{"id":119155,"date":"2022-05-04T12:38:54","date_gmt":"2022-05-04T07:08:54","guid":{"rendered":"https:\/\/www.mapsofindia.com\/my-india\/?p=119155"},"modified":"2022-05-04T12:38:54","modified_gmt":"2022-05-04T07:08:54","slug":"chapter-6-triangles-questions-and-answers-ncert-solutions-for-class-10-mathematics","status":"publish","type":"post","link":"https:\/\/www.mapsofindia.com\/my-india\/education\/chapter-6-triangles-questions-and-answers-ncert-solutions-for-class-10-mathematics","title":{"rendered":"Chapter 6 – Triangles Questions and Answers: NCERT Solutions for Class 10 Mathematics"},"content":{"rendered":"

Exercise 6.1<\/h2>\n

1. Fill in the blanks using correct word given in the brackets:-
\n(i) All circles are __________. (congruent, similar)<\/h2>\n

Answer:\u00a0Similar<\/h3>\n

(ii) All squares are __________. (similar, congruent)<\/h2>\n

Answer:\u00a0Similar<\/h3>\n

(iii) All __________ triangles are similar. (isosceles, equilateral)<\/h2>\n

Answer: Equilateral<\/h3>\n

(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are __________ and (b) their corresponding sides are __________. (equal, proportional)<\/h2>\n

Answer:\u00a0(a) Equal
\n(b) Proportional<\/h3>\n

2. Give two different examples of pair of
\n(i) Similar figures
\n(ii) Non-similar figures<\/h2>\n

Solution:<\/h3>\n

\"\"<\/p>\n

\"\"
\n3. State whether the following quadrilaterals are similar or not:<\/h2>\n

\"\"
\nSolution:
\nFrom the given two figures, we can see their corresponding angles are different or unequal. Therefore they are not similar.<\/h3>\n

Exercise 6.2<\/h2>\n

1. In figure. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).<\/h2>\n

\"\"
\nSolution:
\n(i) Given, in\u00a0\u25b3 ABC, DE\u2225BC
\n\u2234 AD\/DB = AE\/EC [Using Basic proportionality theorem]
\n\u21d21.5\/3 = 1\/EC
\n\u21d2EC = 3\/1.5
\nEC = 3\u00d710\/15 = 2 cm
\nHence, EC = 2 cm.<\/h3>\n

(ii) Given, in\u00a0\u25b3 ABC, DE\u2225BC
\n\u2234 AD\/DB = AE\/EC [Using Basic proportionality theorem]
\n\u21d2 AD\/7.2 = 1.8 \/ 5.4
\n\u21d2 AD = 1.8 \u00d77.2\/5.4 = (18\/10)\u00d7(72\/10)\u00d7(10\/54) = 24\/10
\n\u21d2 AD = 2.4
\nHence, AD = 2.4 cm.<\/h3>\n

2. E and F are points on the sides PQ and PR respectively of a \u0394PQR. 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 = 0.63 cm<\/h2>\n

Solution:
\nGiven, in\u00a0\u0394PQR, E and F are two points on side PQ and PR respectively. See the figure below;<\/h3>\n

\"\"
\n(i) Given, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2,4 cm
\nTherefore, by using Basic proportionality theorem, we get,
\nPE\/EQ = 3.9\/3 = 39\/30 = 13\/10 = 1.3
\nAnd PF\/FR = 3.6\/2.4 = 36\/24 = 3\/2 = 1.5
\nSo, we get, PE\/EQ \u2260 PF\/FR
\nHence, EF is not parallel to QR.<\/h3>\n

(ii) Given, PE = 4 cm, QE = 4.5 cm, PF = 8cm and RF = 9cm
\nTherefore, by using Basic proportionality theorem, we get,
\nPE\/QE = 4\/4.5 = 40\/45 = 8\/9
\nAnd, PF\/RF = 8\/9
\nSo, we get here,
\nPE\/QE = PF\/RF
\nHence, EF is parallel to QR.<\/h3>\n

(iii) Given, PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
\nFrom the figure,
\nEQ = PQ \u2013 PE = 1.28 \u2013 0.18 = 1.10 cm
\nAnd, FR = PR \u2013 PF = 2.56 \u2013 0.36 = 2.20 cm
\nSo, PE\/EQ = 0.18\/1.10 = 18\/110 = 9\/55\u2026\u2026\u2026\u2026. (i)
\nAnd, PE\/FR = 0.36\/2.20 = 36\/220 = 9\/55\u2026\u2026\u2026\u2026 (ii)
\nSo, we get here,
\nPE\/EQ = PF\/FR
\nHence, EF is parallel to QR.<\/h3>\n

3. In the figure, if LM || CB and LN || CD, prove that AM\/AB = AN\/AD<\/h2>\n

\"\"
\nSolution:
\nIn the given figure, we can see, LM || CB,
\nBy using basic proportionality theorem, we get,
\nAM\/AB = AL\/AC\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(i)
\nSimilarly, given, LN || CD and using basic proportionality theorem,
\n\u2234AN\/AD = AL\/AC\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(ii)
\nFrom equation\u00a0(i)\u00a0and\u00a0(ii), we get,
\nAM\/AB = AN\/AD
\nHence, proved.<\/h3>\n

4. In the figure, DE||AC and DF||AE. Prove that BF\/FE = BE\/EC<\/h2>\n

\"\"
\nSolution:
\nIn \u0394ABC, given as, DE || AC
\nThus, by using Basic Proportionality Theorem, we get,
\n\u2234BD\/DA = BE\/EC \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(i)
\nIn \u00a0\u0394BAE, given as, DF || AE
\nThus, by using Basic Proportionality Theorem, we get,
\n\u2234BD\/DA = BF\/FE \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(ii)
\nFrom equation\u00a0(i)\u00a0and\u00a0(ii), we get
\nBE\/EC = BF\/FE
\nHence, proved.<\/h3>\n

5. In the figure, DE||OQ and DF||OR, show that EF||QR.<\/h2>\n

\"\"
\nSolution:
\nGiven,
\nIn \u0394PQO, DE || OQ
\nSo by using Basic Proportionality Theorem,
\nPD\/DO = PE\/EQ\u2026\u2026\u2026\u2026\u2026\u2026\u00a0..(i)
\nAgain given, in \u0394POR, DF || OR,
\nSo by using Basic Proportionality Theorem,
\nPD\/DO = PF\/FR\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u00a0(ii)
\nFrom equation\u00a0(i)\u00a0and\u00a0(ii), we get,
\nPE\/EQ = PF\/FR
\nTherefore, by converse of Basic Proportionality Theorem,
\nEF\u00a0||\u00a0QR, in \u0394PQR.<\/h3>\n

6. In the figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.<\/h2>\n

\"\"
\nSolution:
\nGiven here,
\nIn \u0394OPQ, AB || PQ
\nBy using Basic Proportionality Theorem,
\nOA\/AP = OB\/BQ\u2026\u2026\u2026\u2026\u2026.(i)
\nAlso given,
\nIn \u0394OPR, AC || PR
\nBy using Basic Proportionality Theorem
\n\u2234 OA\/AP = OC\/CR\u2026\u2026\u2026\u2026\u2026(ii)
\nFrom equation\u00a0(i)\u00a0and\u00a0(ii), we get,
\nOB\/BQ = OC\/CR
\nTherefore, by converse of Basic Proportionality Theorem,
\nIn \u0394OQR, BC ||\u00a0QR.<\/h3>\n

7. Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).<\/h2>\n

\"\"
\nSolution:
\nGiven, in \u0394ABC, D is the midpoint of AB such that AD=DB.
\nA line parallel to BC intersects AC at E as shown in above figure such that DE || BC.
\nWe have to prove that E is the mid point of AC.
\nSince, D is the mid-point of AB.
\n\u2234 AD=DB
\n\u21d2AD\/DB = 1 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.\u00a0(i)
\nIn \u0394ABC, DE || BC,
\nBy using Basic Proportionality Theorem,
\nTherefore, AD\/DB = AE\/EC
\nFrom equation (i), we can write,
\n\u21d2 1 = AE\/EC
\n\u2234 AE = EC
\nHence, proved, E is the midpoint of AC.<\/h3>\n

8. Using Converse of basic proportionality 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 have done it in Class IX).<\/h2>\n

Solution:
\nGiven, in \u0394ABC, D and E are the mid points of AB and AC respectively, such that,
\nAD=BD and AE=EC.<\/h3>\n

\"\"
\nWe have to prove that: DE || BC.
\nSince, D is the midpoint of AB
\n\u2234 AD=DB
\n\u21d2AD\/BD = 1\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..\u00a0(i)<\/h3>\n

Also given, E is the mid-point of AC.
\n\u2234 AE=EC
\n\u21d2 AE\/EC = 1
\nFrom equation\u00a0(i)\u00a0and\u00a0(ii), we get,
\nAD\/BD = AE\/EC
\nBy converse of Basic Proportionality Theorem,
\nDE || BC
\nHence, proved.<\/h3>\n

9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO\/BO = CO\/DO.<\/h2>\n

Solution:
\nGiven, ABCD is a trapezium where AB || DC and diagonals AC and BD intersect each other at O.<\/h3>\n

\"\"
\nWe have to prove, AO\/BO = CO\/DO
\nFrom the point O, draw a line EO touching AD at E, in such a way that,
\nEO || DC || AB
\nIn \u0394ADC, we have OE || DC
\nTherefore, By using Basic Proportionality Theorem
\nAE\/ED = AO\/CO\u00a0\u2026\u2026\u2026\u2026\u2026..(i)
\nNow, In \u0394ABD, OE || AB
\nTherefore, By using Basic Proportionality Theorem
\nDE\/EA = DO\/BO\u2026\u2026\u2026\u2026\u2026.(ii)
\nFrom equation\u00a0(i)\u00a0and\u00a0(ii), we get,
\nAO\/CO = BO\/DO
\n\u21d2AO\/BO = CO\/DO
\nHence, proved.<\/h3>\n

10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that\u00a0AO\/BO = CO\/DO.\u00a0Show that ABCD is a trapezium.<\/h2>\n

