Mathematics Textbook PDF

MATHEMATICS Part - STANDARD X 10 2 77.00  The Constitution of India Chapter IV A Fundamental Duties ARTICLE 51A Fundamental Duties- It shall be the duty of every citizen of India- (a) to abide by th...

MATHEMATICS Part - STANDARD X 10 2 77.00  The Constitution of India Chapter IV A Fundamental Duties ARTICLE 51A Fundamental Duties- It shall be the duty of every citizen of India- (a) to abide by the Constitution and respect its ideals and institutions, the National Flag and the National Anthem; (b) to cherish and follow the noble ideals which inspired our national struggle for freedom; (c) to uphold and protect the sovereignty, unity and integrity of India; (d) to defend the country and render national service when called upon to do so; (e) to promote harmony and the spirit of common brotherhood amongst all the people of India transcending religious, linguistic and regional or sectional diversities, to renounce practices derogatory to the dignity of women; (f) to value and preserve the rich heritage of our composite culture; (g) to protect and improve the natural environment including forests, lakes, rivers and wild life and to have compassion for living creatures; (h) to develop the scientific temper, humanism and the spirit of inquiry and reform; (i) to safeguard public property and to abjure violence; (j) to strive towards excellence in all spheres of individual and collective activity so that the nation constantly rises to higher levels of endeavour and achievement; (k) who is a parent or guardian to provide opportunities for education to his child or, as the case may be, ward between the age of six and fourteen years. Fourth Reprint : 2022 Sachin Mehta Asst. Production Officer N/PB/2022-23/90,000 SHRIPAD VALLABHA MUDRANALAYA,KOLHAPUR  1 Similarity Let’s study. Ratio of areas of two triangles Basic proportionality theorem Converse of basic proportionality theorem Tests of similarity of triangles Property of an angle bisector of a triangle Property of areas of similar triangles The ratio of the intercepts made on the transversals by three parallel lines Let’s recall. We have studied Ratio and Proportion. The statement, ‘the numbers a and b are in the ratio m ’ is also written as, ‘the numbers a and b are in proportion m:n.’ n For this concept we consider positive real numbers. We know that the lengths of line segments and area of any figure are positive real numbers. We know the formula of area of a triangle. 1 Area of a triangle = 2 Base ´ Height Let’s learn. Ratio of areas of two triangles Let’s find the ratio of areas of any two triangles. Ex. In D ABC, AD is the height and BC is the base. P In D PQR, PS is the height and A QR is the base 1 ´ BC ´ AD S A( D ABC) 2 B D C Q R = A( D PQR) 1 ´ QR ´ PS Fig. 1.1 Fig. 1.2 2 1  A(D ABC) BC ´ AD \ = A(D PQR) QR ´ PS Hence the ratio of the areas of two triangles is equal to the ratio of the products of their bases and corrosponding heights. Base of a triangle is b1 and height is h1. Base of another triangle is b2 and b1 ´ h1 height is h2. Then the ratio of their areas = b2 ´ h2 Suppose some conditions are imposed on these two triangles, Condition 1 ः If the heights of both triangles are equal then- A P h h B D C Q S R Fig. 1.3 Fig. 1.4 A(D ABC) BC ´ h BC = = A(D PQR) QR ´ h QR A(D ABC) b1 \ = A(D PQR) b2 Propertyः The ratio of the areas of two triangles with equal heights is equal to the ratio of their corresponding bases. Condition 2 ः If the bases of both triangles are equal then - C A(D ABC) AB ´ h1 = A(D APB) AB ´ h2 h1 P A(D ABC) h1 h2 = A(D APB) h2 A D Q B Fig. 1.5 Propertyः The ratio of the areas of two triangles with equal bases is equal to the ratio of their corresponding heights. 2  Activity : Fill in the blanks properly. L (i) A (ii) D B P R Q C M N P Q Fig. 1.6 Fig.1.7 A( D ABC) ´ A(D LMN) ´ = = = = A( D APQ) ´ A(D DMN) ´ (iii) M is the midpoint of C seg AB and seg CM is a median of D ABC \ A(D AMC) = A(D BMC) A M B = = Fig. 1.8 State the reason. ˜˜˜˜˜˜˜˜˜˜˜˜˜ Solved Examples ™™™™™™™™™™™™™ Ex. (1) D A In adjoining figure AE ^ seg BC, seg DF ^ line BC, A( D ABC) AE = 4, DF = 6 , then find A( D DBC). B E C F Fig.1.9 A( D ABC) AE Solution ः =.......... bases are equal, hence areas proportional to A( D DBC) DF heights. 4 2 = 6 = 3 3  In D ABC point D on side BC is such that DC = 6, BC = 15. Ex. (2) Find A(D ABD) : A(D ABC) and A(D ABD) : A(D ADC). Solution ः Point A is common vertex of A D ABD, D ADC and D ABC and their bases are collinear. Hence, heights of these three B C P D triangles are equal Fig. 1.10 BC = 15, DC = 6 \ BD = BC - DC = 15 - 6 = 9 A( D ABD) BD =.......... heights equal, hence areas proportional to A( D ABC) BC bases. 9 3 = 15 = 5 A( D ABD) = BD.......... heights equal, hence areas proportional to A( D ADC) DC bases. 9 3 = 6 = 2 Ex. (3) A D c ABCD is a parallelogram. P is any point on side BC. Find two pairs of triangles with equal areas. B P C Fig. 1.11 Solution ः c ABCD is a parallelogram. \ AD || BC and AB || DC Consider D ABC and D BDC. Both the triangles are drawn in two parallel lines. Hence the distance between the two parallel lines is the height of both triangles. In D ABC and D BDC, common base is BC and heights are equal. Hence, A(D ABC) = A(D BDC) In D ABC and D ABD, AB is common base and and heights are equal. \ A(D ABC) = A(D ABD) 4 Ex.(4) A In adjoining figure in D ABC, point D is P on side AC. If AC = 16, DC = 9 and D BP ^ AC, then find the following ratios. B A( D ABD) A( D BDC) (i) (ii) A( D ABC) A( D ABC) A( D ABD) C (iii) A( D BDC) Fig. 1.12 Solution ः In D ABC point P and D are on side AC, hence B is common vertex of D ABD, D BDC, D ABC and D APB and their sides AD, DC, AC and AP are collinear. Heights of all the triangles are equal. Hence, areas of these triangles are proportinal to their bases. AC = 16, DC = 9 \ AD = 16 - 9 = 7 A( D ABD) AD 7 \ = =........ triangles having equal heights A( D ABC) AC 16 A( D BDC) DC 9 = =........ triangles having equal heights A( D ABC) AC 16 A( D ABD) AD 7 = =........ triangles having equal heights A( D BDC) DC 9 Remember this ! Ratio of areas of two triangles is equal to the ratio of the products of their bases and corresponding heights. Areas of triangles with equal heights are proportional to their corresponding bases. Areas of triangles with equal bases are proportional to their corresponding heights. Practice set 1.1 1. Base of a triangle is 9 and height is 5. Base of another triangle is 10 and height is 6. Find the ratio of areas of these triangles. 5  C 2. In figure 1.13 BC ^ AB, AD ^ AB, A B A( D ABC) BC = 4, AD = 8, then find. A( D ADB) D Fig. 1.13 P T 3. In adjoining figure 1.14 seg PS ^ seg RQ seg QT ^ seg PR. If RQ = 6, PS = 6 and PR = 12, R Q S then find QT. Fig. 1.14 A D 4. In adjoining figure, AP ^ BC, AD || BC, then find A(D ABC) ः A(D BCD). B P C Fig. 1.15 A 5. In adjoining figure PQ ^ BC, AD ^ BC then find following ratios. P A( D PBC) (i) A( D PQB) (ii) A( D PBC) A( D ABC) B Q D C A( D ABC) A( D ADC) (iii) A( D ADC) (iv) A( D PQC) Fig. 1.16 6  Let’s learn. Basic proportionality theorem Theorem : If a line parallel to a side of a triangle intersects the remaining sides in two distinct points, then the line divides the sides in the same proportion. A Given ः In D ABC line l || line BC and line l intersects AB and P Q l AC in point P and Q respectively AP AQ To prove ः = PB QC Construction: Draw seg PC and seg BQ B C Fig. 1.17 Proof ः D APQ and D PQB have equal heights. A(D APQ) AP \ =.......... (I) (areas proportionate to bases) A(D PQB) PB A(D APQ) AQ and =.......... (II) (areas proportionate to bases) A(D PQC) QC seg PQ is common base of D PQB and D PQC. seg PQ || seg BC, hence D PQB and D PQC have equal heights. A(D PQB) = A(D PQC).......... (III) A(D APQ) A(D APQ) =.......... [from (I), (II) and (III)] A(D PQB) A(D PQC) \ AP = AQ.......... [from (I) and (II)] PB QC Converse of basic proportionality theorem Theorem : If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. In figure 1.18, line l interesects the side AB and side AC of D ABC in the AP AQ points P and Q respectively and = , hence line l || seg BC. PB QC 7  A This theorem can be proved by indirect method. Q P l B C Fig. 1.18 Activity : Draw a D ABC. Bisect Ð B and name the point A of intersection of AC and the angle bisector as D. D Measure the sides. AB = cm BC = cm AD = cm DC = cm B C AB AD Find ratios and. Fig. 1.19 BC DC You will find that both the ratios are almost equal. Bisect remaining angles of the triangle and find the ratios as above. You can verify that the ratios are equal. Let’s learn. Property of an angle bisector of a triangle Theorem : The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides. Given : In D ABC, bisector of Ð C interesects D seg AB in the point E. AE CA To prove : = CB C EB Construction : Draw a line parallel to ray CE, passing through the point B. Extend Fig. 1.20 AC so as to intersect it at point D. A E B 8 Proof : ray CE || ray BD and AD is transversal, \ Ð ACE = Ð CDB.......... (corresponding angles)...(I) Now taking BC as transversal Ð ECB = Ð CBD.......... (alternate angle)...(II) But Ð ACE @ Ð ECB.......... (given)...(III) \ Ð CBD @ Ð CDB.......... [from (I), (II) and (III)] In D CBD, side CB @ side CD.......(sides opposite to congruent angles) \ CB = CD...(IV) Now in D ABD, seg EC || seg BD.......... (construction) AE AC \ =....(Basic proportionality theorem)..(V) EB CD AE AC \ =.......... [from (IV) and (V)] EB CB For more information : Write another proof of the theorem yourself. Draw DM ^ AB and DN ^ AC. Use the following properties and write the proof. A (1) The areas of two triangles of equal heights are M proportional to their bases. N B D C Fig. 1.21 (2) Every point on the bisector A of an angle is equidistant M from the sides of the angle. B P N C Fig. 1.22 9 Converse of angle bisector theorem AB BD If in D ABC, point D on side BC such that = , then ray AD bisects AC DC Ð BAC. Property of three parallel lines and their transversals Activity: t1 t2 Draw three parallel lines. Label them as l, m, n. A P l Draw transversals t1 and t2. B Q m AB and BC are intercepts on transversal t1. C R n PQ and QR are intercepts on transversal t2. Fig. 1.23 AB PQ Find ratios and. You will find that they are almost equal. BC QR Theorem ः The ratio of the intercepts made on a transversal by three parallel lines is equal to the ratio of the corrosponding intercepts made on any other transversal by the same parallel lines. Given : line l || line m || line n t1 t2 t1 and t2 are transversals. Transversal t1 intersects the A P l lines in points A, B, C and t2 intersects the lines in points P, B Q m D Q, R. C R n AB PQ To prove : = BC QR Fig. 1.24 Proof : Draw seg PC , which intersects line m at point D. In D ACP, BD || AP AB PD \ =..... (I) (Basic proportionality theorem) BC DC In D CPR, DQ || CR PD PQ \ =..... (II) (Basic proportionality theorem) DC QR AB PD PQ AB PQ \ = =...... from (I) and (II). \ = BC DC QR BC QR 10  Remember this ! A (1) Basic proportionality theorem. P In D ABC, if seg PQ || seg AC AP QC then = BP BQ B Q C Fig. 1.25 (2) Converse of basic proportionality P theorem. PS PT In DPQR, if = S T SQ TR then seg ST || seg QR. Q R Fig. 1.26 A (3) Theorem of bisector of an angle of a triangle. D If in D ABC, BD is bisector of Ð ABC, AB AD then = BC DC B C Fig. 1.27 (4) Property of three parallel lines and their transversals. B C l A If line AX || line BY || line CZ and line l and line m are their X Y Z m AB XY transversals then = BC YZ Fig. 1.28 11  ˜˜˜˜˜˜˜˜˜˜˜˜˜ Solved Examples ™™™™™™™™™™™™™ Ex. (1) In D ABC, DE || BC A If DB = 5.4 cm, AD = 1.8 cm D E EC = 7.2 cm then find AE. Solution : In D ABC, DE || BC AD AE B C \ =..... Basic proportionality theorem Fig. 1.29 DB EC 1.8 AE \ 54 =. 