{"id":12496,"date":"2023-04-05T12:53:10","date_gmt":"2023-04-05T07:23:10","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=12496"},"modified":"2023-04-05T17:44:20","modified_gmt":"2023-04-05T12:14:20","slug":"triangles-class-10-extra-questions","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/triangles-class-10-extra-questions\/","title":{"rendered":"Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers"},"content":{"rendered":"

Here you will find Triangles Class 10 Extra Questions Maths Chapter 6 with Answers Solutions, Extra Questions for Class 10 Maths<\/a> are solved by experts and will guide students in the right direction.<\/p>\n

Extra Questions for Class 10 Maths Triangles with Answers Solutions<\/h2>\n

Extra Questions for Class 10 Maths Chapter 6 Triangles with Solutions Answers<\/strong><\/p>\n

Triangles Class 10 Extra Questions Objective Type<\/h3>\n

Question 1.
\nIn the figure, a line segment PQ is drawn || to the base BC of \u2206ABC If PQ : BC = 1: 3, then the ratio of AP and PB will be :
\n(a) 1 : 2
\n(b) 1 : 3
\n(c) 1 : 4
\n(d) 2 : 3.
\nAnswer:
\n(c) 1 : 4
\nSolution.
\nGiven PQ || BC and PQ : BC = 1 : 3
\n\"Triangles
\n\u2234 AP : PB = 1 : 2
\nHence Choice (c) is correct<\/p>\n

\"Triangles<\/p>\n

Question 2.
\nIn the figure AB = 3 cm, AC = 6 cm, BD = 2 cm and CD = 4 cm. Find the ratio \u2220BAD and \u2220CAD is
\n(a) 2 : 4
\n(b) 1 : 1
\n(c) 3 : 6
\n(d) 6 : 3
\nAnswer:
\n(b) 1 : 1
\nSolution.
\nIn \u2206ADB, sin \u2220BAD = \\(\\frac{B D}{B A}\\) = \\(\\frac{2}{3}\\)
\n\"Triangles
\nIn \u2206ADC, sin \u2220CAD =\\(\\frac{4}{6}\\) = \\(\\frac{2}{3}\\)
\n\u2234 \u2220BAD = \u2220CAD
\n\u2234 ratio is 1 : 1 choice (b) is correct<\/p>\n

Question 3.
\nIn a \u2206ABC, if DE is drawn parallel to BC, cutting AB and ACat Dand E respectively such that AB = 7.2 cm, AC = 6.4 cm and AD = 4.5 cm. Then AE = ?
\n\"Triangles
\n(a) 5.4 cm
\n(b) 4 cm
\n(c) 3.6 cm
\n(d) 3.2 cm
\nAnswer:
\n(b) 4 cm<\/p>\n

Question 4.
\nIn \u2206ABC, DE || BC so that AD = (7x – 4) cm, AE = (5x – 2) cm, DB = (3x + 4) cm and EC = 3x cm. Then, we have :
\n\"Triangles
\n(a) x = 3
\n(b) x = 5
\n(c) x = 4
\n(d) x = 2.5
\nAnswer:
\n(c) x = 4<\/p>\n

\"Triangles<\/p>\n

Question 5.
\nIn \u2206ABC, DE || BC such that \\(\\frac{AD}{DB}\\) = \\(\\frac{3}{5}\\). If AC = 5.6 cm, then AE = ?
\n\"Triangles
\n(a) 4.2 cm
\n(b) 3.1 cm
\n(c) 2.8 cm
\n(d) 2.1 cm
\nAnswer:
\n(d) 2.1 cm<\/p>\n

Question 6.
\nABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of \u2206ABC and \u2206BDE is :
\n(a) 2 : 1
\n(b) 1 : 2
\n(c) 4 : 1
\n(d) 1 : 4.
\nAnswer:
\n(c) 4 : 1<\/p>\n

Question 7.
\nSides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio :
\n(a) 2 : 3
\n(b) 4 : 9
\n(c) 81 : 16
\n(d) 16 : 81.
\nAnswer:
\n(d) 16 : 81.<\/p>\n

Question 8.
\nHari goes 18 m due east and then 24 m due north. The distance from the starting point is :
\n(a) 40 m
\n(b) 30 m
\n(c) 26 m
\n(d) 42 m.
\nAnswer:
\n(b) 30 m<\/p>\n

\"Triangles<\/p>\n

Question 9.
\nThe length of the second diagonal of a rhombus, whose side is 5 cm and one of the diagonals is 6 cm is :
\n(a) 7 cm
\n(b) 8 cm
\n(c) 9 cm
\n(d) 12 cm.
\nAnswer:
\n(b) 8 cm<\/p>\n

Question 10.
\nIn \u2206ABC, AB = 6\u221a3, AC = 12 cm and BC = 6 cm. The angle B is :
\n(a) 120\u00b0
\n(b) 60\u00b0
\n(c) 90\u00b0
\n(d) 45\u00b0
\nAnswer:
\n(c) 90\u00b0<\/p>\n

Triangles Class 10 Extra Questions Very Short Answer Type<\/h3>\n

Question 1.
\nIn the figure, DE || BC. If AD = x, BD = x – 2, AE = x + 2 and EC = x – 1, find the value of x.
\nSolution.
\nIn \u2206ABC, DE || BC
\n\"Triangles
\n\"Triangles
\n\u21d2 (x – 2) (x + 2) = x (x – 1)
\n\u21d2 x2<\/sup> – 4 = x2<\/sup> – x
\n\u2234 x = 4<\/p>\n

Question 2.
\nState whether the following quadrilaterals are similar or not 3 cm C
\n\"Triangles
\nSolution.
\nThe two quadrilaterals, in figure are not similar because their corresponding angles are not equal. It is clear from the figure that, \u2220A is 90\u00b0 but \u2220P is not 90\u00b0.<\/p>\n

\"Triangles<\/p>\n

Question 3.
\nE and Fare points on the sides PQ and PR respectively of a A PQR, for each of the following cases, state whether EF || QR
\n(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm.
\n(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
\n(iii) PQ = 1:28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm.
\nSolution.
\n(i) In figure,
\n\"Triangles
\n\u21d2 EF is not parallel QR because converse of basic proportionality theorem is not satisfied.<\/p>\n

