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Geometry

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65 views26 pages

Geometry

Problem to solve

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singhvik171994
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© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Then L (AB)*(EF) LL (DE) @ Pa ) I< © re be (@ Nothing can be said ‘Which one of the following cannot be the ratio of angles in a right angled triangle? 2995) @ 1:2:3 (by 1:1:2 © 1:3:6 (@)_ None of these in AABC, AB 1 BC and BD 1 AC. And CE bisects the angle C. A = 30°. What is CED? (1995) A D a aS @) &) © 4 @ In the given figure, AB is diameter of the circle and the points C and D are on the circumference such that ZCAD = 30° and ZCBA 70°. What is the measure of ZACD? (1995) pat WS > @ (o) 0° © wv @ vr The length ofa ladder is exactly equal tothe height of the wall itis resting agains. If lower end of the ladder is kept on a stool of height 3 m and the stool is Kept 9 m away ftom the wall the upper end of the ladder coincides with the tip of the wall. Then, the height of the wall is (1999) @ lm () 15m © 18m @ Um Intriangle ABC, angle B is a right angle. If(AC) is 6 em, and D is the mid - point of side AC. The length of BD is (1996) 9 6 a @ tan © Ke © 3m @ 3.5cm Classes | Shorteut workshops | mocks | books | cetking ~ 09504441448 | 09990028086] 09620377300 | www cetkina.com « Gi cali. G 7. The points of intersection ofthree lines , 2X +3Y~5=0, SX-7Y+2=0, 9X-SY-4=0, Om, (®)_ forma triangle (b)_ are on lines perpendicular to each other %, (©) sreon lines parallel to each other (@ are coincident & ABCD ise square and BCE is an equilateral triangle, whatis the measure of the angle DEC? A B 4 @ 1° (o) 30° © 2 @ 4s 9. ABs the diameter ofthe given circle, while points C and D lie onthe circumference as shown. IfAB BDis 9m, find the area ofthe quadrilateral ACBD (yy, iSISeMACiIeag, (9, £ }s D @) S4om* (©) 216cm? © 162cm* (@)_ None of these 10. In the given figure, EADF is a rectangle and ABC is triangle whose vertices lie on the sides of EAD. amy AE=22,BE=6,CF=I6and BF=2 Find the length of the line joining the mid-points of the sides AB and BC. ial 6 16 c. >? @ a2 ®s © 35 (@ None of these M1.” Three circles, ach of radius 20 and centres at P, Q,R. further, AB= 5, CD = 10 and EF. POR? fhe te 2. Whatisthe perimeter oth @ 20 ©) 6 cs oem ee 12 There is acircle of radius mem aanee corer ‘Offa sequence of regular polygons S, (n),n=4,5.6..., where nis the numberof sides the number of sides of the the circle; and each member of the sequece of regular polygons S3(n),n = 4,5,6.... where m is oe, is inscribed in the circle, Let L (n) and L(n) denote the perimeters ofthe corresponding polygons of §(0) and $0). Then {Et3)2223) eae ay (1999) 8 Gecta2 ont (0). Greater than | and less than 2 13. The figure below shows two: (@)_Less than 4 Their ow shor rw connie ces wth cenre PORSisa sae mad in hove iv Italo circumseribes point B, C, D and A. What is the ratio of the perimeter of the outer circle to that of polygon ABCD? (1999) x x x @ > = oF 4 14. Inthe figure below, AB—BC-CD=DE=EF=FG=GA Then, ZDAE is approximately (2000) E c G. A Li_* D @ 1S OA © 2 @ 1s. ache rhombus with diagonais AC and BD intersecting atthe origin on the xy plane. Ifthe equation ofthe line A i+ y= | then the equation ofline BC is (2000) @ x+y=1 @) x+y=4 @ x-y=l (@ None of these vec ere ns raul acagon AB CDE F GH. figs atthe weet A. can jump ont any ofthe eins exo Sas pst cae i eth vertios crac once sod then reaches vertex then how mary times di it jump before eaching (2000) @ 7 @) +1 © 6 (@) can't be determined v. (Pe peer of ingle i 14 andthe sides ae integers. then bow many ifen sangs sr Posse? 