Solution:<\/strong><\/p>\n\n\n\n Steps of construction:<\/p>\n\n\n\n 1. Draw angle BAC = 50o<\/sup> such that AB = 5 cm and AC = 7 cm.<\/p>\n\n\n\n Cut an arc through C at an angle of 50os<\/sup><\/p>\n\n\n\n 2. Draw a straight line passing through C and the arc. This line will be parallel to AB since \u2220CAB =\u2220RCA=50o<\/sup><\/p>\n\n\n\n 3. Alternate angles are equal; therefore the line is parallel to AB.<\/p>\n\n\n\n 4. Again through B, cut an arc at an angle of 50o<\/sup> and draw a line passing through B and this arc and say this intersects the line drawn parallel to AB at D.<\/p>\n\n\n\n 5. \u2220SBA =\u2220BAC = 50o<\/sup>, since they are alternate angles. Therefore BD parallel to AC<\/p>\n\n\n\n 6. Also we can measure BD = 7 cm and CD = 5 cm.<\/p>\n\n\n\n 2. Draw a line PQ. Draw another line parallel to PQ at a distance of 3 cm from it.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Steps of construction:<\/p>\n\n\n\n 1. Draw a line PQ.<\/p>\n\n\n\n 2. Take any two points A and B on the line.<\/p>\n\n\n\n 3. Construct \u2220PBF = 90o<\/sup> and \u2220QAE = 90o<\/sup><\/p>\n\n\n\n 4. With A as center and radius 3 cm cut AE at C.<\/p>\n\n\n\n 5. With B as center and radius 3 cm cut BF at D.<\/p>\n\n\n\n 6. Join CD and produce it on either side to get the required line parallel to AB and at a distance of 3 cm from it.<\/p>\n\n\n\n 3. Take any three non-collinear points A, B, C and draw \u2220ABC. Through each vertex of the triangle, draw a line parallel to the opposite side.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Steps of construction:<\/p>\n\n\n\n 1. Mark three non collinear points A, B and C such that none of them lie on the same line.<\/p>\n\n\n\n 2. Join AB, BC and CA to form triangle ABC.<\/p>\n\n\n\n 3. Parallel line to AC<\/p>\n\n\n\n 4. With A as center, draw an arc cutting AC and AB at T and U, respectively.<\/p>\n\n\n\n 5. With center B and the same radius as in the previous step, draw an arc on the opposite side of AB to cut AB at X.<\/p>\n\n\n\n 6. With center X and radius equal to TU, draw an arc cutting the arc drawn in the previous step at Y.<\/p>\n\n\n\n 7. Join BY and produce in both directions to obtain the line parallel to AC.<\/p>\n\n\n\n Parallel line to AB:<\/p>\n\n\n\n 8. With B as center, draw an arc cutting BC and BA at W and V, respectively.<\/p>\n\n\n\n 9. With center C and the same radius as in the previous step, draw an arc on the opposite side of BC to cut BC at P.<\/p>\n\n\n\n 10. With center P and radius equal to WV, draw an arc cutting the arc drawn in the previous step at Q.<\/p>\n\n\n\n 11. Join CQ and produce in both directions to obtain the line parallel to AB.<\/p>\n\n\n\n Parallel line to BC:<\/p>\n\n\n\n 12. With B as center, draw an arc cutting BC and BA at W and V, respectively (already drawn).<\/p>\n\n\n\n 13. With center A and the same radius as in the previous step, draw an arc on the opposite side of AB to cut AB at R.<\/p>\n\n\n\n 14. With center R and radius equal to WV, draw an arc cutting the arc drawn in the previous step at S.<\/p>\n\n\n\n 15. Join AS and produce in both directions to obtain the line parallel to BC.<\/p>\n\n\n\n 4. Draw two parallel lines at a distance of 5cm apart.