7.2 \ AE ´ 5.4 = 1.8 ´ 7.2 1.8 ´ 7.2 \ AE = 5.4 = 2.4 AE = 2.4 cm Ex. (2) In D PQR, seg RS bisects Ð R. R If PR = 15, RQ = 20 PS = 12 O O then find SQ. P Solution : In D PRQ, seg RS bisects Ð R. PR PS S =...... property of angle bisector RQ SQ 15 12 = Q 20 SQ 12 ´ 20 Fig. 1.30 SQ = 15 \ SQ = 16 Activity : In the figure 1.31, AB || CD || EF A B If AC = 5.4, CE = 9, BD = 7.5 then find DF C D Solution : AB || CD || EF AC =.....( ) E F DF 5.4 Fig. 1.31 = \ DF = 9 DF 12  Activity : A In D ABC, ray BD bisects Ð ABC. D A-D-C, side DE || side BC, A-E-B then E AB AE prove that, = BC EB B C Fig. 1.32 Proof : In D ABC, ray BD bisects Ð B. \ AB = AD.. (I)(Angle bisector theorem) BC DC In D ABC, DE || BC AE = AD..... (II) (........... ) EB DC AB =...... from (I) and (II) EB Practice set 1.2 1. Given below are some triangles and lengths of line segments. Identify in which figures, ray PM is the bisector of Ð QPR. (1) (2) (3) R Q Q 3.5 M 1.5 R 6 3.6 M M 9 8 7 3 7 Q 4 10 R P P P 10 Fig. 1.33 Fig. 1.34 Fig. 1.35 P 2. In D PQR, PM = 15, PQ = 25 PR = 20, NR = 8. State whether line N M NM is parallel to side RQ. Give reason. R Fig. 1.36 Q 13  M 5 2.5 3. In D MNP, NQ is a bisector of Ð N. Q If MN = 5, PN = 7 MQ = 2.5 then N find QP. 7 P Fig. 1.37 A 4. Measures of some angles in the figure 60° P Q are given. Prove that AP AQ = 60° PB QC B C Fig. 1.38 A B 5. In trapezium ABCD, side AB || side PQ || side DC, AP = 15, P Q PD = 12, QC = 14, find BQ. D C Fig. 1.39 M 14 Q 6. Find QP using given information 25 in the figure. N 40 P Fig. 1.40 B 8 D 4 F 7. In figure 1.41, if AB || CD || FE then find x and AE. 12 C x E A Fig. 1.41 14  L 8. In D LMN, ray MT bisects Ð LMN 6 T 8 If LM = 6, MN = 10, TN = 8, N then find LT. M 10 Fig. 1.42 A x-2 9. In D ABC, seg BD bisects Ð ABC. D x If AB = x, BC = x + 5, x+2 AD = x – 2, DC = x + 2, then find B x+5 C the value of x. Fig. 1.43 D 10. In the figure 1.44, X is any point in the interior of triangle. Point X is P joined to vertices of triangle. X Seg PQ || seg DE, seg QR || seg EF. Fill in the blanks to prove that, Q R E F seg PR || seg DF. Fig. 1.44 Proof : In D XDE, PQ || DE.......... XP \ =.......... (I) (Basic proportionality theorem) QE In D XEF, QR || EF.......... \ =..........(II) \ =.......... from (I) and (II) \ seg PR || seg DE.......... (converse of basic proportionality theorem) 11«. In D  ABC, ray BD bisects Ð  ABC and ray CE bisects Ð  ACB. If seg AB @ seg AC then prove that ED || BC. 15  Let’s recall. Similar triangles A In D ABC and D DEF, if Ð A @ Ð D, Ð B @ Ð E, Ð C @ Ð F D AB BC AC and = = DE EF DF B C E F then D ABC and D DEF are similar Fig. 1.45 triangles. ‘D ABC and D DEF are similar’ is expressed as ‘D ABC ~ D DEF’ Let’s learn. Tests of similarity of triangles For similarity of two triangles, the necessary conditions are that their corresponding sides are in same proportion and their corresponding angles are congruent. Out of these conditions; when three specifc conditions are fulfilled, the remaining conditions are automatically fulfilled. This means for similarity of two triangles, only three specific conditions are sufficient. Similarity of two triangles can be confirmed by testing these three conditions. The groups of such sufficient conditions are called tests of similarity, which we shall use. AAA test for similarity of triangles For a given correspondence of P vertices, when corresponding angles of A two triangles are congruent, then the two triangles are similar. In D ABC and D PQR, in the B C Q R correspondence ABC « PQR if Fig. 1.46 Ð A @ Ð P, Ð B @ Ð Q and Ð C @ Ð R then D ABC ~ D PQR. 16  For more information ः Proof of AAA test A Given : In D ABC and D PQR, P Ð A @ Ð P, Ð B @ Ð Q, M N Ð C @ Ð R. Q R To prove ः D ABC ~ D PQR B C Let us assume that D ABC is bigger Fig. 1.47 than D PQR. Mark point M on AB, and point N on AC such that AM = PQ and AN = PR. Show that D AMN @ D PQR. Hence show that MN || BC. Now using basic proportionality theorem, AM = AN MB NC MB NC That is =........... (by invertendo) AM AN MB+AM NC + AN AM =........... (by componendo) AN AB AC \ = AM AN AB AC \ = PQ PR AB BC = Similarly it can be shown that PQ QR AB BC AC \ = = \ D ABC ~ D PQR PQ QR PR A A test for similarity of triangles: We know that for a given correspondence of vertices, when two angles of a triangle are congruent to two corresponding angles of another triangle, then remaining angle of first triangle is congruent to the remaining angle of the second triangle. This means, when two angles of one triangle are congruent to two corresponding angles of another triangle then this condition is sufficient for similarity of two triangles. This condition is called AA test of similarity. 17 SAS test of similarity of triangles For a given correspondence of vertices of two triangles, if two pairs of corresponding sides are in the same proportion and the angles between them are congruent, then the two triangles are similar. For example, if in D KLM and D RST, R K Ð KLM @ Ð RST 1 1.5 KL LM 2 L 2 M S 3 T = = 3 RS ST Fig. 1.48 Therefore, D KLM ~ D RST SSS test for similarity of triangles For a given correspondence of vertices of two triangles, when three sides of a triangle are in proportion to corresponding three sides of another triangle, then the two triangles are similar. X For example, if in D PQR and D XYZ, P Q PQ QR PR If = = R YZ XY XZ Y Z then D PQR ~ D ZYX Fig. 1.49 Properties of similar triangles ः (1) D ABC ~ D ABC - Reflexivity (2) If D ABC ~ D DEF then D DEF ~ D ABC - Symmetry (3) If D ABC ~ D DEF and D DEF ~ D GHI, then D ABC ~ D GHI - Transitivity ˜˜˜˜˜˜˜˜˜˜˜˜˜ Solved Examples ™™™™™™™™™™™™™ Ex. (1) In D XYZ, Ð Y = 100°, Ð Z = 30°, In D LMN, X Ð M = 100°, Ð