then their corresponding angles are
\na) equal
\nb) proportional
\ne) not equal
\nd) not proportional
\nAnswer:
\na) equal<\/p>\n

Question 14.
\nIn the given figure \u2206 ADB ~ \u2206 BDC then BD2<\/sup> is
\n\"KSEEB
\na) AD . AC
\ne) CD . AC
\nb) AD . CD
\nd) AB . BC
\nAnswer:
\nb) AD . CD<\/p>\n

\"KSEEB<\/p>\n

Question 15.
\nIn the given figure \u2206 BDC ~ \u2206 ABC then BC2<\/sup> is
\n\"KSEEB
\na) AC . CD
\nb) AC . AD
\nc) AD . CD
\nd) A13 . BC
\nAnswer:
\na) AC . CD<\/p>\n

Question 16.
\nIn the given figure \u2206 ADB ~ \u2206 BDC then B2<\/sup> is
\n\"KSEEB
\na) AC . AD
\nc) AD . CD
\nb) AC . CD
\nd) AB . BC
\nAnswer:
\na) AC . AD<\/p>\n

Question 17.
\nIf the area of two similar triangles are equal then they are
\na) similar
\nb) congruent
\ne) proportional
\nd) equal
\nAnswer:
\nd) equal<\/p>\n

Question 18.
\nArea of similar triangles are proportional to
\na) squares on their corresponding sides
\nb) squares on their corresponding altitude
\ne) squares on their corresponding medians
\nd) all the above
\nAnswer:
\nd) all the above<\/p>\n

Question 19.
\nThe corresponding altitudes of two similar triangles are 9 cm and 15cm respectively then the ratio between their areas.
\na) 9 : 15
\nb) 100 : 25
\nc) 81 : 225
\nd) 3 : 15
\nAnswer:
\nc) 81 : 225<\/p>\n

Question 20.
\nIn a trapezium ABCD AB || CD and Its diagonals intersect at 0. if AB = 6 cm and DC = 3 cm then the ratio of the area of AOB and COD
\na) 4 : 1
\nb) 1 : 2
\nc) 2 : 1
\nd) 1 : 4
\nAnswer:
\na) 4 : 1<\/p>\n

II. Short Answer Questions:<\/span><\/p>\n

Question 1.
\nState thales theorem (Basic proportionality theorem).
\nAnswer:
\nA straight drawn parallel to one side of triangle then other two sides divides proportionally.<\/p>\n

Question 2.
\nState converse of thales theorem.
\nAnswer:
\nIf two sides divides proportionally then straight line drawn parallel to given side.<\/p>\n

Question 3.
\nWhat is the ratio of areas of two similar triangles whose sides are in the ratio 3 : 4
\nAnswer:
\n\"KSEEB
\n\u2234 Ratio of areas are 9 : 16<\/p>\n

Question 4.
\nIf two triangles are similar such that the ratio of their areas 36 : 121, then what is the ratio of their corresponding medians.
\nAnswer:
\n\"KSEEB
\n\u2234 ratio of their corresponding medians 6 : 11<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nIs the triangle with sides 12 cm, 16 cm and 18 cm a right triangle? give reason.
\nAnswer:
\n122 + 162 = 182
\n144 + 256 = 324
\n400 \u2260 324
\nLHS \u2260 RHS
\n\u2234 Given sides not form right angle \u2206le<\/sup>.<\/p>\n

Question 6.
\nWhat is the name given to the longest side of a right-angled triangle
\nAnswer:
\nhypotenuse<\/p>\n

Question 7.
\nState pythagoras theorem
\nAnswer:
\nIn a right-angled triangle square on the hypotenuse is equal to sum of the squares of other sides.<\/p>\n

Question 8.
\nState converse of pythagoras theorem.
\nAnswer:
\n\u201cIf the square on the longest side of a triangle is equal to the sum of the squares on the other two sides, then those two sides contain a right angle\u201d.<\/p>\n

Question 9.
\nVerify m2<\/sup> – n2<\/sup>, 2mn, m2<\/sup> + n2<\/sup> form the sides of a right-angled triangle.
\nAnswer:
\n(m2<\/sup> + n2<\/sup>)2<\/sup> = (m2<\/sup> – n2<\/sup>)2<\/sup> + (2mn)2<\/sup>
\nm4<\/sup> + 2m2<\/sup>n2<\/sup> + n4<\/sup> = m4<\/sup>\u00a0– 2m2<\/sup>n2<\/sup> + n4<\/sup> + 4m2<\/sup>n2<\/sup>
\nm4<\/sup> + n4<\/sup> + 2m2<\/sup>n2<\/sup> m – n4<\/sup> + 2m2<\/sup>n2<\/sup>
\nLHS = RHS
\ngiven sides form right angle \u2206le<\/sup>.<\/p>\n

