Practical Geometry

10 11 12 l 9 2 10 Construction of a pair of parallel lines using set squares and ruler 1 2 3 4 5 6 7 8 Construction of parallel lined by ruler 1 5 2 6 • 0 cm 1 construction 2 4 Side-Side -Side (SSS) Triangle 5 Triangle construction (Possibilities) 4 • • 3 6 & compasses 7 3 • P 1 • 13 14 CONTENTS 0 15 PRACTICAL GEOMETRY Side-Angle-Side (SAS) Triangle 14 13 1 10 11 12 l Q 4 P 3 CONSTRUCTION OF A PAIR OF PARALLEL LINES USING SET SQUARES AND RULER 4 5 6 7 2 3 when its hypotenuse and one side are given 1 2 3 4 5 6 7 9 Construction of a right angled triangle Q 2 • 1 construction P 8 Angle-Side-Angle (ASA) Triangle 0 • 15 construction l 6 0 cm 1 2 5 Construct a line parallel to a given line l passing through a given point P not on the line : (i) Draw a line l. (ii) Place a set square with the arm of its right angle. (iii) Holding the set square fixed, place a ruler along the other arm of the right angle. (iv) Holding the ruler fixed, slide the set square along the edge of the ruler until the perpendicular edge of the set square passes through the given point P as shown in figure (v) Keeping the set square fixed at same position, draw a line PQ through the point P along the edge of set square. The line PQ is the required line through the point P and parallel to the line l. Construction of a parallel line to a given line m and at a given distance (d) from it : (i) Draw a line m and mark a point P on line m. (ii) Using a set square, construct a line l perpendicular to m at P. (iii) With a ruler or compasses mark a point Q on l such that the segment PQ = d cm. (iv) Using set square construct a line n perpendicular to l at Q. Then, n is the required line parallel to m at a distance of d cm from it. S.No. Steps of Construction Construction 1. Draw a line segment l and mark a point A not lying on it. A 2. 3. 2 3 4 5 6 7 4. 1 1 4 5 3 4 5 3 5. 6 2 6 7 2 8 9 10 11 15 12 13 14 13 14 0 15 l 6. 9 2 10 11 12 1 n Q 8 1 2 3 4 With B as centre and convenient radius draw an arc cutting l at C and AB to D Now with A as centre and the same radius as in step 3 draw an arc EF cutting AB at G m 0 cm 1 1 2 0 A Take any point B on l and join B to A. l B l P l 5 6 7 Place the metal point of the compasses at C and adjust the opening so that the pencil point is at D. D BC l F A G E D C B l F A G E D BC l F H A E G D BC l 7 3 With the same opening as in step 5 and with G as centre draw another arc cutting the arc EF at H. A l 4 1 4 5 6 7. 3 2 2 5 6 0 cm 1 m l Q 90° d 90° P Now join AH and draw a line m. n Now we will show how to construct a line parallel to a given line passing through a given point not lying on it. Given : A line l and a point A not lying on it. To Const. : A line parallel to l through A. m D BC m CONSTRUCTION OF PARALLEL LINES BY RULER AND COMPASSES FA H E G l Then, m is the required line parallel to l and passing through the given point A. Note : In the above figure, l and m are two parallel lines and AB is a transversal. ∠ABC and ∠FAB are alternate interior angles But ∠ABC = ∠FAB [By construction] Hence, l || m [Q alternate interior angles are equal] Alternative Method S.No. Steps of When length of the three sides of a triangle are given, we follow the following steps for constructing a triangle : Construction Construction 1. Draw a line l. 2. Mark a point A which is not lying on l. 3. l l A C l 2. At point A, draw an angle such that it is equal to ∠ABC. A B 5. 