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(Chapter 10) (Practical Geometry) (Class VII) Question 1: Exercise 10.1 Draw a line, say AB, take a point C outside it. Through C, draw a line parallel to AB using ruler and compasses only. Answer 1: To construct: A line, parallel to given line by using ruler and compasses. (a) Draw a line-segment AB and take a point C outside AB. (b) Take any point D on AB and join C to D. (c) With D as centre and take convenient radius, draw an arc cutting AB at E and CD at F. (d) With C as centre and same radius as in step 3, draw an arc GH cutting CD at I. (e) With the same arc EF, draw the equal arc cutting GH at J. (f) Join JC to draw a line l. This the required line AB l. Question 2: Draw a line l. Draw a perpendicular to l at any point on l. On this perpendicular choose a point X, 4 cm away from l. Through X, draw a line m parallel to l. Answer 2: To construct: A line parallel to given line when perpendicular line is also given. (a) Draw a line l and take a point P on it. (b) At point P, draw a perpendicular line n. (c) Take PX = 4 cm on line n. (d) At point X, again draw a perpendicular line m. It is the required construction. Math Class 7 Chapter 10-NCERT Solution 1 Email:info@mywayteaching.com

Question 3: Let l be a line and P be a point not on. l Through P, draw a line m parallel to l. Now join P to any point Q on l. Choose any other point R on m. Through R, draw a line parallel to PQ. Let this meet l at S. What shape do the two sets of parallel lines enclose? Answer 3: To construct: A pair of parallel lines intersecting other part of parallel lines. (a) Draw a line l and take a point P outside of l. (b) Take point Q on line l and join PQ. (c) Make equal angle at point P such that Q = P. (d) Extend line at P to get line m. (e) Similarly, take a point R online m, at point R, draw angles such that P = R. (f) Extended line at R which intersects at S online l. Draw line RS. Thus, we get parallelogram PQRS. Math Class 7 Chapter 10-NCERT Solution 2 Email:info@mywayteaching.com

Exercise 10.2 Question 1: Construct XYZ in which XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm. Answer 1: To construct: XYZ, where XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm. (a) Draw a line segment YZ = 5 cm. (b) Taking Z as centre and radius 6 cm, draw an arc. (c) Similarly, taking Y as centre and radius 4.5 cm, draw another arc which intersects first arc at point X. (d) Join XY and XZ. It is the required XYZ. Question 2: Construct an equilateral triangle of side 5.5 cm. Answer 2: To construct: A ABC where AB = BC = CA = 5.5 cm (a) Draw a line segment BC = 5.5 cm (b) Taking points B and C as centers and radius 5.5 cm, draw arcs which intersect at point A. (c) Join AB and AC. It is the required ABC. Math Class 7 Chapter 10-NCERT Solution 3 Email:info@mywayteaching.com

Question 3: Draw PQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. What type of triangle is this? Answer 3: To construction: PQR, in which PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. (a) Draw a line segment QR = 3.5 cm. (b) Taking Q as centre and radius 4 cm, draw an arc. (c) Similarly, taking R as centre and radius 4 cm, draw an another arc which intersects first arc at P. (d) Join PQ and PR. It is the required isosceles PQR. Question 4: Construct ABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure B. Answer 4: To construct: ABC in which AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. (a) Draw a line segment BC = 6 cm. (b) Taking B as centre and radius 2.5 cm, draw an arc. (c) Similarly, taking C as centre and radius 6.5 cm, draw another arc which intersects first arc at point A. (d) Join AB and AC. (e) Measure angle B with the help of protractor. It is the required ABC where B = 80. Math Class 7 Chapter 10-NCERT Solution 4 Email:info@mywayteaching.com