N = 30°, L Are D XYZ and D LMN similar? If yes, by which test? Y Z M N Fig. 1.50 18 Solution : In D XYZ and D LMN, Ð Y = 100°, Ð M = 100°, \ Ð Y @ Ð M ÐZ = 30°, Ð N = 30°, \ Ð Z @ Ð N \ D XYZ ~ D LMN..... by AA test. Ex. (2) Are two triangles in figure 1.51 similar, according to the P information given? If yes, by 6 U 3 which test? M 10 N V 5 W Solution : In D PMN and D UVW Fig. 1.51 PM 6 2 MN 10 2 = = , = = 1 UV 3 1 VW 5 PM MN \ = UV VW and Ð M @ Ð V......... Given DPMN ~ DUVW......... SAS test of similarity Ex. (3) Can we say that the two triangles in figure 1.52 similar, according to information given? If yes, by M which test ? X 21 Solution : D XYZ and D MNP, 14 XY 14 2 MN = 21 = 3 , Y 20 Z N 30 P YZ 20 2 Fig. 1.52 NP = 30 = 3 It is given that Ð Z @ Ð P. But Ð Z and Ð P are not included angles by sides which are in proportion. \ D XYZ and D MNP can not be said to be similar. 19 Ex. (4) In the adjoining figure BP ^ AC, CQ ^ AB, A A – P- C, A- Q- B , then prove that P D APB and D AQC are similar. Solution : In D APB and D AQC Q ° (I) Ð APB = Ð AQC = ° (II) \ Ð APB @ Ð AQC....from(I) and (II) B C Ð PAB @ Ð QAC.... ( ) Fig. 1.53 \ D APB ~ D AQC..... AA test Ex. (5) Diagonals of a quadrilateral ABCD intersect in point Q. If 2QA = QC , 2QB = QD, then prove that DC = 2AB. B A Given : 2QA = QC 2QB = QD Q To prove : CD = 2AB C D Fig. 1.54 QA 1 Proof : 2QA = QC \ =.......... (I) QC 2 QB 1 2QB = QD \ =.......... (II) QD 2 QA QB \ =..........from (I) and (II) QC QD In D AQB and D CQD, QA QB =.......... proved QC QD Ð AQB @ Ð DQC.......... opposite angles \ D AQB ~D CQD.......... (SAS test of similarity) AQ QB AB \ = =.......... correponding sides are CQ QD CD proportional AQ AB 1 1 But = \ = 2 2 CQ CD \ 2AB = CD 20  Practice set 1.3 A 1. In figure1.55, Ð ABC=75°, Ð EDC=75° state which two triangles D are similar and by which test? Also 75° write the similarity of these two C triangles by a proper one to one 75° E correspondence. B Fig. 1.55 P 2. Are the triangles in figure 1.56 10 L 6 similar? If yes, by which test ? 3 5 Q 8 R M 4 N Fig. 1.56 3. As shown in figure 1.57, two poles of A P height 8 m and 4 m are perpendicular to the ground. If the length of shadow 8 4 of smaller pole due to sunlight is Q R B x C 6 6 m then how long will be the shadow Fig. 1.57 of the bigger pole at the same time ? A Q 4. In D ABC, AP ^ BC, BQ ^ AC B- P-C, A-Q - C then prove that, D CPA ~ D CQB. If AP = 7, BQ=8, BC=12 B P C then find AC. Fig. 1.58 21  P Q 5. Given : In PQRS, trapezium side PQ || side SR, AR = 5AP, A AS = 5AQ then prove that, SR = 5PQ S R Fig. 1.59 C D 6. In trapezium ABCD, (Figure 1.60) O side AB || side DC, diagonals AC and BD intersect in point O. If AB = 20, A DC = 6, OB = 15 then find OD. B Fig. 1.60 A D 7. c ABCD is a parallelogram point E is on side BC. Line DE intersects ray B E C AB in point T. Prove that DE ´ BE = CE ´ TE. Fig. 1.61 T D A 8. In the figure, seg AC and seg BD intersect each other in point P and P AP BP =. Prove that, CP DP C D ABP ~ D CDP B Fig. 1.62 A 9. In the figure, in D ABC, point D on Fig. 1.63 side BC is such that, Ð BAC = Ð ADC. B D C Prove that, CA2 = CB ´ CD 22  Let’s learn. Theorem of areas of similar triangles Theorem : When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides. A P B C Q R D S Fig. 1.64 Given : D ABC ~ D PQR, AD ^ BC, PS ^ QR A(D ABC) AB2 BC2 AC2 To prove: = = = A(D PQR) PQ2 QR2 PR2 A(D ABC) BC ´ AD BC AD Proof : = = ´.......... (I) A(D PQR) QR ´ PS QR PS In D ABD and D PQS, Ð B = Ð Q.......... given Ð ADB = Ð PSQ = 90° \ According to AA test D ABD ~ D PQS AD AB \ =.......... (II) PS PQ But D ABC ~ D PQR AB BC \ =.......... (III) PQ QR From (I), (II) and (III) A(D ABC) BC AD BC BC BC2 AB2 AC2 = ´ = ´ = = = A(D PQR) QR PS QR QR QR2 PQ2 PR2 23  ˜˜˜˜˜˜˜˜˜˜˜˜˜ Solved Examples ™™™™™™™™™™™™™ Ex. (1) ः DABC ~ D PQR , A (DABC) = 16 , A (DPQP) = 25, then find the value of ratio AB. PQ Solution ः DABC ~ D PQR A(D ABC) AB2 \ =.......... theorem of areas of similar triangles A(D PQR) PQ2 16 AB2 AB 4 \ 25 = \ =.......... taking square roots PQ2 PQ 5 Ex. (2) Ratio of corresponging sides of two similar triangles is 2ः5, If the area of the small triangle is 64 sq.cm. then what is the area of the bigger triangle ? Solution ः Assume that D ABC ~ D PQR. D ABC is smaller and D PQR is bigger triangle. A(D ABC) (2) 2 4 \ = =.......... ratio of areas of similar triangles A(D PQR) (5) 2 25 64 4 \ = A(D PQR) 25 4 ´ A(D PQR) = 64 ´ 25 64 ´ 25 A(D PQR) = 4 = 400 \ area of bigger triangle = 400 sq.cm. Ex. (3) In trapezium ABCD, side AB || side CD, diagonal AC and BD intersect A(D ABP) AB2 each other at point P. Then prove that =. A(D CPD) CD2 B A Solution ः In trapezium ABCD side AB || side CD In D APB and DCPD P ÐPAB @ ÐPCD...... alternate angles ÐAPB @ ÐCPD....... opposite angles C D \ DAPB ~ DCPD....... AA test of similarity Fig. 1.65 A(D APB) AB2 =...... theorem of areas of similar triangles A(D CPD) CD2 24  Practice set 1.4 1. The ratio of corresponding sides of similar triangles is 3 ः 5; then find the ratio of their areas. 2. If DABC ~ DPQR and AB: PQ = 2:3, then fill in the blanks. A(D ABC) AB2 22 = = 2 = A(D PQR) 3 3. If D ABC ~ D PQR, A (D ABC) = 80, A (D PQR) = 125, then fill in the blanks. A(D ABC) 80 AB = 125 \ = A(D.... ) PQ 4. D LMN ~ D PQR, 9 ´ A (DPQR ) = 16 ´ A (DLMN). If QR = 20 then find MN. 5. Areas of two similar triangles are 225 sq.cm. 81 sq.cm. If a side of the smaller triangle is 12 cm, then find corresponding side of the bigger triangle. 