Question 10.
\nIn figure \u2220ABC = 90\u00b0, BD \u22a5 AC if BD = 8 cm, AD = 4 cm, find CD
\n\"KSEEB
\nAnswer:
\nBD2<\/sup> = AD \u00d7 CD
\n(8)2<\/sup> = 4 \u00d7 CD
\n64 = 4 CD
\n\"KSEEB
\nIII. Long answer questions:<\/span><\/p>\n

Question 1.
\nIn Fig. \\(\\frac{A O}{O C}=\\frac{B O}{O D}=\\frac{1}{2}\\) and AB = 5 cm. Find the value of DC.
\n\"KSEEB
\nAnswer:
\n\"KSEEB<\/p>\n

Question 2.
\nABC is an equilateral triangle of side 4a. Find each of its altitude.
\nAnswer:
\n\"KSEEB
\nLet ABC be an equilateral triangle of side 4a units, We draw AD \u22a5 BC. Then D is the mid-point of BC.
\n\"KSEEB
\nNow ABD is a right \u2206le <\/sup>right angle at D.
\n\u2234 AB2<\/sup> = AD2<\/sup>\u00a0+ (2a)2<\/sup>. [By Pythagoras theorem]
\n= (4a)2<\/sup> = AD2<\/sup> + 4a2<\/sup>
\n= AD2<\/sup> = 16a2<\/sup> – 4a2<\/sup> = 12a2<\/sup>
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nIf the diagonals of a quadrilateral divide each other proportionally prove that is a trapezium?
\nAnswer:
\n\"KSEEB<\/p>\n

Question 4.
\nABC is a triangle in which AB = AC and D is a point on AC such that BC2<\/sup> = AC \u00d7 CD Prove that BD = BC.
\nAnswer:
\n\"KSEEB
\nGiven: \u2206 ABC in which AB = AC and D is a point on the side AC such that BC2<\/sup> = AC x CD.
\nTo prove: BD = BC
\nConstruction: Join BD
\nProof: We have,
\n\"KSEEB<\/p>\n

Question 5.
\nA door of width 6 meter has an arc above it having a height of 2 meters find the radius of the arc?
\n\"KSEEB
\nAnswer:
\n\"KSEEB
\n\"KSEEB
\nLet the radius of the semi – circle
\nOA = OC = OB
\n= (x + 2)m
\n=> OA = OB. In \u2206 OAB
\n\u2206 OAB is an isosceles \u2206le<\/sup>
\n\u2234 OD \u22a5 AB.
\n=> OD bisects AB
\n\u2234 AD = BD = 3 cm.
\nIn \u2206le<\/sup> OAD, \u2220PDA = 90\u00b0
\nAO2<\/sup> = AD2<\/sup> + OD2<\/sup>
\n(x+2)2<\/sup> = 32<\/sup> + x2<\/sup>
\nx2<\/sup> + 4x + 4 = 9 + x2<\/sup>
\n4x = 9 – 4 = 5
\nx = \\(\\frac{5}{4}\\)
\nHence, radius = x + 2
\n\"KSEEB<\/p>\n

