{"id":47000,"date":"2023-12-20T10:03:34","date_gmt":"2023-12-20T04:33:34","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=47000"},"modified":"2023-12-20T17:20:35","modified_gmt":"2023-12-20T11:50:35","slug":"ml-aggarwal-class-9-solutions-for-icse-maths-chapter-13-chapter-test","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/ml-aggarwal-class-9-solutions-for-icse-maths-chapter-13-chapter-test\/","title":{"rendered":"ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 13 Rectilinear Figures Chapter Test"},"content":{"rendered":"

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 13 Rectilinear Figures Chapter Test<\/h2>\n

P.Q.
\nThe interior angles of a polygon add upto 4320\u00b0. How many sides does the polygon have ?
\nAnswer:
\nSum of interior angles of a polygon
\n= (2n – 4) \u00d7 90\u00b0
\n\u21d2 4320\u00b0 = (2n – 4) \u00d7 90\u00b0
\n\u21d2 \\(\\frac{4320^{\\circ}}{90^{\\circ}}\\) = (2n – 4) \u21d2 \\(\\frac{432}{9}\\) = 2n – 4
\n\u21d2 48 = 2n – 4 \u21d2 48 + 4 = 2n => 52 = 2 n
\n\u21d2 2n = 52 \u21d2 n = \\(\\frac{52}{2}\\) = 26
\nHence, the polygon have 26 sides.<\/p>\n

P.Q.
\nIf the ratio of an interior angle to the exterior angle of a regular polygon is 5 : 1, find the number of sides.
\nAnswer:
\nThe ratio of an interior angle to the exterior angle of a regular polygon = 5 : 1
\n\"ML
\n\"ML
\nHence, number of sides of regular polygon = 12.<\/p>\n

P.Q.
\nIn a pentagon ABCDE, BC || ED and \u2220B: \u2220A : \u2220E = 3 : 4 : 5. Find \u2220A.
\nAnswer:
\n\u2235 BC||ED
\n\u2234 \u2220C +\u2220D= 180\u00b0 (Co-interior angles)
\nBut\u2220A + \u2220B + \u2220C + \u2220D + \u2220E = 540\u00b0
\n\u2234 \u2220A + \u2220B + \u2220E \u2260 180\u00b0 = 540\u00b0
\n\u21d2 \u2220A + \u2220B + \u2220E = 540\u00b0 – 180\u00b0 = 360\u00b0
\nBut \u2220B: \u2220A = \u2220E = 3 : 4 : 5
\nLet \u2220B = 3x, \u2220A = 4x and \u2220E = 5x
\n\"ML
\n\u2234 3x + 4x + 5x = 360\u00b0 \u21d2 12x = 360\u00b0
\n\u21d2 x = \\(\\frac{360^{\\circ}}{12}\\) = 30\u00b0
\n\u2234 A = 4x = 4 \u00d7 30\u00b0= 120\u00b0<\/p>\n

Question 1.
\nIn the given figure, ABCD is a parallelogram. CB is produced to E such that BE = BC. Prove that AEBD is a parallelogram.
\n\"ML
\nSolution:
\nIn the figure, ABCD is a ||gm side CB is produced to E such that BE = BC
\nBD and AE are joined
\nTo prove: AEBD is a parallelogram
\nProof: In \u2206AEB and \u2206BDC
\nEB=BC (Given)
\n\u2220ABE =\u2220DCB (Corresponding angles)
\nAB = DC (Opposite sides of ||gm)
\n\u2234 \u2206AEB = \u2206BDC (SAS axiom)
\n\u2234 AE = DB (c.p.c.t.)
\nBut AD = CB = BE (Given)
\n\u2235 The opposite sides are equal and \u2220AEB = \u2220DBC (c.p.c.t.)
\nBut these are corresponding angle
\n\u2234 AEBD is a parallelogram<\/p>\n

Question 2.
\nIn the given figure, ABC is an isosceles triangle in which AB = AC. AD bisects exterior angle PAC and CD || BA. Show that
\n(i) \u2220DAC = \u2220BCA
\n(ii) ABCD is a parallelogram.
\nSolution:
\nGiven: In isosceles \u2206ABC, AB = AC.
\nAD is the bisector of ext. \u2220PAC and
\nCD || BA
\n\"ML
\nTo prove: (i) \u2220DAC = \u2220BCA
\n(ii) ABCD is a||gm
\nProof: In \u2206ABC
\n\u2235 AB = AC (Given)
\n\u2234 \u2220C = \u2220B (Angles opposite to equal sides)
\n\u2235 Ext. \u2220PAC = \u2220B + \u2220C
\n= \u2220C + \u2220C = 2\u2220C = 2\u2220BCA
\n\u2234 2\u2220DAC = 2\u2220BCA
\n\u2220DAC = \u2220BCA
\nBut these are alternate angles
\n\u2234 AD || BC
\nBut AB || AC (Given)
\n\u2234 ABCD is a||gm<\/p>\n