2000) @ 6 ©) 5. : ome @ 3 te Oy snde are sides ofa triangle a? +b? <= ab + be + then the tangle willbe 2000) (@) equilateral (0) isosceles (©) right angled (obtuse angle 19. cre down below prints AB, and Caetaken onDE,DF and EF respectively such that EC=ACand CF =BC angle D = 40 degrees then what is angle ACB in degrees? 200) . ; / a @ © > (@)_ None of these oa OR ei 20. Based On the © figure below, what isthe value of if , (a) 10 ) yu @ 23 Lvs eee Fy © is a 8 @ 6/> 7 The length of the common chord of two circ! ‘whose centres are 25 cm apart, is (in cm) 0 circles of radii 15 cm and 20 em, whose ) @ » (b) 25 In the figure, ACB isa © 1s 7 Tight angled trian, @ Do th cena tae each naeagl CDi atte Cline win eng ACD.BAD Pay om @5 . ©) 50 o7 @ 8 Instead of walking along two adjacent sides of a rectangular field, boy took a short cut along the dagonslan wd 250 equal to haf the longer side. Then the ratio ofthe shorter side tothe longer side is ome) @ 2 &) 2B © 14 ou Each side of a given polygon is parallel to either the X or the ¥ axis. A comer of such a polygon is sad bene fe em angle is 90° or concave ifthe internal angle is S70" Ite numberof convex comers in sucha pone #25 se concave comers must be () 0 @ 4 - AB =6, BC=8 and AC = 10. ‘Aperpendicular dropped from B. is drawn. Fn eirle cuts AB and BC at Pand Q respectively hen AP ae | ‘ (b) 3:2 @ 41 (@ Ut ‘ABD = ZCDI given below, 4 In the diagram . w 2 neets the side ACD Aveckofmam Bd 6 centre B) i = a stythe avo of CDPOS a 1B = ZPQD = 90°. If AB:CD (a) 1:09 a Gin Cali.» 28. In the figure below, the rectangle at the comer measures 10 cm x 20 em. The comer A of the rectangle is also a point on the circumference of the circle . What is the radius ofthe circle in cm? (20030) (@ 10cm @) 40cm (© 50cm (© _None of the above. 2, Avertical tower OP stands atthe centre O of asquare ABCD. Leth and b denote the length of OP and AB respectively. Suppose _ZAPB = 60° then the relationship between h and b can be expressed as, 20030) @ 2p? =n? © 2h? =0? © 3p? =2? @ 3n? =20? 30. Inthe figure given below, AB is the chord of a circle with centre 0. AB is extended to C such that BC=OB. The straight line CO is produced to meet the circle at D. If ZACD=y degrees and ZAOD = x degrees such that x=ky, then the value of kis (20030) B D| c @ 3 & 2 @1 (@)_None of the above 31. Inthe figure (not drawn to scale) given below, Pis apointon AB such that AP : PB=4: 3. PQ is parallel to AC and QD is parallel toCP.In A ARC, ZARC=90°, and in APQS, ZPSQ=90°. The length of QS is6 cms. WhatisratioAP:PD? (2003) ¢ I~ a P D B (@) 10:3 (b) 2:1 © 7:3 @ 8:3 32. Inthe figure (not drawn to scale) given below, if AD=CD= BC, and ZBCE=96°, how much is 4 DBC? (2003) E “ 96° a D B (bo) (©) (d) Cannot be determined. @ 2 Classes | Shortcut workshops | mocks | books | Cotking — 09594441448 | 09930028086] 09820377380 | www.cetking.com —— ~ ~ a on wee ‘0 The chond WA w onan rac we te pos on se WH : ere TEM L AIC = we wed AE = A fontempe Coan” ‘en, Ja nthe fre veo cv sr wae) A. an pon T uch hat CT be tangent whe cele PO (w 100" (by 19 The larip pert, the Sater M4. Aine and his won are waling at bus ap i the even ‘There is a lamp port behind them Th mnt, de fon tand on the wane strait lin, a vce ta the shadows of hin Wed nd in had a ve ater and hin om ae 6 neren, 1 metre nod reesei pein an the ground Hf de heh of the amp Pt f Pd the father is sanding 2,1 metres away fem” the pont, then hw far (in mecers) 18 the som standing fromn his father? (2004 (by 075 % (a) 0M (a) 09 ‘Directions for Questions 35 t0 371 “Anawer the quevtions on the bnss of the information #1Ve® below, rhe two circle torch each other and have a cortex Inthe adjoining