6. DABC and D DEF are equilateral triangles. If A(DABC) ः A (D DEF) = 1 ः 2 and AB = 4, find DE. 7. In figure 1.66, seg PQ || seg DE, A(D PQF) = 20 units, PF = 2 DP, then find A( c DPQE) by completing the following activity. A(D PQF) = 20 units, PF = 2 DP, Let us assume DP = x. \ PF = 2x D DF = DP + = + = 3x In D FDE and D FPQ, P Ð FDE @ Ð.......... corresponding angles Ð FED @ Ð.......... corresponding angles \ D FDE ~ D FPQ.......... AA test E Q F (3 x) 2 9 \ A(D FDE) = = (2 x) 2 = 4 Fig. 1.66 A(D FPQ ) 9 9 A(D FDE) = 4 A( D FPQ ) = 4 ´ = A(c DPQE) = A( D FDE) - A( D FPQ) = - = 25  Problem set 1 1. Select the appropriate alternative. (1) In D ABC and D PQR, in a one to one correspondence A Q AB BC CA = = then QR PR PQ (A) D PQR ~ D ABC (B) D PQR ~ D CAB B C R P (C) D CBA ~ D PQR Fig. 1.67 (D) D BCA ~ D PQR (2) If in D DEF and D PQR, Q Ð D @ Ð Q, Ð R @ Ð E then which of the following D statements is false ? EF DF DE EF (A) = (B) = PR PQ PQ RP R DE DF EF DE E F P (C) = (D) = QR PQ RP QR Fig. 1.68 (3) In D ABC and D DEF Ð B = Ð E, Ð F = ÐC and AB = 3DE then which of the statements regarding A the two triangles is true ? (A)The triangles are not congruent D and not similar (B) The triangles are similar but B C E F not congruent. Fig. 1.69 (C) The triangles are congruent and similar. (D) None of the statements above is D true. (4) D ABC and D DEF are equilateral A triangles, A (DABC) ः A (DDEF) = 1 ः 2 If AB = 4 then what is length of DE? B C E F (A) 2 2 (B) 4 (C) 8 (D) 4 2 Fig. 1.70 26  (5) In figure 1.71, seg XY || seg BC, then which of the following statements is true? AB AX AX AY A (A) AC = AY (B) XB = AC X Y AX AY AB AC (C) YC = XB (D) YC = XB B C 2. In D ABC, B - D – C and BD = 7, Fig. 1.71 BC = 20 then find following ratios. A (1) A(D ABD) A(D ADC) (2) A(D ABD) A(D ABC) B C (3) A(D ADC) D A(D ABC) Fig. 1.72 3. Ratio of areas of two triangles with equal heights is 2 ः 3. If base of the smaller triangle is 6 cm then what is the corresponding base of the bigger triangle ? D 4. In figure 1.73, ÐABC = ÐDCB = 90° A AB = 6, DC = 8 8 6 A( D ABC) then A( D DCB) = ? B C Fig. 1.73 P 5. In figure 1.74, PM = 10 cm N Q S A(DPQS) = 100 sq.cm M A(DQRS) = 110 sq.cm then find NR. R Fig. 1.74 6. D MNT ~ D QRS. Length of altitude drawn from point T is 5 and length of altitude A( D MNT) drawn from point S is 9. Find the ratio. A( D QRS) 27  A 5 7. In figure 1.75, A – D – C and B – E – C D seg DE || side AB If AD = 5, 3 DC = 3, BC = 6.4 then find BE. B x E 6.4 - x C Fig. 1.75 S D 8. In the figure 1.76, seg PA, seg QB, seg RC and seg SD are perpendicular R C to line AD. Q B AB = 60, BC = 70, CD = 80, PS = 280 P A then find PQ, QR and RS. Fig. 1.76 P 9. In D PQR seg PM is a median. Angle bisectors of ÐPMQ and ÐPMR intersect side PQ and side PR in points X and Y X Y respectively. Prove that XY || QR. O x O x Q M R Complete the proof by filling in the boxes. Fig. 1.77 In D PMQ, ray MX is bisector of ÐPMQ. \ =.......... (I) theorem of angle bisector. In D PMR, ray MY is bisector of ÐPMR. \ =.......... (II) theorem of angle bisector. MP MP But MQ = MR.......... M is the midpoint QR, hence MQ = MR. PX PY \ XQ = YR \ XY || QR.......... converse of basic proportionality theorem. 28  A 10. In fig 1.78, bisectors of Ð B and Ð C of D ABC intersect each other in point X. Line AX intersects side BC in X point Y. AB = 5, AC = 4, BC = 6 then AX B C find XY. Y Fig. 1.78 A D 11. In c ABCD, seg AD || seg BC. P Diagonal AC and diagonal BD intersect each other in point P. Then B C AP PC Fig. 1.79 show that = PD BP 12. In fig 1.80, XY || seg AC. A If 2AX = 3BX and XY = 9. Complete the activity to find the value of AC. X AX Activity : 2AX = 3BX \ = B C BX Y Fig. 1.80 AX +BX + =.......... by componendo. BX AB =.......... (I) BX D BCA ~ D BYX.......... test of similarity. BA AC \ =.......... corresponding sides of similar triangles. BX XY A AC \ = \ AC =...from (I) 9 G F 13«. In figure1.81, the vertices of square DEFG are on the sides of D ABC. Ð A = 90°. Then prove that B D E C DE2 = BD ´ EC Fig. 1.81 (Hint ः Show that D GBD is similar to D CFE. Use GD = FE = DE.) rrr 29  2 Pythagoras Theorem Let’s study. · Pythagorean triplet · Similarity and right angled triangles · Theorem of geometric mean · Pythagoras theorem · Application of Pythagoras theorem · Apollonius theorem Let’s recall. Pythagoras theorem : In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of remaning two sides. R In D PQR Ð PQR = 90° l(PR)2 = l(PQ)2 + l(QR)2 We will write this as, P Q Fig. 2.1 PR = PQ + QR 2 2 2 The lengths PQ, QR and PR of D PQR can also be shown by letters r, p and q. With this convention, refering to figure 2.1, Pythagoras theorem can also be stated as q2 = p2 + r2. Pythagorean Triplet : In a triplet of natural numbers, if the square of the largest number is equal to the sum of the squares of the remaining two numbers then the triplet is called Pythagorean triplet. For Example: In the triplet ( 11, 60, 61 ) , 112 = 121, 602 = 3600, 612 = 3721 and 121 + 3600 = 3721 The square of the largest number is equal to the sum of the squares of the other two numbers. \ 11, 60, 61 is a Pythagorean triplet. Verify that (3, 4, 5), (5, 12, 13), (8, 15, 17), (24, 25, 7) are Pythagorean triplets. Numbers in Pythagorean triplet can be written in any order. 30  For more information Formula for Pythagorean triplet: If a, b, c are natural numbers and a > b, then [(a2 + b2),(a2 - b2),(2ab)] is Pythagorean triplet. (a2 + b2)2 = a4 + 2a2b2 + b4.......... (I) \ (a2 - b2) = a4 - 2a2b2 + b4.......... (II) (2ab)2 = 4a2b2.......... (III) \by (I), (II) and (III) , (a2 + b2)2 = (a2 - b2)2 + (2ab)2 \[(a2 + b2), (a2 - b2), (2ab)] is Pythagorean Triplet. This formula can be used to get various Pythagorean triplets. For example, if we take a = 5 and b = 3, a2 + b2 = 34, a2 - b2 = 16 , 2ab = 30. Check that (34, 16, 30) is a Pythagorean triplet. Assign different values to a and b and obtain 5 Pythagorean triplet. Last year we have studied the properties of right angled triangle with the angles 30° - 60° - 90° and 45° - 45° - 90°. (I)Property of 30°-60°-90° triangle. If acute angles of a right angled triangle are 30° and 60°, then the side opposite 3 30°angle is half of the hypotenuse and the side opposite to 60° angle is times the 2 hypotenuse. See figure 2.2. In D LMN, Ð L = 30°, Ð N = 60°, Ð M = 90° L 1 \side opposite 30°angle = MN = ´ LN 2 30° side opposite 60°angle = LM = 3 ´ LN 2 If LN = 6 cm, we will find MN and LM. 1 3 MN = ´ LN LM = ´ LN 2 2 1 3 = ´ 6 = ´6 90° 60° 2 2 M N Fig. 2.2 = 3 cm =3 3 cm 31 (II ) Property of 45°-45°-90° If the acute angles of a right angled triangle are 45° and 45°, then each of the 1 perpendicular sides is times the hypotenuse. 