1. n Draw a line n, through A which meets the line l at point B Draw line m which is parallel to line l. C A B l m C 3. l TRIANGLE CONSTRUCTION (POSSIBILITIES) There are following possibilities of constructing a triangle : (1) When its three sides are given. (Also known as SSS triangle construction). (2) When its two sides and the included angle is given. (Also known as SAS triangle construction). (3) When its two angles and the included side is given. (Also known as ASA triangle construction). C • The sum of two sides of a triangle is always greater than the third side. • The sum of three angles of a triangle should be equal to 180°. SIDE-SIDE-SIDE (SSS) TRIANGLE CONSTRUCTION Steps of Construction Construction A B 4. S.No. 4. Draw a line segment (with the help of ruler), of a given length equal to one side of a triangle. Let us name it as AB. From one end point A, draw an arc whose distance from A is equal to second side. From second end point B, draw another arc whose distance from B is equal to third side and which cuts the first arc at a point C. Join BC. AC A B 5 cm A B 5 cm C A and B 5 cm C 6 cm 7 cm A 5 cm B The triangle so obtained is the required triangle. It is important to note that the sum of two sides of a triangle is always greater than the third side. Thus we can say that we cannot construct a triangle when the sum of any two sides is less than or equal to the third side. EXAMPLES Ex.1 Construct a triangle ABC such that side AB = 5 cm, BC = 6 cm and AC = 7 cm. sum of any two sides of a triangle is always greater than third side but in the given triangle, the sum of the lengths of the two sides AB (2 cm) and BC (3 cm) is less than the length of the third side AC (6 cm). Sol. C A 1. 2. 3. 4. Ex.2 AB + BC < AC 6 cm 7 cm Therefore, the three given sides of the triangle do not satisfy the triangle inequality. Hence, it is not possible to construct the given triangle. B 5 cm Draw a line segment AB of length 5 cm. With centre A and radius 7 cm draw an arc of the circle. With centre B and radius 6 cm draw another arc intersecting first arc at C. Join AC and BC as shown in figure. ∆ABC is the required triangle. Construct a triangle ABC in which AB = 4.5 cm, AC = 6 cm, BC = 5 cm. C SIDE-ANGLE-SIDE (SAS) TRIANGLE CONSTRUCTION When the length of two sides and the measure of included angle is given, we follow the following steps for constructing a triangle. S.No. Sol. Steps of Construction Construction C 1. 6 cm A 1. 2. 3. 4. Sol. B 4.5 cm Draw a line segment AB of length 4.5 cm. With centre A and radius 6 cm draw an arc of the circle. With centre B and radius 5 cm draw another arc intersecting the first arc at C. Joint AC and BC as shown in figure. ∆ABC is the required triangle. Construct a triangle ABC where AB = 2 cm, BC = 3 cm and AC = 6 cm. Before constructing the required triangle, let us first draw a rough sketch of the triangle. 2. Draw a line segment AB of the triangle with help of ruler. A 5 cm O Draw ∠OBA of measure equal to the given ∠B. 30° 5 cm A 3. O From any point on BO , cut off line segment equal to second side. Mark the cut off point as C. B B C 3 cm 30° B 5 cm A C 4. 6c m Ex.3 5 cm A 2 cm 3 cm Join AC. The triangle so obtained is the required triangle. B By plotting the rough sketch, we find that it is not possible to construct this triangle. Because the EXAMPLES O A C 3 cm 30° B 5 cm Ex.4 Construct ∆ABC, where AB = 5 cm, BC = 3 cm and ∠ABC = 70°. O Ex.6 Sol. C Draw a triangle ABC in which BC = 6 cm, ∠B = 35°, ∠C = 100°. P Q A 3 cm Sol. 70° B 5 cm Draw AB = 5 cm and ∠ABC = 70°. A 1. Ex.5 Then, ∆ABC is the required triangle. It is important to note that 1. 