Example-2 In ∆ABC, ∠A = 60o , ∠B = 90o . Determine the longest and shortest sides. Solution: In ∆ABC, ∠A = 60o , ∠B = 90o . Since ∠A + ∠B + ∠C =180o , we have ∠C =180o – ∠A – ∠B = 180o – 60o – 90o = 30o . Here, ∠B is the greatest angle of ∆ABC, so its opposite side AC is the longest and ∠C is the smallest angle, so its opposite side AB is the shortest. PRACTICE 13.3 1. (a) What is the relation of the sum of any two sides of triangle and its third side? (b) What is the relation between b + c and a in the given triangle? (c) Which is the longest side in the given triangle? (d) Which is the shortest side in the given triangle PQR? (e) Which is the greatest angle in the given triangle LMN? 2. (a) In ∆ABC ∠A = 72o and ∠B = 32o . Find the longest and shortest sides. (b) In ∆PQR ∠Q = 65o and ∠R = 42o . Find the longest and shortest sides. (c) In ∆PQR ∠Q = 90o and ∠R = 25o . Find the longest and shortest sides. 3. (a) In ∆PQR PQ = 6 cm and QR = 8 cm and RP = 7 cm. Determine the greatest and the smallest angles. (b) In ∆XYZ, XY = 5 cm, XZ = 13 cm and YZ = 12 cm. Determine the greatest and the smallest angles. (c) In ∆ABC, AB = 15 cm, AC = 5 cm and BC = 18 cm. Determine the greatest and the smallest angles. 4. (a) In an equilateral triangle, any two sides are together greater than the third side. Verify it by experimental method. (b) In a right-angled triangle, the hypotenuse is the greatest side. Verify it experimentally. 5. (a) Show by measurement that the sum of the medians of a triangle islessthan the perimeter of the triangle. (b) Show by measurement that the perimeter of a quadrilateral is greater than the sum of its diagonals. 2. (a) AB, AC (b) QR, PQ (c) PR, PQ 3. (a) ∠P, ∠R (b) ∠Y, ∠Z (c) ∠A, ∠B Answers B 60° 90° C A Fig. 1(b) Fig. 1(c) a b c B 55° 88° 37° C A Fig. 1(d) Fig. 1(e) Q R P 80° 60° M N L 5 cm 7 cm 6 cm 300 Allied The Leading Mathematics-9 Triangles 301 Geometry Geometry
13.4 Similar Triangles At the end of this topic, the students will be able to: ¾ Define similarity and prove the theorems related to the conditions for similarity of two triangles. Learning Objectives Similarity is one of the simplest and most common geometrical concept that is easily comprehended by everybody. The actual Dharahara and the model Dharahara are the familiar example of similarity. Most of the manufacturing companies and engineers design the models of production items at first and they produce in large quantities. Geometric figures are of different shape and size. Those which are alike in both shape and size, are known as congruent figures. If they are alike but not necessarily equal in size, they are said to be similar. Certain geometrical figures such as circles, squares, equilateral triangles, regular hexagons differ from geometric figures such as ellipses, rectangles, isosceles triangles, etc. We note that all the circles are similar but not all the ellipses. All the squares are similar but not all the rectangles. All equilateral triangles are similar but not all the isosceles triangles. Ample examples of similar figures can be found in everyday life. Some of them are shown below: Similar and dis-similar shapes Square Rectangle Equilateral triangle Isosceles triangle Cube Cuboid Actual Dharahara Model Dharahara 302 Allied The Leading Mathematics-9 Triangles 303 Geometry Geometry
Circle Ellipse To talk about congruence or similarity means to establish how certain parts of a figure are related to certain parts of another figure and conversely. In other words, congruence or similarity between two figures is a one-to-one correspondence between the parts of the two figures. Two polygons are said to be similar to each other, if; i) their corresponding angles are equal, and ii) the lengths of their corresponding sides are proportional. If two polygons ABCDE and A'B'C'D'E' are similar and the correspondence between the vertices are as shown below, A B C D E A' B' C' D' E' We write : Poly (ABCDE) ~ Poly (A'B'C'D'E') Here, the symbol ~ (read ‘tilde’) stands for ‘is similar to’. Since in each regular polygon, the sides as well as angles are all equal, the set of all regular triangles (i. e., all equilateral triangles), the set of all regular quadrilaterals (i.e., all squares), the set of all regular pentagons, the set of all regular hexagons, etc. are separately similar. In what follows, we shall be mainly concerned with similar triangles and properties of similar triangles. Triangles are special kinds of polygons. In the case of triangles, for similarity the two conditions are sufficient but not necessary. We can prove that once one of the two conditions is satisfied the other is automatically satisfied. We can therefore say: Two triangles are said to be similar if three angles of one are equal to the corresponding three angles of the other or if the corresponding sides are proportional. D E D' E' C' A' B' C B A Fig. (a) Fig. (b) 302 Allied The Leading Mathematics-9 Triangles 303 Geometry Geometry
Look at the following pairs of triangles. B C A 60° 90° 30° E F D 60° 90° 30° Here, corresponding angles are equal. B 1 cm C 2 cm 2.5 cm A 4 cm 5 cm E 2 cm F D Here, the corresponding sides are proportional, i.e., AB : DE = BC: EF = CA: FD = 1:2. Statement: If two triangles are equiangular, then they are similar. In other words, two equiangular triangles are similar. *THEOREM 13.12 (AAA Similarity Condition) Since its theoretical proof is not necessary in this section, we present here its experimental verification. Experimental Verification Step 1: Two different pairs oftriangles, each labeled asABCand DEF are drawn in such a way that ∠A = ∠D, ∠B = ∠E, and ∠C = ∠F. B C A E F D B C A E F D Fig. (i) Fig. (ii) Step 2: The corresponding sides of each pair of triangles are measured (in cm) and tabulated and then the ratios of each pair of corresponding sides are calculated as shown below: Fig. In ∆ABC In ∆DEF Ratio of corresponding sides Remarks AB BC AC DE EF DF AB DE BC EF AC DF (i) 1.3 1 1.7 1.7 1.2 2.1 0.8 0.8 0.8 AB DE = BC EF = AC DF (ii) 1.6 1.5 1.3 2 1.9 1.7 0.8 0.8 0.8 AB DE = BC EF = AC DF 304 Allied The Leading Mathematics-9 Triangles 305 Geometry Geometry
Conclusion: ‘The corresponding sides of equiangular triangles are proportional.’ Hence, the equiangular triangles are similar. Note: In what follows, students are advised to draw figures, measure the angles and verify them. Statement: If the corresponding sides of two triangles are proportional, then the triangles are similar. *THEOREM 13.13 (SSS Similarity Condition) Since its theoretical proof is not necessary in this section, we present here its experimental verification. Experimental Verification Step 1: Two different pairs of triangles, each labeled as ABC and DEF are drawn in such a way that AB : DE = BC : EF = CA : FD. Note: Sides of the triangles may be like (i) 1.9 cm, 2 cm, 2.1 cm and 1.5 cm, 1.6 cm, 1.7 cm; (ii) 1.8 cm, 1.3 cm, 2.2 cm and 1.3 cm, 0.9 cm, 1.6 cm; B C A E F D E F D B C A Fig. (i) Fig. (ii) Step 2: The corresponding angles of each pair of triangles are measured and are shown below: Fig In ∆ABC In ∆DEF Remarks ∠A ∠B ∠C ∠D ∠E ∠F (i) 58° 66° 56° 58° 66° 56° ∠A = ∠D, ∠B = ∠E, ∠C = ∠F (ii) 36° 90° 54° 36° 90° 54° ∠A = ∠D, ∠B = ∠E, ∠C = ∠F Conclusion: ‘If the corresponding sides of two triangles are proportional, then the triangles are equiangular’; and hence they are similar. Statement: If two triangles have one angle of the one triangle equal to one angle of the other, and the sides about them are proportional, then the triangles are similar. *THEOREM 13.14 (SAS Similarity Condition) 304 Allied The Leading Mathematics-9 Triangles 305 Geometry Geometry
Experimental Verification Step 1: Two different pairs of triangles, each labeled as ABC and DEF are drawn in such a way that AB:DE = BC: EF and∠B = ∠E. Note: Two adjacent sides of the triangles may be like (i) 2.8 cm, 3.6 cm and 2.1 cm, 2.7 cm ; (ii) 3.2 cm, 2.4 cm and 2.3 cm, 1.8 cm; Constructing any two equal angles as per the above given arms. B A 3.6 cm C 70° 2.8 cm E D 2.7 cm F 2.1 cm 70° B 2.4 cm 3.2 cm C A E F D2.3 cm 1.8 cm Fig. (i) Fig. (ii) Step 2: The remaining pairs of corresponding angles and one pair of corresponding sides of the triangles are measured and tabulated as shown below: Fig. Corresponding angles Corresponding sides Remarks ∠A ∠D ∠C ∠F AC DF AC DF (i) 65° 65° 45° 45° 3.7 2.8 4:3 ∠A = ∠D, ∠C = ∠F, AC DF = AB DE = BC DF (ii) 37° 37° 53° 53° 3.9 2.9 4:3 ∠A = ∠D, ∠C = ∠F, AC DF = AB DE = BC DF Here, two remaining angles A and C are respectively equal to the angles D and F; and the ratio of the remaining sides AC and DF is the same as the ratio of AB and DE. Conclusion: ‘If two triangles have one angle of the one equal to one angle of the other and the sides about them are proportional, then the triangles are similar.’ Before we conclude this section with the following obvious remark: If two angles of a triangle are respectively equal to two angles of the other, the two triangles are similar. 306 Allied The Leading Mathematics-9 Triangles 307 Geometry Geometry
