1. SIMILARITY. 1. Introduction.

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1 1. SIMILARITY 1. Introduction In earlier classes, you have learnt about congruence of two geometric figures, and also some basic theorems and results on the congruence of triangle. Two geometric figures having same shape and size are congruent to each other but two geometric Figures having same shape are called similar. Two congruent geometric Figures are always similar but the converse may or may not be true. All regular polygons of same number of sides such as equilateral triangle, squares, etc. are similar. All circles are similar. In some cases, we can easily notice that two geometric figures are not similar. For example, a triangle and a rectangle can never be similar. In case, we are given two triangles, they may appear to be similar but actually they may not be similar. So, we need some criteria to determine the similarity of two geometric Figures. In particular, we shall discuss similar triangles. Congruent Figures Two geometrical Figures are said to be congruent, provided they must have same shape and same size. Congruent Figures are alike in every respect. Ex. 1. Two squares of the same length.. Two circle of the same radii. 3. Two rectangles of the same dimensions. 4. Two wings of a fan. 5. Two equilateral triangles of same length. Similar Figures Two Figures are said to be similar, if they have the same shape. Similar Figures may differ in size. Thus, two congruent Figures are always similar, but two similar Figures need not be congruent. Ex. 1. Any two line segments are similar.. Any two equilateral triangles are similar.

2 1. Similarity 3. Any two squares are similar. 4. Any two circles are similar. We use the symbol ~ to indicate similarity of Figures. Figure 1.1. Similar Triangles ABC and DEF are said to be similar, if their corresponding angles are equal and the corresponding sides are proportional. i.e., when A D, B E, C F and, we write ABC ~ DEF. The sign ~ is read as is similar to. Theorem 1: (Thales Theorem or Basic Proportionality Theorem): If a line is drawn parallel to one side of a triangle intersecting the other two sides, then the other two sides are divided in the same ratio. Given: A ABC in which line l parallel to BC intersecting AB at D and AC at E. AD AE To prove: DB EC Construction: Join D to C and E to B. Through E drawn EF perpendicular to AB i.e., EF AB and through D draw DG AC. Proof: STATEMENT Figure 1. REASON 1. Area of ( ADE) 1 (AD EF) Area of 1 base altitude Figure 1.3 Area of ( BDE) 1 (BD EF). 1 AD EF Area ( ADE) AD Area ( BDE) 1 DB BD EF By 1

3 Foundation for Mathematics AE DG Area ( ADE) AE Area ( CDE) 1 EC EC DG Similarly 4. Area ( BDE) Area ( CDE) s BDE and CDE are on the same base BC and between the same parallel lines DE and BC. 5. Area( ADE) AE Area( BDE) EC by 3 and 4 6. AD AE by 1 and 5 DB EC Hence, proved. Theorem : (Converse of Basic Proportionality Theorem): If a line divides any two sides of a triangle proportionally, the line is parallel to the third side. Given: A ABC and DE is a line meeting AB and AC at D and E respectively such that AD AE DB EC To prove: DE BC Proof: STATEMENT 1. If possible, let DE be not parallel to BC. REASON Then, draw DF BC. AD AF DB FC by Basic Proportionality Theorem. 3. AD AE DB EC Given AF AE 4. FC EC From and 3. AF 1 AE 1 FC EC Adding 1 on both sides. AF FC AE EC FC EC By addition. AC AC FC EC AF + FC AC and AE + EC AC. FC EC E and F coincide. But, DF BC. Hence DE BC. Hence, proved. Figure 1.4

4 1.4 Similarity Illustration 1: In the adjoining figure, TS QR. (i) If PT 3 cm, PQ 9 cm and PR 1 cm, find PS. (ii) If PT and PR 14 cm, find PS. TQ 5 PT PS Sol: (i) Since TS QR, we have PQ PR 3 PS PS 4 9 Hence, PS 4 cm. PT PS (ii) Since TS QR, we have PS PT (Given) TQ SR SR 5 TQ 5 Let PS x cm. Then, SR (PR PS) (14 x) cm. x 5x (14 x) 5x 8 x 7x 8 x x 5 PS 4 cm. Illustration : In the figure shown above, PT 5.6 cm, PQ 8.4 cm, PS 3.8 cm and PR 5.7 cm. Show that TS QR. Sol: We have, PT 5.6 cm, TQ (PQ PT) ( ) cm.8 cm. PS 3.8 cm, SR (PR PS) ( ) cm 1.9 cm. PT 5.6 And TQ.8 1 Thus, PT PS TQ SR PS 3.8 SR TS divides PQ and PR proportionally. Hence, TS QR. Illustrations 3: In figure, AD AE and ADE ACB. Prove that ABC is an DB EC isosceles triangle. [NCERT] AD AE Sol: It is given that DB EC So, DE BC [Theorem] Therefore, ADE ABC [Corresponding angles] - (1) Also, it is given that ADE ACB () So, ACB ABC [From 1 and ] Therefore AB AC [Sides opposite to the equal angles] i.e., ABC is an isosceles triangle.