(ii) In figure,
\n\"Triangles
\n\u21d2 EF || QR because converse of basic proportionality of theorem is satisfied<\/p>\n

(iii) In figure,
\n\"Triangles
\n\u21d2 EF || QR because converse of basic proportionality theorem is satisfied<\/p>\n

\"Triangles<\/p>\n

Question 4.
\nIn the figure, if \\(\\frac{A O}{O C}=\\frac{B O}{D O}=\\frac{1}{2}\\) and AB = 5 cm find the value of DC.
\n\"Triangles
\nSolution.
\nIn \u2206OAB and \u2206OCD
\n\u2220AOB = \u2220COD (vertically opposite angles)
\n\"Triangles
\n\u2234 DC = 5 \u00d7 2 = 10 cm.<\/p>\n

Question 5.
\nPerimeters of two similar triangles are 40 cm and 60 cm respectively. Find ratio among their areas.
\nSolution.
\nGiven that the triangles are similar
\n\u2234 ratio of the perimeters of triangle = 40 : 60
\n= 2 : 3
\nSo the ratio of the area = (2)2<\/sup> : (3)2<\/sup>
\n= 4 : 9<\/p>\n

\"Triangles<\/p>\n

Question 6.
\nUsing theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (recall that you done it in Class IX).
\nSolution.
\nIn \u2206ABC, D and E are midpoints of side AB and AC, respectively.
\n\"Triangles
\n(By converse of basic proportionality theorem)<\/p>\n

Question 7.
\nIn figure, E is a point on side CB produced of an isosceles \u2206ABC with AB = AC. If AD \u22a5 BC and EF \u22a5 AC, prove that \u2206ABD ~ \u2206ECF
\nSolution.
\nIn figure, we are given that \u2206ABC is isosceles
\n\"Triangles
\nand AB = AC =
\n\u21d2 \u2220B = \u2220C …..(i)
\nFor \u2206ABD and \u2206ECF,
\n\u2220ABD = \u2220ECF [From Eq. (i)]
\nand \u2220ADB = \u2220EFC [Each = 90\u00b0]
\n\u21d2 ABD ~ AECF
\n(AAA similarity criterion)<\/p>\n

\"Triangles<\/p>\n

Question 8.
\nD is point on the side BC of a \u2206ABC such that \u2220ADC = \u2220BAC. Show that CA2<\/sup> = CB \u00b7 CD.
\nSolution.
\nDraw a \u2206ABC such that D is a point on BC and join AD.
\n\"Triangles
\nFor \u2206ABC and \u2206DAC, we have
\n\u2220BAC = \u2220ADC (Given)
\nand \u2220ACB = \u2220DCA (Common \u2220C)
\n\u21d2 \u2206ABC ~ ADAC
\n(AAA similarity criterion)
\n\u21d2 \\(\\frac{A C}{C B}\\) \\(\\frac{C D}{C A}\\)
\n\u21d2 \\(\\frac{C A}{C D}\\) \\(\\frac{C B}{C A}\\)
\n\u21d2 CA \u00d7 CA = CB \u00d7 CD
\n\u21d2 CA2<\/sup> = CB \u00d7 CD<\/p>\n

Question 9.
\nLet \u2206ABC ~ \u2206DEF and their areas be, 64 cm2<\/sup> and 121 cm2<\/sup>, respectively. If EF = 15.4 cm, find BC.
\nSolution.
\n\u2206ABC ~ \u2206DEF (Given)
\n\"Triangles
\n(Using property of area of similar triangles).
\n\"Triangles<\/p>\n

\"Triangles<\/p>\n

Question 10.
\nDiagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD. Find the ratio of the area of \u2206AOB and \u2206COD.
\nSolution.
\n\"Triangles<\/p>\n

Question 11.
\nIf the areas of two similar triangles are equal, prove that they are congruent.
\nSolution.
\nLet \u2206ABC ~ \u2206PQR and ar
\n(\u2206ABC) = ar (\u2206PQR) (Given)
\n\"Triangles
\n(Using property of area of similar triangles)
\n\u21d2 AB = PQ, BC = QR
\nand CA = PR
\n(SSS proportionality criterion)
\n\u21d2 \u2206ABC \u2245 \u2206PQR.<\/p>\n

\"Triangles<\/strong><\/p>\n

Question 12.
\nProve that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.
\nSolution.
\nDraw ABCD is a square having sides of length = a
\n\"Triangles
\nThen, the diagonal, BD = a\u221a2
\nWe construct equilateral \u2206PAB and \u2206QBD.
\n\u2206PAB ~ \u2206QBD
\n(Equilateral triangles are similar)
\n\"Triangles<\/p>\n

Question 13.
\nABC is an isosceles triangle right angled at C. Prove that AB2<\/sup> = 2AC2<\/sup>.
\nSolution.
\nDraw ABC is an isosceles triangle right angled at C.
\n\"Triangles
\nand AC = BC …..(i)
\nBy Pythagoras theorem, we have
\nAB2<\/sup> = AC2<\/sup>+ BC2<\/sup> = AC2<\/sup> + AC2<\/sup> = 2AC2<\/sup>
\n[\u2235 BC = AC by Eq. (i)]<\/p>\n

\"Triangles<\/p>\n

Question 14.
\nA ladder 10 m long reaches a window 8 cm above the ground. Find the distance of the foot of the ladder from base of the wall.
\nSolution.
\nLet B be the position of the window and CB be the length of the ladder.
\n\"Triangles
\nThen, AB = 8 cm (Height of window)
\nBC = 10 cm (Length of ladder)
\nLet AC = x m be the distance of the foot of the ladder from the base of the wall.
\nUsing Pythagoras theorem in \u2206ABC, we get
\nAC2<\/sup> + AB2<\/sup> = BC2<\/sup>
\n\u2234 x2<\/sup> + (8)2<\/sup> = (10)2<\/sup>
\n\u21d2 x2<\/sup> = 100 – 64 = 36
\n\u21d2 x = 6, i.e., AC = 6 cm<\/p>\n