Example-1 Find the values of x and y in the given similar triangles ABC and PQR: Solution: Since ∆ABC ~ ∆PQR so, AB PQ = BC QR = AC PR or, 2 cm 3 cm = x 4.5 cm = 4 cm y Taking the first and the second ratios, we get Taking the first and the third ratios, we get 2 cm 3 cm = x 9 cm 2 cm 3 cm = 4 cm y ∴ x = 6 cm ∴ y = 6 cm. Example-2 In the adjoining ∆CDE, ∠ABD = ∠DCE, AD = 7cm, AB = 5cm, CE = 1.5 cm and BE = 4.2 cm. Find the length of BD and AC. Solution: From the figure, (i) In ∆ADB and ∆CDE, (a) ∠ABD = ∠DCE (A) [ By given] (b) ∠ADB = ∠CDE (A) [ Common angles] (c) ∠BAD = ∠CED (A) [ Remaining angles] (ii) ∆ADB ~ ∆CDE [ By AAA similarity condition] (iii) BD CD = AD DE = AB CE [ Corresponding sides of similar triangles] or, BD 7 + AC = 7 BD + 4.2 = 5 7.5 (iv) 7 BD + 4.2 = 5 7.5 [ Taking the 2nd and the 3rd ratios] or, 5BD + 21 = 52.5 or, 5BD = 52.2 – 21 or, BD = 31.5 5 = 6.3 cm (v) BD 7 + AC = 5 7.5 [ Taking the 1st and the 3rd ratios] or, 6.3 7 + AC = 5 7.5 or, 35 + 5AC = 47.25 or, 5AC = 47.25 – 35 ∴ AC = 12.25 5 = 2.45 cm Q R P 3 cm 4.5 cm y B x 2 cm 4 cm C A D B 5 cm 7 cm 4.2 cm 7.5 cm E A C 306 Allied The Leading Mathematics-9 Triangles 307 Geometry Geometry
Example-3 From the given figure, prove that: (i) PQ2 = PR.PS and (ii) PQ.QR = PR.QS Solution: From the given figure, 1. In ∆PQR and ∆PQS, (i) ∠PQR = ∠PSQ [ Both right angles] (ii) ∠QPR = ∠QPS [ Common angles] (iii) ∠PRQ = ∠PQS [ Remaining angles] 2. ∆PQR ~ ∆PQS [ By AAA similarity condition] 3. PQ PS = PR PQ = QR QS [ Corresponding sides of similar triangles] 4. PQ PS = PR PQ [ Taking the 1st and the 2nd ratios] or, PQ2 = PR.PS 5. PR PQ = QR QS [ Taking the 2nd and the 3rd ratios] or, PQ.QR = PR.QS Proved. PRACTICE 13.4 Keeping Skill Sharp Circle the correct answer. 1. (a) By which condition, the given triangles ABC and DEF are similar? (i) ASA (ii) AAS (iii) AAA (iv) SSS (b) Which is the name of similar condition for ∆PQR and ∆LMN where ∠Q = ∠M and PQ QR = LM MN. (i) ASA (ii) SAS (iii) SSS (iv) AAA (c) What is the similar condition for ∆ABC and ∆PQR when AB QR = BC RP = AC PQ? (i) SSS (ii) SSA (iii) AAA (iv) SAS (d) By which condition ∆PQR and ∆PQS are similar? (i) SSA (ii) AAA (iii) SSS (iv) ASA Q S R P E F D B C A M N L Q R P B C A Q R P Q S R P 308 Allied The Leading Mathematics-9 Triangles 309 Geometry Geometry
2. (a) By which axiom the triangles ABC and PQR are similar? (b) By which condition the adjoining triangles are similar? (c) By which condition the given triangles are similar? 3. (a) The given triangles ABC and PQR are similar. Write a pair of two corresponding angles, other than give angles. (b) If ∆XYZ ~ ∆PQR, write a pair of two corresponding sides. Check Your Performance 4. Solve the following numerical problems: (a) In the following figure AB//DE, find the values of x and y. (b) In the adjoining figure, AC//BD, find the values of x and y. (c) In the figure. DE//BC, find the values of x and y. (d) In the figure, DA//BC and ∠DBA = ∠ACB. Find length of DA and DB. 5. In the following figures, prove that ∆ABE ~ ∆CDE: (a) (b) (c) 6. In the following figures, prove that ∆BAC ~ ∆BDA: (a) (b) (c) 50° 65° 50° 65° 9 cm 6 cm B C 55° A 6 cm 4 cm Q R 55° P Q. No. 2(a) Q. No. 2(b) Q R P Y Z X Q. No. 4(a) Q. No. 4(b) C E 4 cm 12 cm y x B D A 5 cm 13 cm O C A D B x 6 cm 10 cm y 3 cm 5 cm Q. No. 4(d) B C 3 cm D A 4 cm 3.5 cm Q. No. 4(c) B C 4 cm D E A 5 cm y2cm x 7 cm E B A C D E B A C D B D C A E A B D C A D C B B D C A Q. No. 2(c) Q. No. 3(a) 10 cm 14 cm 12 15 cm cm 21 cm 18 cm 2 cm Q 3 cm R P B 4 cm 6 cm C A 308 Allied The Leading Mathematics-9 Triangles 309 Geometry Geometry
7. (a) In the trapezium ABCD, AB//DC and the diagonal AC and BD intersect each other at E. Prove that AB:BE = CD:DE. (b) In a triangle ABC, X and Y are any points on AB and AC respectively and XY//BC. Prove that AX:AB = AY:AC. 8. (a) In ∆ABC, ∠A = 90o and AD ⊥ BC. If BC = a, AC = b, AB = c, BD = x and DC = a – x, prove that a2 = b2 + c2 . (b) In the following figures, Quad. ABCD ~ Quad. EFGH. Find the length of AC, HF, AD and GH. 9. (a) Observe the adjoining picture. Find the height of the window of the tree house. (b) Observe the adjoining picture. Find the height of the building. (c) Observe the adjoining picture at the same time. Find the height of the tree. 4. (a) 4.33 cm, 1.67 cm (b) 6 cm, 10 cm (c) 3.75 cm, 2.67 cm (d) 2.57 cm, 3.43 cm 8. (a) 2.5 cm, 6 cm, 2.5 cm, 4 cm 9. (a) 9 ft (b) 55 ft (c) 37.5 m Answers Project Work Collect the different shapes of triangles and quadrilaterals. Separate the similar triangles and quadrilaterals by using similar conditions. Prepare a report and present it in your classroom. D C E A B A D c x y b a – x B C a C A 1 cm 3 cm 1.5 cm 2 cm B D F E H G 5 cm 2 cm 5 cm 3 cm 5 40 ft ft 72 ft A B F E C 20 ft 200 ft 5.5 ft Mirror 50° 50° 2 m 1.5 m 50 m 310 Allied The Leading Mathematics-9 Triangles PB Geometry Geometry
Geometry Geometry 14.1 Properties of Parallelogram At the end of this topic, the students will be able to: ¾ prove the theorems related on the properties of parallelogram by experimental and theoretical. Learning Objectives Two points in a plane can be joined by a line-segment. Three collinear points (i.e. points in the same straight line can be joined by two line-segments. But three non-collinear (i.e. points not in the same straight line), need three line-segments for joining them two by two. Such three line-segments enclose a certain region called triangular region in the plane. Going further ahead, if we have four noncollinear points, four line-segments are required to join them two by two, it forms a quadrilateral region. Even then none of them represents a quadrilateral. (a) B (AB, BC, CD, AC) C A D (b) (AB, BC, CD, DA) B C D A (c) B D C A (AB, BC, BD, AD) (d) C A D B (AB, CD, BC, AD) We have discussed in detail the various types of quadrilaterals. They may be equiangular (i.e., all angles equal) or regular (i.e., with all sides and angles equal). They may have no side parallel and one or more pairs of sides parallel. (b) Trapezium and Parallelogram A quadrilateral having one pair of parallel sides is called a trapezium and that with both pairs of opposite sides parallel is called a parallelogram. Parallelograms are classified both on the basis of sides and angles. They are as follows: (i) Rectangle A rectangle is a parallelogram in which one of the angles is a right angle. ABCD is a parallelogram in which ∠D = 90°. Hence, ABCD is a rectangle. We can prove that diagonals AC and BD are equal and bisect each other at O. A D B C D C A B D A C B A D B C O CHAPTER 14 PARALLELOGRAM PB Allied The Leading Mathematics-9 Parallelogram 311
Geometry Geometry (ii) Square A square is a rectangle in which adjacent sides are equal. In ABCD is a rectangle, with adjacent sides AB and BC equal. Hence ABCD is a square. Here also diagonals AC and BD are equal, but they bisect each other at right angles at O. (iii) Rhombus A rhombus is a parallelogram whose adjacent sides are equal. In ABCD is a parallelogram with equal adjacent (or consecutive) sides AB and BC. So, ABCD is a rhombus. Diagonals AC and BD bisect each other at right angles at O. Statement: The line segments joining the ends on the same side of equal and parallel line segments are equal and parallel. THEOREM 14.1 Given: Line segments AB and CD are equal and parallel in which A and C are on the same side of the line segment joining BD. To prove: AC = BD, AC//BD Construction: Join A and D. Proof: Statements Reasons 1. In ∆ABD and ∆ACD, (i) AB = CD (S) (ii) ∠BAD = ∠CDA (A) (iii) AD = AD (S) 1. (i) By given (ii) Alternate angles on AB//CD, (iii) Common side 2. ∆ABD ≅ ∆ACD 2. SAS congruent condition 3. (i) AC = BD (ii) ∠ADB = ∠CAD 3. (i) Corresponding sides of congruent triangles (i) Corresponding angles of congruent triangles 4. AC//BD. 4. Being equal alternate angles Hence, it is proved. Corollary: The line segments joining the ends on the opposite side of equal and parallel line segments bisect each other. This corollary can be theoretically proved by the congruence of ∆AOB and ∆COD using ASA congruent condition (∠A = ∠D, AB = CD, ∠B = ∠C) and we obtain AO = DO and BO = CO. B C O A D O D A C B A B C D O A B C D 312 Allied The Leading Mathematics-9 Parallelogram 313
Geometry Geometry Statement: The opposite sides and angles of a parallelogram are equal. THEOREM 14.2 Given: ABCD is a parallelogram in which AD//BC and BA//CD. To prove: (i) AB = DC, AD = BC (ii) ∠A = ∠C, ∠B = ∠D Construction: Join A and C. Proof: Statements Reasons 1. In ∆ABC and ∆ADC, (i) ∠BAC = ∠DCA (A) (ii) AC = AC (S) (iii) ∠ACB = ∠DAC (A) 1. (i) Being are alternate angles on AB//DC (ii) Common side (iii) Being are alternate angles on AD//BC 2. ∆ABC ≅ ∆ADC 2. ASA congruent condition 3. AB = DC, BC = AD 3. Corresponding sides of congruent triangles 4. ∠ABC = ∠ADC 4. Corresponding angles of congruent triangles 5. ∠BAC + ∠DAC = ∠ACB + ∠DCA 5. Adding from (i) and (iii) of statement (1) 6. ∠DAB = ∠BCD 6. Whole part axiom Hence, the theorem is completed. Statement: A quadrilateral is a parallelogram if its opposite sides or angles are equal. THEOREM 14.3 (Converse of Theorem 14.2) Given: ABCD is a quadrilateral in which AB = CD and AD = BC. To prove: ABCD is a parallelogram. Construction: Join A and C. Proof: Statements Reasons 1. In ∆ABC and ∆ADC, (i) AB = CD (S) (ii) AC = AC (S) (iii) AD = BC (S) 1. (i) By given (ii) Common side (iii) By given 2. ∆ABC ≅ ∆ADC 2. SSS congruent condition 3. ∠BAC = ∠ACD and ∠ACB = ∠ CAD 3. Corresponding angles of congruent triangles 4. AB//CD and AD//BC 4. Equal alternate angles in statement (3) 5. ABCD is a parallelogram. 5. Parallel line segments in statement (4) Hence, it is proved. B A C D D A C B 312 Allied The Leading Mathematics-9 Parallelogram 313
