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Mathematics, , (Chapter - 6) (Triangles), (Class 10), Exercise 6.6 (Optional), Question 1: r, In Figure, PS is the bisector of <QPR of 4 PQR, Prove that & = =., , Answer 1:, Aline RT is drawn parallel to SP, which intersects QP produced at T., Given that, SP bisects angle QPR, therefore, ZQPS = zSPR = (i), By construction,, ZSPR = zPRT (As PS || TR) ww (2) pad, ZQPS = ZQTR (As PS || TR) (3), From the above equations, we have, ZPRT = ZQTR %, «PT=PR, By construction, PS || TR, In AQTR, by Thales theorem, , Se. F7.8_N = PT= e e, sn Pr sn OR {= PT=TR], , Question 2:, In Figure, D is a point on hypotenuse AC of 4 ABC, DM 4 BC and DN 4 AB. Prove that:, (i) DM? = DN.MC (i) DM.AN, , Answer 2:, (i) Join B and D., Given that, DN || CB, DM |] AB and 2B = 90°, ADMBN isa rectangle., « DN = MB and DM = NB |, Given that, BD 1 AC, «. 2CDB = 90°, 9 42+23=90° (1) ‘, In ACDM, 21 + 22 + ZDMC = 180°, = 41+22=90° a=» (27, In ADMB, 23 + ZDMB + 24 = 180°, = 23+24=90° @}, From the equations (1) and (2), we have, 21 = 23, From the equations (1) and (3), we have, 22 = 24, , , , , , In ADCM and 4BDM,, , 41223 [Proved above], 42244 [Proved above], « ADCM ~ ABDM [AA similarity], 2 ae Ona |v BM = DN|, => DM? = DNx MC, , (ii) In ADBN, 25 + 27 = 90° w» (4), In ADAN, 26 + 28 = 90° ~ (5), BD 1 AC, -. ZADB = 90°, , = 25+26=90° -- (6), , From the equations (4) and (6), we have, 26 = 27, From the equations (5) and (6), we have, 28 = 25

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In ADNA and SBND,, , 26=27 [Proved above], , 28=25 [Proved above}, , a _ ~ | aa [AA similarity], —_—=— 2 =, , > DN NB = DN* = AN X NB, , = DN? =AN*DM [« NB = DM), , 2, In Figure, ABC is a triangle in which ZABC > 90° and AD 1 CB produced. Prove that AC? = AB? + BC? + 2BC. BD., Answer 3:, In AADB, by Pythagoras theorem, AB? = AD? + DB? (1), In AACD, by Pythagoras theorem, AC? = AD? + DC?, = AC? = AD? + (DB + BC)?, AC? = AD? + DB? + BC? + 2DB « BC, =AC? = AB? + BC? + 2DB « BC [From the equation (1)}, , Question 4:, In Figure, ABC is a triangle in which ZABC < 90° and AD | BC. Prove that, AC? = AB? + BC? + 2BC.BD., Answer 4:, In AADB, by Pythagoras theorem, AD? + DB? = AB?, => AD2 = AB2 - DB2 (1), AADC 4, by Pythagoras theorem, AD? + DC?=AC?, = AB? - BD? + DC? = AC? [From the equation (1)}, => AB? - BD? + (BC - BD)? = AC?, = AC? = AB? - BD? + BC? + BD? -2BC « BD = AB? + BC? - 2BC x BD, , A, xX, ‘, , Question 5: a, , In Figure, AD is a median of a triangle ABC and AM 1 BC. Prove that:, , () AC? = av? + Bc. pM + (*)", , (ii) AB? = AD? - BC.DM + (*)’, , (ii) AC? + AB? = 2AD? +>8C? 8, In, , Answer 5:, (i) In AAMD, by Pythagoras theorem, AM? + MD2 = AD? (1), In AAMC, by Pythagoras theorem, AM? + MC? = AC?, => AM? + (MD + DC)? = AC?, => (AM? + MD®) + DC? + 2MD.DC = AC?, = AD? + DC? + 2MD.DC = AC? [From equation (1)], , 2, , = AD? + (5) +2MD. (5) sac? |v oc= = 6, , 2, = AD? + (=) +MD.BC = AC?