Question 1: Exercise 10.3 Construct DEF such that DE = 5 cm, DF = 3 cm and m EDF = 90. Answer 1: To construct: DEF where DE = 5 cm, DF = 3 cm and m EDF = 90. (a) Draw a line segment DF = 3 cm. (b) At point D, draw an angle of 90 with the help of compass i.e., XDF = 90. (c) Taking D as centre, draw an arc of radius 5 cm, which cuts DX at the point E. (d) Join EF. It is the required right angled triangle DEF. Question 2: Construct an isosceles triangle in which the lengths of each of its equal sides is 6.5 cm and the angle between them is 110. Answer 2: To construct: An isosceles triangle PQR where PQ = RQ = 6.5 cm and Q = 110. (a) Draw a line segment QR = 6.5 cm. (b) At point Q, draw an angle of 110 with the help of protractor, i.e., YQR = 110. (c) Taking Q as centre, draw an arc with radius 6.5 cm, which cuts QY at point P. (d) Join PR It is the required isosceles triangle PQR. Y P 110 Q R X Math Class 7 Chapter 10-NCERT Solution 5 Email:info@mywayteaching.com

Question 3: Construct ABC with BC = 7.5 cm, AC = 5 cm and m C = 60. Answer 3: To construct: ABC where BC = 7.5 cm, AC = 5 cm and m C = 60. (a) Draw a line segment BC = 7.5 cm. (b) At point C, draw an angle of 60 with the help of protractor, i.e., XCB = 60. (c) Taking C as centre and radius 5 cm, draw an arc, which cuts XC at the point A. (d) Join AB It is the required triangle ABC. Math Class 7 Chapter 10-NCERT Solution 6 Email:info@mywayteaching.com

Question 1: Exercise 10.4 Construct ABC, given m A = 60, m B = 30 and AB = 5.8 cm. Answer 1: To construct: ABC where m A = 60, m B = 30 and AB = 5.8 cm. (a) Draw a line segment AB = 5.8 cm. (b) At point A, draw an angle YAB = 60 with the help of compass. (c) At point B, draw XBA = 30 with the help of compass. (d) AY and BX intersect at the point C. It is the required triangle ABC. Question 2: Construct PQR if PQ = 5 cm, m PQR = 105 and m QRP = 40. Answer 2: Given: m PQR = 105 and m QRP = 40 We know that sum of angles of a triangle is 180. m PQR + m QRP + m QPR = 180 105 40 m QPR = 180 145 + m QPR = 180 m QPR = 180 145 m QPR = 35 To construct: PQR where m P = 35, m Q = 105 and PQ = 5 cm. (a) Draw a line segment PQ = 5 cm. (b) At point P, draw XPQ = 35 with the help of protractor. (c) At point Q, draw YQP = 105 with the help of protractor. (d) XP and YQ intersect at point R. It is the required triangle PQR. Math Class 7 Chapter 10-NCERT Solution 7 Email:info@mywayteaching.com

Question 3: Examine whether you can construct DEF such that EF = 7.2 cm, m E = 110 and m F = 80. Justify your answer. Answer 3: Given: In DEF, m E = 110 and m F = 80. Using angle sum property of triangle D + E + F = 180 Which is not possible. D + 110 + 80 = 180 D + 190 = 180 D = 180 190 = 10 Math Class 7 Chapter 10-NCERT Solution 8 Email:info@mywayteaching.com

Question 1: Exercise 10.5 Construct the right angled PQR, where m Q = 90, QR = 8 cm and PR = 10 cm. Answer 1: To construct: A right angled triangle PQR where m Q = 90, QR = 8 cm and PQ = 10 cm. (a) Draw a line segment QR = 8 cm. (b) At point Q, draw QX QR. (c) Taking R as centre, draw an arc of radius 10 cm. (d) This arc cuts QX at point P. (e) Join PQ. It is the required right angled triangle PQR. Question 2: Construct a right angled triangle whose hypotenuse is 6 cm long and one the legs is 4 cm long. Answer 2: To construct: A right angled triangle DEF where DF = 6 cm and EF = 4 cm (a) Draw a line segment EF = 4 cm. (b) At point Q, draw EX EF. (c) Taking F as centre and radius 6 cm, draw an arc. (Hypotenuse) (d) This arc cuts the EX at point D. (e) Join DF. It is the required right angled triangle DEF. Math Class 7 Chapter 10-NCERT Solution 9 Email:info@mywayteaching.com