Geometry Geometry Statement: The diagonals of a parallelogram bisect each other. THEOREM 14.4 Given: In the parallelogram ABCD, the diagonals AC and BD intersect at O. To prove: AO = CO, BO = DO. Proof: Statements Reasons 1. In ∆AOB and ∆DOC, (i) ∠ABO = ∠CDO (A) (ii) AB = DC (S) (iii) ∠BAO = ∠DCO (A) 1. (i) Alternate angles on AB//DC (ii) Opposite sides of parallelogram (iii) Alternate angles on AB//DC 2. ∆ABO ≅ ∆DOC 2. ASA congruent condition 3. AO = CO, BO = DO 3. Corresponding sides of congruent triangles Hence, it is proved. Statement: If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram. THEOREM 14.5 (Converse of Theorem 14.4) Given: In the quadrilateral ABCD the diagonals AC and BD bisect each other at O. i.e. AO = CO and BO = DO. To prove: ABCD is a parallelogram. Proof: Statements Reasons 1. In ∆AOB and ∆DOC (i) AO = CO (S) (ii) ∠AOB = ∠DOC (A) (iii) BO = DO (S) 1. (i) By given (ii) Vertically opposite angles (iii) By given 2. ∆AOB ≅ ∆DOC 2. SAS congruent condition 3. ∠ABO = ∠CDO and ∠BAO = ∠DCO 3. Corresponding sides of congruent triangles 4. AB//DC, AD//BC 4. Alternate angles equal in statement (3) 5. ABCD is a parallelogram. 5. Opposite sides are parallel in statement (4) Hence, it is proved. D O A C B D O A C B 314 Allied The Leading Mathematics-9 Parallelogram 315
Geometry Geometry Example-1 Observe the given parallelogram ABCD. (a) Write the relation between the opposite sides of a parallelogram. (b) Find the value of x. (c) Find the length of AB. Solution: (a) The opposite sides of a parallelogram are equal. (b) In the given parallelogram ABCD, AB = 5x – 3 cm, CD = 2x + 6 cm Now, we have AB = CD [ Opposite sides of ABCD] or, 5x – 3 = 2x + 6 or, 5x – 2x = 6 + 3 or, 3x = 9 or, x = 9 3 = 3 (c) The length of AB = 5x – 3 = 5 × 3 – 3 = 15 – 3 = 12 cm Hence, the length of AB is 12 cm. Example-2 In the given PQRS, ∠PQR = 3x – 5° and ∠PSR = x + 35°. (a) Find the value of x. (b) Find the measure of ∠QRT. Solution: (a) In the given PQRS, ∠PQR = 3x – 5°, ∠PSR = x + 35°. Now, we have (i) ∠PQR = ∠PSR [ Opposite angles of PQRS] or, 3x – 5° = x + 35° or, 3x – x = 35° + 5° or, 2x = 40° or, x = 40° 2 = 20° (b) ∴ ∠PSR = x + 35° = 20° + 35° = 55° ∠QRT = ∠PSR = 55° [ Corresponding angles] A 5x – 3 cm D 2x + 6 cm B C S x + 35° 3x – 5° P T R Q 314 Allied The Leading Mathematics-9 Parallelogram 315
Geometry Geometry Example-3 In the given trapezium ACDE, find the values of x, y and z. Solution: In the given trapezium ACDE, CA//DE, CD//BE, BD//AE, AE = BE and ∠DBE = 68° Now, we have (i) ∠AEB = ∠DBE = 68° [ Alternate angles] (ii) ∠BAE = ∠ABE = x [ Base angles of isosceles ∆ABE] (iii) ∠BAE + ∠ABE + ∠AEB = 180° [ Sum of angles of ∆ABE] or, x + x + 68° = 180° or, 2x = 180° – 68° or, x = 112° 2 = 56° (iv) ∠BDE = ∠BAE [ Opposite angles of ABDE] or, y = 56° (v) ∠BED + ∠DBE + ∠BDE = 180° [ Sum of angles of ∆BDE] or, ∠BED + 68° + 56° = 180° or, ∠BED = 180° – 124° = 56° (vi) ∠BCD = ∠BED [ Opposite angle of BCDE] or, z = 56° Example-4 Show that all the angles of a square are equal. Solution: Given: ABCD is a square in which ∠BAD = 90°, AB = AD. To prove: ∠BAD = ∠ABC = ∠BCD = ∠CDA Proof: Statements Reasons 1. ABCD is a parallelogram. 1. Square is a parallelogram. 2. ∠BAD = ∠BCD = 90° 2. Opposite angles of ABCD. 3. ∠BAD + ∠ABC = 180° or, 90° + ∠ABC = 90° ∴ ∠ABC = 90° 3. Sum of co-interior angles 4. ∠ABC = ∠ADC = 90° 4. Opposite angles of ABCD 5. ∠BAD = ∠ABC = ∠BCD = ∠CDA 5. From statements (2) and (4) Hence, it is proved. Example-5 Prove that if the diagonals of a rhombus are equal, it is a square. Solution: Given: PQRS is a rhombus and PR = QS. To prove: PQRS is a square. Proof: C B A D E y 68° z x A D B C P S Q R 316 Allied The Leading Mathematics-9 Parallelogram 317
Geometry Geometry Statements Reasons 1. In ∆PQS and ∆PQR, i) PQ = PQ (S) ii) PS = QR (S) iii) QS = PR (S) 1. i) Common side ii) Opposite sides of rhombus. iii) By given 2. ∆PQS ≅ ∆PQR 2. By SSS congruent condition 3. ∠QPS = ∠PQR 3. Corresponding angles of congruent triangles. 4. ∠QPS + ∠PQR = 180° 4. Sum of co-interior angles 5. ∠QPS + ∠QPS = 180° or, 2∠QPS = 180° or, ∠QPS = 90° 5. From statements (3) and (4) 6. ∴ PQRS is a square 6. By definition. Proved. Example-6 In the given figure PQRS is a parallelogram. If A and B are the mid-points of PQ and SR respectively, prove that ACBD is a parallelogram. Solution: Given: In the given parallelogram PQRS, PQ//SR and PS//QR. A and B are mid-points of PQ and SR respectively. To prove: ACBD is a parallelogram. Proof: Statements Reasons 1. PQ = SR 1. Opposite sides of PQRS 2. PA = QA and SB = RB 2. By given 3. PA = BR 3. From statements (1) and (2) 4. PA//BR 4. By given PQ//SR. 5. PB//AR and PB = AR 5. From statements (3) and (4) 6. AS//QB and AS = QB 6. Same process as above. 7. ACBD is a parallelogram. 7. From statements (5) and (6) Hence, it is proved. Example-7 In the figure, ABCD is a parallelogram in which its diagonal AC is produced on both sides such that AP = CQ. Prove that PB//DQ. Solution: Given: ABCD is a parallelogram in which PA = CQ. To prove: PB//DQ Proof: P S D A Q B R C A D P B C Q 316 Allied The Leading Mathematics-9 Parallelogram 317
Geometry Geometry Statements Reasons 1. ∠BAC = ∠ACD 1. Alternate angles 2. In ∆APB and ∆CQD, (i) PA = CQ (S) (ii) ∠PAB = ∠DCQ (A) (iii) AB = DC (S) 2. (i) By given (ii) Supplementary angles of statement (1) (iii) Opposite sides of ABCD 3. ∆APB ≅ ∆CQD 3. By SAS congruent condition 4. ∠APB = ∠CQD 4. Corresponding angles of congruent triangles 5. PB//DQ 5. Equal alternate angles in statement (4) Hence, it is proved. PRACTICE 14.1 Keeping Skill Sharp 1. Define the following: (a) Parallelogram (b) Rectangle (c) Rhombus (d) Square 2. Write any two properties of the following quadrilaterals. (a) Parallelogram (b) Rectangle (c) Rhombus (d) Square 3. (a) Write two properties of a rectangle distinct from a parallelogram. (b) What are two properties of a square distinct from a rectangle? (c) Write two properties of a square distinct from a rhombus. (d) Write two properties of a rhombus distinct from a rectangle. 4. (a) In the figure, AB = CD and AB// DC. What is the relation between AD and BC? (b) In the given figure, PQ = RS and PQ// RS. Write the relation between PS and QR. 5. (a) The diagonals of AC and BD in the given parallelogram ABCD ....... to each other. (i) intersect (ii) bisect (iii) trisect (iv) divide (b) In the figure, OS = OQ and OP = OR. The quadrilateral PQRS is a .......... . (i) rectangle (ii) rhombus (iii) parallelogram (iv) square (c) The opposite sides of a parallelogram are ..... (i) equal (ii) parallel (iii) not equal (iv) both (i) and (ii) Fig. 4(a) A D B C Fig. 4(b) P R Q S A D B O C Fig. 5(a) P S Q O R Fig. 5(b) 318 Allied The Leading Mathematics-9 Parallelogram 319
Geometry Geometry (d) The diagonals of a square are ..... (i) equal (ii) parallel (iii) bisected (iv) perpendicularly bisected Check Your Performance 5. (i) Observe the given parallelogram ABCD in which AB = 2x + 3 cm and CD = x + 7 cm. (a) Write the relation between the opposite sides of a parallelogram. (b) Find the value of x in the given parallelogram. (c) Find the length of AB and CD. (ii) Study the given parallelogram PQRS, the diagonals PR and QS intersect at a point O. Also, PO = 3x + 1 cm and RO = x + 7 cm. (a) Write the relation between the opposite angles of a parallelogram. (b) Find the value of x in the given parallelogram PQRS. (c) Find the length of PR. (iii) In given figure, KL = MN and KL//MN. If KM = 3x + 6 cm and LN = 5x – 2 cm, find the length of KM. (iv) In the given figure, AB = CD and A//CD, AO = 5x and DO = x + 8 cm. Find the length of AD. 6. Find the value of x, y and z from the following figures: (i) y x z 150° (ii) x z y 80° 150° (iii) 4x y x z (iv) In the parallelogram, if ∠D = (4x + 28o ) and ∠B = (6x + 20o ). (a) Write the relation between the diagonals of a parallelogram. (b) Find the value of x in the given parallelogram ABCD. (c) Find the measures of ∠B and ∠A of the parallelogram. (v) In the following figure, ABCD is a parallelogram. (a) What is parallelogram? Define it. (b) Find the values of x, y and z. (c) How many degrees added in z will form a rectangle ? (vi) Observe the given figure ABCDE. (a) In which condition does a parallelogram form a rectangle? (b) Find the values of x, y and z. Fig. 5(a) Fig. 5(b) A D 2x + 3 cm B x + 7 cm C P O S Q R x+7 cm 3x+1cm Fig. 5(c) Fig. 5(d) K L M N 3x+6 cm 5x–2 cm A 5x cm x+8 cm C B O D D (4x + 28)° (6x + 20)° A C B Fig. 6(v) D z 88° 58° y G F H A x C E B Fig. 6(vi) E D C z 40° x y A B 40° 318 Allied The Leading Mathematics-9 Parallelogram 319
Geometry Geometry (c) What is the relation between the angles x and z? (d) Write the name of the quadrilateral AECB. 7. (a) Prove that the opposite sides of parallelogram are equal. (b) In the parallelogram RHLA, RH//AL and RA//HL. Prove that ∠ARH = ∠HLA and ∠RHL = ∠RAL. (c) If the opposite sides of a quadrilateral are equal then the quadrilateral is a parallelogram. (d) Ifthe opposite angles of a quadrilateral are equal then the quadrilateral is a parallelogram. (e) Prove that the diagonals of a parallelogram bisect each other. (f) In the given quadrilateral PQRS, PO = RO and QO = SO. Show that PQRS is a parallelogram. 8. (a) Prove that all the angles of a rectangle are right angles. (b) Prove that all the angles and sides of a square are equal. OR Prove that a square is an equilateral and equiangular quadrilateral. (c) Prove that diagonals of a rectangle are equal. (d) Prove that if the diagonals of a parallelogram are equal, then it is a rectangle. 9. (a) Prove that the diagonals of a square are equal. (b) Prove that if the diagonals of a rhombus are equal, then it is a square. (c) Prove that if the adjacent sides of a rectangle are equal, it is a square. 10. (a) ABCD is a rectangle. The mid-point E of AB is joined with D and C. Prove that ∆DEC is an isosceles triangle. (b) In the figure, AB and CD bisect each other at O. Prove that AC//DB. (c) If ABCD is a parallelogram and equal angles are marked similarly, prove that ∠AEB = 90o . 11. (a) In parallelogram WXYZ, XY = 2WX and M is the mid-point of XY. Prove that ∠WMZ = 90o . (b) In the adjoining figure, O is the midpoint of MN of the square KLMN. Prove that ∆LOK is an isosceles triangle. 5. (i) (a) 4 cm, 11 cm (ii) 3 cm, 20 cm (iii) 18 cm (iv) 20 cm 6. (i) (a) 150°, 150°, 30° (ii) 50°, 50o , 30o (iii) 36°, 36°, 30° (iv) 4°, 44o , 136o (v) 34o , 58o , 58°, 32° (vi) 70o , 70o , 70°, equal, isosceles trapezium Answers B C E A D Fig. 10(b) Fig. 10(c) C A O B D E D C A B Fig. 11(a) M Y W Z X K L N O M Fig. 11(b) 320 Allied The Leading Mathematics-9 Parallelogram 321