Triangles Class 10 Extra Questions Short Answer Type<\/h3>\n

Question 1.
\nIn figures, (i) and (ii), DE || BC. Find EC in figure (i) and AD in figure (ii).
\n\"Triangles
\nSolution.
\n\"Triangles<\/p>\n

Question 2.
\nIn \u2206ABC, if the side AD is perpendicular to side BC and AD2<\/sup> = BD \u00d7 CD. Prove that \u2206ABC is a right angle \u2206.
\nSolution.
\nIn \u2206ABD and \u2206ADC, \u2220D = 90\u00b0
\n\u2234 AB2<\/sup> = AD2<\/sup> + BD2<\/sup> …(i)
\nand AC2<\/sup> = AD2<\/sup> + CD2<\/sup> … (ii)
\n\"Triangles
\nAdding (i) & (ii)
\nAB2<\/sup> + AC2<\/sup> = 2AD2<\/sup> +BD2<\/sup>+CD2<\/sup>
\n= 2(BD \u00d7 CD) + BD2<\/sup>+CD2<\/sup>
\n= BD2<\/sup> + 2BD \u00d7 CD + CD2<\/sup>
\n= 2(BD + CD)2<\/sup>
\n(Given AD2<\/sup> = BD \u00d7 CD).
\n= BC2<\/sup>
\n\u2234 \u2220BAC = 90\u00b0
\nHence \u2206BAC is a right angle \u2206.<\/p>\n

\"Triangles<\/p>\n

Question 3.
\nIn figure, AD \u22a5 BC
\n\"Triangles
\nProve that AB2<\/sup> + CD2<\/sup> = BD2<\/sup> + AC2<\/sup>
\nSolution :
\nIn \u2206ADB, \u2220D = 90\u00b0
\n\u2234 AB2<\/sup> = AD2<\/sup> + BD2<\/sup> …(i)
\nagain in \u2206ADC, \u2220D = 90\u00b0
\nAC2<\/sup> = AD2<\/sup> + CD2<\/sup> …(ii)
\nSubtract (ii) from (i)
\nAB2 <\/sup>– AC2<\/sup> = BD2 <\/sup>– CD2<\/sup>
\n\u21d2 AB2<\/sup> + CD2<\/sup> = AC2<\/sup> + BD2<\/sup><\/p>\n

Question 4.
\nUsing theorem, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. :
\nSolution.
\nIn \u2206ABC, D is the mid-point of AB.
\n\"Triangles
\ni.e., \\(\\frac{A D}{D B}\\) = 1 ……(i)
\nAs straight line l || BC.
\nLine l is drawn through D and it meets AC at E.
\nBy basic proportionality theorem,
\n\"Triangles
\n\u21d2 E is the mid-point of AC.<\/p>\n

Question 5.
\nThe diagonals of a quadrilateral ABCD intersect each other at the point O. such \\(\\frac{A O}{B O}=\\frac{C O}{D O}\\) Show that ABCD is a trapezium.
\nSolution.
\n\"Triangles
\n\u21d2 OE || CD(By converse of basic proportionality theorem)
\nNow, we have BA || OE and OE || CD = AB || CD
\n\u21d2 Quadrilateral ABCD is a trapezium.<\/p>\n

\"Triangles<\/p>\n

Question 6.
\nS and T are points on sides PR and QR of \u2206PQR such that \u2220P = \u2220RTS. Show that \u2206RPQ ~ \u2206RTS.
\nSolution.
\nDraw a \u2206RPQ such that S and T are points on PR and QR and joining them.
\n\"Triangles
\nIn figure, we have \u2206RPQ and \u2206RTS in which
\n\u2220RPQ = \u2220RTS (Given)
\n\u2220PRQ = \u2220SRT (Each = \u2220R)
\nThen, by AAA similarity criterion, we have
\n\u2206RPQ ~ \u2206RTS
\nNote: If any two corresponding angles of the triangles are equal, then their third corresponding angles are also equal by AAA.<\/p>\n

Question 7.
\nIn figure, altitudes AD and CE of \u2206ABC intersect each other at the point P. Show that
\n\"Triangles
\n(i) \u2206AEP ~ \u2206CDP
\n(ii) \u2206ABD ~ \u2206CBE
\n(iii) \u2206AEP ~ \u2206ADB
\n(iv) \u2206PDC ~ \u2206BEC
\nSolution.
\n(i) In figure, \u2220AEP = \u2220CDP (Each = 90\u00b0)
\nand \u2220APE = \u2220CPD
\n(Vertically opposite angles)
\n\u21d2 \u2220AEP ~ \u2206CDP
\n(By AAA similarity criterion)<\/p>\n

(ii) In figure,
\n\u2220ADB = \u2220CEB (Each = 90\u00b0)
\nand \u2220ABD = \u2220CBE (Each = \u2220B)
\n\u21d2 \u2206\u0410\u0412D ~ \u2206\u0421\u0412\u0415
\n(By AAA similarity criterion)<\/p>\n

(iii) In figure,
\n\u2220AEP = \u2220ADB (Each = 90\u00b0)
\nand \u2220PAE = \u2220DAB (Common angle)
\n\u21d2 \u2206AEP ~ \u2206ADB
\n(By AAA similarity criterion)<\/p>\n

(iv) In figure,
\n\u2220PDC = \u2220BEC (Each = 90\u00b0)
\nand \u2220PCD = \u2220BCE (Common angle)
\n\u21d2 \u0394PDC ~ \u0394BEC
\n(By AAA similarity criterion)<\/p>\n

\"Triangles<\/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.
\nSolution.
\nDraw a parallelogram ABCD and produce a line AD to AE and joining BE. In parallelogram ABCD,
\n\u2220A = \u2220C …(i)
\nNow, for \u2206ABE and \u2206CFB,
\n\u2220EAB = \u2220BCF [From Eq. (i)],
\n\"Triangles
\n\u2220ABE = \u2220BFC
\n(Alternate angles as AB || FC)
\n\u2206ABE ~ \u2206CFB (AAA similarity)<\/p>\n