figure, Sand tase circles with centers P and Q respectively. T ‘meets the line joining, P anid () at (The dicrese tangent that ches themat points Rand S respectively This conan tangent vardand Tare i he ratio 4 3, al known that the ergs of PO Ie them 1 " 35. What inthe rato ofthe length of PQ to that of QO? am wy 14 yy 3 (a) 30K (4) 34 % eet ; am a) 2am (6) Sem (©) Aan : (4) Sem 37. The length of $0 is enh of 505 a @) #3 om (by 10/5 em to) 12y3 om (4) 1443 om *% In the adjoining figure, chord 14) ia parallel w the diameter AC of the citcle. I ZCHKE,~ 65°, then what isthe value of aon , ue o — am 4 Dt thy 9%" Ie ty ay | Shorten warkehope | mocks | hooks | a) a “FfF = a Ci calls 39. Onasemicircle with diameter AD, chord BC is parallel tothe diameter. Further, each of the chords AB and CD has length 2, while AD has length 8. What is the length of BC? any B c 4A D @ 75 ; ©) 7 © 175 (6) None of the above 40, Ifthe length of diagonals DF, AG and CE ofthe cube shown inthe adjoining figure are equal to the three sides ofa triangle, then the radius ofthe circle circumscribing that triangle will be aoe 4 F ¢ a E > a (2) equal to the side of the cube (b) V5 times the side of the cube (© times the side ofthe cube (@ impossible to find from the given information 41, circle with radius 2is placed against aright angle. Another smaller circle is also placed as shown in the adjoining figure. What is the radius of the smaller circle? (2004) @ 3-22 () 4-22 © 1-4/2 @ 6-42 42. Whatis the distance in cm between two parallel chords of lengths 32 om and 24 em ina circle of radius 20 cm? (2005) (@ 10r7 (b) 2014 (©) 30°21 (@) 40028 43, Inthe following figure, the diameter of the circle is3 cm. AB and MN are two diameters such that MN is perpendicular to AB. In ‘addition, CGis perpendicular to AB such that AE : EB= | : 2, and DF is perpendicular to MN such that NL: LM = 1 :2. The length ofDH in emis (2005) @y2-1) 3 @ 22-1 @ Classes | Shortcut workshops | mocks | books | eet ROMA AAS AA | NEEANDOANAM CAROATTORN | wan colina corn «4 PQ S and R are points on the circumference ofa circle of radius r, such that PQR is an equilateral triangle and PS is « dine of the circle, What is the perimeter of the quadrilateral PQSR? 2003, (® 20+V3) @& 22+43) (© 1+V5) @ ae 7 =9.em,CD=6 emand ZBCD~= ZBAC. shown in the following figure where BC = 12 cm, DB =9 cm, C 4S. Consider the triangle ABC shown : i A D, 9 B 12 c ‘Whaat is the ratio of the perimeter of the triangle ADC to that of the triangle BDC? _ 8 6 5 ary 5 © 5 5 46. Consider a triangle drawn on the X-Y plane with its three vertices at (41,0), (0, 41) and (0, 0), each vertex being represented by its (X, ¥) coordinates. The number of points with integer coordinates inside the triangle (excluding all the points on the boundary) is (2008) @ 780 (b) 800 (©) 0 (@) 741 47. An equilateral triangle BPC is drawn inside a square ABCD. What is the value of the: angle APD in degrees? (2006) @ &) 9 © 120 @ 135 (© 190 48. Two circles with centres P and Q cut each other at two distinct ‘Points A and B. The circles have the same radii and neither P nor Q falls within the intersection of the circles. What is the smallest range that includes all possible values of the angle AQP in degrees? (2007) (@) Between 0 and 45 (&) Between 0 and 90 (©) Between 0 and 30 (@ Between 0 and 60 (©) Between 0 and 75 ‘8. Consider obtuse-anged triangles with sides 8 cm, 15 emand x cm. Ifxis an integer, then how many such triangles exist? (2008) @s5 ® a © 0 @ is © 4 3-em, then what is the radius (in em) of the circle circumscribing the triangle BC? (2008) 2 a © 2785 © 2as (@ 3225 © 51 (2009) ‘What is the length of AB (in em)? 