Geometry Geometry 14.2 Application of Theorems of Parallelogram At the end of this topic, the students will be able to: ¾ prove the relations on the midpoint of the triangle by using theorems of the parallelogram. Learning Objectives The theorem related to the midpoint/s of the sides of a triangle is called the midpoint theorem. The midpoint theorems are mainly used in construction trusts and bridges. The iron steels support to each other for balancing them on the base of midpoint theorems as shown in the picture. This includes two types of theorems below: Statement: The straight line drawn through the midpoint of one side of a triangle and parallel to another side bisects its third side. THEOREM 14.6 A. Observation Method Step 1: Draw a triangle ABC and find out the midpoints D and E of two sides AB and AC, and join them. Step 2: Now, cut out the triangle ABC and also cut along the DE as shown in the figure below. Step 3: Arrange the super triangle with the base remaining part as shown in the figure. What do you see ? Discuss the midpoint theorem on it. B D C A E B D D C A E B D D C A E B. Theoretical Proof Given: In ∆ABC, AD = BD, BC//DE. To prove: AE = CE Construction: Draw a line CF parallel to BA and produce DE upto F. Proof: Statements Reasons 1. BCFD is a parallelogram. 1. By given and construction. 2. BD = CF 2. Being opposite sides of BCFD. 3. BD = AD 3. By given B D Use of midpoint theorem in trust C F A E B C We can discuss the midpoint theorem from this way. A P Q R B D C F A E 320 Allied The Leading Mathematics-9 Parallelogram 321
Geometry Geometry 4. In ∆ADE and ∆CEF, (i) AD = CF (S) (ii) ∠DAE ∠ECF (A) (iii) ∠AED = ∠CEF (A) 4. (i) From statements (2) and (3) (ii) Being alternate angles (iii) Being VOA 5. ∆ ADE ≅ ∆ CEF 5. By SAA congruent condition 6. AE = CE 6. Being corresponding sides of congruent triangles Hence, it is proved. Statement: The line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to half of it. THEOREM 14.7 Theoretical Proof Given: In ∆ABC, AD = BD and AE = CE To prove: DE//BE and DE = 1 2 BC Construction: Draw a line CF parallel to BA and produce DE upto F. Proof: Statements Reasons 1. In ∆ADE and ∆CEF. (i) ∠DAE = ∠ECF (A) (ii) AE = CE (S) (iii) ∠AED = ∠CEF (A) 1. (i) Being alternate angles (ii) By given (iii) Being VOA 2. ∆ADE ≅ ∆CEF 2. By ASA congruent condition 3. AD = CF and DE = EF 3. Being corresponding sides of congruent triangles 4. AD = BD 4. By given 5. ∴ BD = CF 5. From stat. (3) and (4) 6. BD//CF 6. By construction. 7. DF//BC and DF = BC 7. From stat. (5) and (6) 8. DE = 1 2 DF 8. Being mid-point E of DF. 9. DE//BC and DE = 1 2 BC 9. From stat. (7) and (8) Hence, it is proved. PRACTICE 14.2 Keeping Skill Sharp 1. (a) In ∆ABC, D and E are mid-points of AB and AC respectively. Write the relation between DE and BC. (b) In ∆PQR, PS = QS and ST//QR. What is the relation between PT and RT? B D C F A E Q. No. 1 (a) Q. No. 1 (b) S T P B C Q R A D E 322 Allied The Leading Mathematics-9 Parallelogram 323
Geometry Geometry (c) In ∆KMN, P and Q are mid-points of KM and KN respectively. What is the length of PQ? (d) In ∆ABC, AQ = 6cm. What is the length of CQ? 2. (a) In ∆ABC, AB = AC and M is the midpoint of AB and MN parallel to BC. If MN = 2 cm, find BC. (b) In the ∆XYZ, YZ = 8 cm, M is the midpoint of XY and MN is drawn parallel to YZ. Find MN. (c) In the ∆PQR, ∠Q = 50o , ST = 5.7 cm and ∠STR = 110o , find ∠P and QR. Check Your Performance 3. (i) Prove that the straight line drawn through the midpoint of one side of a triangle and parallel to another side bisects the third side. (ii) In the given ∆ABC, AP = BP and AQ = CQ. Prove that PQ//BC and PQ = 1 2 BC. 4. (i) In the given figure, M and N are the midpoints of AB and AC respectively. Prove that AP = PD. (ii) In the parallelogram ABCD, M and N are the midpoints of AB and CD respectively. Prove that: (a) AMCN is a parallelogram. (b) BF = FE = ED. (iii) In the fig, BF = FC and AG = GC. If FG produced to meet AD at E, prove that AE = ED. (iv) From the adjoining figure, prove that: (a) BX = 1 2 AP (b) AP + CR = 2 BQ. 5. (i) Prove that the lines joining the midpoints of the opposite sides of a parallelogram bisect each other. (ii) Show that a quadrilateral joining the midpoints of the sides of a quadrilateral is a parallelogram. (iii) Prove that the line segmentsjoining the midpoints of the opposite sides of a quadrilateral bisect each other. 2. (a) 4 cm (b) 4 cm (c) 60°, 11.4 cm Answers Q. No. 1 (c) Q. No. 1 (d) 6 cm P Q A B C P Q K M 10 cm N Q. No. 2 (b) Q. No. 2 (c) 8 cm M N X Y Z 50° 110° 5.7 cm S T P Q R B C A P Q B C A N P M Q. No. 4 (i) Q. No. 4 (ii) D D N C E F A M B A P Q B C R D C A B E G F Q. No. 4 (iii) Q. No. 4 (iv) 322 Allied The Leading Mathematics-9 Parallelogram 323
Geometry Geometry 15.1 Construction of Scalene Quadrilaterals At the end of this topic, the students will be able to: ¾ construct a quadrilateral from the given measurements. Learning Objectives High school geometry, on one hand, makes us think, collect reasons and then argue systematically or logically; and, on the other hand, it helpsin developing enough skill in drawing and constructing geometrical figures. This skill is indispensable in all walks of life. Science, engineering and technology cannot be imagined in the absence of such a process of thinking, arguing, and drawing and reproducing exact copies. Drawing geometrical figures differs basically from the geometrical constructions of the same geometrical figures. Geometrical figures can be drawn freely or with the help of any instrument that is convenient. But a geometrical construction is usually done with the help of a ruler and a compass. We can begin with anything and at any time. What we need is : (i) a rough sketch of the figure we want to construct, (ii) appropriate labeling of the rough sketch with given data (or measurement), and (iii) basic idea necessary before stepwise construction. To minimize the errors that may occur during the construction of a geometrical figure, the following points should be kept in mind. (i) Every point should be observed vertically, (ii) Observed point or points should be marked by sharply pointed pencil, and (iii) Accurate instruments should be used. In what follows, we shall be concerned with geometrical constructions under three subheadings: (a) Triangles (b) Parallelograms (c) Rectangle (d) Squares (e) Rhombus (f) Trapeziums (g) Quadrilaterals We discuss the following types of constructions of quadrilateral: CHAPTER 15 CONSTRUCTION 324 Allied The Leading Mathematics-9 Construction 325
Geometry Geometry (I) Construction of a quadrilateral when four sides and one diagonal of a quadrilateral are given: Given: In quadrilateral ABCD, AB = 4 cm, BC = 5 cm, CD = 4 cm, AD = 4.5 cm and the diagonal BD = 3 cm. Basic idea/Rough sketch: a) Sides AB, BC, CD, DA and diagonal BD are given. b) The two triangles ABD & BCD are thus fixed; and so the four vertices A, B, C and D of the required parallelogram ABCD are fixed. Actual construction: 1. Draw AB = 4 cm. 2. With centre at A and radius AD = 4.5 cm draw an arc. 3. With centre at B and radius BD = 3 cm draw an arc intersecting the arc in step 2 at D. 4. Join AD and BD to get the triangle ABD. 5. With centre at D and radius CD = 4 cm, draw an arc on the side of BD opposite to A. 6. With centre at B and radius BC = 5 cm, draw an arc to intersect the arc in step 5 at C. 7. Join DC and BC to obtain the required quadrilateral ABCD. (II) Construction of a quadrilateral when four sides and one angle are given: Given: In the quad. ABCD, AB = 4 cm, BC = 3 cm, CD = 3.5 cm, DA = 3.5 cm and ∠ABC = 60o . Basic idea/Rough sketch: a) Four sides AB, BC, CD, DA and one angle ABC are given. Once the vertex B of the angle ABC is chosen, the other end-points A and C of the arms BA and BC are also fixed. b) The fourth point D lies at a fixed point determined by the given distances from A and C. Actual construction: 1. Draw BC = 3 cm. 2. With centre at B and any radius, draw an arc cutting it at P. 3. With centre at P and same radius as in step 2 draw an arc intersecting the arc in step 2 at Q. 4. Join BQ and produce it to A so that BA = 4 cm. 5. With centre at A and radius 3.5 cm draw an arc on the same side of AB as D. 6. With centre at C and radius 3.5 cm draw an arc on cutting the arc in step 5 at D. 7. Join AD and CD to obtain the required quadrilateral ABCD. Rough Sketch A B D C 4 cm 3 cm 4.5 cm 4 cm 5 cm A D C 5 cm 4 cm 3 cm 4.5 cm 4 cm B B D A 3.5 cm 4 cm 3.5 cm 60° 3 cm C Rough Sketch B 3 cm C D A 3.5 cm 4 cm 3.5 cm 60° 324 Allied The Leading Mathematics-9 Construction 325