Solution:
\nGiven, Quadrilateral ABCD where AC and BD intersects each other at O such that,
\nAO\/BO = CO\/DO.<\/h3>\n

\"\"
\nWe have to prove here, ABCD is a trapezium
\nFrom the point O, draw a line EO touching AD at E, in such a way that,
\nEO || DC || AB
\nIn \u0394DAB, EO || AB
\nTherefore, By using Basic Proportionality Theorem
\nDE\/EA = DO\/OB \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(i)
\nAlso, given,
\nAO\/BO = CO\/DO
\n\u21d2 AO\/CO = BO\/DO
\n\u21d2 CO\/AO = DO\/BO
\n\u21d2DO\/OB = CO\/AO\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(ii)
\nFrom equation\u00a0(i)\u00a0and\u00a0(ii), we get
\nDE\/EA = CO\/AO
\nTherefore, By using converse of Basic Proportionality Theorem,
\nEO || DC also EO || AB
\n\u21d2 AB || DC.
\nHence, quadrilateral ABCD is a trapezium with AB || CD.<\/h3>\n

Exercise 6.3<\/h2>\n

1. State which pairs of triangles in Figure, are similar. Write the similarity criterion used by\u00a0you for answering the question and also write the pairs of similar triangles in the symbolic\u00a0form:<\/h2>\n

\"\"
\nSolution:
\n(i) Given, in \u0394ABC and \u0394PQR,
\n\u2220A =\u00a0\u2220P = 60\u00b0
\n\u2220B = \u2220Q = 80\u00b0
\n\u2220C = \u2220R = 40\u00b0
\nTherefore by AAA similarity criterion,
\n\u2234 \u0394ABC ~ \u0394PQR<\/h3>\n

(ii) Given, in \u00a0\u0394ABC and \u0394PQR,
\nAB\/QR = 2\/4 = 1\/2,
\nBC\/RP = 2.5\/5 = 1\/2,
\nCA\/PA = 3\/6 = 1\/2
\nBy SSS similarity criterion,
\n\u0394ABC ~ \u0394QRP<\/h3>\n

(iii) Given, in \u0394LMP and \u0394DEF,
\nLM = 2.7, MP = 2, LP = 3, EF = 5, DE = 4, DF = 6
\nMP\/DE = 2\/4 = 1\/2
\nPL\/DF = 3\/6 = 1\/2
\nLM\/EF = 2.7\/5 = 27\/50
\nHere , MP\/DE = PL\/DF \u2260 LM\/EF
\nTherefore, \u0394LMP and \u0394DEF are not similar.<\/h3>\n

(iv) In \u0394MNL and \u0394QPR, it is given,
\nMN\/QP = ML\/QR = 1\/2
\n\u2220M = \u2220Q = 70\u00b0
\nTherefore, by SAS similarity criterion
\n\u2234 \u0394MNL ~ \u0394QPR<\/h3>\n

(v) In \u0394ABC and \u0394DEF, given that,
\nAB = 2.5, BC = 3, \u2220A = 80\u00b0, EF = 6, DF = 5, \u2220F = 80\u00b0
\nHere , AB\/DF = 2.5\/5 = 1\/2
\nAnd, BC\/EF = 3\/6 = 1\/2
\n\u21d2 \u2220B \u2260 \u2220F
\nHence, \u0394ABC and \u0394DEF are not similar.<\/h3>\n

(vi) In \u0394DEF, by sum of angles of triangles, we know that,
\n\u2220D\u00a0+\u00a0\u2220E\u00a0+\u00a0\u2220F = 180\u00b0
\n\u21d2 70\u00b0\u00a0+ 80\u00b0\u00a0+ \u2220F = 180\u00b0
\n\u21d2 \u2220F = 180\u00b0 \u2013 70\u00b0 \u2013 80\u00b0
\n\u21d2 \u2220F = 30\u00b0
\nSimilarly, In \u0394PQR,
\n\u2220P\u00a0+\u00a0\u2220Q\u00a0+\u00a0\u2220R = 180 (Sum of angles of \u0394)
\n\u21d2 \u2220P\u00a0+ 80\u00b0\u00a0+ 30\u00b0 = 180\u00b0
\n\u21d2 \u2220P = 180\u00b0 \u2013 80\u00b0 -30\u00b0
\n\u21d2 \u2220P = 70\u00b0
\nNow, comparing both the triangles, \u0394DEF and \u0394PQR, we have
\n\u2220D = \u2220P = 70\u00b0
\n\u2220F = \u2220Q = 80\u00b0
\n\u2220F = \u2220R = 30\u00b0
\nTherefore, by AAA similarity criterion,
\nHence, \u0394DEF ~ \u0394PQR<\/h3>\n

2.\u00a0\u00a0In the figure, \u0394ODC \u221d \u00bc \u0394OBA, \u2220 BOC = 125\u00b0 and \u2220 CDO = 70\u00b0. Find \u2220 DOC, \u2220 DCO and \u2220 OAB.<\/h2>\n

\"\"
\nSolution:
\nAs we can see from the figure, DOB is a straight line.
\nTherefore, \u2220DOC + \u2220 COB = 180\u00b0
\n\u21d2 \u2220DOC = 180\u00b0 \u2013 125\u00b0 (Given, \u2220 BOC = 125\u00b0)
\n= 55\u00b0
\nIn \u0394DOC, sum of the measures of the angles of a triangle is 180\u00ba
\nTherefore, \u2220DCO + \u2220 CDO + \u2220 DOC = 180\u00b0
\n\u21d2 \u2220DCO + 70\u00ba + 55\u00ba = 180\u00b0(Given, \u2220 CDO = 70\u00b0)
\n\u21d2 \u2220DCO = 55\u00b0
\nIt is given that, \u0394ODC \u221d \u00bc \u0394OBA,
\nTherefore, \u0394ODC ~ \u0394OBA.
\nHence, Corresponding angles are equal in similar triangles
\n\u2220OAB = \u2220OCD
\n\u21d2 \u2220 OAB = 55\u00b0
\n\u2220OAB = \u2220OCD
\n\u21d2 \u2220OAB = 55\u00b0<\/h3>\n

3. Diagonals 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 AO\/OC = OB\/OD<\/h2>\n

Solution:<\/h3>\n

\"\"
\nIn \u0394DOC and \u0394BOA,
\nAB || CD, thus alternate interior angles will be equal,
\n\u2234\u2220CDO = \u2220ABO
\nSimilarly,
\n\u2220DCO = \u2220BAO
\nAlso, for the two triangles \u0394DOC and \u0394BOA, vertically opposite angles will be equal;
\n\u2234\u2220DOC = \u2220BOA
\nHence, by AAA similarity criterion,
\n\u0394DOC ~ \u0394BOA
\nThus, the corresponding sides are proportional.
\nDO\/BO = OC\/OA
\n\u21d2OA\/OC = OB\/OD
\nHence, proved.<\/h3>\n

4.\u00a0In the fig.6.36, QR\/QS = QT\/PR and\u00a0\u22201 =\u00a0\u22202. Show that\u00a0\u0394PQS ~\u00a0\u0394TQR.<\/h2>\n

\"\"
\nSolution:
\nIn \u0394PQR,
\n\u2220PQR = \u2220PRQ
\n\u2234 PQ = PR \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(i)
\nGiven,
\nQR\/QS = QT\/PRUsing\u00a0equation\u00a0(i), we get
\nQR\/QS = QT\/QP\u2026\u2026\u2026\u2026\u2026\u2026.(ii)
\nIn \u0394PQS and \u0394TQR, by equation (ii),
\nQR\/QS = QT\/QP
\n\u2220Q = \u2220Q
\n\u2234 \u0394PQS ~ \u0394TQR [By SAS similarity criterion]<\/h3>\n

5. S and T are point on sides PR and QR of \u0394PQR such that \u2220P = \u2220RTS. Show that \u0394RPQ ~ \u0394RTS.<\/h2>\n

Solution:
\nGiven, S and T are point on sides PR and QR of \u0394PQR
\nAnd \u2220P = \u2220RTS.<\/h3>\n

\"\"
\nIn \u0394RPQ and \u0394RTS,
\n\u2220RTS = \u2220QPS (Given)
\n\u2220R = \u2220R (Common angle)
\n\u2234 \u0394RPQ ~ \u0394RTS (AA similarity criterion)<\/h3>\n

6. In the figure, if \u0394ABE \u2245 \u0394ACD, show that \u0394ADE ~ \u0394ABC.<\/h2>\n

\"\"
\nSolution:
\nGiven, \u0394ABE \u2245 \u0394ACD.
\n\u2234 AB = AC [By CPCT] \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(i)
\nAnd, AD = AE [By CPCT] \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(ii)
\nIn \u0394ADE and \u0394ABC, dividing eq.(ii) by eq(i),
\nAD\/AB = AE\/AC
\n\u2220A = \u2220A [Common angle]
\n\u2234 \u0394ADE ~ \u0394ABC [SAS similarity criterion]<\/h3>\n

7.\u00a0In the figure, altitudes AD and CE of \u0394ABC intersect each other at the point P. Show that:<\/h2>\n

\"\"
\n(i) \u0394AEP ~ \u0394CDP
\n(ii) \u0394ABD ~ \u0394CBE
\n(iii) \u0394AEP ~ \u0394ADB
\n(iv) \u0394PDC ~ \u0394BEC<\/h2>\n

Solution:
\nGiven, altitudes AD and CE of \u0394ABC intersect each other at the point P.
\n(i) In \u0394AEP and \u0394CDP,
\n\u2220AEP = \u2220CDP (90\u00b0 each)
\n\u2220APE = \u2220CPD (Vertically opposite angles)
\nHence, by AA similarity criterion,
\n\u0394AEP ~ \u0394CDP<\/h3>\n

(ii) In \u0394ABD and \u0394CBE,
\n\u2220ADB = \u2220CEB ( 90\u00b0 each)
\n\u2220ABD = \u2220CBE (Common Angles)
\nHence, by AA similarity criterion,
\n\u0394ABD ~ \u0394CBE<\/h3>\n

