{"id":44193,"date":"2022-05-29T02:00:35","date_gmt":"2022-05-28T20:30:35","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=44193"},"modified":"2023-11-09T16:08:54","modified_gmt":"2023-11-09T10:38:54","slug":"ml-aggarwal-class-8-solutions-for-icse-maths-chapter-14-ex-14-1","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-14-ex-14-1\/","title":{"rendered":"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 14 Constructions of Quadrilaterals Ex 14.1"},"content":{"rendered":"
Question 1.
\nConstruct a quadrilateral PQRS where PQ = 4.5 cm, QR = 6 cm, RS = 5.5 cm, PS = 5 cm and PR = 6.5 cm.
\nSolution:
\nSteps of constructions :
\n(i) Draw a line segment PR = 6.5 cm.
\n(ii) With centre P and radius 4.5 cm and with centre R
\nand radius 6 cm draw arcs intersecting each other at Q.
\n(iii) Join PQ and QR.
\n(iv) Similarly with centre P and radius 5 cm and with centre R
\nand radius 5.5 cm, draw arcs intersecting each other at S.
\n(v) Join PS and SR.
\nPQRS is the required quadrilateral.
\n<\/p>\n
Question 2.
\nConstruct a quadrilateral ABCD in which AB = 3\u00b75 cm, BC = 5 cm, CD = 5\u00b76 cm, DA = 4 cm and BD = 5\u00b74 cm
\nSolution:
\nSteps of construction :
\n(i) Draw AB = 3\u00b75 cm.
\n(ii) With A as centre and radius = 4 cm,
\ndraw an arc with B as centre and radius = 5\u00b74 cm
\ndraw an arc to meet the previous arc at D. Join AD andBD.
\n(iii) With B as centre and radius = 5 cm,
\ndraw an arc With D as centre and radius = 5\u00b76 cm,
\ndraw an arc to meet the previous arc at C.
\n(iv) Join BC and CD, then ABCD is the required quadrilateral.
\n<\/p>\n
Question 3.
\nConstruct a quadrilateral PQRS in which PQ = 3 cm, QR = 2\u00b75 cm, PS = 3\u00b75 cm, PR = 4 cm and QS = 5 cm.
\nSolution:
\nSteps of construction :
\n(i) Draw PQ = 3 cm.
\n(ii) With P as centre and radius = 4 cm,
\ndraw an arc with Q as centre and radius = 2\u00b75 cm
\ndraw an arc to meet the previous arc at R.
\nJoin PR and QR.
\n(iii) With P as centre and radius = 3\u00b75 cm,
\ndraw an arc. With Q as centre and radius = 5 cm,
\ndraw an arc to meet the previous arc at S.
\n(iv) Join PS, QS and SR.
\n(v) Hence, PQRS is the required quadrilateral.
\n<\/p>\n
Question 4.
\nConstruct a quadrilateral ABCD such that BC = 5 cm, AD = 5.5 cm, CD = 4.5 cm, AC = 7 cm, and BC = 5.5 cm.
\nSolution:
\nSteps of construction :
\n(i) Draw a line segment CD = 4.5 cm.
\n(ii) With centre C and radius of 5.5 cm and with centre D
\nand radius 7 cm draw arcs intersecting each other at B.
\n(iii) Join BC and BD.
\n(iv) Similarly with centre C and radius 5.5 cm
\nand with centre D and radius 5.5 cm,
\ndraw arcs intersecting each other at A.
\n(v) Join AC and AD.
\n(vi) Join AB.
\nABCD is the required quadrilateral.
\n<\/p>\n
Question 5.
\nConstruct a quadrilateral ABCD given that BC = 6 cm, CD = 4 cm, \u2220B = 45\u00b0, \u2220C = 90\u00b0 and \u2220D = 120\u00b0.
\nSolution:
\nSteps of construction :
\n(i) Draw BC = 6 cm.
\n(ii) At B, construct \u2220CBP = 45\u00b0.
\n(iii) At C, construct \u2220BCQ = 90\u00b0
\n(iv) From CQ, cut off CD = 4 cm.