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Adding equations (ii) and (iii), we have, , DA? + AB? + 2EA = AB + AD? + DC? - 2DC = FD = DB? + AC?, , => DA? + AB? + AD? + DC? + 2EA « AB - 2DC = FD = DB? + AC?, , = BC? + AB? + AD? + DC? + 2EA « AB ~ 2AB « EA = DB? + AC? [From the equation (iv) and (vi)}, => AB? + BC? + CD? + DA? = AC? + BD®, , Question 7: a, , In Figure, two chords AB and CD intersect each other at the point P. Prove that: f/ J, , {i) AAPC ~ ADPB (ii) AP.BP = CP_DP T wl, Answer 7: \ 4, , Join CB. x JS, , (i) In AAPC and DPB, —, , ZAPC = ZDPB [Vertically Opposite Angles] pti, , ZCAP = ZBDP [Angles in the same segment} - \, , AAPC ~ ADPB [AA similarity] / X, , (ii) We have already proved that AAPC ~ ADPB. ‘t —— "|, , We know that the corresponding sides of similar triangles are proportional. So, \é : — ;, , AP PC CA AP PC ‘ J, , DP PB” BD "Dp PB = AP. PB = PC.DP he ail, , Question 8:, In Figure, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the, circle. Prove that, (i) APAC ~ APDB, (ii) PAPB = PC.PD, Answer 8:, (i) In APAC and SPDB,, 2P=2P [Common], ZPAC = ZPDB, , [The exterior angle of cyclic quadrilateral is equal to opposite interior angle] —, « APAC ~ APDB [AA similarity] ,, , (ii) We know that the corresponding sides of similar triangles are proportional. Therefore,, PA AC PC PA PC, = PA.PB = PC.DP, , PD BD PB PD PB, Question 9:, , In Figure, D is a point on side BC of A ABC such that => =. Prove that AD is the |, bisector of 2 BAC. \, Answer 9:, , Produce BA to P, such that AP = AC and join P to C., , Given that:, , BD AB BD AP, , —_—_—Te o_—rs:-—, , cD AC CD AC, , By the converse of Thales theorem, we have x, AD||PC = ZBAD=ZAPC [Corresponding angle] (1) {/\, and, ZDAG = ZACP [Alternate angle] ww (2)

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By construction,, , AP = AC /, = ZAPC = ZACP w (3) f, From the equations (1), (2) and (3), we have /, ZBAD = ZAPC Ae, , = AD, bisects angle BAC. \ N, , Question 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?, Answer 10: ee, , Let AB be the height of rod tip from the surface of water and BC is the horizontal distance between fly to tip, of the rod., , Then, the length of the string is AC., , In AABC, by Pythagoras theorem, , AC? = AB? + BC?, , = AB? = (1.8 m)? + (2.4 m)?, , = AB? = (3.24 + 5.76) m?, , = AB? = 9.00 m?, , = AB=V9=3m fy, Hence, the length of string, which is out, is'3 m. ‘ le ., , , , Tea, , If she pulls in the string at the rate of S cm/s, then the distance travelled by fly in 12 seconds, =12*5=60cm=06m, , Let, D be the position of fly after 12 seconds., , Hence, AD is the length of string that is out after 12 seconds. A, The length of the string pulls in by Nazima = AD = AC ~ 12, = (3.00 - 0.6) m, , =24m, , In AADB,, AB? + BD? = AD? ' D, = (1.8m) + BD? = (2.4 m)? le, = BD? = (5.76 - 3.24) m?= 2.52 m?, , = BD = 1.587 m, , Horizontal distance travelled by Fly, =BD+12m, , = (1.587 + 1.2)m, , = 2.787 m, , =2.79m