(iii)\u00a0In \u0394AEP and \u0394ADB,
\n\u2220AEP = \u2220ADB (90\u00b0 each)
\n\u2220PAE = \u2220DAB (Common Angles)
\nHence, by AA similarity criterion,
\n\u0394AEP ~ \u0394ADB<\/h3>\n

(iv) In \u0394PDC and \u0394BEC,
\n\u2220PDC = \u2220BEC (90\u00b0 each)
\n\u2220PCD = \u2220BCE (Common angles)
\nHence, by AA similarity criterion,
\n\u0394PDC ~ \u0394BEC<\/h3>\n

8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that \u0394ABE ~ \u0394CFB.<\/h2>\n

Solution:
\nGiven, E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Consider the figure below,<\/h3>\n

\"\"<\/p>\n

In \u0394ABE and \u0394CFB,
\n\u2220A = \u2220C (Opposite angles of a parallelogram)
\n\u2220AEB = \u2220CBF (Alternate interior angles as AE || BC)
\n\u2234 \u0394ABE ~ \u0394CFB (AA similarity criterion)<\/h3>\n

9.\u00a0In the figure, ABC and AMP are two right triangles, right angled at B and M respectively, prove that:<\/h2>\n

\"\"
\n(i) \u0394ABC ~ \u0394AMP
\n(ii) CA\/PA = BC\/MP<\/h2>\n

Solution:
\nGiven, ABC and AMP are two right triangles, right angled at B and M respectively.
\n(i) In \u0394ABC and \u0394AMP, we have,
\n\u2220CAB = \u2220MAP (common angles)
\n\u2220ABC = \u2220AMP = 90\u00b0 (each 90\u00b0)
\n\u2234 \u0394ABC ~ \u0394AMP (AA similarity criterion)<\/h3>\n

(ii) As, \u0394ABC ~ \u0394AMP (AA similarity criterion)
\nIf two triangles are similar then the corresponding sides are always equal,
\nHence, CA\/PA = BC\/MP<\/h3>\n

10. CD and GH are respectively the bisectors of \u2220ACB and \u2220EGF such that D and H lie on sides AB and FE of \u0394ABC and \u0394EFG respectively. If \u0394ABC ~ \u0394FEG, Show that:
\n(i) CD\/GH = AC\/FG
\n(ii) \u0394DCB\u00a0~\u00a0\u0394HGE
\n(iii) \u0394DCA ~ \u0394HGF<\/h2>\n

Solution:
\nGiven, CD and GH are respectively the bisectors of \u2220ACB and \u2220EGF such that D and H lie on sides AB and FE of \u0394ABC and \u0394EFG respectively.<\/h3>\n

\"\"
\n(i) From the given condition,
\n\u0394ABC ~ \u0394FEG.
\n\u2234 \u2220A = \u2220F, \u2220B = \u2220E, and \u2220ACB = \u2220FGE
\nSince, \u2220ACB = \u2220FGE
\n\u2234 \u2220ACD = \u2220FGH (Angle bisector)
\nAnd, \u2220DCB = \u2220HGE (Angle bisector)
\nIn \u0394ACD and \u0394FGH,
\n\u2220A = \u2220F
\n\u2220ACD = \u2220FGH
\n\u2234 \u0394ACD ~ \u0394FGH (AA similarity criterion)
\n\u21d2CD\/GH = AC\/FG<\/h3>\n

(ii) In \u0394DCB and \u0394HGE,
\n\u2220DCB = \u2220HGE (Already proved)
\n\u2220B = \u2220E (Already proved)
\n\u2234 \u0394DCB ~ \u0394HGE (AA similarity criterion)<\/h3>\n

(iii) In \u0394DCA and \u0394HGF,
\n\u2220ACD = \u2220FGH (Already proved)
\n\u2220A = \u2220F (Already proved)
\n\u2234 \u0394DCA ~ \u0394HGF (AA similarity criterion)<\/h3>\n

11. In 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 \u0394ABD ~ \u0394ECF.<\/h2>\n

\"\"
\nSolution:
\nGiven, ABC is an isosceles triangle.
\n\u2234 AB = AC
\n\u21d2 \u2220ABD = \u2220ECF
\nIn \u0394ABD and \u0394ECF,
\n\u2220ADB = \u2220EFC (Each 90\u00b0)
\n\u2220BAD = \u2220CEF (Already proved)
\n\u2234 \u0394ABD ~ \u0394ECF (using AA similarity criterion)<\/h3>\n

12.\u00a0Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of \u0394PQR (see Fig 6.41). Show that \u0394ABC ~ \u0394PQR.<\/h2>\n

\"\"
\nSolution:
\nGiven, \u0394ABC and \u0394PQR, AB, BC and median AD of \u0394ABC are proportional to sides PQ, QR and median PM of \u0394PQR
\ni.e. AB\/PQ = BC\/QR = AD\/PM
\nWe have to prove: \u0394ABC ~ \u0394PQR
\nAs we know here,
\nAB\/PQ = BC\/QR = AD\/PM<\/h3>\n

\"\"
\n\u21d2AB\/PQ = BC\/QR = AD\/PM (D is the midpoint of BC. M is the midpoint of QR)
\n\u21d2 \u0394ABD ~ \u0394PQM [SSS similarity criterion]
\n\u2234 \u2220ABD = \u2220PQM [Corresponding angles of two similar triangles are equal]
\n\u21d2 \u2220ABC = \u2220PQR
\nIn \u0394ABC and \u0394PQR
\nAB\/PQ = BC\/QR \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(i)
\n\u2220ABC = \u2220PQR \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(ii)
\nFrom equation\u00a0(i)\u00a0and\u00a0(ii), we get,
\n\u0394ABC ~ \u0394PQR [SAS similarity criterion]<\/h3>\n

13. D is a point on the side BC of a triangle ABC such that \u2220ADC = \u2220BAC. Show that CA2\u00a0= CB.CD<\/h2>\n

Solution:
\nGiven, D is a point on the side BC of a triangle ABC such that \u2220ADC = \u2220BAC.<\/h3>\n

\"\"
\nIn \u0394ADC and \u0394BAC,
\n\u2220ADC = \u2220BAC (Already given)
\n\u2220ACD = \u2220BCA (Common angles)
\n\u2234 \u0394ADC ~ \u0394BAC (AA similarity criterion)
\nWe know that corresponding sides of similar triangles are in proportion.
\n\u2234 CA\/CB = CD\/CA
\n\u21d2\u00a0CA2\u00a0= CB.CD.
\nHence, proved.<\/h3>\n

14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that\u00a0\u0394ABC ~ \u0394PQR.<\/h2>\n

Solution:
\nGiven: Two triangles \u0394ABC and \u0394PQR in which AD and PM are medians such that;
\nAB\/PQ = AC\/PR = AD\/PM
\nWe have to prove, \u0394ABC ~ \u0394PQR
\nLet us construct first: Produce AD to E so that AD = DE. Join CE, Similarly produce PM to N such that PM = MN, also Join RN.<\/h3>\n

\"\"
\nIn \u0394ABD and \u0394CDE, we have
\nAD = DE \u00a0[By Construction.]
\nBD = DC [Since, AP is the median]
\nand, \u2220ADB = \u2220CDE [Vertically opposite angles]
\n\u2234 \u0394ABD\u00a0\u2245\u00a0\u0394CDE [SAS criterion of congruence]
\n\u21d2 AB = CE [By CPCT] \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(i)
\nAlso, in \u0394PQM and \u0394MNR,
\nPM = MN [By Construction.]
\nQM = MR [Since, PM is the median]
\nand, \u2220PMQ = \u2220NMR [Vertically opposite angles]
\n\u2234 \u0394PQM = \u0394MNR [SAS criterion of congruence]
\n\u21d2 PQ = RN [CPCT] \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(ii)
\nNow, AB\/PQ = AC\/PR = AD\/PM
\nFrom equation\u00a0(i)\u00a0and\u00a0(ii),
\n\u21d2CE\/RN = AC\/PR = AD\/PM
\n\u21d2 CE\/RN = AC\/PR = 2AD\/2PM
\n\u21d2 CE\/RN = AC\/PR = AE\/PN [Since 2AD = AE and 2PM = PN]
\n\u2234 \u0394ACE ~ \u0394PRN [SSS similarity criterion]
\nTherefore, \u22202 = \u22204
\nSimilarly, \u22201 = \u22203
\n\u2234 \u22201\u00a0+\u00a0\u22202\u00a0=\u00a0\u22203\u00a0+\u00a0\u22204
\n\u21d2 \u2220A = \u2220P \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(iii)
\nNow, in \u0394ABC and \u0394PQR, we have
\nAB\/PQ = AC\/PR (Already given)
\nFrom equation (iii),
\n\u2220A = \u2220P
\n\u2234 \u0394ABC ~ \u0394PQR [ SAS similarity criterion]<\/h3>\n

15. A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.<\/h2>\n

Solution:
\nGiven, Length of the vertical pole = 6m
\nShadow of the pole = 4 m
\nLet Height of tower =\u00a0h\u00a0m
\nLength of shadow of the tower = 28 m<\/h3>\n

\"\"
\nIn \u0394ABC and \u0394DEF,
\n\u2220C = \u2220E (angular elevation of sum)
\n\u2220B = \u2220F = 90\u00b0
\n\u2234 \u0394ABC ~ \u0394DEF (AA similarity criterion)
\n\u2234 AB\/DF = BC\/EF (If two triangles are similar corresponding sides are proportional)
\n\u2234 6\/h = 4\/28
\n\u21d2h = (6\u00d728)\/4
\n\u21d2\u00a0h\u00a0=\u00a06 \u00d7 7
\n\u21d2\u00a0h\u00a0= 42 m
\nHence, the height of the tower is 42 m.<\/h3>\n

16. If AD and PM are medians of triangles ABC and PQR, respectively where\u00a0\u0394ABC ~\u00a0\u0394PQR prove that AB\/PQ = AD\/PM.<\/h2>\n