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Steps of construction:<\/p>\n\n\n\n 1. Draw a line PQ.<\/p>\n\n\n\n 2. Take any two points A and B on the line.<\/p>\n\n\n\n 3. Construct \u2220PBF = 90o<\/sup> and \u2220QAE = 90o<\/sup><\/p>\n\n\n\n 4. With A as center and radius 5 cm cut AE at C.<\/p>\n\n\n\n 5. With B as center and radius 5 cm cut BF at D.<\/p>\n\n\n\n 6. Join CD and produce it on either side to get the required line parallel to AB and at a distance of 5 cm from it.<\/p>\n\n\n\n Exercise 17.2 Page No: 17.3<\/p>\n\n\n\n 1. Draw \u25b3ABC in which AB = 5.5 cm. BC = 6 cm and CA = 7 cm. Also, draw perpendicular bisector of side BC.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Steps of construction:<\/p>\n\n\n\n 1. Draw a line segment AB of length 5.5 cm.<\/p>\n\n\n\n 2. From B, cut an arc of radius 6 cm.<\/p>\n\n\n\n 3. With center A, draw an arc of radius 7 cm intersecting the previously drawn arc at C.<\/p>\n\n\n\n 4. Join AC and BC to obtain the desired triangle.<\/p>\n\n\n\n 5. With center B and radius more than half of BC, draw two arcs on both sides of BC.<\/p>\n\n\n\n 6. With center C and the same radius as in the previous step, draw two arcs intersecting the arcs drawn in the previous step at X and Y.<\/p>\n\n\n\n 7. Join XY to get the perpendicular bisector of BC.<\/p>\n\n\n\n 2. Draw \u2206PQR in which PQ = 3 cm, QR = 4 cm and RP = 5 cm. Also, draw the bisector of \u2220Q<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Steps of construction:<\/p>\n\n\n\n 1. Draw a line segment PQ of length 3 cm.<\/p>\n\n\n\n 2. With Q as center and radius 4 cm, draw an arc.<\/p>\n\n\n\n 3. With P as center and radius 5 cm, draw an arc intersecting the previously drawn arc at R.<\/p>\n\n\n\n 4. Join PR and QR to obtain the required triangle.<\/p>\n\n\n\n 5. From Q, cut arcs of equal radius intersecting PQ and QR at M and N, respectively.<\/p>\n\n\n\n 6. From M and N, cut arcs of equal radius intersecting at point S.<\/p>\n\n\n\n 7. Join QS and extend to produce the angle bisector of angle PQR.<\/p>\n\n\n\n 8. Verify that angle PQS and angle SQR are equal to 45o<\/sup> each.<\/p>\n\n\n\n 3. Draw an equilateral triangle one of whose sides is of length 7 cm.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Steps of construction:<\/p>\n\n\n\n 1. Draw a line segment AB of length 7 cm.<\/p>\n\n\n\n 2. With center A, draw an arc of radius 7 cm.<\/p>\n\n\n\n 3. With center B, draw an arc of radius 7 cm intersecting the previously drawn arc at C.<\/p>\n\n\n\n 4. Join AC and BC to get the required triangle.<\/p>\n\n\n\n 4. Draw a triangle whose sides are of lengths 4 cm, 5 cm and 7 cm. Draw the perpendicular bisector of the largest side.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Steps of construction:<\/p>\n\n\n\n 1. Draw a line segment PR of length 7 cm.<\/p>\n\n\n\n 2. With center P, draw an arc of radius 5 cm.<\/p>\n\n\n\n 3. With center R, draw an arc of radius 4 cm intersecting the previously drawn arc at Q.<\/p>\n\n\n\n 4. Join PQ and QR to obtain the required triangle.<\/p>\n\n\n\n 5. From P, draw arcs with radius more than half of PR on either sides.<\/p>\n\n\n\n 6. With the same radius as in the previous step, draw arcs from R on either sides of PR intersecting the arcs drawn in the previous step at M and N.