1 sin oe Ni-2 ain ABCD isa parallelogram. Eis point on AB such that AE : BE = 2: 3. EF eo UpSRACcaTGn Ce tert i en afipee amen plelio AD nd itneea CDEC @ 3:5 (b) 10:3, TSEC ADEC tothe area of AEFG? (2009) (d)_ None of these ‘Classes | Shortcut workshops Cetking — 09504441440 | ners | Mocks: Gin Cail, aw Grovarmny $B Ay As, As,Au As, Agare 6 points in clockwise order on the circumference ofa circle of radius 4 om: js one-third the ‘Set ‘OFA js, Aen forall i= 1 104 Ifthe sum of the Tengiha of the ara Se ee cmkeance ot he Girele, what isthe angle (in radians) subtended by AA, atthe center ofthe circle? (2009) ® 3n © aa © a4 © 38 o A, Intriangle ABC, the intemal angle bisector of ZA meets BC at point D. IFAB= Sem, AD= 6emand BAC =120% then was he length of AC? (2009) @ wom ®) 12cm © 6y3em (@)_ none of these point Eis on BC, such that AD= 2CD and CE= EB. If we draw perpendiculars for D and E individually 55, ABC isan equilateral triangle. Point Dis on AC and (2009) fom D end E to other two sides and find the sumof' ‘and denote them as per (D) and per (E) respectively, then which A /\ G B E (@ None of these (b) per(D) per(E) 56, Misthe centre ofthe circle £(QS)= 10 2. £( () 11459, units (@) 100sq. uni ) 57. Twocircles, touching each other, ‘are drawn inside a square of side 10 cm. Each ‘What is the maximum possible value (in cm) of the sum of ir radii? (2010) @ 10(2-V2) © 10(2+v2) © 102 @ 10 58. The figure given below shows 8 circle with center O and radius 8m. BD isa chord of the circle ‘and A is a point on the minor Cis perpendicular to BD. The length of AC is 4 em ‘and BC is 12 cm. What = length 2010) arc BD. Cis point on BD such that A (in em) of CD? D © af @ 3 = 8m, BD = 6 emand DC 3 em. Find AC. (2010) @) 2 w 4 59. Ins AABC, AD isthe bisector of BAC, AB A /o™~ 2 6D 7 © (e) Sem (@) Sem 4om (o) Gem ° Classes | Shortcut workshope | mocks | Books | cclng ~ ovsgeae4® | 09090028086) 00820977980 | worw.cakina any Cc D = ZBAC ZABD= ZACD (@ ZACD ; (@) ZADB=ZACB ) : ide. Ate ide of a road 50m wide. poll a. ‘Two vertical lampposts of equal height stand on ete oft road oe amp pos (4) ZADB= ZCAD int P on the road between them, the hich makes angle of 6°." (2011) tthe tops ofthe lamp-posts are 60° and 30” ors () 25m © ee eee their radii is 15, find the 2. ‘The sum of the areas of two circles which touch each other externally is 153x, Ifthe sum of oe larger tothe smaller radius oa ee ! 4 (b) 2 (c) : ° 8. (© cb ina rectangle. The points p and Q lie on AD and AB respectively Ife wiangles PAQ, OBC and PCD all hve hag areas and BQ = 2, then AQ= 7 I) @® 1+V5 () 1-v5 © V7 (4) 27 it i 146 m respectively. Ramu and Shams ‘wo cidos with conta Aan B toch each ther at. Thera of eto cra are ee “ To Sie we and pi and wave ng he cle Wilh = sian fam Cv insane Ra's stat would hey e separated by a dsance of 18m (2012) @7 &) 10 © 125 (d) Never 65. Froma point P, the tangents PQ and PT are drawn to acircle with centre O and radius 2 units. From the centre O, OA and OB = {drawn parallel to PQ and PT respectively. The length of the chord TQ is 2 units, Find the measure of the AC = om. AD=EF=8 AC=10 (By Pythagorous theorem) ‘Also since line joining mid-points of two sides is half 23 - the length of the third side. 1 (&) S=|5 -7 2 . " : 9 5 4 Hence, required length = 1L,(17) Now in APED, 1807-54749 1 le and at the same time circumference of the circ! circumscribing the polygon will be always greater than the perimeter of the respective polygon. Therefore we can say that the value of the above equation has to be greater than 2. 