Question 6.
\nIn an equilateral \u2206le<\/sup> ABC, AD is drawn \u22a5r<\/sup> to BC meeting BC in D, prove that AD2<\/sup> = 3BD2<\/sup>.\"KSEEB
\nAnswer:
\nIn \u2206 ABD, \u2220ADB = 90\u00b0
\nThen by pythagoras theorem
\n=> AB2<\/sup> = AD2<\/sup> + BD2<\/sup>
\n=> BC2<\/sup> = AD2<\/sup> + BD2<\/sup>, [BC = 2BD]
\n\u2234 AB = BC = CA
\n=> (2BD)2<\/sup> = AD2<\/sup> + BD2<\/sup>
\n\u2234 (\u22a5 is the median in an equilateral \u2206le<\/sup>)
\n3BD2<\/sup> = AD2<\/sup>.<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nIn the given figure, AD \u22a5 BC and BD = \\(\\frac{1}{3}\\) CD prove that 2AC2<\/sup> = 2AB2<\/sup> + BC2<\/sup>.\"KSEEB
\nAnswer:
\nIn right \u2206 ABD, by pythogoras theorem.
\n\u2234 AB2<\/sup> = AD2<\/sup> + BD2<\/sup>
\nIn right \u2206 ABC, by pythagoras theorem.
\n\u2234 AC2<\/sup> = AD2<\/sup> + DC2<\/sup>
\nSubtracting (ii) from (i) we get,
\n\u2234 AB2<\/sup> – AC2<\/sup> = BD2<\/sup> – DC2
\n<\/sup>\"KSEEB
\n\"KSEEB
\nQuestion 8.
\nTriangle ABC is right angles at B and D is the mid-point of BC? Prove that AC2<\/sup> = 4AD2<\/sup> – 3AB2<\/sup>.
\n\"KSEEB
\nAnswer:
\nGiven: ABC is right angled at B and D is the mid-point of B C
\nBD = DC = \\(\\frac{1}{2}\\) BC
\nIn \u2206 ABD, AD2<\/sup> = AB2<\/sup> + BD2<\/sup>
\n[pythogoras theorem] – (1)
\nIn \u2206 ABC, AC2<\/sup> = AB2<\/sup> + BC2<\/sup>
\n[pythogoras theorem] – (2)
\nFrom eqn (1) and (2)
\n\"KSEEB
\n=> 4AD2<\/sup> = 4AB2<\/sup> + BC2<\/sup>
\n=> BC2<\/sup> = 4AD2<\/sup> – 4AB2<\/sup> – (3)
\nUsing this eqn in (2)
\nAC2<\/sup> = AB2<\/sup> + 4AD2<\/sup> – 4AB2<\/sup>
\nAC2<\/sup> = 4AD2<\/sup> – 3AB2<\/sup>.<\/p>\n

Question 9.
\nThrough the midpoint M of the sides of parallelogram ABCD. The line BM is drawn intersecting AC at L and AD produced to E. Prove that EL = 2BL.
\nAnswer:
\n\"KSEEB
\n\"KSEEB<\/p>\n

Question 10.
\nProve that any two medians of a \u2206le<\/sup> divide each other in the ratio 2:1.
\nAnswer:
\n\"KSEEB
\nIn \u2206le<\/sup> ABC, BD, CE are two medians intersecting at G.
\n\u2234 AE = EB, AD = DC [medians divides equally]
\n\"KSEEB<\/p>\n

Question 11.
\nA ladder of length 2.6 m is lended against a wall. When it is at a distance of 2.4 m from the foot of the wall, the top of the ladder touches the bottom edge of the window in the wall, if the foot of the ladder is moved 1.4 m towards the wall it touches the top edge of the window, find the height of the window.
\nAnswer:
\n\"KSEEB
\nLength of ladder AC = DE = 2.6 m
\nHeight of window AD =?
\nBC = 2.4 m
\nEC = 1.4 m
\nBE = 1 m
\nIn \u2206le<\/sup> ABC
\nAC2<\/sup> = AB2<\/sup> + BC2<\/sup>
\nAB2<\/sup> = AC2<\/sup> – BC2<\/sup>
\nAB2<\/sup> = (2.6)2<\/sup> – (2.4)2<\/sup>
\n= (2.6 + 2.4) x (2.6 – 2.4)
\n= 5 x 0.2 = lm
\nAB = 1m
\nIn \u2206le<\/sup> DBE
\nDE2<\/sup> = BD2<\/sup> + BE2<\/sup>
\n(2.6)2<\/sup> – 12<\/sup> = BD2<\/sup>
\nBD2 = (2.6 + 1) (2.6 – 1)
\n= 3.6 x 1.6
\nBD = \"KSEEB
\nBD = 2.4 m
\n\u2234 Height of window
\nAD = BD – AB
\n= 2.4 – 1 = 1.4
\nHeight of window = 1.4m<\/p>\n

\"KSEEB<\/p>\n

Question 12.
\nIn a \u2206le<\/sup> ABC, D and E be two points on side AB such that AD = BE : IF DE || BC and EQ || AC, then prove that PQ || AB.
\nAnswer:
\n\"KSEEB
\nFrom fig, given BE = AD
\nEA = AD + DE
\n= BE + ED
\n= BD
\nThen (2) becomes
\n\"KSEEB
\n\u2234 PQ || AB [converse of BPT]<\/p>\n