Question 9.
\nIn figure, ABC and AMP are two right triangles, right angled at B and M, respectively. Prove that
\n\"Triangles
\n(i) \u2206ABC ~ \u2206AMP
\n(ii) \\(\\frac{C A}{P A}=\\frac{B C}{M P}\\)
\nSolution.
\n(i) In figure, we have \u2220ABC = \u2220AMP (Each = 90\u00b0)
\nBecause the \u2206ABC and \u2206AMP are right angled at B and M, respectively.
\nAlso, \u2220BAC = \u2220PAM
\n(Common angle \u2220A)
\n\u21d2 ABC – \u2206AMP
\n(By AAA similarity criterion)<\/p>\n

(ii) As \u2206ABC ~ \u2206AMP,
\n\\(\\frac{A C}{A P}=\\frac{B C}{M P}\\)
\n(Ratio of the corresponding sides of similar triangles)
\n\u21d2 \\(\\frac{C A}{P A}=\\frac{B C}{M P}\\)<\/p>\n

\"Triangles<\/p>\n

Question 10.
\nIn a triangle ABC, AD is the median of BC and E is mid-point of AD. If BE produced, it meets AC in figure. Prove that AF = \\(\\frac {1}{3}\\) AC.
\nSolution.
\nGiven: In \u2206ABC, AD is the median and E is the mid-point of AD.BE is joined and produced intersect AC at F.
\nT.P.T. – AF = \\(\\frac {1}{3}\\) AC
\n\"Triangles
\nConstruction-Draw a line parallel to BF through B, which intersect AC at G. EF || DG
\nProof – In \u2206ADC, EF || DG
\n\u2234 AF = FG (Mid-point theorem) …(i)
\nIn ABFC, EF || DG
\n\u2234 FG = GC (Mid-point theorem) ….(ii)
\nFrom Eq. (i) & (ii)
\nAF = FG = GC Now
\nAC = AF + FG + GC
\n= AF + AF + AF
\n= 3AF
\nAF = \\(\\frac {1}{3}\\)AC.<\/p>\n

Question 11.
\nBL and CM are the medians of a right triangle ABC right angle at A. Prove that 4(BL2<\/sup> + CM)2<\/sup> = 5 BC2<\/sup>.
\nSolution.
\nGiven,
\nIn \u2206BAC, \u2220A = 90\u00b0 and BL and CM are the medians.
\n\"Triangles
\nT.P.T. – 4(BL2<\/sup>+cm2<\/sup>) = 5BC2<\/sup>
\nProof – In \u2206BAC, \u2220A = 90\u00b0
\nAB2<\/sup> + AC2<\/sup> = BC2<\/sup> ….(i)
\nIn \u2206BAL, \u2220A = 90\u00b0
\nBL2<\/sup> = AB2<\/sup> + AL2<\/sup>
\n\"Triangles
\nIn \u2206MAC,\u00a0 \u2220A = 90\u00b0
\nCM2<\/sup> = AM2<\/sup> = AC2<\/sup>
\n\"Triangles
\n\u21d2 4(BL2<\/sup> + cm2<\/sup>) = 5(AB2<\/sup> + AC2<\/sup>)
\n= 5BC2<\/sup> From Eq. (i)
\n(AB2<\/sup> + AC2<\/sup> = BC2<\/sup>)<\/p>\n

\"Triangles<\/p>\n

Question 12.
\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.
\nSoultion.
\nIn figure (i), AB is a pole behind it a Sun is risen which casts a shadow of length BC = 4 cm and makes an angle \u03b8 to the horizontal and in figure it, PM is a height of the tower and behind a Sun risen which casts a shadow of length, NM = 28 cm.
\n\"Triangles
\n\"Triangles<\/p>\n

Question 13.
\nIn figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O,
\nshow that
\n\"Triangles
\nSolution.
\nDraw AL \u22a5 BC and DM \u22a5 BC (See figure)
\n\u2220ALO = \u2220DMO = 90\u00b0
\nand \u2220AOL = \u2220DOM
\n(Vertically opposite angle)
\n\u2234 \u0394OLA ~ \u0394\u039f\u039cD
\n(AAA similarity criterion)
\n\"Triangles<\/p>\n

Question 14.
\nD, E and Fare respectively the midpoint of sides AB, BC and CA of \u2206ABC. Find the ratio of the areas of \u2206DEF and \u2206ABC.
\nSolution.
\nDraw a \u2206ABC taking mid-points D, E and F on AB, BC and AC. Join them.
\nHere, DF = \\(\\frac {1}{2}\\)BC, DE = \\(\\frac {1}{2}\\)CA
\nand EF = \\(\\frac {1}{2}\\)AB …….(i)
\n(\u2235 D, E and Fare mid-points of sides AB, BC and CA respectively)
\n\"Triangles
\n[\u2235 ar (\u2206CAB)= ar (\u2206ABC)]
\nHence, the required ratio is 1 : 4.<\/p>\n

\"Triangles<\/p>\n

Question 15.
\nEquilateral triangles are drawn on the sides of a right angled triangle. Show that the area of the triangle on the hypotenuse is equal to the sum of the areas of triangles on the other two sides.
\nSolution.
\nGiven. A right angled triangle ABC, with right angle at B. Equilateral triangles PAB, QBC and RAC are described on sides AB, BC and CA respectively.
\n\"Triangles
\nTo prove. Area (\u2206PAB) + area (\u2206QBC) = area (\u2206RAC)
\nProof. Since, triangles PAB, QBC and RAC are equilateral. Therefore, they are equiangular and hence similar.
\n\"Triangles
\n\"Triangles<\/p>\n