Question 3.
\nProve that the quadrilateral obtained by joining the mid-points of an isosceles trapezium is a rhombus.
\nAnswer:
\nGiven. ABCD is an isosceles trapezium in which AB || DC and AD = BC
\nP, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively PQ, QR, RS and SP are joined.
\n\"ML
\nTo Prove. PQRS is a rhombus.
\nConstructions. Join AC and BD.
\nProof, \u2235 ABCD is an isosceles trapezium
\n\u2234 Its diagnoals are equal
\n\u2234 AC = BD
\nNow in \u2206ABC,
\nP and Q are the mid-points of AB and BC
\n\u2234 PQ || AC and PQ = \\(\\frac{1}{2}\\) AC …(i)
\nSimilarly in \u2206ADC,
\nS and R mid-points of CD and AD
\n\u2234 SR || AC and SR = \\(\\frac{1}{2}\\) AC …(ii)
\nfrom (i) and (ii)
\nPQ | | SR and PQ = SR
\n\u2234 PQRS is a parallelogram
\nNow in \u2206APS and \u2206BPQ,
\nAP = BP (P is mid-point of AB)
\nAS = BQ Half of equal sides)
\n\u2220A = \u2220B ( \u2235 ABCD is isosceles trapezium)
\n\u2234 \u2206APS \u2245 BPQ
\n\u2234 PS = PQ
\nBut there are the adjacent sides of a parallelogram
\n\u2234 Sides of PQRS are equal
\nHence PQRS is a rhombus.
\nHence proved.<\/p>\n

Question 4.
\nFind the size of each lettered angle in the following figures:
\n\"ML
\nAnswer:
\n(i) \u2235 CDE is a st. line
\n\u2234 \u2220ADE + \u2220ADC = 180\u00b0
\n\"ML
\n122\u00b0+ \u2220ADC = 180\u00b0
\n\u2220ADC = 180\u00b0 – 122\u00b0\u00b0
\n\u2220ADC = 58\u00b0 ….(1)
\n\u2220ABC = 360\u00b0 – 140\u00b0 = 220\u00b0
\n(At any point the angle is 360\u00b0) …(2)
\nNow, in quadrilateral ABCD,
\n\u2220ADC + \u2220BCD + \u2220B AD + \u2220ABC = 360\u00b0
\n\u21d2 58\u00b0 + 53\u00b0 + x + 220\u00b0 = 36o\u00b0 [using (1) and (2)]
\n\u21d2 331\u00b0 + x = 360\u00b0 \u21d2 x = 360\u00b0 – 331\u00b0
\n\u21d2 x = 29\u00b0 Answer:<\/p>\n

(ii) \u2235 DE || AB (given)
\n\u2234 \u2220ECB = \u2220CBA (Alternate angles)
\n\u21d2 75\u00b0 = \u2220CBA
\n\u2234 \u2220CBA =75\u00b0
\n\u2235 AD || BC (given)
\n\u2234 (x + 66\u00b0) + (75\u00b0) =180\u00b0 (co-interior angles are supplementary)
\n\u21d2 x + 66\u00b0 + 75\u00b0 = 180\u00b0 \u21d2 x + 141\u00b0 = 180\u00b0
\n\u21d2 x = 180\u00b0- 141\u00b0
\nA = 39\u00b0 …(1)
\nNow, in \u2206AMB,
\nx + 30\u00b0 + \u2220AMB = 180\u00b0 (sum of all angles in a triangle is 180\u00b0)
\n\u21d2 39\u00b0 + 30\u00b0 + \u2220AMB = 180\u00b0 [From (1)]
\n\u21d2 69\u00b0 + \u2220AMB = 180\u00b0
\n\u21d2 \u2220AMB = 180\u00b0- 69\u00b0
\n\u21d2 \u2220AMB = 111\u00b0 ….(2)
\n\u2235 \u2220AMB = y (vertically opposite angles)
\n\u21d2 111\u00b0 = y [From (2)]
\n\u2234 y = 111\u00b0
\nHence, A = 39\u00b0 and y =111\u00b0<\/p>\n