Question 13.
\nA tree 32 m tall broke due to gale and Its top fell at a distance of 16m from its foot at what height above the ground did the tree break?
\nAnswer:
\n\"KSEEB
\nHeight of the tree AB = 32 m
\nAD is broken part of tree = x
\nBC is distance between foot to the top oftreeBC = 16m
\nBD is the remaining height of tree = 32 – x
\nIn Ale BDC \u2220B = 90\u00b0
\n(CD)2<\/sup> = BD2<\/sup> + BC2<\/sup>
\nx2<\/sup> = (32 -x)2<\/sup> + (16)2<\/sup>
\nx2<\/sup> = 1024 + x2<\/sup> – 64x + 256
\nx2<\/sup> – x2<\/sup> = 1280 – 64x
\n64x = 1280
\nx = \\(\\frac{1280}{64}\\)
\nx = 20m
\nRemaining part of a tree
\nBD = (32 – x) = 32 – 20 = 12m<\/p>\n

Question 14.
\nIn \u2206le<\/sup> ABC, i) \u2220ACB = 60\u00b0, ii) AD \u22a5ar<\/sup> BC. Derive an expression for AB in terms of AC and BC.
\nAnswer:
\n\"KSEEB
\nIn \u2206le<\/sup> ABC
\ni) \u2220ACB = 60\u00b0 and
\nii) AD \u22a5r<\/sup> BC
\nConstruction:
\nDraw DE such that \u2220EDC = 60\u00b0
\n\u2234 Ale DEC is an equilateral
\n(Data and construction)
\nDE = DC = EC
\nAle DEA is an isosceles Ale [\u2234 AED = 120 and \u2220 ADE = 30\u00b0 = \u2220 DAE]
\nDE = EA
\n=> DC = EA = EC [\u2234 Axiom – 1]
\n\u2234 AC = 2DC
\nNow in Ale ADB, AB2<\/sup> = AD2<\/sup> + BD2<\/sup> [pythagoras theorem]
\nAB2<\/sup> = (AC2<\/sup> – DC2<\/sup>) + (BC -CD)2<\/sup>
\nAB2<\/sup> = 4DC2<\/sup> – DC2<\/sup> + BC2<\/sup> – 2BC DC + DC2<\/sup> [\u2234 AC = 2DC]
\nAB2<\/sup> = 4DC2<\/sup> + BC2<\/sup> – 2BC . DC
\nAB2<\/sup> = 4DC2<\/sup> – 2BC . DC + BC2<\/sup>
\nAB2<\/sup> = 2DC (2DC – BC) + BC2<\/sup>
\n= AC (AC – BC) + BC2<\/sup> [2DC = AC]
\nAB2 = AC2<\/sup> – AC . BC + BC2<\/sup>
\nAB2<\/sup> = AC2<\/sup> + BC2<\/sup> – AC.BC<\/p>\n

Question 15.
\nABCD is a rhombus. Prove that AC2<\/sup> + BD2<\/sup> = 4AB2<\/sup>
\nAnswer:
\n\"KSEEB
\nGiven: In rhombus ABCD, AC and BD are diagonals, AB = BC = CD = DA
\n\u2234 AO = 1\/2 AC and BO = 1\/2 BD
\nTo prove: AC2<\/sup> + BD2<\/sup> = 4AB2<\/sup>
\nIn right angle \u2206le<\/sup> AOB, \u2220O = 90\u00b0
\nAB2<\/sup> = OA2<\/sup> + OB2<\/sup>
\nAB2<\/sup> = (1\/2 AC)2<\/sup> + (1\/2 BD)2<\/sup>
\n\"KSEEB
\n4AB2<\/sup> = AC2<\/sup> + BD2<\/sup><\/p>\n

Question 16.
\nAD is the altitude from A to BC in the \u2206 ABC and DB:CD = 3:1 prove that BC2<\/sup> = 2(AB2<\/sup> – AC2<\/sup>)
\nAnswer:
\n\"KSEEB
\nIn \u2206le <\/sup>ABD \u2220D = 90\u00b0
\nAB2<\/sup> = AD2<\/sup> + BD2<\/sup>
\nAD2<\/sup> = AB2<\/sup> – BD2<\/sup> – (1)
\nIn \u2206le<\/sup> ADC \u2220D = 90\u00b0
\nAC2<\/sup> = AD2<\/sup> + CD2<\/sup>
\nAD2<\/sup> = AC2<\/sup> – CD2<\/sup> – (2)
\nFrom (1) and (2)
\nAB2<\/sup> – BD2<\/sup> = AC2<\/sup> – CD2<\/sup>
\nAB2<\/sup> – AC2<\/sup> = BD2<\/sup> – CD2<\/sup>
\n\"KSEEB
\n\"KSEEB
\nBC2<\/sup> = 2(AB2<\/sup> – AC2<\/sup>)<\/p>\n