<\/p>\n\n\n\n 7. MN is the required perpendicular bisector of the largest side.<\/p>\n\n\n\n 5. Draw a triangle ABC with AB = 6 cm, BC = 7 cm and CA = 8 cm. Using ruler and compass alone, draw (i) the bisector AD of \u2220A and (ii) perpendicular AL from A on BC. Measure LAD.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Steps of construction:<\/p>\n\n\n\n 1. Draw a line segment BC of length 7 cm.<\/p>\n\n\n\n 2. With center B, draw an arc of radius 6 cm.<\/p>\n\n\n\n 3. With center C, draw an arc of radius 8 cm intersecting the previously drawn arc at A.<\/p>\n\n\n\n 4. Join AC and AB to get the required triangle.<\/p>\n\n\n\n Angle bisector steps:<\/p>\n\n\n\n 5. From A, cut arcs of equal radius intersecting AB and AC at E and F, respectively.<\/p>\n\n\n\n 6. From E and F, cut arcs of equal radius intersecting at point H.<\/p>\n\n\n\n 7. Join AH and extend to produce the angle bisector of angle A, meeting line BC at D.<\/p>\n\n\n\n 8. Perpendicular from Point A to line BC steps:<\/p>\n\n\n\n 9. From A, cut arcs of equal radius intersecting BC at P and Q, respectively (Extend BC to draw these arcs).<\/p>\n\n\n\n 10. From P and Q, cut arcs of equal radius intersecting at M.<\/p>\n\n\n\n 11. Join AM cutting BC at L.<\/p>\n\n\n\n 12. AL is the perpendicular to the line BC.<\/p>\n\n\n\n 13. Angle LAD is 15o<\/sup>.<\/p>\n\n\n\n 6. Draw \u25b3DEF such that DE= DF= 4 cm and EF = 6 cm. Measure \u2220E and \u2220F.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Steps of construction:<\/p>\n\n\n\n 1. Draw a line segment EF of length 6 cm.<\/p>\n\n\n\n 2. With E as center, draw an arc of radius 4 cm.<\/p>\n\n\n\n 3. With F as center, draw an arc of radius 4 cm intersecting the previous arc at D.<\/p>\n\n\n\n 4. Join DE and DF to get the desired triangle DEF.<\/p>\n\n\n\n 5. By measuring we get, \u2220E= \u2220F= 40o<\/sup>.<\/p>\n\n\n\n 7. Draw any triangle ABC. Bisect side AB at D. Through D, draw a line parallel to BC, meeting AC in E. Measure AE and EC.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Steps of construction:<\/p>\n\n\n\n We first draw a triangle ABC with each side = 6 cm.<\/p>\n\n\n\n Steps to bisect line AB:<\/p>\n\n\n\n 1. Draw an arc from A on either side of line AB.<\/p>\n\n\n\n 2. With the same radius as in the previous step, draw an arc from B on either side of AB intersecting the arcs drawn in the previous step at P and Q.<\/p>\n\n\n\n 3. Join PQ cutting AB at D. PQ is the perpendicular bisector of AB.<\/p>\n\n\n\n Parallel line to BC:<\/p>\n\n\n\n 4. With B as center, draw an arc cutting BC and BA at M and N, respectively.<\/p>\n\n\n\n 5. With center D and the same radius as in the previous step, draw an arc on the opposite side of AB to cut AB at Y.<\/p>\n\n\n\n 6. With center Y and radius equal to MN, draw an arc cutting the arc drawn in the previous step at X.<\/p>\n\n\n\n 7. Join XD and extend it to intersect AC at E.<\/p>\n\n\n\n 8. DE is the required parallel line.<\/p>\n\n\n\n Exercise 17.3 Page No: 17.5<\/p>\n\n\n\n 1. Draw \u25b3ABC in which AB = 3 cm, BC = 5 cm and \u2220B = 70o<\/sup>.