Geometry Geometry (III) Construction of a quadrilateral when three sides and two diagonals are given: Given: In quad. ABCD, AB = 5 cm, BC = 4.5 cm, AD = 3.8 cm, AC = 6 cm and BD = 5.5 cm. Basic idea a) Two adjacentsidesAB andAD, and the diagonal BD determine the triangle ABD. In the same way, two adjacent sides AB & BC and the diagonal AC determine the triangle ABC. b) The end-points A and B of the common base AB and the vertices D and C of the triangles ABD and ABC determine the required quadrilateral. Actual construction 1. Draw AB = 5 cm. 2. With centre at A and radius AD = 3.8 cm draw an arc. 3. With centre at B and radius BD = 5.5 cm, draw an arc to intersect the arc in step 2 at D. 4. Join AD and BD to get the triangle ABD. 5. With centre at A and radius AC = 6 cm, draw an arc on the same side of AD as B. 6. With centre at B and radius BC = 4.5 cm, draw an arc cutting the arc in step 5 at C. 7. Join BC and CD to get the required quadrilateral ABCD. (IV) Construction of a quadrilateral when three sides and two angles are given: Given: In quad. ABCD, AB = 4.5 cm, BC = 3 cm, AD = 4 cm, ∠ABC = 60o and ∠BAD = 75o . Basic idea/Rough sketch: a) One side and two angles at its ends and side joining fixed points on their arms determine a quadrilateral ABCD. Actual construction 1. Draw AB = 4.5 cm. 2. With centre at A and any radius on compass, construct ∠BAD = 75o . 3. With centre at B and any radius on compass, construct ∠ABC = 60o . 4. Cut the arms AD = 4 cm and BC = 3 cm from A and B respectively by using compass. 5. Join BC to get the required quadrilateral ABCD. A B 5.5 cm 5 cm 6 cm 3.8 cm D C 4.5 cm Rough Sketch A 5 cm B C D 6 cm 5.5 3.8 cm cm 4.5 cm 4.5 cm 75° 60° 3 cm 4 cm A B C D Rough Sketch 4.5 cm 75° 60° 3 cm 4 cm A B C D 326 Allied The Leading Mathematics-9 Construction 327
Geometry Geometry (v) Construction of a quadrilateral when two adjacent sides and three angles are given: Given: In quad. ABCD, AB = 5 cm, BC = 2.5 cm, ∠ABC = 60o , ∠BCD = 120o and ∠BAD = 75o . Basic idea/Rough sketch: a) Three angles on two adjacent sides determine a quadrilateral ABCD. Actual construction 1. Draw AB = 5 cm. 2. With centre at A and any radius on compass, construct ∠ABC = 60o . 3. With centre at B and any radius on compass, construct ∠BAD = 75o . 4. Cut the arm BC = 2.5 cm from B by using compass. 5. With centre at C and any radius on compass, construct ∠BCD = 120o . 6. The arms of 60o and 75o intersect at D and we get the required quadrilateral ABCD. PRACTICE 15.1 1. Construct a quadrilateral ABCD from the following data: (a) AB = 5.5 cm, BC = DA = 4.2 cm, CD = 5 cm and diagonal AC = 6.5 cm. (b) AB = 4 cm, BC = 3.2 cm, CD = 4.6 cm, DA = 3.9 cm and diagonal AC = 5.7 cm. (c) AB = 6 cm, BC = 4.4 cm, CD = 5.2 cm, DA = 4.2 cm and diagonal AC = 7 cm. 2. Construct a quadrilateral PQRS from the following information: (a) PQ = 5.9 cm, QR = 4.1 cm, RS = ST = 4.5 cm, ∠PQR = 60o . (b) PQ = 4.6 cm = QR= 3.6 cm, RS = 5.2 cm, PS = 4.2 cm, ∠P = 75o . (c) PQ = 5 cm = QR= 4 cm, RS = PS = 4.5 cm, ∠QRS = 75o . 3. Construct a quadrilateral KITE from the following data: (a) KI = 5 cm, IT = 4.5 cm, KE = 4.8 cm and diagonals KT = 6 cm and IE = 6.5 cm. (b) KI = IT = 4.5 cm, TE = 5.1 cm and diagonals KT = 6.3 cm, IE = 5.8 cm. (c) KE = IT = 5 cm, TE = 4.5 cm and diagonals KT = 6.5 cm, IE = 6 cm. 4. (a) Construct a quadrilateral KLMN with KL = 4 cm, MN = 5 cm, LM = 6 cm, ∠M = ∠L = 60o . (b) Construct a quadrilateral PQRS with PQ = QR = 3.6 cm, PS = 4.2 cm, ∠P = ∠Q = 120o . (c) Construct a quadrilateral WXYZ with YZ = XY = 5 cm, WZ = 3.6 cm, ∠Y = 60o , ∠Z = 75o . 5. (a) Construct a quadrilateral ABCD in which AB = 4.3, BC = 5.5 cm, ∠A = 45o , ∠B = ∠C = 60o . (b) Construct a quadrilateral BIKE having BI = 4.3, IK = 5.5 cm, ∠B = 45o , ∠I = ∠K = 60o . (c) Construct a quadrilateral PRST with PR = RS = 4.8 cm, ∠P = 60o , ∠R = 105o , ∠S = 75o . 6. (a) Construct a quadrilateral ABCD with AB = BD = 4.5 cm, AD = BC = 5 cm , ∠BAC = 45o . (b) Construct a quadrilateral HERA with HE = 5 cm = EA, ER = 4.5 cm, ∠HEA = 60o , ∠EAR = 45o . 5 cm 2.5 cm A B D C 75° 120° 60° Rough Sketch 5 cm 2.5 cm A B C D 75° 120° 60° 326 Allied The Leading Mathematics-9 Construction 327
Geometry Geometry 15.2 Construction of Trapeziums At the end of this topic, the students will be able to: ¾ construct a trapezium from the given information. Learning Objectives We construct the following types of constructions of trapeziums: (I) Construction of a trapezium when three sides and one angle are given Given: In the trapezium ABCD, AB = 6 cm, BC = 3 cm, CD = 3.5 cm ∠ABC = 60° and AB//DC. Basic idea/Rough sketch: a) In the quadrilateral ABCD, ∠ABC = 60° and AB//DC, then ∠BCD = 180° – 60° = 120°. b) We can easily construct the required trapezium ABCD as scalene quadrilateral. Actual construction: 1. Draw a line segment AB = 6 cm. 2. Construct the angle ABX 60° at B by taking any radius. 3. With centre at B and radius 3 cm, cut the arm AX of 60° at C. 4. Construct the angle BCY 120° at C by taking any radius. 5. With centre at C and radius 3.5 cm, cut the arm CY of 120° A at D. 6. Join A and D. Then we get the required trapezium ABCD. (II) Construction of a trapezium when four sides are given and the parallel sides are indicated Given: In the trapezium ABCD, AB//DC, AB = 6 cm, BC = CD = 4 cm, DA = 5 cm. Basic idea/Rough sketch: a) Given AB//DC. If AD//BC, the required quadrilateral would be parallelogram. b) Take a point E on AB = 6 cm such that AE = CD = 4 cm and form a parallelogram AECD. c) Then join B and C. We get the required trapezium ABCD. 6 cm 3.5 cm 3 cm A B C D 120° 60° Rough Sketch 6 cm 3.5 cm 3 cm A B C X D Y 120° 60° A B D C E 6 cm 5 cm 4 cm 4 cm 4 cm Rough Sketch 328 Allied The Leading Mathematics-9 Construction 329
Geometry Geometry Actual construction: 1. Draw a line segment AB = 6 cm. 2. With centre at A with radius 4 cm, cut AB at E. 3. With centre at E with radius 5 cm, draw an arc upper side of B. 4. With centre at B with radius 4 cm, cut the arc of step 2, say C. 5. With centre at C with radius 4 cm, draw an arc left side of E and upper side of AB. 6. With centre at A with radius 5 cm, cut the arc of step 5, say D. 7. Join B, C, D and A. Then we get the required trapezium ABCD. (III) Construction of a trapezium when three sides and one diagonal are given. Given: In the trapeziumABCD,AB//DC,AB = 4 cm, BC = 4 cm, CD = 3.5 cm, a diagonalAC = 5 cm. Basic idea/Rough sketch: a) Two sides AB and BC, and one diagonal AC form a triangle. b) Since AB//DC so, the alternate angles BAC andACD are equal. c) After determining the point D. Then A, D and C can be joined to get the required trapezium. Actual construction: 1. Draw a line segment AB = 4 cm. 2. With centre at B and radius 4 cm, draw an arc above AB. 3. With centre at A and radius 5 cm, cut the previous arc drawn from B, say C. Join C with A and B. 4. Construct ∠ACX at C equal with ∠BAC because of showing parallel lines AB and DC. 5. With centre at C and radius 3.5 cm, cut the arm CX at D. Join D with A. Then we get the required trapezium ABCD. (IV) Construction of a trapezium when two sides and two angles are given. Given: In the trapeziumABCD,AB//DC,AB = 5 cm, BC = 3.5 cm, ∠ABC = 60o and ∠BAD = 120o . Basic idea/Rough sketch: a) One side AB and two angles at A and B determine three sides of the trapezium. b) Since AB//DC so, the sum of co-interior anglesABC and BCD is 180o . A B E C D 6 cm 5 cm 4 cm 4 cm 4 cm 5 cm 4 cm 3.5 cm A B D C Rough Sketch 4 cm 5 cm 4 cm 3.5 cm A B C Y D Z X 3.5 cm 5 cm A B D C 60° 120° 120° Rough Sketch 328 Allied The Leading Mathematics-9 Construction 329
Geometry Geometry c) After constructing ∠BCD = 120o at they form the required trapezium ABCD. Actual construction: 1. Draw a line segment AB = 5 cm. 2. Construct ∠ABX = 60o and ∠BAY = 120o at B and A respectively. 3. With centre at B and radius 3.5 cm, draw an arc that cuts BX at C. 4. Construct ∠BCZ = 120o at C whose arm intersects AZ at D. Hence, the required trapezium ABCD forms. (V) Construction of a trapezium when three sides and the height are given Given: In the trapeziumABCD,AB//DC,AB = 5.5 cm, CD = 2.5 cm,AD = 3.5 cm and height = 2.5 cm. Basic idea/Rough sketch: a) Given the height and two parallel sides, a rectangle ABEF with the longer of the parallel sides (AB) and height (AF) as adjacent sides can be constructed. b) The given oblique side AD can be drawn from the end A of the longer side to the opposite side; and from this point, the shorter of the parallel sides DC can be cut off along FE. c) Four points A, B, C and D of the trapezium are then fixed. They can be joined to get the required trapezium. Actual construction: 1. Draw a line segment AB = 5.5 cm. 2. With centre at A and any radius, draw an arc cutting AB at P. 3. With centre at P and same radius as in step 2, draw an arc cutting the arc in step 2 at Q. 4. With centre at Q and same radius as in step 3, draw an arc cutting the arc in step 2 at R a portion of arc drawn above the arc PQR. 5. With centre at R and same radius as in step 4, draw an arc cutting the arc in step 4 at S and join AS. 6. Cut AE = 2.5 cm. With E and B as centers and radii ET = AB = 5.5 cm and BT = AE = 2.5 cm respectively, draw arcs intersecting each other at T. 7. Join ET. Then, cut AD = 3.5 cm and DC = 2.5 cm along ET. 8. Join AD and CB. Then we get the required trapezium ABCD. 3.5 cm 5 cm A B C D 60° 120° 120° Z Y X A 5.5 cm B F D 2.5cm C E 3.5 cm 2.5 cm Rough Sketch A E S R Q P C T 5.5 cm 2.5 cm 2.5 cm B 3.5 cm D 330 Allied The Leading Mathematics-9 Construction 331