Solution:
\nGiven, \u0394ABC ~ \u0394PQR<\/h3>\n

\"\"
\nWe know that the corresponding sides of similar triangles are in proportion.
\n\u2234AB\/PQ = AC\/PR = BC\/QR\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(i)
\nAlso, \u2220A = \u2220P, \u2220B = \u2220Q, \u2220C = \u2220R \u2026\u2026\u2026\u2026.\u2026..(ii)
\nSince AD and PM are medians, they will divide their opposite sides.
\n\u2234 BD = BC\/2 and QM = QR\/2 \u2026\u2026\u2026\u2026\u2026..\u2026\u2026\u2026\u2026.(iii)
\nFrom equations\u00a0(i)\u00a0and\u00a0(iii), we get
\nAB\/PQ = BD\/QM\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(iv)
\nIn \u0394ABD and \u0394PQM,
\nFrom equation (ii), we have
\n\u2220B = \u2220Q
\nFrom equation\u00a0(iv), we have,
\nAB\/PQ = BD\/QM
\n\u2234 \u0394ABD ~ \u0394PQM (SAS similarity criterion)
\n\u21d2AB\/PQ = BD\/QM = AD\/PM<\/h3>\n

Exercise 6.4<\/h2>\n

1. Let \u0394ABC ~ \u0394DEF and their areas be, respectively, 64 cm2\u00a0and 121 cm2. If EF = 15.4 cm, find BC.<\/h2>\n

Solution:\u00a0Given, \u0394ABC ~ \u0394DEF,
\nArea of \u0394ABC = 64 cm2
\nArea of \u0394DEF =\u00a0121 cm2
\nEF = 15.4 cm<\/h3>\n

\"\"
\nAs we know, if two triangles are similar, ratio of their areas are equal to the square of the ratio of their corresponding sides,
\n= AC2\/DF2\u00a0= BC2\/EF2
\n\u2234 64\/121 = BC2\/EF2
\n\u21d2 (8\/11)2\u00a0= (BC\/15.4)2
\n\u21d2 8\/11 = BC\/15.4
\n\u21d2 BC = 8\u00d715.4\/11
\n\u21d2 BC = 8 \u00d7 1.4
\n\u21d2 BC = 11.2 cm<\/h3>\n

2. Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.<\/h2>\n

Solution:
\nGiven, ABCD is a trapezium with AB || DC. Diagonals AC and BD intersect each other at point O.<\/h3>\n

\"\"
\nIn \u0394AOB and \u0394COD, we have
\n\u22201 = \u22202 (Alternate angles)
\n\u22203 = \u22204 (Alternate angles)
\n\u22205 = \u22206 (Vertically opposite angle)
\n\u2234 \u0394AOB ~ \u0394COD [AAA similarity criterion]
\nAs we know, If two triangles are similar then the ratio of their areas are equal to the square of the ratio of their corresponding sides. Therefore,
\nArea of (\u0394AOB)\/Area of (\u0394COD) = AB2\/CD2
\n= (2CD)2\/CD2\u00a0[\u2234 AB = 2CD]
\n\u2234 Area of (\u0394AOB)\/Area of (\u0394COD)
\n= 4CD2\/CD2\u00a0= 4\/1
\nHence, the required ratio of the area of \u0394AOB and \u0394COD = 4:1<\/h3>\n

3. In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that area (\u0394ABC)\/area (\u0394DBC) = AO\/DO.<\/h2>\n

\"\"
\nSolution:
\nGiven, ABC and DBC are two triangles on the same base BC. AD intersects BC at O.
\nWe have to prove: Area (\u0394ABC)\/Area (\u0394DBC) = AO\/DO
\nLet us draw two perpendiculars AP and DM on line BC.<\/h3>\n

\"\"
\nWe know that area of a triangle = 1\/2\u00a0\u00d7 Base\u00a0\u00d7 Height<\/h3>\n

\"\"
\nIn \u0394APO and \u0394DMO,
\n\u2220APO = \u2220DMO (Each 90\u00b0)
\n\u2220AOP = \u2220DOM (Vertically opposite angles)
\n\u2234 \u0394APO ~ \u0394DMO (AA similarity criterion)
\n\u2234 AP\/DM = AO\/DO
\n\u21d2 Area (\u0394ABC)\/Area (\u0394DBC) = AO\/DO.<\/h3>\n

4. If the areas of two similar triangles are equal, prove that they are congruent.<\/h2>\n

Solution:
\nSay \u0394ABC and \u0394PQR are two similar triangles and equal in area<\/h3>\n

\"\"
\nNow let us prove \u0394ABC \u2245 \u0394PQR.
\nSince, \u0394ABC ~ \u0394PQR
\n\u2234 Area of (\u0394ABC)\/Area of (\u0394PQR) = BC2\/QR2
\n\u21d2 BC2\/QR2\u00a0=1 [Since, Area(\u0394ABC) = (\u0394PQR)
\n\u21d2 BC2\/QR2
\n\u21d2 BC = QR
\nSimilarly, we can prove that
\nAB = PQ and AC = PR
\nThus, \u0394ABC \u2245 \u0394PQR [SSS criterion of congruence]<\/h3>\n

5. D, E and F are respectively the mid-points of sides AB, BC and CA of \u0394ABC. Find the ratio of the area of \u0394DEF and \u0394ABC.<\/h2>\n

Solution:
\nGiven, D, E and F are respectively the mid-points of sides AB, BC and CA of \u0394ABC.<\/h3>\n

\"\"
\nIn \u0394ABC,
\nF is the mid-point of AB (Already given)
\nE is the mid-point of AC (Already given)
\nSo, by the mid-point theorem, we have,
\nFE || BC and FE = 1\/2BC
\n\u21d2 FE || BC and FE || BD [BD = 1\/2BC]
\nSince, opposite sides of parallelogram are equal and parallel
\n\u2234 BDEF is parallelogram.
\nSimilarly, in \u0394FBD and \u0394DEF, we have
\nFB = DE (Opposite sides of parallelogram BDEF)
\nFD = FD (Common sides)
\nBD = FE (Opposite sides of parallelogram BDEF)
\n\u2234 \u0394FBD\u00a0\u2245 \u0394DEF
\nSimilarly, we can prove that
\n\u0394AFE \u2245 \u0394DEF
\n\u0394EDC \u2245 \u0394DEF
\nAs we know, if triangles are congruent, then they are equal in area.
\nSo,
\nArea(\u0394FBD) = Area(\u0394DEF) \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(i)
\nArea(\u0394AFE) = Area(\u0394DEF) \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(ii)
\nand,
\nArea(\u0394EDC) = Area(\u0394DEF) \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(iii)
\nNow,
\nArea(\u0394ABC) = Area(\u0394FBD)\u00a0+ Area(\u0394DEF)\u00a0+ Area(\u0394AFE)\u00a0+\u00a0Area(\u0394EDC) \u2026\u2026\u2026(iv)
\nArea(\u0394ABC) = Area(\u0394DEF)\u00a0+\u00a0Area(\u0394DEF)\u00a0+\u00a0Area(\u0394DEF)\u00a0+\u00a0Area(\u0394DEF)
\nFrom equation\u00a0(i),\u00a0(ii)\u00a0and\u00a0(iii),
\n\u21d2 Area(\u0394DEF) = (1\/4)Area(\u0394ABC)
\n\u21d2 Area(\u0394DEF)\/Area(\u0394ABC) = 1\/4
\nHence, Area(\u0394DEF): Area(\u0394ABC) = 1:4<\/h3>\n

6. Prove that the ratio of the areas of two similar triangles is equal to the square\u00a0of the ratio of their corresponding medians.<\/h2>\n

Solution:
\nGiven: AM and DN are the medians of triangles ABC and DEF respectively and \u0394ABC ~ \u0394DEF.<\/h3>\n

\"\"
\nWe have to prove: Area(\u0394ABC)\/Area(\u0394DEF) = AM2\/DN2
\nSince, \u0394ABC ~ \u0394DEF (Given)
\n\u2234 Area(\u0394ABC)\/Area(\u0394DEF) = (AB2\/DE2) \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(i)
\nand, AB\/DE = BC\/EF = CA\/FD \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(ii)<\/h3>\n

\"\"<\/p>\n

In \u0394ABM and \u0394DEN,
\nSince \u0394ABC ~ \u0394DEF
\n\u2234 \u2220B = \u2220E
\nAB\/DE = BM\/EN [Already Proved in equation\u00a0(i)]
\n\u2234 \u0394ABC ~ \u0394DEF [SAS similarity criterion]
\n\u21d2 AB\/DE = AM\/DN \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(iii)
\n\u2234 \u0394ABM ~ \u0394DEN
\nAs the areas of two similar triangles are proportional to the squares of the corresponding sides.
\n\u2234 area(\u0394ABC)\/area(\u0394DEF) = AB2\/DE2\u00a0= AM2\/DN2
\nHence, proved.<\/h3>\n

7.\u00a0Prove 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.<\/h2>\n

Solution:<\/h3>\n

\"\"
\nAC. \u0394APC and \u0394BQC are two equilateral triangles described on the diagonals AC and side BC of the square ABCD.
\nArea(\u0394BQC) = \u00bd Area(\u0394APC)
\nSince, \u0394APC and \u0394BQC are both equilateral triangles, as per given,
\n\u2234 \u0394APC ~ \u0394BQC [AAA similarity criterion]
\n\u2234 area(\u0394APC)\/area(\u0394BQC) = (AC2\/BC2) = AC2\/BC2
\nSince, Diagonal = \u221a2 side = \u221a2 BC = AC<\/h3>\n

\"\"
\n\u21d2 area(\u0394APC) = 2\u00a0\u00d7 area(\u0394BQC)
\n\u21d2 area(\u0394BQC) = 1\/2area(\u0394APC)
\nHence, proved.
\nTick the correct answer and justify:<\/h3>\n

8. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of triangles ABC and BDE is
\n(A) 2 : 1
\n(B) 1 : 2
\n(C) 4 : 1
\n(D) 1 : 4<\/h2>\n

Solution:
\nGiven,\u00a0\u0394ABC and \u0394BDE are two equilateral triangle. D is the midpoint of BC.<\/h3>\n