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n\n\n Steps of construction:<\/p>\n\n\n\n 1. Draw a line segment AB of length 3 cm.<\/p>\n\n\n\n 2. Draw \u2220XBA=70o<\/sup>.<\/p>\n\n\n\n 3. Cut an arc on BX at a distance of 5 cm at C.<\/p>\n\n\n\n 4. Join AC to get the required triangle.<\/p>\n\n\n\n 2. Draw \u25b3ABC in which \u2220A<\/em>=70o<\/sup>. AB = 4 cm and AC= 6 cm. Measure BC.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Steps of construction:<\/p>\n\n\n\n 1. Draw a line segment AC of length 6 cm.<\/p>\n\n\n\n 2. Draw \u2220XAC=70o<\/sup>.<\/p>\n\n\n\n 3. Cut an arc on AX at a distance of 4 cm at B.<\/p>\n\n\n\n 4. Join BC to get the desired triangle.<\/p>\n\n\n\n 5. We see that BC = 6 cm.<\/p>\n\n\n\n 3. Draw an isosceles triangle in which each of the equal sides is of length 3 cm and the angle between them is 45o<\/sup>.<\/strong><\/p>\n\n\n\n Steps of construction:<\/strong><\/p>\n\n\n\n 1. Draw a line segment PQ of length 3 cm.<\/p>\n\n\n\n 2. Draw \u2220QPX=45o<\/sup>.<\/p>\n\n\n\n 3. Cut an arc on PX at a distance of 3 cm at R.<\/p>\n\n\n\n 4. Join QR to get the required triangle.<\/p>\n\n\n\n 4. Draw \u25b3ABC in which \u2220A = 120o<\/sup>, AB = AC = 3 cm. Measure \u2220B and \u2220C.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Steps of construction:<\/p>\n\n\n\n 1. Draw a line segment AC of length 3 cm.<\/p>\n\n\n\n 2. Draw \u2220XAC = 120o<\/sup>.<\/p>\n\n\n\n 3. Cut an arc on AX at a distance of 3 cm at B.<\/p>\n\n\n\n 4. Join BC to get the required triangle.<\/p>\n\n\n\n 5. By measuring, we get \u2220B = \u2220C = 30o<\/sup>.<\/p>\n\n\n\n 5. Draw \u25b3ABC in which \u2220C <\/em>= 90o<\/sup> and AC = BC = 4 cm.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Steps of construction:<\/p>\n\n\n\n 1. Draw a line segment BC of length 4 cm.<\/p>\n\n\n\n 2. At C, draw \u2220BCY=90\u00b0.<\/p>\n\n\n\n 3. Cut an arc on CY at a distance of 4 cm at A.<\/p>\n\n\n\n 4. Join AB. ABC is the required triangle.<\/p>\n\n\n\n 6. Draw a triangle ABC in which BC = 4 cm, AB = 3 cm and \u2220B <\/em>= 45o<\/sup>. Also, draw a perpendicular from A on BC.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Steps of construction:<\/p>\n\n\n\n 1. Draw a line segment AB of length 3 cm.<\/p>\n\n\n\n 2. Draw an angle of 45o<\/sup> and cut an arc at this angle at a radius of 4 cm at C.<\/p>\n\n\n\n 3. Join AC to get the required triangle.<\/p>\n\n\n\n 4. With A as center, draw intersecting arcs at M and N.<\/p>\n\n\n\n 5. With center M and radius more than half of MN, cut an arc on the opposite side of \u2220A.<\/p>\n\n\n\n 6. With N as center and as same radius taken in the previous step, cut an arc intersecting the previous arc at E.<\/p>\n\n\n\n 7. Join AE, it meets BC at D, then AE is the required perpendicular.<\/p>\n\n\n\n 7. Draw a triangle ABC with AB = 3 cm, BC = 4 cm and \u2220B <\/em>= 60o<\/sup>. Also, draw the bisector of angles C and A of the triangle, meeting in a point O. Measure \u2220COA.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Steps of construction:<\/p>\n\n\n\n 1. Draw a line segment BC = 4 cm.<\/p>\n\n\n\n 2. Draw \u2220CBX = 60o<\/sup>.<\/p>\n\n\n\n 3. Draw an arc on BX at a radius of 3 cm cutting BX at A.<\/p>\n\n\n\n 4. Join AC to get the required triangle.