Geometry Geometry PRACTICE 15.2 1. Construct a trapezium ABCD in which; (a) AB = 5 cm, BC = 4.2 cm, CD = 4 cm ∠ABC = 45° and AB//CD. (b) BC = 6.5 cm, CD = 5 cm, DA = 4.5 cm ∠BCD = 60° and BC//AD. (c) BC = 4.8 cm, AB = 5.2 cm, AD = 6 cm ∠BAD = 75° and CB//DA. 2. (a) Construct a trapezium PQRS in which AB = 5.5 cm, BC = 4.2 cm, CD = 4.5 cm, DA = 5 cm and AB//DC. (b) Construct a trapezium PREM in which PR = 4 cm, RE = 4.5 cm, EM = 6 cm, PM = 4.6 cm and PR//EM. (c) Construct a trapezium SOVA in which SO = 6 cm, VO = 7 cm, VA = SA = 5 cm and VO//AS. 3. (a) Construct a trapeziumABCD,AB//DC,AB= 6.2 cm,BC= 4.8 cm,CD = 5.3 cm, a diagonal AC = 6 cm. (b) Construct a trapezium WXYZ, ZY//WX, WX = 4.4 cm, WZ = 5.6 cm, ZY = 5.5 cm, a diagonal ZX = 6.5 cm. (c) Construct a trapeziumABCD,AD//BC,AD = 4.5 cm,BC= 6.3 cm,CD = 4.3 cm, a diagonal BD = 7 cm. 4. Construct a trapezium ABCD in which; (a) AB = 5.2 cm, BC = 4.2 cm, ∠B = 60o , ∠A = 90o , AB//DC. (b) AB = 4.2 cm, BC = 5.2 cm, ∠B = 45o , CD = 5.3, AB//CD. (c) AB//DC,AB = 5 cm,AD = 3.8 cm, ∠BAD = 75o , ∠ABC = 105o . 5. (a) Construct a trapezium ABCD with AB = 6.2 cm, BC = 4.3 cm, AD = 4.2 cm, AB//CD and distance between the parallel sides 3.2 cm. (b) Construct a trapezium PQRS with PQ = 4.5 cm, QR = 5.2 cm, PS = 4 cm, PQ//RS and distance between the parallel sides 3 cm. (c) Construct a trapezium CDEF with CD = 4.5 cm, DE = 5.2 cm, CF = 4 cm, CD//EF and distance between the parallel sides 3.4 cm. Consult with your teacher. Answers 330 Allied The Leading Mathematics-9 Construction 331
Geometry Geometry 15.3 Construction of Rhombus At the end of this topic, the students will be able to: ¾ construct a rhombus from the given information. Learning Objectives We construct the following types of constructions of rhombus: (I) Construction of a rhombus when its one side and one angle are given: Given: Construct a rhombus ABCD in which AB = 6 cm and ∠A = 60o . Basic idea/Rough sketch: When a side of a rhombus ABCD is given, its all sides are equal. Then AB = BC = CD = AD = 6 cm. At first, we construct ∠BAD with arms AB = AD = 6 cm and then cut down C using compass with the same radius of 6 cm from B and D. Actual construction: 1. Draw line segment AB = 6 cm. 2. Construct ∠BAX = 60o . 3. With A as the centre and radius equal to 6 cm, draw an arc on AX and mark the point of intersection as D. 4. From D and B as the centre and radius equal to 6 cm, draw an arc, which intersect at C. 5. Join points C, D and points C, B. Then, ABCD is the required rhombus. (II) Construction of a rhombus given one side and one diagonal: Given: Construct a rhombus ABCD in which AB = 5.4 cm and AC = 7.4 cm. Basic idea: When a side of a rhombus ABCD is given, its all sides are equal. Then AB = BC = CD = AD = 5.4 cm. At first, we construct ∆ABC with AB = BC = 5.4 cm and AC = 7.4 cm and then cut down D using compass with the same radius of 5.4 cm from A and C. Actual construction: 1. Draw a line segment AB = 5.4 cm . 2. Draw the arcs from B and A with the radii 5.4 cm and 7.4 cm respectively, which intersect at C and join C with A and B. 3. Draw the arcs from A and C with the same radius 5.4 cm, which intersect at D and join D with A and C. Then ABCD is the required rhombus. A D B 60° 6 cm 6 cm 6 cm 6 cm C Rough Sketch A 60° B C D X 6 cm A D 5.4 cm B 5.4 cm 5.4 cm 7.4 cm 5.4 cm C Rough Sketch A D 5.4 cm 7.4 cm 5.4 cm 5.4 cm 5.4 cm B C 332 Allied The Leading Mathematics-9 Construction 333
Geometry Geometry (III) Construction of a rhombus when two diagonals are given: This construction is same as the construction of the parallelogram with given two diagonals. Given: Construct a rhombus ABCD in which AC = 6 cm and BD = 7.4 cm. Basic idea/Rough sketch: In a rhombus, its diagonals are perpendicularly bisected. So, at first, draw one diagonal and it is perpendicularly bisected by using compass. Take half of the next diagonal on up and down of the bisector and then join the points. Actual construction: 1. Draw a line segment AC = 6 cm . 2. Draw a perpendicular bisector XY of AC from A and C by taking the radius of the compass more than half of 6 cm that intersects at a point, say O. 3. From O, cut the lines OX and OY with radius 2.5 cm at the points D and B. 4. Join A, B, C and D, we get the required rhombus ABCD. (IV) Construction of a rhombus when a side and an angle between the side and diagonal are given: Given: Construct a rhombus ABCD in which AB = 4.5 cm, AC = 5 cm and ∠BAC = 60o . Basic idea/Rough sketch: Two sides and an angle between them determine a triangle and other two sides determine another triangle on the same diagonal. Actual construction: 1. Draw a line segment AB = 4.5 cm . 2. At A, construct ∠BAX = 60o . 3. From B, cut the line AX with radius 4.5 cm at a point, say C. 4. From A and C, draw the arcs with radius 4.5 cm that intersect at a point, say D. 5. Join D with A and C and we get the required rhombus ABCD. Rough Sketch 6 cm 5 cm 6 cm 2.5 cm 2.5 cm A X Y B C O D 5 cm A B D C 60° 4.5 cm Rough Sketch 4.5 cm A B C X D 60° 5 cm 332 Allied The Leading Mathematics-9 Construction 333
Geometry Geometry PRACTICE 15.3 1. Construct a rhombus ABCD having; (a) Side = 6 cm, ∠BAD = 60o (b) BC = 5.4 cm, ∠D = 30o (c) AD = 5 cm, ∠B = 120o 2. Construct a rhombus PQRS having; (a) PQ = 5.5 cm, PR = 7 cm (b) RS = 6 cm, QS = 8 cm (c) PS = 5 cm, RP = 7.4 cm 3. Construct a rhombus SERA having; (a) SR = 6 cm, EA = 7.4 cm (b) EA = 7 cm, RS = 8 cm (c) SR = 4.6 cm, RP = 6 cm 4. Construct a rhombus RAME having; (a) ME = 4.5 cm, RM = 7.2 cm and ∠EMR = 45o (b) MA = 5 cm, EA = 6 cm and ∠AMR = 30o (c) RE = 6.2 cm, EA = 8 cm and REA = 60° 334 Allied The Leading Mathematics-9
Geometry Geometry 16.1 Circle At the end of this topic, the students will be able to: ¾ Define the different parts of a circle and prove the theorems related to the chords of a circle. Learning Objectives We are all familiar with objects such as ring , bicycle wheels, coins and so on. They all have the same type of boundary, i.e. their shapes are circular. Now we study first two of all such shapes under four sub-headings: I. Basic concepts II. Circle and chords III. Circle and arcs IV. Circle and tangents I Basic Concepts (a) Circle, Centre, Radius and Circumference A circle is the set of all points in a plane that lie at a constant distance from a fixed point on the plane. The fixed point is called the centre of the circle and the line segment joining the centre to any point on the circle is called its radius. In circle, a circle with centre at O and radius r is denoted by (O, ABC) or simply by C(ABC) or ABC where A, B, and C are any three points on the circle. Obviously, each of the line segments OA, OB and OC is a radius of the circle. The circumference of a circle is the perimeter or boundary of the circle. In a sense, it may be considered as the path traced out by a point moving at a constant distance from a fixed point. (b) Interior and Exterior Points A point P in the same plane as a circle C(O, r) is said: a) to lie inside the circle or be an interior point, if OP < r. b) to lie outside the circle or be an exterior point, if OP > r. and c) to lie on the circle or be a point of the circle, if OP = r. Circumference O C B A r Interior point O P A r Exterior point O A P r CHAPTER 16 CIRCLE PB Allied The Leading Mathematics-9 Circle 335
Geometry Geometry II Circle and Chords (a) Chord, Diameter and Semi-circle of a circle A line segment joining any two points on a circle is called the chord of a circle. In the following figure, the line segment AB joining the two given points A and B is the chord AB of the circle. If a chord passes through the centre of a circle, it is called a diameter. In the above figure, MN passes through the centre O and is therefore a diameter of the circle. A diameter divides a circle into two equal parts, each of which is known as a semi-circle. Obviously, a circle can have many diameters. It can be shown that the diameter isthe largest chord of a circle; and all diameters of the same circle are equal. Furthermore, a diameter of a circle is twice as long as its radius. i.e. d = 2r where, d = length of diameter and r = length of radius. (b) Segment of a circle Achord of a circle dividesthe region enclosed by a circle into two portions. Each such part is a segment. The segment APB containing the diameter MN is called the major segment and the other AQB not containing the diameter MN is called the minor segment shown in the adjoining figure. Statement: The perpendicular drawn from the centre of a circle to a chord bisects the chord. THEOREM 16.1 A. Experimental Verification Step 1: Two circles of different radii are drawn and in each circle the centre is labeled as O. Step 2: A chord BC is drawn in each circle. Step 3: From the centre O, OD is drawn perpendicular to BC in each circle by using compass and ruler or a pair of set squares. Step 4: The lengths of BD and DC in each figure are measured in cm. and tabulated below O A D B C Fig. (ii) O A D B C Fig. (i) O C D A B Chord Chord O P M N Diameter Chord A B Q Semi-circle O P Q M N A B Minor segment Major segment 336 Allied The Leading Mathematics-9 Circle 337
Geometry Geometry Fig. BD DC Remarks (i) 1.4 cm 1.4 cm BD = DC (ii) 1.1 cm 1.1 cm BD = DC Conclusion: From the above table, we can conclude that 'The perpendicular drawn from the centre of a circle to a chord bisects the chord.' B. Theoretical Proof Given: BC is a chord of the circle C(O, r) and OD ⊥ BC. To prove: OD bisects BC or DB = DC. Construction: Join OB and OC. Proof: Statements Reasons 1. In ∆ ODB and ∆ODC, (i) ∠ODB = ∠ODC (R) (ii) OB = OC (H) (iii) OD = OD( S) 1. (i) OD ⊥ BC (ii) Radii of the same circle (iii) Common side 2. ∆ODB ≅ ∆ODC 2. RHS congruent condition 3. BD = CD 3. Corresponding sides of congruent triangles 4. D is the mid point of BC. i.e. OD bisects BC. 4. BD = CD Hence, it is proved. Statement: The line joining the centre of a circle to the midpoint of a chord is perpendicular to the chord. THEOREM 16.2 (Converse of Theorem 16.1) A. Experimental Verification Step 1: Two circles of different radii are drawn and the centre is denoted by O in each circle. Step 2: A chord BC is drawn in each circle. Step 3: The mid-point of BC is found with the help of compass and ruler or otherwise in each figure. B D C O A O D B C Fig. (i) Fig. (ii) O D B C 336 Allied The Leading Mathematics-9 Circle 337