\"\"
\n\u2234 BD = DC = 1\/2BC
\nLet each side of triangle is 2a.
\nAs, \u0394ABC ~ \u0394BDE
\n\u2234 Area(\u0394ABC)\/Area(\u0394BDE) = AB2\/BD2\u00a0= (2a)2\/(a)2\u00a0= 4a2\/a2\u00a0= 4\/1 = 4:1
\nHence, the correct answer is (C).<\/h3>\n

9. Sides 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<\/h2>\n

Solution:
\nGiven, Sides of two similar triangles are in the ratio 4 : 9.<\/h3>\n

\"\"
\nLet ABC and DEF are two similar triangles, such that,
\n\u0394ABC ~ \u0394DEF
\nAnd AB\/DE = AC\/DF = BC\/EF = 4\/9
\nAs, the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides,
\n\u2234 Area(\u0394ABC)\/Area(\u0394DEF) = AB2\/DE2
\n\u2234 Area(\u0394ABC)\/Area(\u0394DEF) =\u00a0(4\/9)2\u00a0= 16\/81 = 16:81
\nHence, the correct answer is (D).<\/h3>\n

Exercise 6.5<\/h2>\n

1. \u00a0Sides of triangles are given below. Determine which of them are right triangles? In case of a right triangle, write the length of its hypotenuse.
\n(i) 7 cm, 24 cm, 25 cm
\n(ii) 3 cm, 8 cm, 6 cm
\n(iii) 50 cm, 80 cm, 100 cm
\n(iv) 13 cm, 12 cm, 5 cm<\/h2>\n

Solution:
\n(i) Given, sides of the triangle are 7 cm, 24 cm, and 25 cm.
\nSquaring the lengths of the sides of the, we will get 49, 576, and 625.
\n49 + 576 = 625
\n(7)2\u00a0+ (24)2\u00a0= (25)2
\nTherefore, the above equation satisfies, Pythagoras theorem. Hence, it is right angled triangle.
\nLength of Hypotenuse = 25 cm<\/h3>\n

(ii) Given, sides of the triangle are 3 cm, 8 cm, and 6 cm.
\nSquaring the lengths of these sides, we will get 9, 64, and 36.
\nClearly, 9 + 36 \u2260 64
\nOr, 32\u00a0+ 62\u00a0\u2260 82
\nTherefore, the sum of the squares of the lengths of two sides is not equal to the square of the length of the hypotenuse.
\nHence, the given triangle does not satisfies Pythagoras theorem.<\/h3>\n

(iii)\u00a0Given, sides of triangle\u2019s are 50 cm, 80 cm, and 100 cm.
\nSquaring the lengths of these sides, we will get 2500, 6400, and 10000.
\nHowever, 2500 + 6400 \u2260 10000
\nOr, 502\u00a0+ 802\u00a0\u2260 1002
\nAs you can see, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.
\nTherefore, the given triangle does not satisfies Pythagoras theorem.
\nHence, it is not a right triangle.<\/h3>\n

(iv) Given, sides are 13 cm, 12 cm, and 5 cm.
\nSquaring the lengths of these sides, we will get 169, 144, and 25.
\nThus, 144 +25 = 169
\nOr, 122\u00a0+ 52\u00a0= 132
\nThe sides of the given triangle are satisfying Pythagoras theorem.
\nTherefore, it is a right triangle.
\nHence, length of the hypotenuse of this triangle is 13 cm.<\/h3>\n

2. PQR is a triangle right angled at P and M is a point on QR such that PM \u22a5 QR. Show that PM2\u00a0= QM\u00a0\u00d7\u00a0MR.<\/h2>\n

Solution:
\nGiven, \u0394PQR is right angled at P is a point on QR such that PM \u22a5QR<\/h3>\n

\"\"
\nWe have to prove, PM2\u00a0= QM\u00a0\u00d7\u00a0MR
\nIn \u0394PQM, by Pythagoras theorem
\nPQ2\u00a0= PM2\u00a0+ QM2
\nOr, PM2\u00a0= PQ2\u00a0\u2013 QM2\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(i)
\nIn \u0394PMR, by Pythagoras theorem
\nPR2\u00a0= PM2\u00a0+ MR2
\nOr, PM2\u00a0= PR2\u00a0\u2013 MR2\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(ii)
\nAdding\u00a0equation,\u00a0(i)\u00a0and\u00a0(ii), we get,
\n2PM2\u00a0=\u00a0(PQ2\u00a0+ PM2) \u2013 (QM2\u00a0+ MR2)
\n= QR2\u00a0\u2013 QM2\u00a0\u2013 MR2 \u00a0 \u00a0 \u00a0\u00a0\u00a0[\u2234 QR2\u00a0= PQ2\u00a0+ PR2]
\n= (QM\u00a0+ MR)2\u00a0\u2013 QM2\u00a0\u2013 MR2
\n= 2QM \u00d7 MR
\n\u2234 PM2\u00a0= QM\u00a0\u00d7\u00a0MR<\/h3>\n

3. In Figure, ABD is a triangle right angled at A\u00a0and AC \u22a5 BD. Show that
\n(i) AB2\u00a0= BC \u00d7\u00a0BD
\n(ii) AC2\u00a0= BC \u00d7\u00a0DC
\n(iii) AD2\u00a0= BD\u00a0\u00d7\u00a0CD<\/h2>\n

\"\"
\nSolution:
\n(i) In \u0394ADB and \u0394CAB,
\n\u2220DAB = \u2220ACB (Each 90\u00b0)
\n\u2220ABD = \u2220CBA (Common angles)
\n\u2234 \u0394ADB ~ \u0394CAB [AA similarity criterion]
\n\u21d2 AB\/CB = BD\/AB
\n\u21d2 AB2\u00a0= CB \u00d7\u00a0BD<\/h3>\n

(ii) Let \u2220CAB = x
\nIn \u0394CBA,
\n\u2220CBA = 180\u00b0 \u2013 90\u00b0 \u2013 x
\n\u2220CBA = 90\u00b0 \u2013 x
\nSimilarly, in \u0394CAD
\n\u2220CAD = 90\u00b0 \u2013 \u2220CBA
\n= 90\u00b0 \u2013\u00a0x
\n\u2220CDA = 180\u00b0 \u2013 90\u00b0 \u2013 (90\u00b0 \u2013\u00a0x)
\n\u2220CDA =\u00a0x
\nIn \u0394CBA and \u0394CAD, we have
\n\u2220CBA = \u2220CAD
\n\u2220CAB = \u2220CDA
\n\u2220ACB = \u2220DCA (Each 90\u00b0)
\n\u2234 \u0394CBA ~ \u0394CAD [AAA similarity criterion]
\n\u21d2 AC\/DC = BC\/AC
\n\u21d2 AC2\u00a0= \u00a0DC \u00d7 BC<\/h3>\n

(iii) In \u0394DCA and \u0394DAB,
\n\u2220DCA = \u2220DAB (Each 90\u00b0)
\n\u2220CDA = \u2220ADB (common angles)
\n\u2234 \u0394DCA ~ \u0394DAB [AA similarity criterion]
\n\u21d2 DC\/DA = DA\/DA
\n\u21d2 AD2\u00a0= BD\u00a0\u00d7\u00a0CD<\/h3>\n

4. ABC is an isosceles triangle right angled at C. Prove that AB2\u00a0= 2AC2\u00a0.<\/h2>\n

Solution:
\nGiven, \u0394ABC is an isosceles triangle right angled at C.<\/h3>\n

\"\"<\/p>\n

In \u0394ACB, \u2220C = 90\u00b0
\nAC = BC (By isosceles triangle property)
\nAB2\u00a0= AC2\u00a0+ BC2\u00a0[By Pythagoras theorem]
\n= AC2\u00a0+\u00a0AC2\u00a0[Since, AC = BC]
\nAB2\u00a0= 2AC2<\/h3>\n

5.\u00a0ABC is an isosceles triangle with AC = BC. If\u00a0AB2\u00a0= 2AC2, prove that ABC is a right triangle.<\/h2>\n

Solution:
\nGiven, \u0394ABC is an isosceles triangle having AC = BC and AB2\u00a0= 2AC2<\/h3>\n

\"\"
\nIn \u0394ACB,
\nAC = BC
\nAB2\u00a0= 2AC2
\nAB2\u00a0= AC2\u00a0+ AC2
\n= AC2\u00a0+ BC2\u00a0[Since, AC = BC]
\nHence, by Pythagoras theorem\u00a0\u0394ABC is right angle triangle.<\/h3>\n

6.\u00a0ABC is an equilateral triangle of side 2a. Find each of its altitudes.<\/h2>\n

Solution:
\nGiven, ABC is an equilateral triangle of side 2a.<\/h3>\n

\"\"
\nDraw, AD\u00a0\u22a5\u00a0BC
\nIn \u0394ADB and \u0394ADC,
\nAB = AC
\nAD = AD
\n\u2220ADB = \u2220ADC [Both are 90\u00b0]
\nTherefore, \u0394ADB\u00a0\u2245\u00a0\u0394ADC by RHS congruence.
\nHence, BD = DC [by CPCT]
\nIn right angled \u0394ADB,
\nAB2\u00a0= AD2\u00a0+ BD2
\n(2a)2\u00a0= AD2\u00a0+\u00a0a2
\n\u21d2\u00a0AD2 =\u00a04a2\u00a0\u2013\u00a0a2
\n\u21d2\u00a0AD2 =\u00a03a2
\n\u21d2\u00a0AD\u00a0=\u00a0\u221a3a<\/h3>\n

7. Prove that the sum of the squares of the sides of rhombus is equal to the sum of the squares of its diagonals.<\/h2>\n

Solution:
\nGiven, ABCD is a rhombus whose diagonals AC and BD intersect at O.<\/h3>\n