2 See Figure 2.3. In D XYZ, 1 Z XY = ´ ZY 2 45° 1 XZ = ´ ZY 2 1 \ XY = XZ = ´ ZY 2 If ZY = 3 2 cm then we will find XY and 45° X Y ZX 1 Fig. 2.3 XY = XZ = ´ 3 2 2 XY = XZ = 3cm In 7th standard we have studied theorem of Pythagoras using areas of four right angled triangles and a square. We can prove the theorem by an alternative method. Activity: Take two congruent right angled triangles. Take another isosceles right angled triangle whose congruent sides are equal to the hypotenuse of the two congruent right angled triangles. Join these triangles to form a trapezium 1 Area of the trapezium = ´ (sum of the lengths of parallel sides) ´ height 2 Using this formula, equating the area of trapezium with the sum of areas of the three right angled triangles we can prove the theorem of Pythagoras. z x z y y x Fig. 2.4 32  Let’s learn. Now we will give the proof of Pythagoras theorem based on properties of similar triangles. For this, we will study right angled similar triangles. Similarity and right angled triangle Theorem : In a right angled triangle, if the altitude is drawn to the hypotenuse, then the two triangles formed are similar to the original triangle and to each other. A ´ Given : In D ABC, Ð ABC = 90°, seg BD ^ seg AC, A-D-C To prove : D ADB ~ D ABC D D BDC ~ D ABC D ADB ~ D BDC B ´ C Fig. 2.5 Proof : In D ADB and D ABC In D BDC and D ABC Ð DAB @ Ð BAC...(common angle) Ð BCD @ Ð ACB.....(common angle) Ð ADB @ Ð ABC... (each 90°) Ð BDC @ Ð ABC..... (each 90°) D ADB ~ D ABC... (AA test)... (I) D BDC ~ D ABC..... (AA test)... (II) \ D ADB ~ D BDC from (I) and (II)........(III) \ from (I), (II) and (III), D ADB ~ D BDC ~ D ABC....(transitivity) Theorem of geometric mean In a right angled triangle, the perpendicular segment to the hypotenuse from the opposite vertex, is the geometric mean of the segments into which the hypotenuse is divided. R Proof ः In right angled triangle PQR, seg QS ^ hypotenuse PR D QSR ~ D PSQ.......... ( similarity of right triangles ) QS SR S = PS SQ QS SR = P Q PS QS Fig. 2.6 QS2 = PS ´ SR \ seg QS is the ‘geometric mean’ of seg PS and SR. 33  Pythagoras Theorem In a right angled triangle, the square of the hypotenuse is equal to the sum of A the squares of remaining two sides. Given : In D ABC, ÐABC = 90° Fig. 2.7 To prove : AC2 = AB2 + BC2 D Construction :Draw perpendicular seg BD on side AC. A-D-C. B C Proof : In right angled D ABC, seg BD ^ hypotenuse AC..... (construction) \ D ABC ~ D ADB ~ D BDC..... (similarity of right angled triangles) D ABC ~ D ADB Similarly, D ABC ~ D BDC AB BC AC AB BC AC = = - corresponding BD = = -corresponding AD DB AB DC BC sides sides AB AC AC = BC = AD AB DC BC AB2 = AD ´ AC.......... (I) BC2 = DC ´ AC.......... (II) Adding (I) and (II) AB2 + BC2 = AD ´ AC + DC ´ AC = AC (AD + DC) = AC ´ AC.......... (A-D-C) \ AB2 + BC2= AC2 \ AC2 = AB2 + BC2 Converse of Pythagoras theorem In a triangle if the square of one side is equal to the sum of the squares of the remaining two sides, then the triangle is a right angled triangle. Given ः In D ABC, AC2 = AB2 + BC2 To prove ः Ð ABC = 90° A P B C Q R Fig. 2.8 Fig. 2.9 34 Construction : Draw D PQR such that, AB = PQ, BC = QR , Ð PQR = 90°. Proof : In D PQR, Ð Q = 90° PR2 = PQ2 + QR2.......... (Pythagoras theorem) = AB2 + BC2.......... (construction)......(I) = AC2.......... (given)......(II) \ PR2 = AC2 \ PR = AC.......... (III) \ D ABC @ D PQR.......... (SSS test) \ Ð ABC = Ð PQR = 90° Remember this ! (1) (a) Similarity and right angled triangle P In DPQR Ð Q = 90° , seg QS ^ seg PR, S DPQR ~ DPSQ ~ DQSR. Thus all the right angled triangles in the figure are similar to one another. Q R Fig. 2.10 (b) Theorem of geometric mean In the above figure, DPSQ ~ DQSR \ QS2 = PS ´ SR \ seg QS is the geometric mean of seg PS and seg SR (2) Pythagoras Theorem: In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of remaining two sides. (3) Converse of Pythagoras Theorem: In a triangle, if the square of one side is equal to the sum of the squares of the remaining two sides, then the triangle is a right angled triangle (4) Let us remember one more very useful property. In a right angled triangle, if one side is half of the hypotenuse then the angle opposite to that side is 30°. This property is the converse of 30°-60°-90° theorem. 35  ˜˜˜˜˜˜˜˜˜˜˜˜˜ Solved Examples ™™™™™™™™™™™™™ Ex. (1) See fig 2.11. In D ABC, Ð B= 90°, Ð A= 30°, AC=14, then find AB and BC Solution : In D ABC, A ÐB = 90°, ÐA = 30°, \ ÐC = 60° 30° By 30°- 60°- 90° theorem, 14 BC = 1 ´ AC AB = 3 ´ AC 2 2 1 3 BC = ´ 14 AB = ´ 14 2 2 60° BC = 7 AB = 7 3 B C Fig. 2.11 Ex. (2) See fig 2.12, In D ABC, seg AD ^ seg BC, ÐC = 45°, BD = 5 and AC = 8 2 then find AD and BC. Solution : In D ADC A ÐADC = 90°, ÐC = 45°, \ ÐDAC = 45° 1 8 AD = DC = ´ 8 2... by 45°-45°-90° theorem 2 2 DC = 8 \ AD = 8 BC = BD + DC B 5 D C 8 = 5 + 8 Fig. 2.12 BC = 13 Ex. (3) In fig 2.13, Ð PQR = 90°,seg QN ^ seg PR, PN = 9, NR = 16. Find QN. Solution : In D PQR, seg QN ^ seg PR NQ2 = PN ´ NR... theorem of geometric P 9 mean N \NQ = PN ´ NR 16 = 9 ´16 =3´4 Q R = 12 Fig. 2.13 36 Ex. (4) See figure 2.14.In D PQR, Ð PQR = 90°, seg QS ^ seg PR then find x, y, z. Solution : In D PQR, Ð PQR = 90°, seg QS ^ seg PR P QS = PS ´ SR.......... (theorem of geometric mean) = 10 ´ 8 10 = 5´ 2´8 