\"KSEEB<\/p>\n

Question 17.
\nIn the given figure in \u2206 ABC, AD \u22a5 BC and AD2<\/sup> = BD \u00d7 CD. Prove that the \u2206 ABC is a right angle triangle.
\nAnswer:
\n\"KSEEB<\/p>\n

\u2206 ABC, AD \u22a5 BC and AD2<\/sup> = BD \u00d7 CD
\nTo prove: \u2206 ABC is a right angle triangle
\nProof: In \u2206le<\/sup> ABD \u2220D = 90\u00b0
\nAB2 = BD2<\/sup> + AD2<\/sup> – (1)
\nIn \u2206le<\/sup> ADC \u2220D = 90\u00b0
\nAC2<\/sup> = AD2<\/sup> + DC2<\/sup> – (2)
\nAdding eqn (1) and (2)
\nAB2<\/sup> + AC2<\/sup> = BD2<\/sup> + AD2<\/sup> + AD2<\/sup> + DC2<\/sup>
\n= 2 AD2<\/sup> + BD2<\/sup> + DC2<\/sup>
\n= 2(BD x CD) + BD2<\/sup> + DC2<\/sup>
\n= (BD + DC)2<\/sup>
\n\u2234 AB2<\/sup> + AC2<\/sup> = BC2<\/sup>
\n\u2206 ABC is a right triangle, right angled at A.<\/p>\n

Question 18.
\nIn trapezium ABCD, AB || CD and DC = 2AB. EF || AB, where E and F lie on BC and AD respectively such \\(\\frac{B E}{E C}=\\frac{4}{3}\\) . Diagonal BD intersect EF at G. Prove that 7EF = 11 AB.
\nAnswer:
\nIn trapezium ABCD, AB || CD and
\n\"KSEEB
\n\"KSEEB
\n\"KSEEB<\/p>\n

Question 19.
\nA peacock on a pillar of 9 feet height on seeing a snake coming towards is hole situated just below the pillar from a distance 27 feet away from the pillar will fly to catch it. If both possess the same speed, how far from the pillar they are going to meet?
\nAnswer:
\n\"KSEEB<\/p>\n

AB is the height of pillar = 9ft
\nBC is the distance b\/w pillar and the snake hole = 27 ft
\nAD = CD = x ft
\nBD = 27 – x
\nIn \u2206 ABD \u2220D = 90\u00b0
\nAD2<\/sup> = AB2<\/sup> + BD2 [By pythagoras theorem]
\n(x)2<\/sup> = (9)2<\/sup> + (27 – x)2<\/sup>
\nx2<\/sup> = 81 + 729 + x2<\/sup> – 54x
\n54x = 810
\nx = \\(\\frac{810}{54}\\)
\nx = 15 ft
\nBD = 27 – x = 27 – 15 = 12 ft
\n\u2234 Both of these going to meet at a distance. 12ft from the pillar.<\/p>\n

Question 20.
\nIn \u2206 MGN, MP \u22a5 GN. If MG = a units, MN = b units, GP = c units and PN = d unit. Prove that (a + b) (a – b) = (c + d) (c – d).
\nAnswer:
\n\"KSEEB
\nIn \u2206le <\/sup>MGP \u2220P = 90\u00b0
\nMG2<\/sup> = GP2<\/sup> + MP2<\/sup>
\n(a)2<\/sup> = (c)2<\/sup> + MP2<\/sup>
\nMP2<\/sup> = a2<\/sup> – c2<\/sup> – (1)
\nIn \u2206le<\/sup> MNP \u2220P = 90\u00b0
\nMN2<\/sup> = PM2<\/sup> + PN2<\/sup>
\n(b)2<\/sup> = d2<\/sup> + MP2<\/sup>
\nMP2<\/sup> = b2<\/sup> – d2<\/sup> – (2)
\nFrom (1) and (2)
\na2 – c2<\/sup> = b2<\/sup> – d2<\/sup>
\na2<\/sup> – b2<\/sup> = c2 – d2<\/sup>
\n(a + b) (a – b) = (c + d) (c – d).<\/p>\n

\"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Class 10 Maths Chapter 2 Triangles Additional Questions Questions, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your in examinations Karnataka State Syllabus Class 10 Maths Chapter 2 Triangles Additional Questions I Multiple choice questions: Question 1. …<\/p>\n","protected":false},"author":5,"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\/25288"}],"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\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=25288"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/25288\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=25288"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=25288"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=25288"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}