<\/p>\n\n\n\n Angle bisector for angle A:<\/p>\n\n\n\n 5. With A as center, cut arcs of the same radius cutting AB and AC at P and Q, respectively.<\/p>\n\n\n\n 6. From P and Q cut arcs of same radius intersecting at T.<\/p>\n\n\n\n 7. Join AT to get the angle bisector of angle A.<\/p>\n\n\n\n Angle bisector for angle C:<\/p>\n\n\n\n 8. With A as center, cut arcs of the same radius cutting CB and CA at M and N, respectively.<\/p>\n\n\n\n 9. From M and N, cut arcs of the same radius intersecting at R<\/p>\n\n\n\n 10. Join CR to get the angle bisector of angle C.<\/p>\n\n\n\n 11. Mark the point of intersection of CR and AT as O.<\/p>\n\n\n\n 12. Angle \u2220COA = 120o<\/sup>.<\/p>\n\n\n\n Exercise 17.4 Page No: 16.3<\/p>\n\n\n\n 1. Construct \u2206ABC in which BC = 4 cm, \u2220B = 50o<\/sup> and \u2220C = 70o<\/sup>.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Steps of construction:<\/p>\n\n\n\n 1. Draw a line segment BC of length 4 cm.<\/p>\n\n\n\n 2. Draw \u2220CBX such that \u2220CBX=50o<\/sup>.<\/p>\n\n\n\n 3. Draw \u2220BCY with Y on the same side of BC as X such that \u2220BCY=70o<\/sup>.<\/p>\n\n\n\n 4. Let CY and BX intersects at A.<\/p>\n\n\n\n 5. ABC is the required triangle.<\/p>\n\n\n\n 2. Draw \u2206ABC in which BC = 8 cm, \u2220B = 50o<\/sup> and \u2220A = 50o<\/sup>.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Steps of construction:<\/p>\n\n\n\n 1. Draw a line segment BC of length 8 cm.<\/p>\n\n\n\n 2. Draw \u2220CBX such that \u2220CBX = 50o<\/sup>.<\/p>\n\n\n\n 3. Draw \u2220BCY with Y on the same side of BC as X such that \u2220BCY = 80o<\/sup>.<\/p>\n\n\n\n 4. Let CY and BX intersects at A.<\/p>\n\n\n\n 3. Draw \u2206ABC in which \u2220Q = 80o<\/sup>, \u2220R = 55o<\/sup> and QR = 4.5 cm. Draw the perpendicular bisector of side QR.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Steps of construction:<\/p>\n\n\n\n 1. Draw a line segment QR = 4.5 cm.<\/p>\n\n\n\n 2. Draw \u2220RQX = 80\u00b0 and \u2220QRY = 55o<\/sup>.<\/p>\n\n\n\n 3. Let QX and RY intersects at P so that PQR is the required triangle.<\/p>\n\n\n\n 4. With Q as center and radius more than 4.5 cm, draw arcs on either sides of QR.<\/p>\n\n\n\n 5. With R as center and radius more than 4.5 cm, draw arcs intersecting the previous arcs at M and N.<\/p>\n\n\n\n 6. Join MN<\/p>\n\n\n\n 7. MN is the required perpendicular bisector of QR.<\/p>\n\n\n\n 4. Construct \u2206ABC in which AB = 6.4 cm, \u2220A = 45o<\/sup> and \u2220B = 60o<\/sup><\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Steps of construction:<\/p>\n\n\n\n 1. Draw a line segment AB = 6.4 cm.<\/p>\n\n\n\n 2. Draw \u2220BAX = 45o<\/sup>.<\/p>\n\n\n\n 3. Draw \u2220ABY with Y on the same side of AB as X such that \u2220ABY = 60o<\/sup>.<\/p>\n\n\n\n 4. Let AX and BY intersects at C.<\/p>\n\n\n\n 5. ABC is the required triangle.<\/p>\n\n\n\n 5. Draw \u2206ABC in which AC = 6 cm, \u2220A = 90o<\/sup> and \u2220B = 60o<\/sup><\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Steps of construction:<\/p>\n\n\n\n 1. Draw a line segment AC = 6 cm.<\/p>\n\n\n\n 2. Draw \u2220ACX = 30o<\/sup>.<\/p>\n\n\n\n 3. Draw \u2220CAY with Y on the same side of AC as X such that \u2220CAY = 90o<\/sup>.<\/p>\n\n\n\n 4. Join CX and AY. Let these intersects at B.