Question 3: Construct an isosceles right angled triangle ABC, where m ACB = 90 and AC = 6 cm. Answer 3: To construct: An isosceles right angled triangle ABC where m C = 90, AC = BC = 6 cm. (a) Draw a line segment AC = 6 cm. (b) At point C, draw XC CA. (c) Taking C as centre and radius 6 cm, draw an arc. (d) This arc cuts CX at point B. (e) Join BA. It is the required isosceles right angled triangle ABC. Math Class 7 Chapter 10-NCERT Solution 10 Email:info@mywayteaching.com

(Chapter 10) (Practical Geometry) (Class VII) Questions: Miscellaneous Questions Below are given the measures of certain sides and angles of triangles. Identify those which cannot be constructed and say why you cannot construct them. Construct rest of the triangle. Triangle Given measurements 1. ABC m A = 85 ; m B = 115 ; AB = 5 cm 2. PQR m Q = 30 ; m R = 60 ; QR = 4.7 cm 3. ABC m A = 70 ; m B = 50 ; AC = 3 cm 4. LMN m L = 60 ; m N = 120 ; LM = 5 cm 5. ABC BC = 2 cm; AB = 4 cm; AC = 2 cm 6. PQR PQ = 3.5 cm; QR = 4 cm; PR = 3.5 cm 7. XYZ XY = 3 cm; YZ = 4 cm; XZ = 5 cm 8. DEF DE = 4.5 cm; EF = 5.5 cm; DF = 4 cm Answer 1: In ABC, m A = 85,m B = 115, AB = 5 cm Construction of ABC is not possible because m A = 85 m B = 200, and we know that the sum of angles of a triangle should be 180. Answer 2: To construct: PQR where m Q = 30, m R = 60 and QR = 4.7 cm. (a) Draw a line segment QR = 4.7 cm. (b) At point Q, draw XQR = 30 with the help of compass. (c) At point R, draw YRQ = 60 with the help of compass. (d) QX and RY intersect at point P. It is the required triangle PQR. Math Class 7 Chapter 10-NCERT Solution 11 Email:info@mywayteaching.com

Answer 3: We know that the sum of angles of a triangle is 180. m A + m B + m C = 180 70 50 m C = 180 120 + m C = 180 m C = 180 120 m C = 60 To construct: ABC where m A = 70, m C = 60 and AC = 3 cm. (a) Draw a line segment AC = 3 cm. (b) At point C, draw YCA = 60. (c) At point A, draw XAC = 70. (d) Rays XA and YC intersect at point B It is the required triangle ABC. Answer 4: In LMN, m L = 60, m N = 120, LM = 5 cm This LMN is not possible to construct because m L + m N = 60 120 180 which forms a linear pair. Answer 5: ABC, BC = 2 cm, AB = 4 cm and AC = 2 cm This ABC is not possible to construct because the condition is Sum of lengths of two sides of a triangle should be greater than the third side. AB < BC + AC 4 < 2 + 2 4 = 4, Math Class 7 Chapter 10-NCERT Solution 12 Email:info@mywayteaching.com

Answer 6: To construct: PQR where PQ = 3.5 cm, QR = 4 cm and PR = 3.5 cm (a) Draw a line segment QR = 4 cm. (b) Taking Q as centre and radius 3.5 cm, draw an arc. (c) Similarly, taking R as centre and radius 3.5 cm, draw an another arc which intersects the first arc at point P. It is the required triangle PQR. Answer 7: To construct: A triangle whose sides are XY = 3 cm, YZ = 4 cm and XZ = 5 cm. (a) Draw a line segment ZY = 4 cm. (b) Taking Z as centre and radius 5 cm, draw an arc. (c) Taking Y as centre and radius 3 cm, draw another arc. (d) Both arcs intersect at point X. It is the required triangle XYZ. Math Class 7 Chapter 10-NCERT Solution 13 Email:info@mywayteaching.com

Answer 8: To construct: A triangle DEF whose sides are DE = 4.5 cm, EF = 5.5 cm and DF = 4 cm. (a) Draw a line segment EF = 5.5 cm. (b) Taking E as centre and radius 4.5 cm, draw an arc. (c) Taking F as centre and radius 4 cm, draw an another arc which intersects the first arc at point D. It is the required triangle DEF. Math Class 7 Chapter 10-NCERT Solution 14 Email:info@mywayteaching.com

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