(iii) In \u2206ABD
\nAB = AD (given)
\n\u2220ABD = \u2220ADB (\u2235 equal sides have equal angles opposite to them)
\n\u21d2 \u2220ABD = 42\u00b0 [\u2235 \u2220ADB = 42\u00b0 (given)]
\n\u2235 \u2220ABD + \u2220ADB + \u2220BAD = 180\u00b0
\n\"ML
\n(Sum of all angles in a triangle is 180\u00b0)
\n\u21d2 42\u00b0 + 42\u00b0 + y = 180\u00b0 \u21d2 84\u00b0 + y = 180\u00b0
\n\u21d2 y = 180\u00b0 – 84\u00b0 \u21d2 y = 96\u00b0
\n\u2220BCD = 2 \u00d7 26\u00b0 = 52\u00b0
\nIn \u2220BCD
\n\u2235 BC = CD (given)
\n\u2234 \u2220CBD = \u2220CDB = x [equal side have equal angles opposite to them]
\n\u2234 \u2220CBD + \u2220CDB + \u2220BCD = 180\u00b0
\n\u21d2 x + x + 52\u00b0 = 180\u00b0 \u21d2 2A = 180\u00b0 – 52\u00b0
\n\u21d2 2x = 128\u00b0 \u21d2 x = \\(\\frac{128^{\\circ}}{2}\\) \u21d2 x = 64\u00b0
\nHence, x = 64\u00b0 and y = 90\u00b0<\/p>\n

Question 5.
\nFind the size of each lettered angle in the following figures :
\n\"ML
\nAnswer:
\n(i) Here AB || CD and BC II AD (given)
\n\u2234 ABCD is a || gm
\n\u2234 y = 2 \u00d7 \u2220ABD
\n\u21d2 y = 2 \u00d7 53\u00b0 = 106\u00b0 ….(1)
\nAlso, y + \u2220DAB = 180\u00b0
\n\u21d2 106\u00b0 + \u2220DAB = 180\u00b0
\n\u21d2 \u2220DAB = 180\u00b0 – 106\u00b0 \u21d2 \u2220DAB = 74\u00b0
\n\u2234 x = \\(\\frac{1}{2}\\) \u2220DAB (\u2235 AC bisect \u2220DAB)
\n\"ML
\n\u21d2 x = \\(\\frac{1}{2}\\) \u00d7 74\u00b0 = 37\u00b0
\nand \u2220D AC = x = 37 ….(2)
\n\u2234 \u2220DAC = z (Alternate angles) ….(3)
\nFrom (2) and (3),
\nz = 37\u00b0
\nHence,x = 37\u00b0, y = 106\u00b0, z = 37\u00b0<\/p>\n

(ii) \u2235 ED is a st. line
\n\"ML
\n\u2234 60\u00b0 + \u2220AED = 180\u00b0 (linear pair)
\n\u21d2 \u2220AED = 180\u00b0 – 60\u00b0
\n\u21d2 \u2220AED = 120\u00b0 …(1)
\n\u2235 CD is a st. line
\n\u2234 50\u00b0 +\u2220BCD = 180\u00b0 (linear pair)
\n\u21d2 \u2220BCD = 180\u00b0 – 50\u00b0
\n\u21d2 \u2220BCD = 130\u00b0 ….(2)
\nIn pentagon ABCDE
\n\u2220A + \u2220B + \u2220AED + \u2220BCD + x = 540\u00b0 (Sum of interior angles in pentagon is 540\u00b0)
\n\u21d2 90\u00b0 + 90\u00b0 + 120\u00b0 + 130\u00b0 + x = 540\u00b0
\n\u21d2 430\u00b0 + x = 540\u00b0 \u21d2 x = 540\u00b0 – 430\u00b0
\n\u21d2 x = 110\u00b0
\nHence, value of x = 110\u00b0<\/p>\n

(iii) In given figure , AD || BC (given)
\n\"ML
\n\u2234 60\u00b0 + y = 180\u00b0and x + 110\u00b0 = 180\u00b0
\n\u21d2 y = 180\u00b0 – 60\u00b0 and x = 180\u00b0 – 110\u00b0
\n\u21d2 y = 120\u00b0 and x = 70\u00b0
\n\u2235 CD||AF (given)
\n\u2234 \u2220FAD = x (Alternate angles)
\n\u21d2 \u2220FAD = 70\u00b0 ….(1)
\nIn quadrilateral ADEF,
\n\u2220FAD + 75\u00b0 + z + 130\u00b0 = 360\u00b0
\n\u21d2 70\u00b0 + 75\u00b0 + z + 130\u00b0 = 360\u00b0 [using(1)]
\n\u21d2 275\u00b0 + z = 360\u00b0 \u21d2 z = 85\u00b0
\nHence, x = 10\u00b0, y = 120\u00b0 and z = 85\u00b0<\/p>\n