z = 5 ´16 S x = 4 5 8 \ x = 4 5 Q y R Fig. 2.14 In D QSR, by Pythagoras theorem In D PSQ, by Pythagoras theorem QR2 = QS2 + SR2 PQ2 = QS2 + PS2 ( ) 2 = ( 4 5 ) + 8 2 2 = 4 5 + 102 = 16 ´ 5 + 100 = 16 ´ 5 + 64 = 80 + 100 = 80 + 64 = 180 = 144 = 36 ´ 5 \ QR = 12 \ PQ = 6 5 Hence x = 4 5, y = 12, z = 6 5 Ex. (5) In the right angled triangle, sides making right angle are 9 cm and 12 cm. Find the length of the hypotenuse Solution: In D PQR ,Ð Q = 90° P PR2 = PQ2 + QR2 (Pythagoras theorem) = 92 + 122 9 = 81 + 144 PR2 = 225 Q R PR = 15 12 Hypotenuse= 15 cm Fig. 2.15 37 Ex. (6) In D LMN ,l = 5, m = 13, n = 12. State whether D LMN is a right angled triangle or not. Solution : l = 5, m = 13, n = 12 l2= 25, m2 = 169, n2 = 144 \ m2 = l2 + n2 \ by converse of Pythagoras theorem D LMN is a right angled triangle. Ex. (7) See fig 2.16. In D ABC, seg AD ^ seg BC. Prove that: AB2 + CD2 = BD2 + AC2 C Solution : According to Pythagoras theorem, in D ADC AC2 = AD2 + CD2 \ AD2 = AC2 - CD2... (I) In D ADB D AB2 = AD2 + BD2 \ AD2 = AB2 - BD2... (II) B A \ AB2 - BD2 = AC2 - CD2..........from I and II Fig. 2.16 \ AB2 + CD2 = AC2 + BD2 Practice set 2.1 1. Identify, with reason, which of the following are Pythagorean triplets. (i)(3, 5, 4) (ii)(4, 9, 12) (iii)(5, 12, 13) (iv) (24, 70, 74) (v)(10, 24, 27) (vi)(11, 60, 61) M 2. In figure 2.17, Ð MNP = 90°, seg NQ ^seg MP, MQ = 9, Q QP = 4, find NQ. P N P Fig. 2.17 10 3. In figure 2.18, Ð QPR = 90°, seg PM ^ seg QR and Q-M-R, Q M R 8 PM = 10, QM = 8, find QR. Fig. 2.18 38  S 30° P 4. See figure 2.19. Find RP and PS using the information given in 6 Fig. 2.19 D PSR. R 5. For finding AB and BC with the A help of information given in figure 2.20, complete following activity. 8 AB = BC.......... \ Ð BAC = B C \ AB = BC = ´ AC Fig. 2.20 = ´ 8 = ´ 2 2 = 6. Find the side and perimeter of a square whose diagonal is 10 cm. 8 D G 7. In figure 2.21, Ð DFE = 90°, 12 FG ^ ED, If GD = 8, FG = 12, find (1) EG (2) FD and (3) EF E F Fig. 2.21 8. Find the diagonal of a rectangle whose length is 35 cm and breadth is 12 cm. P « 9. In the figure 2.22, M is the midpoint of QR. Ð PRQ = 90°. Prove that, PQ2 = 4PM2 - 3PR2 Q R M Fig. 2.22 10«. Walls of two buildings on either side of a street are parellel to each other. A ladder 5.8 m long is placed on the street such that its top just reaches the window of a building at the height of 4 m. On turning the ladder over to the other side of the street , its top touches the window of the other building at a height 4.2 m. Find the width of the street. 39  Let’s learn. Application of Pythagoras theorem In Pythagoras theorem, the relation between hypotenuse and sides making right angle i.e. the relation between side opposite to right angle and the remaining two sides is given. In a triangle, relation between the side opposite to acute angle and remaining two sides and relation of the side opposite to obtuse angle with remaining two sides can be determined with the help of Pythagoras theorem. Study these relations from the following examples. Ex. (1) In D ABC, Ð C is an acute angle, seg AD ^ seg BC. Prove that: AB2 = BC2 + AC2 - 2BC ´ DC In the given figure let AB = c, AC = b, AD = p, BC = a, DC = x, A \ BD = a - x In D ADB , by Pythagoras theorem c b c2 = (a-x)2 + p c2 = a2 - 2ax + x2 +.......... (I) In D ADC, by Pythagoras theorem B a-x D x C b2 = p2 + Fig. 2.23 p2 = b2 -.......... (II) Substituting value of p2 from (II) in (I), c2 = a2 - 2ax + x2 + b2 - x2 \ c2 = a2 + b2 - 2ax \ AB2 = BC2+ AC2 - 2BC ´ DC Ex. (2) In D ABC, Ð ACB is obtuse angle, seg AD ^ seg BC. Prove that: AB2 = BC2 + AC2 + 2BC ´ CD A In the figure seg AD ^ seg BC Let AD = p, AC = b, AB = c, c p b BC = a and DC = x. DB = a + x D x C a B In D ADB, by Pythagoras theorem, c2 = (a + x)2 + p2 Fig. 2.24 c2 = a2 + 2ax + x2 + p2.......... (I) 40  Similarly, in D ADC b2 = x2 + p2 \ p2 = b2 - x2.......... (II) \ substituting the value of p2 from (II) in (I) \ c2 = a2 + 2ax + b2 \ AB2 = BC2+ AC2 + 2BC ´ CD Apollonius theorem In D ABC, if M is the midpoint of side BC, then AB2 + AC2 = 2AM2 + 2BM2 A Given ःIn D ABC, M is the midpoint of side BC. To prove ः AB2 + AC2 = 2AM2 + 2BM2 B M D C Constructionः Draw seg AD ^ seg BC Fig. 2.25 Proof ः If seg AM is not perpendicular to seg BC then out of Ð AMB and Ð AMC one is obtuse angle and the other is acute angle In the figure, Ð AMB is obtuse angle and Ð AMC is acute angle. From examples (1) and (2) above, AB2 = AM2 + MB2 + 2BM ´ MD..... (I) and AC2 = AM2 + MC2 - 2MC ´ MD \ AC2 = AM2 + MB2 - 2BM ´ MD ( BM = MC)..........(II) \ \ adding (I) and (II) AB2 + AC2 = 2AM2 + 2BM2 Write the proof yourself if seg AM ^ seg BC. From this example we can see the relation among the sides and medians of a triangle. This is known as Apollonius theorem. ˜˜˜˜˜˜˜˜˜˜˜˜˜ Solved Examples ™™™™™™™™™™™™™ Ex. (1) In the figure 2.26, seg PM is a median of D PQR. PM = 9 and PQ2 + PR2 = 290, then find QR. Solution ः In D PQR , seg PM is a median. M is the midpoint of seg QR. 41  P 1 QM = MR = QR 2 PQ + PR = 2PM + 2QM2 (by Apollonius 2 2 2 theorem) 290 = 2 ´ 92 + 2QM2 9 290 = 2 ´ 81 + 2QM2 290 = 162 + 2QM2 2QM2 = 290 - 162 Q M R 2QM2 = 128 Fig. 2.26 QM2 = 64 QM = 8 \ QR = 2 ´ QM = 2 ´ 8 = 16 Ex. (2) Prove that, the sum of the squares of the diagonals of a rhombus is equal to the sum of the squares of the sides. P S Given : c PQRS is a rhombus. Diagonals PR and SQ intersect each other at T point T To prove : PS2 + SR2 + QR2 + PQ2 = PR2 + QS2 Q R Fig. 2.27 Proof : Diagonals of a rhombus bisect each other. \ by Apollonius’ theorem, PQ2 + PS2 = 2PT2 + 2QT2.......... (I) QR2 + SR2 = 2RT2 + 2QT2.......... (II) \ adding (I) and (II) , PQ2 + PS2+ QR2 + SR2 = 2(PT2 + RT2) + 4QT2 = 2(PT2 + PT2) + 4QT2.......... (RT = PT) = 4PT2 + 4QT2 = (2PT)2 + (2QT)2 = PR2 + QS2 (The above proof can be written using Pythagoras theorem