Question 16.
\nIn adjoining Fig. ABC and BCD are two triangles on the same base BC. If AD intersects BC at O, show that
\n\"Triangles
\nSolution.
\nGiven. \u2206ABC and ABCD are on the same base and AD intersects BC at O.
\n\"Triangles
\nConstruction. Draw AM \u22a5 BC and DN
\nProof. In \u2206AMO and \u2206DNO,
\nM = \u2220N
\n[Each 90\u00b0, by construction]
\n\u2220AOM – \u2220DON [same angles]
\n\u2206AMO ~ \u2206DNO [AA corollary]
\n\u21d2 \\(\\frac{A O}{D O}=\\frac{A M}{D N}\\)
\n[Corresponding sides are proportional in similar triangles]
\n\"Triangles<\/p>\n

Question 17.
\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 \u00d7 MR.
\nSolution.
\nIn \u2206PQR and \u2206MPQ,
\n\"Triangles
\n\u22201 + \u22202 = \u22202 + \u22204 (Each = 90\u00b0)
\n\u21d2 \u22201 = \u22204
\nSimilarly, \u22202 = \u22203
\nand \u2220PMR = \u2220PMQ (each 90\u00b0)
\n\u2206QPM ~ \u2206PRM (AAA criterion)
\n\"Triangles
\n\u21d2 PM2<\/sup> = QM \u00d7 RM
\nor PM2<\/sup> = QM \u00d7 MR<\/p>\n

\"Triangles<\/p>\n

Question 18.
\nABC is an isosceles triangle with AC = BC. If AB2<\/sup> = 2AC?, prove that ABC is a right triangle.
\nSolution.
\nDraw an isosceles \u2206ABC with AC =BC.
\n\"Triangles
\nIn \u2206ABC, we are given that
\nAC = BC …..(i)
\nand AB2<\/sup> = 2AC2<\/sup> …(ii)
\nNow, AC2<\/sup> + BC2<\/sup>-AC2<\/sup> + AC2<\/sup> [By Eq. (i)]
\n= 2AC2<\/sup> = AB2<\/sup> (By Eq. (ii)]
\ni.e., AC2<\/sup> + BC2<\/sup> = AB2<\/sup>
\nHence, by the converse of the Pythagoras theorem, we have \u2206ABC is right angled at C.<\/p>\n

Question 19.
\nABC is an equilateral triangle of side 2a. Find each of its altitude.
\nSolution.
\nDraw equilateral \u2206ABC, each side is 2a.
\n\"Triangles
\nAlso, draw AD \u22a5 BC.
\nWhere AD is an altitude.
\nIn \u2206ADB and \u2206ADC
\nAD = AD (Common)
\nand \u2220ADB = \u2220ADC = 90\u00b0
\n\u2206ADB \u2245 \u2206ADC (RHS congruency)
\nBD = CD = \\(\\frac {1}{2}\\)BC = a
\n(\u2235 In an equilateral triangle altitude AD is the perpendicular bisector of BC).
\nNow, from \u2206ABD by Pythagoras theorem, we get
\nAB2<\/sup> = AD2<\/sup> + BD2<\/sup>
\n\u21d2 (2a)2<\/sup> = AD2<\/sup> + a2<\/sup>
\n\u21d2 AD2<\/sup> = 3a2<\/sup>
\n\u21d2 AD = \u221a3a.<\/p>\n

\"Triangles<\/p>\n

Question 20.
\nProve that the sum of the square of the sides of a rhombus is equal to the sum of the squares of its diagonals.
\nSolution.
\nDraw ABCD is a rhombus in which AB = BC = CD = DA = a (Say)
\n\"Triangles
\nIts diagonal AC and BD are right angled bisector of each other at O.
\nIn \u2206OAB, \u2220AOB = 90\u00b0,
\nOA = \\(\\frac {1}{2}\\)AC and OB = \\(\\frac {1}{2}\\)BD
\nIn \u2206AOB, use Pythagoras theorem, we have
\nOA2<\/sup> + OB2<\/sup> = AB2<\/sup>
\n\"Triangles
\n\u21d2 AC2<\/sup> + BD2<\/sup> = 4AB2<\/sup>
\nor 4AB2<\/sup> = AC2<\/sup> + BD2<\/sup>
\n\u21d2 AB2<\/sup> + BC2<\/sup> + CD2<\/sup> + DA2<\/sup> = AC2<\/sup> + BD2<\/sup>
\n(\u2235 AB = BC = CD = DA)
\nHence proved.<\/p>\n

Question 21.
\nIn figure, O is a point in the interior of a \u2206ABC, OD \u22a5 BC, OE \u22a5 AC and OF \u22a5 AB. Show that
\n\"Triangles
\n(i) OA2<\/sup> + OB2<\/sup>+OC2<\/sup>-OD2<\/sup>-OE2<\/sup>-OF2<\/sup> – AF2<\/sup> + BD2<\/sup> + CE2<\/sup>
\n(ii) AF2<\/sup>+ BD2<\/sup> + CE2<\/sup> = AE2<\/sup> + CD2<\/sup> + BF2<\/sup>
\nSolution.
\nIn \u2206ABC, from point O join lines OB, OC and OA.
\n(i) In right angled \u2206OFA,
\nOA2<\/sup> = OF2<\/sup> + AF2<\/sup>
\n(By Pythagoras therorem)
\n\u21d2 OA2<\/sup> – OF2<\/sup> = AF2<\/sup> …..(i)
\nSimilarly, in \u2206OBD,
\nOB2<\/sup> – OD2<\/sup> = BD2<\/sup> …(ii)
\nand in \u2206OCE,
\nOC2<\/sup> – OE2<\/sup> = CE2<\/sup> …(iii)
\nOn adding Eqs. (i), (ii) and (iii), we get
\nOA2<\/sup> + OB2<\/sup> + OC2<\/sup> – OD2<\/sup> – OE2<\/sup> – OF2<\/sup>
\n= AF2<\/sup> + BD2<\/sup> + CE2<\/sup><\/p>\n