Geometry Geometry Step 4: OD is joined and ∠ODB and ∠ODC are measured and tabulated below: Fig. ∠ODB ∠ODC Remarks (i) 90° 90° ∠ODB = ∠ODC = 90° (ii) 90° 90° ∠ODB = ∠ODC = 90° Conclusion: 'The line joining the centre of a circle and the midpoint of its chord is perpendicular to the chord.' B. Theoretical Proof Given: D is the mid-point of the chord BC of a circle with centre O. O and D are joined by OD. To prove: OD ⊥ BC. Construction: Join OB and OC. Proof: Statements Reasons 1. In ∆ODB and ∆ODC, (i) BD = DC (S) (ii) OB = OC (S) (iii) OD = OD (S) 1. (i) By given (ii) Radii of the same circle (iii) Common side 2. ∆ODB ≅ ∆ODC 2. SSS congruent condition 3. ∠ODB = ∠ODC 3. Corresponding angles of congruent triangles 4. ∠ODB + ∠ODC = 180° 4. Linear pair of angles 5. 2 ∠ODB = 180° 5. From statement (4) 6. ∠ODB = ∠ODC = 90°, i.e., OD ⊥ BC 6. From statement (5) Hence, it is proved. Statement: The perpendicular bisector of a chord of a circle passes through the centre of the circle. THEOREM 16.3 A. Experimental Verification Step 1: Two circles of different radii are drawn. In each case, the centre is denoted by O. Step 2: In each figure, a chord BC is drawn. B D C O A 338 Allied The Leading Mathematics-9 Circle 339
Geometry Geometry Step 3: Perpendicular bisector of the chord BC is drawn in each circle by using a compass and ruler. Step 4: After drawing the perpendicular bisector of the chord BC in each circle whether the perpendicular bisector passes through the centre O or not is examined by visual observation. O B D C Fig. (i) O D B Fig. (ii) C Conclusion: 'The perpendicular bisector of a chord of a circle passes through the centre of the circle.' B. Theoretical Proof Given: AD is the perpendicular bisector of the chord BC of a circle with centre O. To prove: AD passes through the centre O. Construction: Join OB and OC. Proof: Statements Reasons 1. AD is the locus of points equidistant from B and C. 1. AD is given to be the perpendicular bisector of BC. 2. OB = OC 2. Radii of the same circle 3. ∴ O must lie on AD. 3. OA = OB 4. ∴ CD passes through O. 4. From statement (3) Hence, it is proved. Statement: Equal chords of a circle are equidistant from the centre. THEOREM 16.4 A. Experimental Verification Step 1: Two circles of different radii are drawn and the centre is labeled as O in each circle. Step 2: Two equal chords AB and CD are drawn. Step 3: OX ⊥ AB and OY ⊥ CD are drawn by using set squares or compass. B D C O A 338 Allied The Leading Mathematics-9 Circle 339
Geometry Geometry O A C B X Y D O D B C Y X A Fig. (i) Fig. (ii) Step 4: The lengths of OX and OY are measured in cm and the results are tabulated below. Fig. OX OY Remarks (i) 1.1 cm 1.1 cm OX = OY (ii) 0.9 cm 0.9 cm OX = OY Conclusion: 'Equal chords AB and CD of a circle are equidistant from the centre O.' A. Theoretical Proof Given: AB and CD are two equal chords of a circle with centre at O. OX is the distance of AB from O and OY is the distance of CD from O. To prove: OX = OY Construction: Join OA and OC. Proof: Statements Reasons 1. (i) OX ⊥ AB and (ii) OY ⊥ CD 1. OX and OY being the distances of AB and CD from O respectively. 2. ∴ AX = 1 2 AB and CY = 1 2 CD 2. By statement (1) 3. AB = CD 3. By given 4. ∴ AX = CY 4. By statements (2) and (3). 5. In ∆OXA and ∆OYC, (i) ∠OXA = ∠OYC (R) (ii) OA = OC (H) (iii) AX = CY (S) 5. (i) OX ⊥ AB and OY ⊥ CD (ii) Radii of the same circle (iii) From statement (5). 6. ∆OXA ≅ ∆ OYC 6. RHS congruent condition 7. ∴ OX = OY 7. Corresponding sides of congruent triangles Hence, it is proved. B O X Y A C B D 340 Allied The Leading Mathematics-9 Circle 341
Geometry Geometry Statement: Chords equidistant from the centre of a circle are equal in length. THEOREM 16.5 A. Experimental Verification Step 1: Two circles of different radii are drawn and in each case the centre is labeled as O. Step 2: By using compass, lengths OX and OY making OX = OY are cut off from any two radii. Step 3: Two chords AB ⊥ OX and CD ⊥ OY are drawn by using set squares or compass. O A C B X Y D Fig. (i) O A C B X Y D Fig. (ii) Step 4: The lengths of the chords AB and CD are measured in cm and the results are tabulated below: Fig. AB CD Remarks (i) 2.5 cm 2.5 cm AB = CD (ii) 1.9 cm 1.9 cm AB = CD Conclusion: 'Chords equidistant from the centre of a circle are equal in length.' B. Theoretical Proof Given: AB and CD are two chords of a circle whose centre is at O such that OX ⊥ AB, OY ⊥ CD and OX = OY. To prove: AB = CD. Construction: Join OA and OC. Proof: Statements Reasons 1. In ∆ OXA and ∆OYC, (i) ∠OXA = ∠OYC (R) (ii) OA = OC (H) (iii) OX = OY (S) 1. (i) OX ⊥ AB and OY ⊥ CD. (ii) Radii of the same circle (iii) By given 2. ∆OXA ≅ ∆OYC 2. By statement (1) 3. AX = CY 3. By given 4. AX = 1 2 AB and CY = 1 2 CD 4. OX ⊥ AB and OY ⊥ CD. 5. ∴ AB = CD 5. From statements (3) and (4) Hence, it is proved. O X Y A C B D 340 Allied The Leading Mathematics-9 Circle 341
Geometry Geometry Example-1 In a circle, the length of a chord BC = 24 cm. Its distance from the centre OE = 5 cm. Find the radius of the circle. Solution: Here, in the right-angled ∆OEB, OB2 = OE2 + BE2 or, r2 = (5)2 + (12)2 [ BE = 1 2 BC = 1 2 × 24 cm = 12 cm] = 25 + 144 = 169 = 132 => r = 13 cm. Example-2 In the given figure, AB and CD are two chords of a circle with centre at O. OE⊥AB and OF ⊥ CD. AB = 16 cm, CD = 12 cm and OE = 6 cm. Find OF. Solution: Here, AE = 1 2 AB = 1 2 × 16 = 8 cm [ OE ⊥ AB and OF ⊥ CD] and CF = 1 2 CD = 1 2 × 16 = 6 cm Now, in ∆OEA, OA2 = OE2 + AE2 [ By Pythagoras theorem] = 62 + 82 = 36 + 64 = 100 = 102 ∴ OA = 10 cm ∴ OC = OA = 10 cm [ OC = OA, radii of the same circle] Hence, from ∆OFC, OF2 = OC2 – CF2 = 100 – 36 = 64 = 82 ∴ OF = 8 cm. Example-3 In the given circle with center O, AS = AK, OS⊥AH and OK⊥AD. Prove that AH = AD. Solution: Given: In the give circle with center O, AS = AK, OS⊥AH and OK⊥AD. To prove: AH = AD Construction: Join A and O. Proof: Statements Reasons 1. In ∆ASO and ∆AKO, (i) ∠ASO = ∠AKO (R) (ii) OA = OA (H) (iii) OS = OK (S) 1. (i) OX ⊥ AB and OY ⊥ CD. (ii) Common side (iii) By given 2. ∆ASO ≅ ∆AKO 2. RHS congruent condition 3. OS = OK 3. Corresponding sides of congruent triangles 4. ∴ AB = CD 4. Chords equidistant from the origin Hence, it is proved. B C E 5 r cm O A B E 6 cm 16 cm O C F D 12 cm H D O S K A 342 Allied The Leading Mathematics-9 Circle 343
Geometry Geometry Example-4 In the given concentric circles with common center D, prove that AS = OK. Solution: Given: In the given intersecting circles. D is the common center. To prove: AS = OK Construction: Draw DH⊥AK. Proof: Statements Reasons 1. AH = KH 1. DH⊥AK at H 2. SH = OH 2. DH⊥SO at H 3. AH – SH = KH – OH 3. Subtracting statement (2) from (1) 4. AS = OK 4. Remaining parts of statement (3) Hence, it is proved. Example-5 In the given intersecting circles with centers R and M, prove that RM is a perpendicular bisector of HI. Solution: Given: The centres of the given intersecting circles are R and M. To prove: RM is a perpendicular bisector of HI. i.e., RM⊥HI and HA = IA. Construction: Join R and M with H and I. Proof: Statements Reasons 1. In ∆RMH and ∆RIM, (i) RH = RI (S) (ii) RM = RM (S) (iii) HM = MI (S) 1. (i) Radii of the same circle with centre R (ii) Common side (iii) Radii of the same circle with centre M 2. ∆RMH ≅ ∆RIM 2. SSS congruent condition 3. ∠HRM = ∠IRM 3. Corresponding angles of congruent triangles 4. In ∆RAH and ∆RAI, (i) RH = RI (S) (ii) ∠HRA = ∠IRA (A) (iii) RA = RA (S) 4. (i) Radii of the same circle with centre R (ii) From statement (3) (iii) Common side 5. ∆RAH ≅ ∆RAI 5. SAS congruent condition 6. ∠RAH = ∠RAI, HA = AI 6. Corresponding parts of congruent triangles 7. RM⊥HI 7. Equal supplementary angles in Stat. (6) Hence, it is proved. D H S O A K R M H I A 342 Allied The Leading Mathematics-9 Circle 343
Geometry Geometry PRACTICE 16.1 Keeping Skill Sharp 1. Define the following terms. (a) Circle (b) Centre (c) Radius (d) Circumference (e) Chard (f) Diameter (g) Area (h) Semi-circle 2. Write the relation between the following terms about the circle. (a) Diameter and Radius (b) Circumference and Diameter (c) Circumference and Radius (d) Area and Radius 3. (a) What do O, OP, AB, BC and AD represent in the adjoining figure? (b) Write the name of the radius, diameter and chord in the given figure. 4. Circle the correct answer. (a) What is the name of the shaded portion in the given circle? (i) radii (ii) segment (iii) sector (iv) semi-circle (b) What is the name of the shaded region in the given circle? (i) sector (ii) semi-circle (iii) radii (iv) segment (c) What is the length of the diameter in a circle with radius 7 cm? (i) 3.5 cm (ii) 14 cm (iii) 44 cm (iv) 21 cm (d) What is the length of circumference of circle having diameter 21 cm? (i) 66 cm (ii) 10.5 cm (iii) 132 cm (iv) 42 cm (e) In the adjoining figure, AB = 10 cm. What is the length of CD? (i) 5 cm (ii) 20 cm (iii) 7 cm (iv) 10 cm (f) What is the length of QT if RP = 6 cm? (i) 12 cm (ii) 6 cm (iii) 24 cm (iv) 3 cm 5. (i) What is the relation between the chord and the perpendicular drawn from the centre of a circle to the chord? (ii) Write the relation between PR and RQ in the given circle. (iii) Write the relation between OC and AB in the adjoining circle. O D A B C Q. No. 4(a) A O B Q. No. 4(b) O Q. No. 4(e) O M N A C B D Q. No. 4(f) S O T R P Q Q. No. 5(ii) Q. No. 5(iii) P R Q O A C B O 344 Allied The Leading Mathematics-9 Circle 345