\"\"
\nWe have to prove, as per the question,
\nAB2\u00a0+ BC2\u00a0+ CD2\u00a0+ AD2\u00a0= AC2\u00a0+ BD2
\nSince, the diagonals of a rhombus bisect each other at right angles.
\nTherefore, AO = CO and BO = DO
\nIn \u0394AOB,
\n\u2220AOB =\u00a090\u00b0
\nAB2\u00a0= AO2\u00a0+ BO2\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..\u00a0(i)\u00a0[By Pythagoras theorem]
\nSimilarly,
\nAD2\u00a0= AO2\u00a0+ DO2\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..\u00a0(ii)
\nDC2\u00a0= DO2\u00a0+ CO2\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..\u00a0(iii)
\nBC2\u00a0= CO2\u00a0+ BO2\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..\u00a0(iv)
\nAdding equations\u00a0(i) + (ii)\u00a0+ (iii)\u00a0+ (iv), we get,
\nAB2\u00a0+ AD2\u00a0+\u00a0DC2\u00a0+\u00a0BC2\u00a0= 2(AO2\u00a0+ BO2\u00a0+ DO2\u00a0+ CO2)
\n= 4AO2\u00a0+ 4BO2\u00a0[Since, AO = CO and BO =DO]
\n= (2AO)2\u00a0+ (2BO)2\u00a0= AC2\u00a0+ BD2
\nAB2\u00a0+ AD2\u00a0+\u00a0DC2\u00a0+\u00a0BC2\u00a0= AC2\u00a0+ BD2
\nHence, proved.<\/h3>\n

8. In Fig. 6.54, O is a point in the interior of a triangle.<\/h2>\n

\"\"
\nABC, OD \u22a5 BC, OE\u00a0\u22a5\u00a0AC and OF \u22a5 AB. Show that:
\n(i) OA2\u00a0+ OB2\u00a0+ OC2\u00a0\u2013 OD2\u00a0\u2013 OE2\u00a0\u2013 OF2\u00a0= AF2\u00a0+ BD2\u00a0+ CE2\u00a0,
\n(ii) AF2\u00a0+ BD2\u00a0+ CE2\u00a0= AE2\u00a0+ CD2\u00a0+ BF2.<\/h2>\n

Solution:
\nGiven, in \u0394ABC, O is a point in the interior of a triangle.
\nAnd OD \u22a5 BC, OE\u00a0\u22a5\u00a0AC and OF \u22a5 AB.
\nJoin OA, OB and OC<\/h3>\n

\"\"
\n(i) By Pythagoras theorem in \u0394AOF, we have
\nOA2\u00a0= OF2\u00a0+\u00a0AF2
\nSimilarly, in\u00a0\u0394BOD
\nOB2\u00a0= OD2\u00a0+ BD2
\nSimilarly,\u00a0in\u00a0\u0394COE
\nOC2\u00a0= OE2\u00a0+ EC2
\nAdding these equations,
\nOA2\u00a0+ OB2\u00a0+ OC2\u00a0= OF2\u00a0+ AF2\u00a0+ OD2\u00a0+ BD2\u00a0+ OE2\u00a0+ EC2
\nOA2\u00a0+ OB2\u00a0+ OC2\u00a0\u2013 OD2\u00a0\u2013 OE2\u00a0\u2013 OF2\u00a0= AF2\u00a0+ BD2\u00a0+ CE2.<\/h3>\n

(ii) AF2\u00a0+ BD2\u00a0+ EC2\u00a0= (OA2\u00a0\u2013 OE2)\u00a0+ (OC2\u00a0\u2013 OD2)\u00a0+ (OB2\u00a0\u2013 OF2)
\n\u2234 AF2\u00a0+ BD2\u00a0+ CE2\u00a0= AE2\u00a0+ CD2\u00a0+ BF2.<\/h3>\n

9. A 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.<\/h2>\n

Solution:
\nGiven, a ladder 10 m long reaches a window 8 m above the ground.<\/h3>\n

\"\"
\nLet BA be the wall and AC be the ladder,
\nTherefore, by Pythagoras theorem,
\nAC2\u00a0=\u00a0AB2\u00a0+ BC2
\n102\u00a0= 82\u00a0+ BC2
\nBC2\u00a0= 100 \u2013 64
\nBC2\u00a0= 36
\nBC\u00a0= 6m
\nTherefore, the distance of the foot of the ladder from the base of the wall is\u00a06 m.<\/h3>\n

10.\u00a0A 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?<\/h2>\n

Solution:
\nGiven, a guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end.<\/h3>\n

\"\"
\nLet AB be the pole and AC be the wire.
\nBy Pythagoras theorem,
\nAC2\u00a0=\u00a0AB2\u00a0+ BC2
\n242\u00a0= 182\u00a0+ BC2
\nBC2\u00a0= 576 \u2013 324
\nBC2\u00a0= 252
\nBC\u00a0= 6\u221a7m
\nTherefore, the distance from the base is 6\u221a7m.<\/h3>\n

11. An aeroplane leaves an airport and flies due north at a speed of 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. How far apart will be the two planes after<\/h2>\n

\"\"<\/p>\n

hours (OA) = 1000 \u00d7 3\/2 km = 1500 km<\/h3>\n

Speed of second aeroplane = 1200 km\/hr<\/h3>\n

Distance covered by second aeroplane flying due west in<\/h3>\n

\"\"hours (OB) = 1200 \u00d7 3\/2 km = 1800 km<\/h3>\n

\"\"
\nIn right angle \u0394AOB, by Pythagoras Theorem,
\nAB2\u00a0=\u00a0AO2\u00a0+ OB2
\n\u21d2 AB2\u00a0=\u00a0(1500)2\u00a0+ (1800)2
\n\u21d2 AB = \u221a(2250000\u00a0+ 3240000)
\n= \u221a5490000
\n\u21d2 AB = 300\u221a61\u00a0km
\nHence, the distance between two aeroplanes will be 300\u221a61\u00a0km.<\/h3>\n

12. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.<\/h2>\n

Solution:
\nGiven, Two poles of heights 6 m and 11 m stand on a plane ground.
\nAnd distance between the feet of the poles is 12 m.<\/h3>\n

\"\"
\nLet AB and CD be the poles of height 6m and 11m.
\nTherefore, CP = 11 \u2013 6 = 5m
\nFrom the figure, it can be observed that AP = 12m
\nBy Pythagoras theorem for \u0394APC, we get,
\nAP2\u00a0=\u00a0PC2\u00a0+ AC2
\n(12m)2\u00a0+ (5m)2\u00a0= (AC)2
\nAC2\u00a0= (144+25) m2\u00a0= 169 m2
\nAC = 13m
\nTherefore, the distance between their tops is 13 m.<\/h3>\n

13. D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE2\u00a0+ BD2\u00a0= AB2\u00a0+ DE2.<\/h2>\n

Solution:
\nGiven, D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.<\/h3>\n

\"\"
\nBy Pythagoras theorem in \u0394ACE, we get
\nAC2\u00a0+\u00a0CE2\u00a0= AE2\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(i)
\nIn \u0394BCD, by Pythagoras theorem, we get
\nBC2\u00a0+\u00a0CD2\u00a0= BD2\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(ii)
\nFrom equations\u00a0(i)\u00a0and\u00a0(ii), we get,
\nAC2\u00a0+\u00a0CE2\u00a0+ BC2\u00a0+\u00a0CD2\u00a0= AE2\u00a0+ BD2\u00a0\u2026\u2026\u2026\u2026..(iii)
\nIn \u0394CDE, by Pythagoras theorem, we get
\nDE2\u00a0=\u00a0CD2\u00a0+ CE2
\nIn \u0394ABC, by Pythagoras theorem, we get
\nAB2\u00a0=\u00a0AC2\u00a0+ CB2
\nPutting the above two values in equation\u00a0(iii), we get
\nDE2\u00a0+ AB2\u00a0= AE2\u00a0+ BD2.<\/h3>\n

14. The perpendicular from A on side BC of a\u00a0\u0394 ABC intersects BC at D such that DB = 3CD\u00a0(see Figure). Prove that 2AB2\u00a0= 2AC2\u00a0+ BC2.<\/h2>\n

\"\"
\nSolution:
\nGiven, the perpendicular from A on side BC of a\u00a0\u0394 ABC intersects BC at D such that;
\nDB = 3CD.
\nIn \u0394 ABC,
\nAD \u22a5BC and BD = 3CD
\nIn right angle triangle, ADB and ADC, by Pythagoras theorem,
\nAB2\u00a0=\u00a0AD2\u00a0+ BD2\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(i)
\nAC2\u00a0=\u00a0AD2\u00a0+ DC2\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(ii)
\nSubtracting equation\u00a0(ii)\u00a0from equation\u00a0(i), we get
\nAB2\u00a0\u2013 AC2\u00a0= BD2\u00a0\u2013 DC2
\n= 9CD2\u00a0\u2013 CD2\u00a0[Since, BD = 3CD]
\n= 8CD2
\n= 8(BC\/4)2\u00a0[Since, BC = DB\u00a0+ CD = 3CD\u00a0+ CD = 4CD]
\nTherefore, AB2\u00a0\u2013 AC2\u00a0= BC2\/2
\n\u21d2 2(AB2\u00a0\u2013 AC2) = BC2
\n\u21d2 2AB2\u00a0\u2013 2AC2\u00a0= BC2
\n\u2234 2AB2\u00a0= 2AC2\u00a0+ BC2.<\/h3>\n

15. \u00a0In an equilateral triangle ABC, D is a point on side BC such that BD = 1\/3BC. Prove that 9AD2\u00a0= 7AB2.<\/h2>\n

Solution:
\nGiven, ABC is an equilateral triangle.
\nAnd D is a point on side BC such that BD = 1\/3BC<\/h3>\n

\"\"
\nLet the side of the equilateral triangle be\u00a0a, and AE be the altitude of \u0394ABC.
\n\u2234 BE = EC = BC\/2 = a\/2
\nAnd, AE =\u00a0a\u221a3\/2
\nGiven, BD = 1\/3BC
\n\u2234 BD =\u00a0a\/3
\nDE = BE \u2013 BD =\u00a0a\/2 \u2013\u00a0a\/3 =\u00a0a\/6
\nIn \u0394ADE, by Pythagoras theorem,
\nAD2\u00a0= AE2\u00a0+ DE2<\/h3>\n