(ii) From part Eqs. (i), we get
\nOA2<\/sup> + OB2<\/sup> + OC2<\/sup> – OD2<\/sup> – OE2<\/sup> – OF2<\/sup>
\n= AF2<\/sup> + BD2<\/sup> + CE2<\/sup> …(iv)
\nSimilarly,
\nOA2<\/sup> + OB2<\/sup> + OC2<\/sup> – OD2<\/sup> – OE2<\/sup> – OF2<\/sup>
\n= BF2<\/sup> + CD2<\/sup> + AE2<\/sup> …(v)
\nFrom Eqs. (iv) and (v), we have
\nAF2<\/sup> + BD2<\/sup> + CE2<\/sup> = AE2<\/sup> + CD2<\/sup> + BF2<\/sup><\/p>\n

\"Triangles<\/p>\n

Question 22.
\nA guy wire attached to a vertical pole of height 18m is 24m 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?
\nSolution.
\nLet AB be the vertical pole of height 18 m, A guy wire is of length BC = 24 m.
\n\"Triangles
\nLet AC = x m be the distance of the stake from the base of the pole.
\nUsing Pythagoras theorem in \u2206ABC, we get
\ni.e., AC2<\/sup> + AB2<\/sup> = BC2<\/sup>
\n\u2234 x2<\/sup> + (18)2<\/sup> = (24)2<\/sup>
\n\u21d2 x2<\/sup> = (24)2<\/sup> – (18)2<\/sup>
\n= 576 – 324 = 252
\n\u21d2 x = \u221a252 m
\n(\u2235 We take positive sign because cannot be negative)
\n\u21d2 x = 6 \u221a7 m<\/p>\n

Question 23.
\nAn aeroplane leaves an airport and flies due North at a speed of 1000 kmh-1<\/sup>. At the same time, another aeroplane leaves the same airport and flies due West at a speed of 1200 kmh-1<\/sup>. How far apart will be two planes after 1\\(\\frac {1}{2}\\)h?
\nSolution.
\nThe first plane travels distance BC in the direction of North in 1\\(\\frac {1}{2}\\)h at speed of 1000 km\/h.
\n\"Triangles<\/p>\n

Question 24.
\nTwo poles of heights 6 m and 11 m stand on a plane ground. If the distance between the f00t of the poles is 12 m, find the distance between their tops.
\nSolution.
\nLet BC and AD be the two poles of heights 11 m and 6 m.
\nThen, CE = BC – AD
\n= 11 – 6
\n= 5 cm
\n\"Triangles
\nLet distance between tops of two poles DC m = x cm
\nUsing Pythagoras theorem in \u2206DEC, we get
\ni.e., DC2<\/sup> = DE2<\/sup> + CE2<\/sup>
\n\u21d2 x2<\/sup> = (12)2<\/sup> + (5)2<\/sup> = 169
\n\u21d2 x = 13
\nHence, distance between their tops = 13 m.<\/p>\n

\"Triangles<\/p>\n

Question 25.
\nD and E are points on the sides CA and CB, respectively of a \u2206ABC right angled at C. Prove that AE2<\/sup> + BD2<\/sup> = AB2<\/sup> + DE2<\/sup>.
\nSolution.
\nDraw a right \u2206ABC at C. Take D and E points on the sides CA and BC and join ED, BD and EA.
\n\"Triangles
\nIn right angled \u2206ACE,
\nAE2<\/sup> = CA2<\/sup> + CE2<\/sup> ……(i)
\n(By pythagoras theorem)
\nand in right angled ABCD,
\nBD2<\/sup> = BC2<\/sup> + CD2<\/sup> …(ii)
\nOn adding Eqs. (i) and (ii), we get
\nAE2<\/sup> + BD2<\/sup> = (Ca2<\/sup> + CE2<\/sup>) + (BC2<\/sup> + CD2<\/sup>)
\n= (BC2<\/sup> + Ca2<\/sup>) + (CD2<\/sup> + CE2<\/sup>)
\n(\u2235 In \u2206ABC, Ba2<\/sup> = BC2<\/sup> + CA2<\/sup> and in \u2206ECD, DE2<\/sup> = CD2<\/sup> + CE2<\/sup>)
\n= BA2<\/sup> + DE2<\/sup>
\n(By Pythagoras theorem),
\n\u2234 AE2<\/sup> + BD2<\/sup> = AB2<\/sup> + DE2<\/sup>
\nHence proved.<\/p>\n

\"Triangles<\/p>\n

Question 26.
\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.
\nSolution.
\nDraw \u2206ABC is an equilateral triangle of side a.(Say)
\n\"Triangles
\nand AD \u22a5 BC
\nLet AD = x
\nNow, BD = CD = \\(\\frac {1}{2}\\)BC = \\(\\frac {1}{2}\\)a
\n(In an equilateral triangle altitude AD is a perpendicular bisector of BC)
\nIn right angled \u2206ABD,
\nAB2<\/sup> = AD2<\/sup> + BD2<\/sup>
\n\u21d2 a2<\/sup> = x2<\/sup> + (\\(\\frac {1}{2}\\)a)2<\/sup>
\n\u21d2 a2<\/sup> = x2<\/sup> + \\(\\frac {1}{4}\\)a2<\/sup>
\n\u21d2 4a2<\/sup> = ax2<\/sup> + a2<\/sup>
\n\u21d2 3a2<\/sup> = 4x2<\/sup>
\nHence proved.<\/p>\n

Triangles Class 10 Extra Questions Long Answer Type<\/h3>\n

Question 1.
\nIn figure, if LM || CB and LN || CD,
\nprove that \\(\\frac{A M}{A B}=\\frac{A N}{A D}\\)
\n\"Triangles
\nSolution.
\n\"Triangles<\/p>\n

Question 2.
\nABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that \\(\\frac{A O}{B O}=\\frac{C O}{D O}\\).
\nSolution.
\nWe draw,
\nEOF ||AB (Also || CD)
\nIn \u2206ACD, OE || CD
\n\"Triangles<\/p>\n