also.) 42  Practice set 2.2 1. In D PQR, point S is the midpoint of side QR.If PQ = 11,PR = 17, PS =13, find QR. 2. In D ABC, AB = 10, AC = 7, BC = 9 then find the length of the median drawn from point C to side AB P 3. In the figure 2.28 seg PS is the median of D PQR and PT ^ QR. Prove that, 2  QR  (1) PR = PS + QR ´ ST + 2 2  2     QR  2 Q T S R ii) PQ = PS - QR ´ ST + 2 2  2  Fig. 2.28   A 4. In D ABC, point M is the midpoint of side BC. If, AB2 + AC2 = 290 cm2, AM = 8 cm, find BC. B M C Fig. 2.29 P Q « 5. In figure 2.30, point T is in the interior of rectangle PQRS, B A Prove that, TS + TQ = TP + TR 2 2 2 2 T (As shown in the figure, draw S R seg AB || side SR and A-T-B) Fig. 2.30 Problem set 2 1. Some questions and their alternative answers are given. Select the correct alternative. (1) Out of the following which is the Pythagorean triplet? (A) (1, 5, 10) (B) (3, 4, 5) (C) (2, 2, 2) (D) (5, 5, 2) (2) In a right angled triangle, if sum of the squares of the sides making right angle is 169 then what is the length of the hypotenuse? (A) 15 (B) 13 (C) 5 (D) 12 43  (3) Out of the dates given below which date constitutes a Pythagorean triplet ? (A) 15/08/17 (B) 16/08/16 (C) 3/5/17 (D) 4/9/15 (4) If a, b, c are sides of a triangle and a2 + b2 = c2, name the type of triangle. (A) Obtuse angled triangle (B) Acute angled triangle (C) Right angled triangle (D) Equilateral triangle (5) Find perimeter of a square if its diagonal is 10 2 cm. (A)10 cm (B) 40 2 cm (C) 20 cm (D) 40 cm (6) Altitude on the hypotenuse of a right angled triangle divides it in two parts of lengths 4 cm and 9 cm. Find the length of the altitude. (A) 9 cm (B) 4 cm (C) 6 cm (D) 2 6 cm (7) Height and base of a right angled triangle are 24 cm and 18 cm find the length of its hypotenus (A) 24 cm (B) 30 cm (C) 15 cm (D) 18 cm (8) In D ABC, AB = 6 3 cm, AC = 12 cm, BC = 6 cm. Find measure of Ð A. (A) 30° (B) 60° (C) 90° (D) 45° 2. Solve the following examples. (1) Find the height of an equilateral triangle having side 2a. (2) Do sides 7 cm , 24 cm, 25 cm form a right angled triangle ? Give reason. (3) Find the length a diagonal of a rectangle having sides 11 cm and 60cm. (4) Find the length of the hypotenuse of a right angled triangle if remaining sides are 9 cm and 12 cm. (5) A side of an isosceles right angled triangle is x.Find its hypotenuse. (6) In D PQR; PQ = 8 , QR = 5 , PR = 3. Is D PQR a right angled triangle ? If yes, which angle is of 90° ? 3. In D RST, Ð S = 90°, Ð T = 30°, RT = 12 cm then find RS and ST. 4. Find the diagonal of a rectangle whose length is 16 cm and area is 192 sq.cm. 5«. Find the length of the side and perimeter of an equilateral triangle whose height is 3 cm. 6. In D ABC seg AP is a median. If BC = 18, AB2 + AC2 = 260 Find AP. 44  1 7«. D ABC is an equilateral triangle. Point P is on base BC such that PC = BC, 3 if AB = 6 cm find AP. P 8. From the information given in the figure 2.31, prove that a a PM = PN = 3 ´a M N a Q S R a a Fig. 2.31 9. Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides. 10. Pranali and Prasad started walking to the East and to the North respectively, from the same point and at the same speed. After 2 hours distance between them was 15 2 km. Find their speed per hour. B 11«. In D ABC, Ð BAC = 90°, seg BL and seg CM are medians M of D ABC. Then prove thatः 4(BL2 + CM2) = 5 BC2 C L A Fig. 2.32 12. Sum of the squares of adjacent sides of a parallelogram is 130 sq.cm and length of one of its diagonals is14 cm. Find the length of the other diagonal. A 13. In D ABC, seg AD ^ seg BC DB = 3CD. Prove that ः 2AB2 = 2AC2 + BC2 C B D Fig. 2.33 14«. In an isosceles triangle, length of the congruent sides is 13 cm and its base is 10 cm. Find the distance between the vertex opposite the base and the centroid. 45 15. In a trapezium ABCD, D C seg AB || seg DC seg BD ^ seg AD, 15 15 seg AC ^ seg BC, If AD = 15, BC = 15 and AB = 25. Find A(c ABCD) A B 25 Fig. 2.34 P 16«. In the figure 2.35, D PQR is an equilatral triangle. Point S is on seg QR such that 1 QS = QR. 3 Prove that : 9 PS2 = 7 PQ2 60° 60° Q S T R Fig. 2.35 17«. Seg PM is a median of D PQR. If PQ = 40, PR = 42 and PM = 29, find QR. 18. Seg AM is a median of D ABC. If AB = 22, AC = 34, BC = 24, find AM ICT Tools or Links Obtain information on ‘the life of Pythagoras’ from the internet. Prepare a slide show. rrr 46  3 Circle Let’s study. · Circles passing through one, two,three points · Secant and tangent · Circles touching each other · Arc of a circle · Inscribed angle and intercepted arc · Cyclic quadrilateral · Secant tangent angle theorem · Theorem of intersecting chords Let’s recall. You are familiar with the concepts regarding circle, like - centre, radius, diameter, chord, interior and exterior of a circle. Also recall the meanings of - congruent circles, concentric circles and intersecting circles. congruent circles concentric circles intersecting circles Recall the properties of chord studied in previous standard and perform the activity below. Activity I : In the adjoining figure, seg DE is a chord of a circle with centre C. C seg CF ^ seg DE. If diameter of the D E F circle is 20 cm, DE =16 cm find CF. Fig. 3.1 Recall and write theorems and properties which are useful to find the solution of the above problem. (1) The perpendicular drawn from centre to a chord (2) (3) Using these properties, solve the above problem. 47  Activity II : In the adjoining figure, seg QR is a chord of the circle with O centre O. P is the midpoint of the chord QR. If QR = 24, OP = 10 , Q P R find radius of the circle. Fig. 3.2 To find solution of the problem, write the theorems that are useful. (1) (2) Using these theorems solve the problems. Activity III : In the adjoining figure, M D is the centre of the circle and S seg AB is a diameter. M A B seg MS ^ chord AD seg M