13. (©) Letthe diagonal of PQRS be 2 ‘Therefore, side = 1/2 2 Now, ABCD is a square. And side nf V2=r Perimeter of ABCD = 4r. (Circumference of bigger circle = nr. Therefore, required ratio = = wm Let ZEAD= a, then, ZAFG= a andalso ZACB=a. Hence 14,a+b> cd 6 So, the possible measures of sides 2+ laa =18P of Bate FE.9203.5,6), (6,4, 9550) ™” a Cmmse8 | Sheree Thus, there are four pairs pos Pas 18 (a) a+bP+c?=ab+ be tac Put a=b=c=k we get 3k? = 3k, which satisfies the above equation. Thus the triangle is equilateral. 1% © In ADEF, 0+ 4+ 40 =180 or 6+9=140 In AEC, ZACE = 180-26 InABCF, ZBCF = 180-20 ZACE+a+ ZBCF =180 180 +180-2(8 + 6) +0=180 = a=100° 2 0) a Zz 7 4S +4 Pg Biexe (x44)? + (x-3)? = 2x? +254 2x ‘Since solving this equation is very difficult. So, itis @ better approach (Time saving) to put the values given in the options and try to find out a solution. Hence, trying out we get 1] as the value of x 21.) A 4) B. D. —_—_'— Let BC=x and AD=y Using the theorem of angle of bisector, BD AB_4 pe” Ac 3 4 3 = BD = Fx and DE => ‘on | Shortcut workshops 2 @ Given AB = 25 cm, AN = 20 em, BM= 15cm Then, AM=AB-BM =25-15=10 BN = AB-AN=25-20=5 AM+MN+NB=25 => 10+MN+5=28 = MN=10 Let MO=x,ON = 10~ x, (since we know PQ LAB) In AAOP OP? = (207 -(10+x? = 400-(100+ x? + 20x) = 300-x? - 20% 0) In BOP, op? =(15)? -(0B =225-(10-x+5)? = 225-(1S-x)? = 225-225-x? +30x aii) By (i) and (i), we have 300-x?- 20x = =x? +30x = 300=50x => x=6 : MO=6 hence, AO 16 oP = y(2 @ led ABA‘ In right ang 2+ 208 => BC= 25 In right angled SADB, AB?= x? + AD? > iste xp tad? =) In Fatt ang x2 + 4xy = 4x? = 20?= x} +AD* (iil) xy +xz=25 Subsracting (i) from (i), (© The number of convex corners is always less thn aon 2252 x} xf = (82-4100 +) i concave corners by 4. Hence 25—4= 21. 26. (@)_ From the question, we get the following diagram: => 175 = (x2 —%1)*25 2x)-K=7 oniV) A Adding, (iii) and (iv), we get, SS A x) =16 and x, =9 [AN S10, ‘Using x,, x, in (ii), we get, 6 D> Ww 8 400 = 256+ AD? => AD | ay a g Cp c In AADB, using the formula for inradius <—_ 3 —> (FexxaD) (Fxox12}2 In ABC and ABDC, yw Ama, i2 _J 2s ZABC = ZBDC = 90°; s (2) 15+12+9 ‘ZC is common; 2 One side is common Where S is semi-perimeter 2. SABC ~ ABDC In AADC, AB LAC, 6 100 py (tessa0) (Jaton) Bo" BC BD 8 case ee 24_6 24_16 i AP =6-— =< and QC =8-—=— 20+12+16 35 Qca8 sas Rene ( +AD+CD_ 2 cof thy =34+4=7 and ry - In APQR, PQ? = QR?+PR? cp _cD 27. (b)_ Using the quality of simitar triangles, === 3° = PQ= Vly +)? +(-9)? = V7? 41? = 150 ae D c In ABPQ and BCD, 2 BC 4 Lh :075 PQ BP 3 y 2B x ql Let x and y be the longer and shorter sid Cpepstbbedr ide respectively ‘According to the question, (ery yet ay? o% 2 Gry Sadat yy? In the right angled 4, r? = 20)? +(r-10)? 2 60r + 500 Solving we get r= lansee | Shortcut workshops | mone Teo Grom ee vo Pp Again, QD || CP, Dy JN ic A 1B . Given that ZAPB = 60° ‘Then, 4 APB is an equilateral A ( ZA = ZB) 2 AP=AB=b. ‘Now in right angled 4 AOP, AP? = AO? + PO? 2 > » (Ss) +h? 2 ©) = pe =n? oan? =o? iS D B x @ — Let ZCAD= ZACD=x At point C, c x+(180°—4x) + 96° = 180° = 180°—3x +96" =180° axe Hence, ZDBC =2 *32= 64° InaBOC, 2B0C= ZBCO=y [since BC= OB] Then, ZABO=2y {extemal angles} InAAOB, ZABO = ZOAB=2y __ [isosceles triangle] ZAOB =(180-4y) —_ [sumofangles= 180] ‘Then, ZDOA + ZAOB+ ZBOC = 180 x+(180-4y)+y =180 => x=3y Hence k=3 31. (@) From figure c sa to SN tare > Q IN ; ¢ J fs, a cen, PQIIAC, eC QB PBS Classes | Shortcut workshops | mocks | books | Cetking ~ 09594441448 | 09930028086] 09820377380 | werw.cetking.com wt VAS en AI? a lamp-post (6 metres), In the above diagram, AB is the I Scere te) DBs father (1.8 metres), FGis the son the shadows meet at C. We have to find DF. Using the porperties of similar triangles, we get, CF © cD Also CB Now we have, CD = 2CF and CD =0.3 CB. Let DEC= ZACE=x Hence, 2CF=0.3 CB. [Since AC || ED] But CB=BD+ DF + CF =2.1 +2CF (since CF = DF) Complete the ACAE =2.1+03CB ZCBE= ZCAE (angles in same sp = 0.7CB=2.1 ZAEC=90° [angle subtended in se, ce => CB=3 and CD=3-2.1=09. .. INACAE, ZCAE+ ZAEC+x= 180° or 65+90+x= 180 {Since FD= CF, then DI = = x25" 38.37. 