\"\"
\n\u21d2 9 AD2\u00a0= 7 AB2<\/h3>\n

16.\u00a0In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.<\/h2>\n

Solution:
\nGiven, an equilateral triangle say ABC,<\/h3>\n

\"\"<\/p>\n

Let the sides of the equilateral triangle be of length a, and AE be the altitude of \u0394ABC.
\n\u2234 BE = EC = BC\/2 =\u00a0a\/2
\nIn \u0394ABE, by Pythagoras Theorem, we get
\nAB2\u00a0= AE2\u00a0+ BE2<\/h3>\n

\"\"
\n4AE2\u00a0= 3a2
\n\u21d2 4 \u00d7 (Square of altitude) = 3 \u00d7 (Square of one side)
\nHence, proved.<\/h3>\n

17. Tick the correct answer and justify: In \u0394ABC, AB = 6\u221a3\u00a0cm, AC = 12 cm and BC = 6 cm.
\nThe angle B is:
\n(A) 120\u00b0
\n(B) 60\u00b0
\n(C) 90\u00b0
\n(D) 45\u00b0<\/h2>\n

Solution:
\nGiven, in \u0394ABC, AB = 6\u221a3\u00a0cm, AC = 12 cm and BC = 6 cm.<\/h3>\n

\"\"<\/p>\n

We can observe that,
\nAB2\u00a0= 108
\nAC2\u00a0= 144
\nAnd, BC2\u00a0= 36
\nAB2\u00a0+ BC2\u00a0= AC2
\nThe given triangle, \u0394ABC, is satisfying Pythagoras theorem.
\nTherefore, the triangle is a right triangle, right-angled at B.
\n\u2234 \u2220B = 90\u00b0
\nHence, the correct answer is (C).<\/h3>\n

Exercise 6.6<\/h2>\n

1. In Figure, PS is the bisector of \u2220 QPR of \u2206 PQR. Prove that QS\/PQ = SR\/PR<\/h2>\n

\"\"
\nSolution:
\nLet us draw a line segment RT parallel to SP which intersects extended line segment QP at point T.
\nGiven, PS is the angle bisector of \u2220QPR. Therefore,
\n\u2220QPS = \u2220SPR\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(i)<\/h3>\n

\"\"
\nAs per the constructed figure,
\n\u2220SPR=\u2220PRT(Since, PS||TR)\u2026\u2026\u2026\u2026\u2026(ii)
\n\u2220QPS = \u2220QRT(Since, PS||TR) \u2026\u2026\u2026\u2026..(iii)
\nFrom the above equations, we get,
\n\u2220PRT=\u2220QTR
\nTherefore,
\nPT=PR
\nIn \u25b3QTR, by basic proportionality theorem,
\nQS\/SR = QP\/PT
\nSince, PT=TR
\nTherefore,
\nQS\/SR = PQ\/PR
\nHence, proved.<\/h3>\n

2. In Fig. 6.57, D is a point on hypotenuse AC of \u2206ABC, such that BD \u22a5AC, DM \u22a5 BC and DN \u22a5 AB. Prove that: (i) DM2\u00a0= DN . MC (ii) DN2\u00a0= DM . AN.<\/h2>\n

\"\"
\nSolution:
\n1. Let us join Point D and B.<\/h3>\n

\"\"
\nGiven,
\nBD \u22a5AC, DM \u22a5 BC and DN \u22a5 AB
\nNow from the figure we have,
\nDN || CB, DM || AB and \u2220B = 90 \u00b0
\nTherefore, DMBN is a rectangle.
\nSo, DN = MB and DM = NB
\nThe given condition which we have to prove, is when D is the foot of the perpendicular drawn from B to AC.
\n\u2234 \u2220CDB = 90\u00b0 \u21d2 \u22202 + \u22203 = 90\u00b0 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (i)
\nIn \u2206CDM, \u22201 + \u22202 + \u2220DMC = 180\u00b0
\n\u21d2 \u22201 + \u22202 = 90\u00b0 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (ii)
\nIn \u2206DMB, \u22203 + \u2220DMB + \u22204 = 180\u00b0
\n\u21d2 \u22203 + \u22204 = 90\u00b0 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (iii)
\nFrom equation (i) and (ii), we get
\n\u22201 = \u22203
\nFrom equation (i) and (iii), we get
\n\u22202 = \u22204
\nIn \u2206DCM and \u2206BDM,
\n\u22201 = \u22203 (Already Proved)
\n\u22202 = \u22204 (Already Proved)
\n\u2234 \u2206DCM \u223c \u2206BDM (AA similarity criterion)
\nBM\/DM = DM\/MC
\nDN\/DM = DM\/MC (BM = DN)
\n\u21d2 DM2\u00a0= DN \u00d7 MC
\nHence, proved.<\/h3>\n

(ii) In right triangle DBN,
\n\u22205 + \u22207 = 90\u00b0 \u2026\u2026\u2026\u2026\u2026\u2026.. (iv)
\nIn right triangle DAN,
\n\u22206 + \u22208 = 90\u00b0 \u2026\u2026\u2026\u2026\u2026\u2026\u2026 (v)
\nD is the point in triangle, which is foot of the perpendicular drawn from B to AC.
\n\u2234 \u2220ADB = 90\u00b0 \u21d2 \u22205 + \u22206 = 90\u00b0 \u2026\u2026\u2026\u2026.. (vi)
\nFrom equation (iv) and (vi), we get,
\n\u22206 = \u22207
\nFrom equation (v) and (vi), we get,
\n\u22208 = \u22205
\nIn \u2206DNA and \u2206BND,
\n\u22206 = \u22207 (Already proved)
\n\u22208 = \u22205 (Already proved)
\n\u2234 \u2206DNA \u223c \u2206BND (AA similarity criterion)
\nAN\/DN = DN\/NB
\n\u21d2 DN2\u00a0= AN \u00d7 NB
\n\u21d2 DN2\u00a0= AN \u00d7 DM (Since, NB = DM)
\nHence, proved.<\/h3>\n

3. In Figure, ABC is a triangle in which \u2220ABC > 90\u00b0 and AD \u22a5 CB produced. Prove that
\nAC2= AB2+ BC2+ 2 BC.BD.<\/h2>\n

\"\"
\nSolution:
\nBy applying Pythagoras Theorem in \u2206ADB, we get,
\nAB2\u00a0= AD2\u00a0+ DB2\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (i)
\nAgain, by applying Pythagoras Theorem in \u2206ACD, we get,
\nAC2\u00a0= AD2\u00a0+ DC2
\nAC2\u00a0= AD2\u00a0+ (DB + BC)\u00a02
\nAC2\u00a0= AD2\u00a0+ DB2\u00a0+ BC2\u00a0+ 2DB \u00d7 BC
\nFrom equation (i), we can write,
\nAC2\u00a0= AB2\u00a0+ BC2\u00a0+ 2DB \u00d7 BC
\nHence, proved.<\/h3>\n

4. In Figure, ABC is a triangle in which \u2220 ABC < 90\u00b0 and AD \u22a5 BC. Prove that
\nAC2= AB2+ BC2\u00a0\u2013 2 BC.BD.<\/h2>\n

\"\"
\nSolution:
\nBy applying Pythagoras Theorem in \u2206ADB, we get,
\nAB2\u00a0= AD2\u00a0+ DB2
\nWe can write it as;
\n\u21d2 AD2\u00a0= AB2\u00a0\u2212 DB2\u00a0\u2026\u2026\u2026\u2026\u2026\u2026.. (i)
\nBy applying Pythagoras Theorem in \u2206ADC, we get,
\nAD2\u00a0+ DC2\u00a0= AC2
\nFrom equation (i),
\nAB2\u00a0\u2212 BD2\u00a0+ DC2\u00a0= AC2
\nAB2\u00a0\u2212 BD2\u00a0+ (BC \u2212 BD)\u00a02\u00a0= AC2
\nAC2\u00a0= AB2\u00a0\u2212 BD2\u00a0+ BC2\u00a0+ BD2\u00a0\u22122BC \u00d7 BD
\nAC2\u00a0= AB2\u00a0+ BC2\u00a0\u2212 2BC \u00d7 BD
\nHence, proved.<\/h3>\n

5. In Figure, AD is a median of a triangle ABC and AM \u22a5 BC. Prove that :
\n(i) AC2\u00a0= AD2\u00a0+ BC.DM + 2 (BC\/2)\u00a02
\n(ii) AB2\u00a0= AD2\u00a0\u2013 BC.DM + 2 (BC\/2)\u00a02
\n(iii) AC2\u00a0+ AB2\u00a0= 2 AD2\u00a0+ \u00bd BC2<\/h2>\n

\"\"
\nSolution:
\n(i) By applying Pythagoras Theorem in \u2206AMD, we get,
\nAM2\u00a0+ MD2\u00a0= AD2\u00a0\u2026\u2026\u2026\u2026\u2026\u2026. (i)
\nAgain, by applying Pythagoras Theorem in \u2206AMC, we get,
\nAM2\u00a0+ MC2\u00a0= AC2
\nAM2\u00a0+ (MD + DC)\u00a02\u00a0= AC2
\n(AM2\u00a0+ MD2\u00a0) + DC2\u00a0+ 2MD.DC = AC2
\nFrom equation(i), we get,
\nAD2\u00a0+ DC2\u00a0+ 2MD.DC = AC2
\nSince, DC=BC\/2, thus, we get,
\nAD2\u00a0+ (BC\/2)\u00a02\u00a0+ 2MD.(BC\/2)\u00a02\u00a0= AC2
\nAD2\u00a0+ (BC\/2)\u00a02\u00a0+ 2MD \u00d7 BC = AC2
\nHence, proved.<\/h3>\n

(ii) By applying Pythagoras Theorem in \u2206ABM, we get;
\nAB2\u00a0= AM2\u00a0+ MB2
\n= (AD2\u00a0\u2212 DM2) + MB2
\n= (AD2\u00a0\u2212 DM2) + (BD \u2212 MD)\u00a02
\n= AD2\u00a0\u2212 DM2\u00a0+ BD2\u00a0+ MD2\u00a0\u2212 2BD \u00d7 MD
\n= AD2\u00a0+ BD2\u00a0\u2212 2BD \u00d7 MD
\n= AD2\u00a0+ (BC\/2)2\u00a0\u2013 2(BC\/2) MD
\n= AD2\u00a0+ (BC\/2)2\u00a0\u2013 BC MD
\nHence, proved.<\/h3>\n