<\/p>\n\n\n\n 5. ABC is the required triangle where angle \u2220ABC = 60o<\/sup>.<\/p>\n\n\n\n Exercise 17.5 Page No: 16.3<\/p>\n\n\n\n 1. Draw a right triangle with hypotenuse of length 5 cm and one side of length 4 cm.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Steps of construction:<\/p>\n\n\n\n 1. Draw a line segment QR = 4 cm.<\/p>\n\n\n\n 2. Draw \u2220QRX of measure 90o<\/sup>.<\/p>\n\n\n\n 3. With center Q and radius PQ = 5 cm, draw an arc of the triangle to intersect ray RX at P.<\/p>\n\n\n\n 4. Join PQ to obtain the desired triangle PQR.<\/p>\n\n\n\n 5. PQR is the required triangle.<\/p>\n\n\n\n 2. Draw a right triangle whose hypotenuse is of length 4 cm and one side is of length 2.5 cm.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Steps of construction:<\/p>\n\n\n\n 1. Draw a line segment QR = 2.5 cm.<\/p>\n\n\n\n 2. Draw \u2220QRX of measure 90o<\/sup>.<\/p>\n\n\n\n 3. With center Q and radius PQ = 4 cm, draw an arc of the triangle to intersect ray RX at P.<\/p>\n\n\n\n 4. Join PQ to obtain the desired triangle PQR.<\/p>\n\n\n\n 5. PQR is the required triangle.<\/p>\n\n\n\n 3. Draw a right triangle having hypotenuse of length 5.4 cm, and one of the acute angles of measure 30o<\/sup><\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Let ABC be the right triangle at A such that hypotenuse BC = 5.4 cm. Let C = 30o<\/sup>.<\/p>\n\n\n\n Therefore \u2220A + \u2220B + \u2220C = 180o<\/sup><\/p>\n\n\n\n \u2220B = 180o <\/sup>\u2212 30o<\/sup>\u2212 90o<\/sup> = 60o<\/sup><\/p>\n\n\n\n Steps of construction:<\/p>\n\n\n\n 1. Draw a line segment BC = 5.4 cm.<\/p>\n\n\n\n 2. Draw angle CBY = 60o<\/sup><\/p>\n\n\n\n 3. Draw angle BCX of measure 30o<\/sup> with X on the same side of BC as Y.<\/p>\n\n\n\n 4. Let BY and CX intersects at A.<\/p>\n\n\n\n 5. Then ABC is the required triangle.<\/p>\n\n\n\n 4. Construct a right triangle ABC in which AB = 5.8 cm, BC = 4.5 cm and \u2220C = 90o<\/sup>.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Steps of construction:<\/p>\n\n\n\n 1. Draw a line segment BC = 4.5 cm.<\/p>\n\n\n\n 2. Draw \u2220BCX of measure 90o<\/sup><\/p>\n\n\n\n 3. With center B and radius AB = 5.8 cm, draw an arc of the triangle to intersect ray CX at A.<\/p>\n\n\n\n 4. Join AB to obtain the desired triangle ABC.<\/p>\n\n\n\n 5. ABC is the required triangle.<\/p>\n\n\n\n 5. Construct a right triangle, right angled at C in which AB = 5.2 cm and BC= 4.6 cm.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Steps of construction:<\/p>\n\n\n\n 1. Draw a line segment BC = 4.6 cm.<\/p>\n\n\n\n 2. Draw \u2220BCX of measure 90o<\/sup><\/p>\n\n\n\n 3. With center B and radius AB = 5.2 cm, draw an arc of the triangle to intersects ray CX at A.<\/p>\n\n\n\n 4. Join AB to obtain the desired triangle ABC.<\/p>\n\n\n\n 5. ABC is the required triangle.<\/p>\n\n\n\n RD Sharma Solutions for Class 7 Maths Chapter 17\u2013Constructions<\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/rd-sharma-solutions-for-class-7-maths-chapter-17-constructions-image-1.png\" "\\"RD")
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<\/figure>\n\n\n\nRD Sharma Solutions for Class 7 Maths Chapter 17: Download PDF<\/strong><\/h2>\n\n\n\n