2 2DEC=25° R ®. 0) Bec V4 Ey SD k Fi mT A E Oo FD bate) 35. (b) In ASOQand AROP AO=BO=radius=4 Oi common a oe 1 4S= R= 90" (angentat cle Aenresaan Bef) 31 *-A$0Q~AROP 2x24) 42 RP_ OP _PQ+0Q =o BE OF _ PQs : SQ_0Q BC=AD-AE-FD a4 ae 40. @ Letthe side of the cube be 1. Then dag! > 0g Now we ha rewith sie O* - : Wve an equilateral triangle wit = PQ=7and 0Q=21 its altitude will be 3/2. : Circumradius = 2/3 « altitude =2/3 «32°! “ jired = x le: oe ae Thus, the radius of the circle circunseite® % ©) PQ=1, 41,27 ‘Tiangle will be equal to the side of the As the ratio of radii is 4 : 3, 4. @ So, the only value whi circle = 3)” “MME Which satisfies the redii of 37. ©) mnasog, = 802+ Sq2= 092 A PEO A121 43)— 1B 24n 439 /| >S80=12)3 D Gi. cavllio. Gooey _ A113 ‘OABCis square with side=2 Hence, EOHL isa square of side 0.5 em. op= VP oF =23 eta ioreospens? Letr be the radius of smaller circle = (DH+05P=(15+05)(1.5-0.5)=2 op=202 =oD+r+OB=2+r+8y7 =n J? +1)=2/2-1) r= 20=D 20-9! (2+) 2-1 =2(2+1-2V2) =6-4y2 Remember that a perpendicular from the centre to a chord divides it into two equal parts. 2 @ => OB= 1207-167 =12 ‘Similarly in A OAA’, OA'= 20? ~12? =16 :. Distance between the two parallel chords =16-12=4em or 16+12=28em a © IM 2 => EB=2AE and LM=2NL = EB+AE=3 and LM+LN=3 => AE=1&EB=2 and LN=1 &LM=) Now, OL= LM-OM=2-1.5=0.5 And EO=EB-OB=2-1.5=0.5 46. @) 3-1 = DH=V2-05=2-— 7 ‘As PQR is an equilateral triangle, hence PS will be Perpendicularto QR and will divide it into 2 equal pas. 2 & ZS will be supplementary so ZS = 120° and ‘ZQSA= ZRSA= 6, InA0AQ, OA =0Qsin30°= > => AS=OS-OA= Now in APAQ, PA=PQcos30" Similarly in AAQS, 2 AS=QS.c0s60" => Qs= "2 Seas = Sar, Perimeter ofPQSR=2(PQ+QS) = 2r(V3 +1) (common) (given) AD=16-9=7 Perimeter of AADC _ AD+DC+AC Perimeter of ABDC BD+DC+BC = 2168 21. 946412 27 9 27 Classes | Shortcut workshops | mocks | books | etking — 09594444448 | 09930028086) 09820377380 | wwww.cetking.com 46, (@) The equation of the line becomes 2 + 47, (s) y a1 41 0.40) 49, (41,0) The (x. y) co-ordinates (x and y € 1) satisfying this equation are (1, 40) (2,39)....(40, 1) Total required points are ,1(,2).. DQ,2).. = 38 points and so on. So total integral points = 39 + 38 + 37 +... +1 which is an A.P. Hence Sap = 32 +39) = 780 ZPBA = LABC- ZPBC =90°-60°=30° Furtherin AABP PB=AB=a=> ZBPA = BAP Further 2(ZBPA)+ ZPBA = 180° = 2ZBPA =180°—30°=150° => ZBPA=75° = ZBAP Similarly ZPAD = 90°— ZPAB = 90°75 ‘Again in right angled A APE, ZEPA = 90°~ ZPAE = 90° 15°=75° Similarly we can calculate that ZDPE = 75° DPA © 15° + 75°= 150° 3. Since P and Q are outside the intersection of the two circles, A, v Q © therefore r 2r, ZAQP 5 ge Hence, in both the cases, 0° ZAQP Z6q0 If ‘a’, *b’ and ‘c' are the length of sidey triangle and ‘a’ be the length of lon ‘Then a? > b? +c? Care (1): If length of one longest side 225>64+x2 161 =8,9, 10, 11,12 {Since, the value of x is less than 8, because length of any two sides ofa triangle is greater longest side. Case (Ii): If length of longest side be xm, then x2>225+64 x2>289 x=18, 19,20, 21 and 22 [Since the value of x is less than 23] Therefore, total number of values of x is 10 and total number of triangles is 10. of; IBESt Side, ts be 15cm (bo) B Classes | Shortcut workshops | mocks | books | Radius of circum circle ofa trianlge, _axbxe 4a R xbxe _175x9* BC aR | 175x9x BC tan R=26.28 ZBAD= 90" Cetking — 09594441448 | 09930028086! 