\n(v) At D, construct \u2220CDR = 120\u00b0.
\n(vi) Let BP and DR meet at A.
\nThen ABCD is the required quadrilateral.
\n<\/p>\n
Question 6.
\nConstruct a quadrilateral PQRS where PQ = 4 cm, QR = 6 cm, \u2220P = 60\u00b0, \u2220Q = 90\u00b0 and \u2220R = 120\u00b0.
\nSolution:
\nSteps of construction :
\n(i) Draw a line segment QR = 6 cm.
\n(ii) At Q, draw a ray QX making an angle of 90\u00b0
\nand cut off QP = 4 cm.
\n
\n(iii) At P, draw a ray making an angle of 60\u00b0 and at R,
\na ray making an angle 120\u00b0 which meet each other at S.
\nPQRS is the required quadrilateral.<\/p>\n
Question 7.
\nConstruct a quadrilateral ABCD such that AB = 5 cm, BC = 4\u00b72 cm, AD = 3\u00b75 cm, \u2220A = 90\u00b0, and \u2220B = 60\u00b0
\nSolution:
\nSteps of construction :
\n(i) Draw AB = 5 cm.
\n(ii) At A, construct angle A = 90\u00b0
\n(iii) At B, construct angle B = 60\u00b0
\n(iv) With B as centre and 4\u00b72 cm as radius, cut off \u2220B atC.
\n(v) With A as centre and 3\u00b75 cm as radius, cut off \u2220A at D.
\n(vi) Join CD. Then ABCD is the required quadrilateral.
\n<\/p>\n
Question 8.
\nCosntruct a quadrilateral PQRS where PQ = 4 cm, QR = 5 cm, RS = 4.5 cm, \u2220Q = 60\u00b0 and \u2220R = 90\u00b0.
\nSolution:
\nSteps of construction :
\n(i) Draw a line segment QR = 5 cm.
\n(ii) At Q, draw a ray QX making an angle of 60\u00b0
\nand cut off QP = 4 cm.
\n(iii) At R, draw a ray RY making an angle of 90\u00b0
\nand cut off RS. = 4.5 cm.
\n(iv) Join PS.
\nPQRS is the required quadrilateral.
\n<\/p>\n
Question 9.
\nConstruct a quadrilateral BEST where BE = 3.8 cm, ES = 3.4 cm, ST = 4.5 cm, TB = 5 cm and \u2220E = 80\u00b0.
\nSolution:
\nSteps of construction :
\n(i) Draw a line segment BE = 3.8 cm.
\n(ii) At E, draw a ray EX making an angle of 80\u00b0
\nand cut off ES = 3.4 cm.
\n(iii) With centre B and radius 5 cm and
\nwith centre S with radius 4.5 cm,
\ndraw arcs intersecting each other at T.
\n(iv) Join TB and TS.
\nBEST is the required quadrilateral.
\n<\/p>\n
Question 10.
\nConstruct a quadrilateral ABCD where AB = 4\u00b75 cm, BC = 4 cm, CD = 3\u00b79 cm, AD = 3\u00b72 cm and \u2220B = 60\u00b0.
\nSolution:
\nSteps of construction :
\n(i) Draw AB = 4\u00b75 cm.
\n(ii) At B, construct \u2220ABP = 60\u00b0.
\n(iii) From BP, cut of \u2220BC = 4 cm.
\n(iv) With C as centre, and 3\u00b79 cm as radius draw an arc.
\n(v) With A as centre and 3\u00b72 cm as radius,
\ndraw an arc to meet the previous arc at D.
\n(vi) Join AD and CD.
\n<\/p>\n
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 14 Constructions of Quadrilaterals Ex 14.1 Question 1. Construct a quadrilateral PQRS where PQ = 4.5 cm, QR = 6 cm, RS = 5.5 cm, PS = 5 cm and PR = 6.5 cm. Solution: Steps of constructions : (i) Draw a line segment PR = … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3034],"tags":[],"yoast_head":"\n