Question 3.
\nProve that If the corresponding sides of two triangles are proportional then their corresponding angles are equal, and hence the two triangles are similar.
\nSolution.
\nGiven. \u2206ABC and \u2206DEF in which
\n\\(\\frac{A B}{D E}\\) = \\(\\frac{B C}{B F}\\) = \\(\\frac{A C}{D F}\\)
\nTo Prove. \u2206ABC ~ \u2206DEF.
\nConstruction. Let us take \u2206ABC and \u2206DEF such that
\n\\(\\frac{A B}{D E}\\) = \\(\\frac{B C}{B F}\\) = \\(\\frac{A C}{D F}\\)(<1).
\nCut DP = AB and DQ = AC. Join PQ.
\n\"Triangles
\n\"Triangles
\n\u21d2 BC = PQ
\nThus, AB = DP, AC = DQ and DC = PQ
\n\u2234 \u2206ABC \u2245 \u2206DPQ
\n[By SSS-congruence]
\n\u2234 \u2220A = \u2220D, \u2220B = P = \u2220E
\nand C = \u2220Q = \u2220F
\n\u21d2 \u2220A = \u2220D, \u2220B = \u2220E and
\n\u2220C = \u2220F.
\nThus, the given triangles are equiangular and hence similar.<\/p>\n

\"Triangles<\/p>\n

Question 4.
\nProve that If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional then the two triangles are similar.
\nAnswer:
\nGiven. \u2206ABC and \u2206DEF in which
\n\u2220A = \u2220D and \\(\\frac{A B}{D E}\\) = \\(\\frac{A C}{D F}\\)
\nTo prove. \u2206ABC ~ \u2206DEF.
\nConstruction. Let us take \u2206ABC and \u2206DEF such that
\n\\(\\frac{A B}{D E}\\) = \\(\\frac{A C}{D F}\\) (<1) and \u2220A = \u2220D.
\nCut DP = AB and DQ = AC. Join PQ.
\n\"Triangles
\nProof. In \u2206ABC and ADPQ, we have :
\nAB = DP [By construction]
\n\u2220A = \u2220D (Given)
\nAC = DQ [By construction]
\n\u2206ABC \u2245 \u2206DPQ [By SAS-congruence]
\n\u2234 \u2220A = \u2220D, \u2220B = \u2220P
\nand \u2220C = \u2220Q.
\n\"Triangles
\n[\u2235 AB = DP and AC = DQ)
\n\u21d2 PQ || EF
\n[By the converse of Thales’ theorem)
\n\u21d2 \u2220P = \u2220E and \u2220Q = \u2220F
\n[Corresponding \u2220s]
\n\u2234 \u2220A = \u2220D, \u2220B = P = \u2220E
\nand C = 2Q = \u2220F.
\nThus, \u2220A = \u2220D, \u2220B = \u2220E
\nand C = \u2220F.
\nSo, the given triangles are equiangular and hence similar.<\/p>\n

\"Triangles<\/p>\n

Question 5.
\nProve that If two triangles are equiangular, prove that the ratio of their corresponding sides is the same as the ratio of the corresponding medians.
\nSolution.
\nGiven. \u2206ABC and ADEF in which \u2220A = \u2220D, \u2220B = \u2220E and \u2220C = \u2220F, and AL and DM are the medians.
\n\"Triangles
\n\"Triangles<\/p>\n

Question 6.
\nDiagonals AC and BD of a trapezium ABCD with AB || DC intersect each other the point O. Using a similarity criterion for two triangles, show that
\n\\(\\frac{O A}{O C}=\\frac{O B}{O D}\\)
\nSolution.
\nDraw ABCD is a trapezium and AC and BD are diagonals intersect at O.
\n\"Triangles
\nIn figure, AB || DC (Given)
\n\u21d2 \u22201 = \u22203, \u22202 = 24
\n(Alternate interior angles)
\nAlso, \u2220DOC = \u2220BOA
\n(Vertically opposite angles)
\n\u21d2 \u2206OCD ~ \u2206QAB (Similar triangle)
\n\u21d2 \\(\\frac{O C}{O A}=\\frac{O D}{O B}\\)
\n(Ratios of the corresponding sides of the similar triangles)
\n\u21d2 \\(\\frac{O A}{O C}=\\frac{O B}{O D}\\) (Taking reciprocals)
\nHence proved.<\/p>\n

\"Triangles<\/p>\n

Question 7.
\nSides AB and AC and median AD of a \u2206ABC are respectively proportional to sides PQ and PR and median PM of another \u2206PQR. Show that \u2206ABC ~ \u2206PQR.
\nSolution.
\nGiven, in \u2206ABC and \u2206PQR,
\nAD and PM are their medians, respectively.
\n\"Triangles
\nTo Prove. \u2206ABC ~ \u2206PQR
\nConstruction. Produce AD to E such that AD = DE and produce PM to N such that PM = MN. Join BE, CE, QN and RN.
\n\"Triangles
\nProof. Quadrilaterals ABEC and PQNR are parallelograms because their diagonals bisect each other at D and M, respectively.
\n\"Triangles
\nFrom Eqs. (ii) and (iii), we have
\n\\(\\frac{A B}{P Q}=\\frac{B E}{Q N}=\\frac{A E}{P N}\\)
\n\u21d2 \u0394\u0391\u0392\u0395 ~ \u0394\u03a1\u03a9\u039d
\n\u21d2 \u22201 = \u22202 ….(iv)
\nSimilarly, we can prove that
\n\u2206ACE ~ \u2206PRN
\n\u22203 = \u22204 …..(v)
\nOn adding Eqs. (iv) and (v), we have
\n\u22201 + \u22203 = \u22202 + \u22204
\n\u21d2 \u2220A = \u2220P
\n\u21d2 \u2206BC ~ \u2206PQR
\n(SAS similarity criterion)<\/p>\n

\"Triangles<\/p>\n

Question 8.
\nIn figure, if \u2206ABE \u2245 \u2206ACD, show that \u2206ADE ~ \u2206ABC.
\n\"Triangles
\nSolution.
\nIn figure,
\n\u2206ABE \u2245 \u2206ACD (Given)
\n\u21d2 AB = AC and AE = AD (CPCT)
\n\"Triangles
\nand also,
\n\u2220DAE = \u2220BAC (Each = \u2220A)
\n\u2206ADE ~ \u2206ABC (By SAS similarity criterion)
\nHence proved.<\/p>\n