09820377380 | www .cetkinc cam Extend BA and CD to meet at E. AEBC becomes an equilateral triangle, SoBE=4cn In AADE,ZADE = 30° ae tan 30° => G ° yar B 2 _Av3-2 — AB = BE-AE= 4-2 SS 2om : . ‘Area of triangle ABC= = AB x AC * sin Z BAC...(i) D FE Cc 2 @ 4 Area oftiangle = 3» AB AD *sin 2BAD ii) is 1 Area of triangle ACD => x AC x AD x sin ZCAD A? E— 3 B (4’sbetween same il) ‘Area of ADEC~Areaof ADBC parallel line and same base) Also, 2xBAD =CAD + BAC=60° and @=Gi)+ Gi) Hence 7 - ‘Area of parallelogram ABCD 1 ‘Area of AEFG = > * Area of parallelogram EBCF Bo fi 1 1 Ax 8x ACx SS = — x 8x 6x +2 x ACK 6x 2eexACx 22 2 = = AC=24cm 55. (©) Sumoflength ofall perpendiculars drawn on the sides of any equilateral triangle is constant. Perpendicular (D)=Perpendicular (E) 56. () ZPQS=ZPRS=90". [Diameter subtends 90? at the centre] PR=RS and PR || QS ;, Rectangle PQRS is a square. PS= 2 xQS (Diagonal = 2 * side] = V2 «10/2 =20 ©. Radius =10 =3.14=3(approx) Areaofeircle= xr=3 x 10% 10=3000 Area of square =(10 /2 P= 200 ©. Area of shaded region = Area of parallelogram ABCD ale 1 2 pxarea(ABCD) So, aI 3% gxarea( ABCD) 1 6 = 7 (Area of circle — Area of square) Are length = 355% 2" As the radius is fixed, the arc length is directly Proportional to the angle subtended by the arc at the . So ifthe are length is getting tripled the angleis $7. (a) ™& also getting tripled. ) Therefore (0+30-+90+270+810)= 22x 1 3 * 100= 5059. units. => 1210=2 o-— 4 484 ‘Angle subtended by A,A; at the centre is 38 yn. x SR From the above figure 484 484 Classes | Shortcut workshops | mocks | books | nena sed sae | ORANOANAR NORAATTEA | arene catkinn crm @ © Given, AC=4and BC= 12. LaDC=2x. ‘Now, ACx CE=BC CDor4 x CE= Hence, CE= 6x. AE=AC+CE=4+ 6x 122k. EG & 243K. BD=BC+CD=12 + 2xand cF=06 =B-co-6-x ‘As OGis perpendicular to AE, in right triangle EGO: BG+0G=EO? (243x) +(6-x)' = (2+3x) +(6-x) = 12x) =64 (44 9x? +12x) + 6427 10x? +40 = 64+ 2 em. 3 Hence, CD =2x= CD. Bs >< + Cc Line AD bisects ZCAB ABE ACE ene £8 CE AC sac mem a B é D Classes | Shortcut worksh Cetking - 09594441448 | 09930028086) ncannon e. 6 0) @ nA ADB and ABCA, we have AD=BC, BD=AC. ‘AB is the common side By 85S congruencys ADB (ZADB= ZACB @ {iso, ABCD is acyclic quadrilateral ZABD= ZACD Ail) In ‘SACD and ABDC, we have ‘AD=BC, BD=AC, it ZBAC «. ZACD= ZBAC, TTAB be the road between the lamp-pods 4 BM.AB=50m. angle LAP, SE =tan60" 7 ‘From trig le LAP, AP =tan Saree BM romiringle MPB, “pry =18" 30°= PB=b f PA+PB=AB =50 [Given] h 3 =50=>h B > Let the radii ber, andr, arp +r} = 1530 af +r} =153 nth =15 =r tr} + 2qn = 225 Wy = 72 > Hy =36 Ifr, and ry are roots of equation then, r?—-15r+36=0 = P12 343620 2 (r-12Kr-9°? deg % 08 | mocks | books | “a @ 6. 6, Gromer 6. © ) (@) As areas of APQ, QBC and PDC one equal, so —2—. A P| x D 7 c 1 1 1 SxDx x= 5 xAPx(y-2)= 2 PDX. 5 7 -2)=5PDxy 1 1 x=5(y-2)xAP=t yx or = 50-2) 27xPD 2x => PD=%and AP y v2 = p+ ap=220=2)+ 297 ¥O-2) > YY-2)=2y-2)42y oF y? rym 4y—4 Again, AD = ay -byt4= P 4x1 yot er 1 4.69 eg = AQ=3+V5-2=1+V5 This is never possible, because data is not sufficient. Since, OQ = TQ = 2 units, therefore AOTQ is an equilateral, P 2. ZT0Q= 60" Since, PQ isa tangent to the circle, therefore ZOQP Since, PQ is a parallel to OA therefore ZAOQ= 90° For the same reason ZBOT = 90°. :. LAOB = 360° -(ZTOQ+ ZAOQ+ ZBOT) =120°. E 5 A ‘|p Cw Cena AS © & "ZAEB = 90° (Angle in a semicircle) ZBAE = 35° (Given) :. ZABE = 180-(90+35)= 55° ‘ZBAC = 55° (Alternate interior angles) Now, ZCDB= 180-55 = 125° (Sum ofopposite angles of acyclic quadrilateral is 180°.) . ZABD= 180-125" 55° (Sum of interior opposite angles is 180°). Hence required difference ZCDB~ ZABD =125-55=70°. Let the two sides of the rectangle be x and 2x and ZAED be 8. al7 *. ZDEC=0 2 D c Ne o x als 8 Z a OS B Now, in ABCE 20 = 90° + 60° (Exterior