50° 2. Cut off DF = 4 cm from DO . 3. Join EF. ∆DEF is the required triangle. ANGLE-SIDE-ANGLE (ASA) TRIANGLE CONSTRUCTION When the measure of two angles and length of included side is given, we follow the following steps for constructing a triangle : S.No. Steps of Construction Construction 1. Draw the given A B 4 cm line segment say AB O 2. Draw ∠OAB of measure equal to ∠A. 40° B A 4 cm O 3. Draw ∠PBA of P measure equal to C ∠B. Let the ray 50° B AO and Ray BP A 40° 4 cm intersect at C. Then ∆ABC is the required triangle. EXAMPLES Ex.7 Sol. Construct a ∆ABC in which ∠A = 100°, ∠B = 20° and CA = 5.4 cm. Q O B ° 20 C ∠B + ∠C = 35° + 100° = 135° Which is less than 180°. In other words, the sum of two angles of a triangle is always less than 180°. 4 cm E 5 cm Draw DE = 5 cm and ∠EDF = 50°. The two rays BQ and CP intersect at A. 2. F D 4. 1. 2. Cut off BC = 3 cm from BO . 3. Join AC. ∆ABC is the required triangle. Construct a ∆DEF in which DE = 5 cm, DF = 4 cm, and ∠D = 50°. O Sol. 3. 35° 100° C 6 cm Draw BC = 6 cm as shown in figure. Draw ∠CBQ = 35° Draw ∠BCP = 100°. B A 100° 5.4 cm C 1. Draw AC = 5.4 cm as shown in figure 2. 3. Draw ∠CAQ = 100°. We know the sum of three angles in a triangle = 180° So, ∠C = 180° – (∠A + ∠B) = 180° – (100 + 20) = 180° – 120° = 60° 4. 5. Draw ∠ACO = 60°. The two rays AQ and CO intersect at B. Then ∆ABC is the require triangle. Note : We cannot construct a triangle if the sum of two angles is equal to 180º or greater than 180º. CONSTRUCTION OF RIGHT ANGLED TRIANGLE WHEN ITS HYPOTENUSE AND ONE SIDE ARE GIVEN We follow the following steps for RHS triangle construction : S.No. Steps of Construction Construction B 1. Draw a line A 4 cm segment of a given length. 2. Draw an angle of 90° on one end of the segment. A 3. 4. Ex.8 With other end as centre and radius equal to the hypotenuse draw an arc intersecting the perpendicular line at C. Join the intersecting point with the other end of the line segment. The triangle so obtained is the required triangle. 4 cm X Sol. P 6 cm Q Draw a line segment QR = 4 cm. 2. 3. At Q draw ∠RQX = 90°. With R as the centre and radius PR = 6 cm (i.e., hypotenuse) draw an arc to intersect ray B QX at P. C Ex.10 90° 4 cm Join PR as in figure, Thus, ∆PQR is the required triangle. Construct an isosceles right triangle ABC in which ∠C = 90° and AC = BC = 5 cm. B X Sol. A 5 cm C A 90° 4 cm 90° B 5 cm B 5 cm 1. Draw a line segment BC of length = 5 cm. 2. 3. Draw ∠C = 90°. With centre C and radius equal to 5 cm, draw 4. Ex.11 Join BA. Thus, ∆ABC is the required triangle. Draw a right triangle whose hypotenuse is of length 5 cm and one side of length 3 cm. O C 1. 2. 3. 4. 90° 3.8 cm X C Sol. 4.2 cm A C an arc intersecting ray CX at A. EXAMPLES Construct a right triangle when its hypotenuse is of length 4.2 cm and one of its side is of length 3.8 cm. Sol. R 4 cm 1. 4. A 90° 5 cm B Draw a line segment of length 3.8 cm, as shown in figure. Draw ∠BAO of measure 90°. With centre B and radius equal to 4.2 cm, draw an arc intersecting ray AO at C. Join BC. Thus, ∆BAC is the required triangle. A 90° 3 cm B 1. Draw a line segment AB = 3 cm as in figure. 2. 3. Draw ∠A = 90°. With centre B and radius equal to 5 cm (hypotenuse) draw an arc intersecting ray AX at C. 4. Join BC. Thus, ∆BAC is the required triangle. Ex.9 Construct a right triangle PQR in which ∠Q = 90°, PR = 6 cm and QR = 4 cm.