Geometry Geometry (iv) From which point does the line segment CD when CD ⊥ AB and AD = BD in the given circle? Check Your Performance 6. (i) The length of the chord AB of a circle with centre at O and radius 10 cm is 12 cm and ∠OAB = 45°. (a) Write the relation between AE and BE when OE ⊥ AB. (b) Write the length of AE. (c) Find the distance of the chord AB from the centre O. (ii) The radius of a circle with centre at P is 13 cm. The length of the chord AB of this circle is 24 cm. Determine the distance from P to the chord AB. (iii) Find the length of a chord which is at a distance of 4 cm from the centre of a circle with radius 5 cm. (iv) Find the length of the radius of a circle in which 32 cm long chord is 12 cm far from its origin. 7. In the adjoining figure, P is the centre of the circle and its radius is 8 cm, ∠APB = 60o and PM ⊥ AB. (a) Name the types of triangle APB. (b) Calculate the length of AB. (c) Find the length of AM. (d) Justify that the length of PM is 4 3. 8. (i) Prove that the perpendicular drawn from the centre of a circle to a chord bisects the chord. (ii) O is the centre of a circle and R is the midpoint of PQ. The centre O and the midpoint are joined. (a) Write the relation between PQ and OR. (b) Prove the above relation by statements and reasons. (iii) Verify experimentally that the perpendicular bisector of a chord of a circle passes through the centre of the circle. 9. (i) In the given figure, ABC is an isosceles triangle and a circle with A as the centre cuts BC at D and E. (a) If AF BC, write the relation between DF and EF. (b) Prove that BD = CE. A D B C Q. No. 5(iv) O A E B 12 cm P Q 13 cm A B 24 cm P A M B 60° R O P Q Q. No. 10(i) A B C D E 344 Allied The Leading Mathematics-9 Circle 345
Geometry Geometry (ii) In the adjoining figure, OD is perpendicular to the chord AB of a circle whose centre is O. (a) Is the point D midpoint of AB ? Why ? Justify your answer with reason. (b) If BC is a diameter, show that AC = 2 OD and AC // DO. (iii) In the given figure, P and Q are the centres of two intersecting circles and AC // PQ. (a) Prove that AC = 2PQ. (b) If AC = 12 cm, what is the length of PQ? 10. (i) Prove that the diameter of a circle perpendicular to one of the two parallel chords of a circle is perpendicular to the other and bisects it. (ii) If a line intersects two concentric circles at the points A, B, C and D as shown in the figure, prove that AB = CD. (iii) In the figure, XY is a diameter and it is also perpendicular to AB. Prove that XA = XB. 11. (i) In the given figure, the chords AB and CD of a circle with centre at O intersect at X so that ∠AXO = ∠OXD. Prove that BX = CX. (ii) In the given figure, X and Y are the centres of two circles which intersect at A and C. XA and XC are produced to meet the other circle at B and D. Prove that AB = CD. 6. (i) 6 cm, 8 cm (ii) 5 cm (iii) 6 cm (iv) 20 cm 7. (i) 8 cm (ii) 4 cm (iii) 4 3 cm Answers Project Work Cut down the circle from a paper and draw its different parts with labels. O A C D B A B P Q C A B Y X D C B A Q. No. 11 (ii) Q. No. 11 (iii) X O C B A D Y B D X A C Q. No. 12(i) Q. No. 12(ii) 346 Allied The Leading Mathematics-9 Circle 347
Geometry Geometry CONFIDENCE LEVEL TEST - V Unit V : Geometry Class: 9, The Leading Maths Time: 45 mins. FM: 20 Attempt all questions. 1. Answer the given questions. (a) Find the value of x and y. [2] (b) Draw two triangle RAM and produce AM to X then examine 'the exterior angle of a triangle is equal to the sum of two opposite interior angle' experimentally? Also write the conclusion of above experiment. [3] 2. In the given figure BO = 5 cm and OD = x + 1 cm. (a) Find the value of x [1] (b) Recall about the relation of diagonals. [1] (c) Look at the given parallelogram ABCD and answer the following: (i) Prove that: ∆ACD ≅ ABD. [3] (ii) Write the relation between triangle ABD and parallelogram ABCD. [1] 3. Study the given figure answer the given questions. (a) Name the two parallel line [1] (b) Write any one condition similarity for two triangles [1] (c) Find the value of AP [3] 4. In a circle, O is the center of the circle. OA = 10 cm and AB = 12 cm. (a) Write the length of AP. [1] (b) Prove that: OP = 8 cm [2] (c) Prove that: ΔOAP ≅ ΔOBP. [2] Best of Luck 30° x y 30° 50° A C B D E A D B O C A D B C P A B 3 cm 5 cm 4 cm 10 cm Q R A P B O 346 Allied The Leading Mathematics-9 Circle 347
Geometry Geometry Additional Practice – V 1. In triangle ABC, a side BC is produced to D. (a) What would be the value of ∠ABC when ∠ACD = 120° and ∠BAC = 50°. (b) Experimentally verify that the relation of exterior angle of the triangle with sum of two non-adjacent interior angles of the triangle ABC by making different size of two triangles. (c) If ABC is an equilateral triangle, calculate and mention the ratio between the ∠ABC and ∠CAD formed by joining the vertex A and the point D in such a way that AC = CD of ∆ACD. 2. In the figure of an isosceles triangle ABC, the straight line AD is drawn perpendicular from the vertex A to the base BC. (a) If ∠BAC = 68°, and BD = CD, what is the value of ∠BAD? (b) Experimentally verify that the straight line drawn perpendicular from the vertex to the base of an isosceles triangle ABC bisects the vertical angle and the base by making two triangles of different size. (c) In which axiom, ΔABD is congruent to ΔADC? 3. Two triangles QPR and QSR, right angled at P and S respectively are drawn on the same base QR and on the same side of QR. (a) Prove that: Δ QPT ~ Δ RST. (b) Prove that PT × TR = ST × TQ. 4. In the given figure, a girl looks the reflection of the top of the lamppost on the mirror which is 6.6 m away from the foot of the lamppost. The girl whose height is 1.25 m is standing 2.5 m away from the mirror. (Assuming the mirror is placed on the ground facing the sky and the girl, mirror and the lamppost are in a same line.) (a) Prove that: ΔABD ~ ΔECD. (b) Find the height of the lamppost. 5. The figure of rhombus MNOP is given below. (a) Write the relationship between the diagonals of rhombus. (b) If MN = (4m – 7 ) cm and MP = (m + 2) cm, find the value of m. (c) Construct a rhombus MNOP having diagonal MO = 6 cm and diagonal PN = 8 cm. A B C D A B C D R S Q T P P B C 2.5 m 6.6 m 1.25 m E x D O N P M 348 Allied The Leading Mathematics-9 Circle 349
Geometry Geometry 6. Parallelograms BCDE and ABDE with AE = BE are given in the trapezium ACDE as shown in the figure with measurement. (a) Write the relationship between opposite angles of parallelogram. (b) Find the value of x, y and z. (c) Construct a trapezium in which AB = 6 cm, BC = CD = 4 cm and DA = 5 cm. Also, AB || CD. 7. In the given figure, O is the center of the circle. OP is the perpendicular distance from the center to the chord. (a) What type of triangle is OAB? (b) Find the radius of circle having chord AB = 16 cm and OP = 6 cm. (c) Can we draw a circle through the points A and B having radius 5 cm? Write with reason. 8. In a circle, O is the center of the circle. OA = 10 cm and AB = 12 cm. (a) Write the length of AP. (b) Prove that: OP = 8 cm (c) Prove that: ΔOAP ≅ ΔOBP. 9. In the given circle with center O, AS = AK, OS ⊥ AH and OK ⊥ AD. (a) Prove that: ΔOAS ≅ ΔOAK (b) Are AH equal to AD? Justify with reason. (c) If AH = 12 cm, what should be the value of AS. 10. In the given figure, the length of two parallel chords AB and CD are 12 cm and 6 cm respectively. Both chords are lying in the same side from the center O of the circle. (a) Write the relation of chords which are equidistant from the center of a circle. [1] (b) If the distance between two chords AB and CD is 3 cm, find the radius of the circle. [3] D 12 cm 6 cm A C B O 1. 70° 2. 34° 4. 3.3 m 5. 3 cm 6. 60°, 55°, 55° 7. 10 cm 8. 6 cm 9. 6 cm 10. 6.7 cm Answers D E y x z 65° 55° C B A A P B O A P B O A D H K O S 348 Allied The Leading Mathematics-9 Circle 349
Statistics and Probability Statistics and Probability 350 Allied The Leading Mathematics-9 Classification and Representation of Data 351 development and demonstration of skills of collection, tabulation, presentation and analysis. demonstration of skills of basic concept, addition and multiplication of probability in problem solving of daily life. Estimated Teaching Hours : Competency Learning Outcomes At the end of this unit, the students will be able to: classify and tabulate the data. construct the frequency table from the data of discrete and continuous series. construct histogram, frequency polygon and frequency ogive from collected data. find the mean, median, mode, quartiles and range of the data. develop the basic concept of probability. solve the problems on daily life related to priory probability. 17. Classification and Representation of Data 17.1 Collection of Data and Frequency Table 17.2 Histogram 17.3 Frequency Polygon 17.4 Cumulative Frequency Curve (Ogive) 18. Central Tendency of Ungrouped Data 18.1 Arithmetic Mean 18.2 Median 18.3 Quartiles 18.4 Mode 18.5 Range 19. Probability 19.1 Introduction to Probability 19.2 Empirical and Classical Probabilities Chapters / Lessons STATISTICS AND PROBABILITY VI UNIT 24 [Th. + Pr.] Specification Grid Unit Areas Total working hour Knowledge Understanding Application Higher ability Total number of items Total number of questions Total Marks No. of items Marks No. of items Marks No. of item Marks No. of item Marks VI Statistics and Probability 24 2 2 2 3 2 4 2 2 8 2 11 FEE Concept: "Maths is Fun, Maths is Easy and Maths is Everywhere"