\"Triangles<\/p>\n

Question 9.
\nIf AD and PM are medians of \u2206ABC and \u2206PQR, respectively, where \u2206ABC ~ \u2206PQR,
\nprove that \\(\\frac{A B}{P Q}=\\frac{A D}{P M}\\)
\nSolution.
\nDraw two \u2206ABC and APQR taking D and M points on BC and QR such that AD and PM are the medians of the \u2206ABC and \u2206PQR.
\n\"Triangles
\n\u2206ABC – \u2206PQR (Given)
\n\u21d2 \\(\\frac{A B}{P Q}=\\frac{B C}{Q R}=\\frac{A C}{P R}\\) ; \u2220A = \u2220P,
\n\u2220B = \u2220Q, \u2220C = \u2220R …(i)
\nNow, BD = CD = \\(\\frac {1}{2}\\)BC
\nandQM = RM = \\(\\frac {1}{2}\\)QR …..(ii)
\n(\u2235 D is mid-point of BC and M is mid point of QR)
\nFrom Eq. (i),
\n\"Triangles<\/p>\n

Question 10.
\nUsing the properties of similar triangles, prove that sum of square of two sides is equal to square of its hypotenuse in a right triangle.
\nSolution.
\nGiven : In right angled \u2206ABC, \u2220B= 90\u00b0
\n\"Triangles
\nTo Prove: AC2<\/sup> = AB2<\/sup> + BC2<\/sup>
\nConstruction : Draw BD \u22a5 AC.
\nPr00f : \u2206ADB ~ \u2206ABC(by A similarly criterion)
\n\u2234 \\(\\frac{A D}{A B}\\) = \\(\\frac{A B}{A C}\\)
\n\u21d2 AB2<\/sup> = AD \u00d7 AC
\n\u2206BDC ~ \u2206ABC(by A similarty criterion)
\n\u2234 \\(\\frac{B C}{D C}\\) + \\(\\frac{A C}{B C}\\)
\n\u21d2 BC2<\/sup> = AC \u00d7 DC …(ii)
\nOn adding (i) & (ii)
\nAB2<\/sup> + BC2<\/sup> = AD \u00d7 AC + AC \u00d7 DC
\n= AC(AD + DC)
\n= AC \u00d7 AC
\n= AC2<\/sup><\/p>\n

\"Triangles<\/p>\n

Question 11.
\nProve that the ratio of the area of two similar triangles is equal to the square of the ratio of their corresponding medians.
\nSolution.
\n\"Triangles
\nIn figure, AD is a median of \u2206ABC and PM is a median of \u2206PQR. Here, D is mid-point of BC and M is mid-point of QR. Now, we have,
\n\u2206ABC ~ APQR
\n\u21d2 \u2220B = \u2220Q
\n(Corresponding angles are equal) …(i)
\nAlso, \\(\\frac{A B}{P Q}\\) + \\(\\frac{B C}{Q R}\\)
\n(Ratio of corresponding sides are equal)
\n\"Triangles<\/p>\n

Question 12.
\nThe perpendicular from A on side BC of a \u2206ABC intersects BC at D such that DB = 3 CD (see in figure). Prove that 2AB2<\/sup> = 2AC2<\/sup> + BC2<\/sup>.
\n\"Triangles
\nSolution.
\nGiven, DB = 3CD
\n\"Triangles
\n= CD = \\(\\frac{1}{4}\\)BC …(i)
\nand DB = \\(\\frac{3}{4}\\) BC
\nIn \u2206ABD, AB2<\/sup> = DB2<\/sup> + AD2<\/sup> …(ii)
\nIn \u2206ACD, AC2<\/sup> = CD2<\/sup> + AD2<\/sup> …(iii)
\n(By Pythagoras as theorem)
\nOn subtracting Eq. (iii) from Eq. (ii), we get
\nAB2<\/sup> – AC2<\/sup> = DB2<\/sup> – CD2<\/sup>
\n\"Triangles
\n\u21d2 2AB2<\/sup> – 2AC2<\/sup> = BC2<\/sup>
\n\u21d2 2AB2<\/sup> = 2AC2<\/sup> + BC2<\/sup>
\nHence proved.<\/p>\n

\"Triangles<\/p>\n

Question 13.
\nIn an equilateral \u2206ABC, D is a point on side BC such that BD = \\(\\frac{1}{3}\\)BC. Prove that 9AD2<\/sup> = 7AB2<\/sup>
\nSolution.
\nDraw ABC is an equilateral triangle, D is a point on side BC such that BD = \\(\\frac{1}{3}\\)BC. Draw a line AE is perpendicular to BC.
\n\"Triangles
\nAB = BC = CA = a (Say)
\n(By property of equilateral triangle)
\nBD = \\(\\frac{1}{3}\\)BC = \\(\\frac{1}{3}\\)a
\n\u21d2 CD = \\(\\frac{2}{3}\\)BC = \\(\\frac{2}{3}\\)a
\n\u2235 AE \u22a5 BC
\n\u21d2 BE = EC = \\(\\frac{1}{2}\\)a
\n(\u2235 In an equilateral triangle altitude AE is perpendicular bisector of BC.)
\nDE = BE – BD = \\(\\frac{1}{2}\\)a – \\(\\frac{1}{3}\\)a – \\(\\frac{1}{6}\\)a
\nUsing pythagoras theorem in \u2206ADE,
\nAD2<\/sup> = AE2<\/sup> + DE2<\/sup>
\n= AB2<\/sup> – BE2<\/sup> + DE2<\/sup>
\n(\u2235 Right \u2206ABE, AE2<\/sup> = AB2<\/sup> – BE2<\/sup>)
\n\"Triangles
\n\u21d2 9AD2<\/sup> = 7AB2<\/sup> Hence proved.<\/p>\n

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

Here you will find Triangles Class 10 Extra Questions Maths Chapter 6 with Answers Solutions, Extra Questions for Class 10 Maths are solved by experts and will guide students in the right direction. Extra Questions for Class 10 Maths Triangles with Answers Solutions Extra Questions for Class 10 Maths Chapter 6 Triangles with Solutions Answers …<\/p>\n

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