(iii) By applying Pythagoras Theorem in \u2206ABM, we get,
\nAM2\u00a0+ MB2\u00a0= AB2\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026.\u2026 (i)
\nBy applying Pythagoras Theorem in \u2206AMC, we get,
\nAM2\u00a0+ MC2\u00a0= AC2\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026..\u2026 (ii)
\nAdding both the equations (i) and (ii), we get,
\n2AM2\u00a0+ MB2\u00a0+ MC2\u00a0= AB2\u00a0+ AC2
\n2AM2\u00a0+ (BD \u2212 DM)\u00a02\u00a0+ (MD + DC)\u00a02\u00a0= AB2\u00a0+ AC2
\n2AM2+BD2\u00a0+ DM2\u00a0\u2212 2BD.DM + MD2\u00a0+ DC2\u00a0+ 2MD.DC = AB2\u00a0+ AC2
\n2AM2\u00a0+ 2MD2\u00a0+ BD2\u00a0+ DC2\u00a0+ 2MD (\u2212 BD + DC) = AB2\u00a0+ AC2
\n2(AM2+ MD2) + (BC\/2)\u00a02\u00a0+ (BC\/2)\u00a02\u00a0+ 2MD (-BC\/2 + BC\/2)\u00a02\u00a0= AB2\u00a0+ AC2
\n2AD2\u00a0+ BC2\/2 = AB2\u00a0+ AC2<\/h3>\n

6. Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.<\/h2>\n

Solution:
\nLet us consider, ABCD be a parallelogram. Now, draw perpendicular DE on extended side of AB, and draw a perpendicular AF meeting DC at point F.<\/h3>\n

\"\"
\nBy applying Pythagoras Theorem in \u2206DEA, we get,
\nDE2\u00a0+ EA2\u00a0= DA2\u00a0\u2026\u2026\u2026\u2026\u2026\u2026.\u2026 (i)
\nBy applying Pythagoras Theorem in \u2206DEB, we get,
\nDE2\u00a0+ EB2\u00a0= DB2
\nDE2\u00a0+ (EA + AB)\u00a02\u00a0= DB2
\n(DE2\u00a0+ EA2) + AB2\u00a0+ 2EA \u00d7 AB = DB2
\nDA2\u00a0+ AB2\u00a0+ 2EA \u00d7 AB = DB2\u00a0\u2026\u2026\u2026\u2026\u2026. (ii)
\nBy applying Pythagoras Theorem in \u2206ADF, we get,
\nAD2\u00a0= AF2\u00a0+ FD2
\nAgain, applying Pythagoras theorem in \u2206AFC, we get,
\nAC2\u00a0= AF2\u00a0+ FC2\u00a0= AF2\u00a0+ (DC \u2212 FD)\u00a02
\n= AF2\u00a0+ DC2\u00a0+ FD2\u00a0\u2212 2DC \u00d7 FD
\n= (AF2\u00a0+ FD2) + DC2\u00a0\u2212 2DC \u00d7 FD AC2
\nAC2= AD2\u00a0+ DC2\u00a0\u2212 2DC \u00d7 FD \u2026\u2026\u2026\u2026\u2026\u2026\u2026 (iii)
\nSince ABCD is a parallelogram,
\nAB = CD \u2026\u2026\u2026\u2026\u2026\u2026\u2026.\u2026(iv)
\nAnd BC = AD \u2026\u2026\u2026\u2026\u2026\u2026. (v)
\nIn \u2206DEA and \u2206ADF,
\n\u2220DEA = \u2220AFD (Each 90\u00b0)
\n\u2220EAD = \u2220ADF (EA || DF)
\nAD = AD (Common Angles)
\n\u2234 \u2206EAD \u2245 \u2206FDA (AAS congruence criterion)
\n\u21d2 EA = DF \u2026\u2026\u2026\u2026\u2026\u2026 (vi)
\nAdding equations (i) and (iii), we get,
\nDA2\u00a0+ AB2\u00a0+ 2EA \u00d7 AB + AD2\u00a0+ DC2\u00a0\u2212 2DC \u00d7 FD = DB2\u00a0+ AC2
\nDA2\u00a0+ AB2\u00a0+ AD2\u00a0+ DC2\u00a0+ 2EA \u00d7 AB \u2212 2DC \u00d7 FD = DB2\u00a0+ AC2
\nFrom equation (iv) and (vi),
\nBC2\u00a0+ AB2\u00a0+ AD2\u00a0+ DC2\u00a0+ 2EA \u00d7 AB \u2212 2AB \u00d7 EA = DB2\u00a0+ AC2
\nAB2\u00a0+ BC2\u00a0+ CD2\u00a0+ DA2\u00a0= AC2\u00a0+ BD2<\/h3>\n

7. In Figure, two chords AB and CD intersect each other at the point P. Prove that :
\n(i) \u2206APC ~ \u2206 DPB
\n(ii) AP . PB = CP . DP<\/h2>\n

\"\"
\nSolution:
\nFirstly, let us join CB, in the given figure.
\n(i) In \u2206APC and \u2206DPB,
\n\u2220APC =\u00a0\u2220DPB (Vertically opposite angles)
\n\u2220CAP =\u00a0\u2220BDP (Angles in the same segment for chord CB)
\nTherefore,
\n\u2206APC\u00a0\u223c\u00a0\u2206DPB (AA similarity criterion)<\/h3>\n

(ii) In the above, we have proved that \u2206APC\u00a0\u223c\u00a0\u2206DPB
\nWe know that the corresponding sides of similar triangles are proportional.
\n\u2234 AP\/DP = PC\/PB = CA\/BD
\n\u21d2AP\/DP = PC\/PB
\n\u2234AP. PB = PC. DP
\nHence, proved.<\/h3>\n

8. In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that:
\n(i) \u2206 PAC ~ \u2206 PDB
\n(ii) PA . PB = PC . PD.<\/h2>\n

\"\"
\nSolution:
\n(i) In \u2206PAC and \u2206PDB,
\n\u2220P = \u2220P (Common Angles)
\nAs we know, exterior angle of a cyclic quadrilateral is \u2220PCA and \u2220PBD is opposite interior angle, which are both equal.
\n\u2220PAC = \u2220PDB
\nThus, \u2206PAC \u223c \u2206PDB(AA similarity criterion)<\/h3>\n

(ii) We have already proved above,
\n\u2206APC \u223c \u2206DPB
\nWe know that the corresponding sides of similar triangles are proportional.
\nTherefore,
\nAP\/DP = PC\/PB = CA\/BD
\nAP\/DP = PC\/PB
\n\u2234 AP. PB = PC. DP<\/h3>\n

9. In Figure, D is a point on side BC of \u2206 ABC such that BD\/CD = AB\/AC. Prove that AD is the bisector of \u2220 BAC.<\/h2>\n

\"\"
\nSolution:
\nIn the given figure, let us extend BA to P such that;
\nAP = AC.
\nNow join PC.<\/h3>\n

\"\"
\nGiven, BD\/CD = AB\/AC
\n\u21d2 BD\/CD = AP\/AC
\nBy using the converse of basic proportionality theorem, we get,
\nAD || PC
\n\u2220BAD = \u2220APC (Corresponding angles) \u2026\u2026\u2026\u2026\u2026\u2026.. (i)
\nAnd, \u2220DAC = \u2220ACP (Alternate interior angles) \u2026\u2026.\u2026 (ii)
\nBy the new figure, we have;
\nAP = AC
\n\u21d2 \u2220APC = \u2220ACP \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (iii)
\nOn comparing equations (i), (ii), and (iii), we get,
\n\u2220BAD = \u2220APC
\nTherefore, AD is the bisector of the angle BAC.
\nHence, proved.<\/h3>\n

10. Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see Figure)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?<\/h2>\n

\"\"
\nSolution:
\nLet us consider, AB is the height of the tip of the fishing rod from the water surface and BC is the
\nhorizontal distance of the fly from the tip of the fishing rod. Therefore, AC is now the length of the string.<\/h3>\n

\"\"
\nTo find AC, we have to use Pythagoras theorem in \u2206ABC, is such way;
\nAC2\u00a0= AB2+ BC2
\nAB2\u00a0= (1.8 m)\u00a02\u00a0+ (2.4 m)\u00a02
\nAB2\u00a0= (3.24 + 5.76) m2
\nAB2\u00a0= 9.00 m2
\n\u27f9 AB = \u221a9 m = 3m
\nThus, the length of the string out is 3 m.
\nAs its given, she pulls the string at the rate of 5 cm per second.
\nTherefore, string pulled in 12 seconds = 12 \u00d7 5 = 60 cm = 0.6 m<\/h3>\n

\"\"
\nLet us say now, the fly is at point D after 12 seconds.
\nLength of string out after 12 seconds is AD.
\nAD = AC \u2212 String pulled by Nazima in 12 seconds
\n= (3.00 \u2212 0.6) m
\n= 2.4 m
\nIn \u2206ADB, by Pythagoras Theorem,
\nAB2\u00a0+ BD2\u00a0= AD2
\n(1.8 m)\u00a02\u00a0+ BD2\u00a0= (2.4 m)\u00a02
\nBD2\u00a0= (5.76 \u2212 3.24) m2\u00a0= 2.52 m2
\nBD = 1.587 m
\nHorizontal distance of fly = BD + 1.2 m
\n= (1.587 + 1.2) m = 2.787 m
\n= 2.79 m<\/h3>\n","protected":false},"excerpt":{"rendered":"

Chapter 6 – Triangles Questions and Answers: NCERT Solutions for Class 10 Mathematics<\/p>\n","protected":false},"author":21830,"featured_media":117421,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[7],"tags":[],"class_list":{"0":"post-119155","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-education"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/119155","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/users\/21830"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/comments?post=119155"}],"version-history":[{"count":1,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/119155\/revisions"}],"predecessor-version":[{"id":119295,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/119155\/revisions\/119295"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media\/117421"}],"wp:attachment":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media?parent=119155"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/categories?post=119155"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/tags?post=119155"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}