Question 6.
\nIn the adjoining figure, ABCD is a rhombus and DCFE is a square. If \u2220ABC = 56\u00b0, find
\n(i) \u2220DAG
\n(ii) \u2220FEG
\n(iii) \u2220GAC
\n(iv) \u2220AGC
\n\"ML
\nAnswer:
\nHere ABCD and DCFE is a rhombus and square respectively.
\n\"ML
\n\u2234 AB = BC = DC = AD ….(1)
\nAlso DC = EF = FC = EF ….(2)
\nFrom (1) and (2),
\nAB = BC = DC = AD = EF = FC = EF ….(3)
\n\u2220ABC =56\u00b0 (given)
\n\u2220ADC = 56\u00b0 (opposite angle in rhombus are equal)
\n\u2234 \u2220EDA = \u2220EDC +\u2220ADC = 90\u00b0 + 56\u00b0 = 146\u00b0
\nIn \u2206ADE,
\nDE = AD [From (3)]
\n\u2220DEA =\u2220DAE (equal sides have equal opposite angles)
\n\u21d2 DEA = DAE = \\(\\frac{180^{\\circ}-\\angle \\mathrm{EDA}}{2}\\)
\n= \\(\\frac{180^{\\circ}-146^{\\circ}}{2}=\\frac{34^{\\circ}}{2}\\) = 17\u00b0
\n\u21d2 \u2220DAG = 17\u00b0
\nAlso, \u2220DEG =17\u00b0
\n\u2234 \u2220FEG = \u2220E – \u2220DEG
\n= 90\u00b0- 17\u00b0 = 73\u00b0
\nIn rhombus ABCD,
\n\u2220DAB = 180\u00b0 – 56\u00b0 = 124\u00b0
\n\u2220DAC = \\(\\frac{124^{\\circ}}{2}\\) (\u2235 AC diagonals bisect the \u2220A)
\n\u2220DAC =62\u00b0
\n\u2234 \u2220GAC = \u2220DAC – \u2220DAG
\n= 62\u00b0 – 17\u00b0 = 45\u00b0
\nIn \u2206EDG,
\n\u2220D + \u2220DEG + \u2220DGE = 180\u00b0 (Sum of all angles in a triangle is 180\u00b0)
\n\u21d2 90\u00b0 + 17\u00b0 + \u2220DGE = 180\u00b0
\n\u21d2 \u2220DGE = 180\u00b0 – 107\u00b0 = 73\u00b0 ….(4)
\nHence, \u2220AGC = \u2220DGE ….(5) (vertically opposite angles)
\nFrom (4) and (5)
\n\u2220AGC = 73\u00b0<\/p>\n

Question 7.
\nIf one angle of a rhombus is 60\u00b0 and the length of a side is 8 cm, find the\/lengths of its diagonals.
\nAnswer:
\nEach side of rhombus ABCD is 8 cm. AB = BC = CD = DA = 8cm.
\n\"ML
\nLet \u2220A = 60\u00b0
\n\u2234 \u2206ABD is an equilateral triangle
\n\u2234 AB = BD = AD = 8cm.
\n\u2235 Diagonals of a rhombus bisect each other eight angles.
\n\u2234 AO = OC, BO = OD=4 cm.
\nand \u2220AOB = 90\u00b0
\nNow in right \u2206AOB,
\nAB = AO2<\/sup> + OB2<\/sup> (Pythagoras Theorem)
\n\u21d2 (8)2<\/sup> = AO2<\/sup> + (4)2<\/sup>
\n\u21d2 64 = AO2<\/sup> + 16 ,
\n\u21d2 AO2<\/sup> = 64 – 16 = 48 = 16 + 3
\n\u2234 AO = \\(\\sqrt{16 \\times 3}=4 \\sqrt{3}\\) cm.
\nBut AC = 2 AO
\n\u2234 AC = 2 x \\(4 \\sqrt{3}=8 \\sqrt{3}\\) cm<\/p>\n

Question 8.
\nUsing ruler and compasses only, construct a parallelogram ABCD with AB = 5 cm, AD = 2.5 cm and \u2220BAD = 45\u00b0. If the bisector of \u2220BAD meets DC at E, prove that \u2220AEB is a right angle.
\nAnswer:
\nGiven : AB = 5 cm, AD = 2.5 cm and \u2220BAD =45\u00b0.
\nRequired : (i) To construct a parallelogram ABCD.
\n(ii) If the bisector of \u2220BAD meets DC at E then prove that \u2220AEB = 90\u00b0.
\n\"ML
\nSteps of Construction:<\/p>\n