angle is equal to the sum of ‘opposite interior angles.) = 0=75° Here, Two cases are possible Case I: When B lies in greater arc AC. Then, B 1 ZABC= 5 ZAOC=70° (Case II: When B lies in smaller arc AC, Then, Classes | Shortcut workshops | mocks | books | Cotking ~ 09594441448 | 09930028086) 09820377380 | www.cetking.com Let D be a point 0 is ZOAC= ZOCA®20 = ZA0C= 140° », ZADC* ; ZAOC= 10° cyclic quadrilateral 180° 70°= 110°. ~The length of AE is #h because E and F is mi ABCD i atways equa tothe point of side AB ZAB ‘Option a) length of CF and CD. ‘So, statement ~ [is always true, Option (b) the Length of BC is equal length of BF so statement Il is always False. Option (c): Length of AE is equal because of EF is parallel to BC Option (4): ZAEF = ZEFC=90° So statement Ils also true, tothe length of DF Rn © Given that PT]SR| QU PT=aunits, QU=b units, APTQand ASRQ are similar. . We have FQ_SQ PT SR PQ_SQ oR A) AUQP and ARSP are similar. PQ_ PS Wehave 72 = PS. ete QU” SR ] roEre ob SR ‘Combining (i) and (i) PQ, PQ _sQ+Ps ab. SR orro{tot) 0 ab)” sp FA8SQ+ PS ~ pay ra[s4] Fo ab)” SR @ i) ab ors asp Here, ZEOQ=85° and ZB0D= 15° ZEOD = 180° (85° + 15°)=80" hh 20) OE =OD (radius), COED = ZODE=6° (lt) £.0+0+80°= 180 => 26°=180—80= 100° . ZOED= ZODE= 50° in ZE0C, ZEOC=80°+ 15°=95° and ZOEC= 50° “ ZECA=180° — (95° + 50°)=35° : x PQis perpendicular to line y= Jj i +: Slop of PQ= B Let equation of line PQ be Atpoint M,when x= J3,¥= esd Se Vix +4 ix+o . 4 ~: Co-ordinates of point Q = (¥ ‘) - Co-ordinates of point P= (0, 4). Se Hence, PQ= (A) -#- et -F (18+ 2r— 15P +915? abe 12 as re Shem Radius of smallercircle= 4.5 em. n. ‘Area of the shaded region 1 L x30.30-34{ 510510) — 30059 magTr, ZQTR+ ZORT+ ZRQ ZRQT = 180° ~(55°+25%)= 100° ZORT=90° (TRis a tangent to the circle at R) ZQRT+ ZORQ=90° 3s @ ZQRQ=90°-55°= 35° 0Q=OR (Both are radius of circle) ~ ZOQR= ZORQ=35° ZRQT + ZPQR= 180° (POT isa straight line) 100° +35°+ ZPQO= 180° 2. 2QO=45° Tn 4OQP,0Q=OP (radius of circle) AC: LOPQ= ZPQO=45° nd 20-1)+(21- 70-90" 0, 2r= 78 (@) Let Gea point on AC such that DG is parallel to BF. =r nN ‘AOEC = AOFC(RHS) ‘Area (AFOEC)=2 x Area (AOEC) f = 25.156 =90 cn? g Similarly, AAOD = AAOF (RES) AREA (AADOF)=2 * Area (AAOD) B D c Here, BD =8cm_ = 24.146 =84em? bakers DC=6cmand so,Required ‘Area(AFOEC) _ 90 _ 15 pe , ratio= <— anon 8414 =12cm Requit '0= ‘area(AADOF) 84 14 ear AB AE 3 FG BD 84 FG ED 4°GC DC 6 3 AF: FG:GC=3:4:3, 10, 4-2 n2yeme AC= (AF) = (12) em= 40em Let the radius of the bigger circle be R and the smaller i circle be r and the side of the square is 2a. 79. (b) AreaofGBC = 3 (area of AABC) = OE=R-EF eee ‘Area of renining part= 3» Arenof AABC ive (2a+2r-RP +a? = R? 2 a=9(-; 2a=18);R=15 = 5 Vats—an(s= 5-9 Classes | Shortcut workshops | mocks | books | Cetking - 09594441448 | 09930028086) 09820377380 | www.cetking.com Gi, cali. A120 2 = 5 v50(50-25,50—35)50-40) _ $00 e 80. (b) 81. tee. A 6 B Enclosed Area = Area of semi circle BQC with centre 0 Area of Arc BPC with O 1 = 512)? -BPOD (Are) fi) Now, Area of triangle ABC + BPCO (Are) = Area of Quarter circle (ABPC) 1 1 2 +%6x6+BPCO = + 36% 6+ BPCO = 7 n(6) ‘Area of BPCO=9n-18 Put the value in equation (i) Enclosed Area = _on-18) => 9-9 +18 =18sqem (24) Ina 3, 4, 5 triangle the length of the altitude to the Hypotenuse = at =24 therefore ina 15,20, 25 triangle itis 1.2. This isthe shortest distance from A to BC. At 60 knvhr i.e. 1 km/min It would be take 24 min to cover 24 km. Grower, D a kK F J B A E H 1 ic Total 11 point formed triangle ; ‘ u = 11C,—colinear point group= 3755 —5 _ 11x10x9x8! 3x28! => 165-5=160 shortest distance according graph = 1 yabeod+lr+1| letx,-1

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