5 Foundation for Mathematics 1.5 Illustration 4: Prove that any line parallel to parallel sides of a trapezium divides the non-parallel sides proportionally (i.e., in the same ratio). OR PQRS is a trapezium with SR parallel to PQ. T and U are points on PS and QR respectively such that TU is parallel to PQ. Show that PT QU [NCERT] TS UR Sol: We are given trapezium PQRS such that RS QP, TU PQ and RS both. We join PR. It meets TU at O. In PRS, OT RS PO PT... (i) [Basic Proportionality OR TS Theorem] In RPQ, OU PQ RO RU [B.P.T] OP UQ PO QU...(ii) OR UR PT QU From (i) and (ii) TS UR Hence, proved. Illustration 5: Prove that the internal bisector of an angle of a triangle divides the opposite side in the ratio of the sides containing the angle. (Internal Angle Bisector Theorem) Sol: Given: a PQR in which PS is the internal bisector of P. To Prove: QS PQ SR PR Construction: DrawRT SP, meeting QP produced at T. Proof: STATEMENT REASON 1. 1 PS is the bisector of P. 3 Alt. s are equal, as RT is parallel to SP and PR is the transversal Corres. s are equal, as RT is parallel to SP and QT is the transversal From 1, and PT PR Sides opposite to equal angles are equal 6. In QRT, PRT QS QP SR PT By B.P.T. (Basic Proportionality Theorem) QS PQ SR PR Using 5 Hence, Proved.

6 1.6 Similarity Remark: The external bisector of an angle of a triangle divides the opposite side externally in the ratio of the sides containing the angle. i.e., if in a ABC, AD is the bisector of the exterior of angle A and intersect BC produced in D, BD AB CD AC 3. Axioms of Similarity of Triangles 3.1 AA (Angle-Angle) Axiom of Similarity If two triangles have two pairs of corresponding angles equal, then the triangles are similar. In the given figure, ABC and DEF are such that A D and B E. ABC ~ DEF 3. SAS (Side-Angle-Side) Axiom of Similarity If two triangles have a pair of corresponding angles equal and the sides s including them proportional, then the triangles are similar. In the given figure, ABC and DEF are such that A D and AB AC ABC ~ DEF. DE DF 3.3 SSS (Side-Side-Side) Axiom of Similarity If two triangles have three pairs of corresponding sides proportional, then the triangles are similar. If in ABC and DEF we have: AB AC BC, then ABC ~ DEF DE DF EF Illustration 6: In figure shown alongside, find C. Figure 1.5 Figure 1.6 Figure 1.7 Sol: In PQR and ABC, PQ 4.4 CA 11 5 QR 4 RP 3.6 and AB 10 5 BC 9 5 PQ QR RP CA AB CB PQR ~ CAB (SSS Similarity) C P 180 Q R

7 Foundation for Mathematics C 60 Illustration 7: In the figure, PQ QR, ST PR, and VU QR. Prove that PST ~ VRU. Sol: (each side 90 ) 4 P V... (i) Also T U... (ii) (each equal to 90 ) From (i) and (ii), we get AA similarity for triangles PST and VRU. PST ~ VRU. Illustration 8: In figure, BE BC and 1. Prove that ABD ~ EBC. AC BD Sol: 1 (Given) AC AB... (i) (Sides opposite to equal angles in BCA) Also BE BC (Given)... (ii) AC BD BE BC BA BD Form (i) and (ii), we have... (iii) AC BD BE BC Now, in triangles ABD and EBC, we have ABD EBC (each 1) and AB BD EB BC (from (3)) ABD ~ EBC (SAS Similarity) Illustration 9: In figure, RS and VW are respectively, the medians of PQR and TUV, If PQR ~ RS PQ TUV, prove that (i) PSR ~ UWV (ii) NCERT) VW UT Sol: (i) PQR ~ UTV (given) P U, (i) ( The corresponding angles of the similar triangles are equal) Also, PR PQ (Corresponding sides are proportional) UV UT PR PS S ismid-point of PQ UV UW W ismid-point of UT PR PS PR UV... (ii) UV UW PS UW Now, in PSR and UWV, we have P U and PR UV (By (i) and (ii)) PS UW

8 1.8 Similarity PSR ~ UWV (SAS similarity) (ii) PSR ~ UWV RS PS (Corresponding sides proportional) VW UW RS PS RS PQ VW UW VW UT Illustration 10: PQR is a right triangle, right angled at Q. If QS is the length of the perpendicular drawn from Q to PR. Prove that: (i) PSQ ~ PQR and hence PQ PS PR (ii) QSR ~ PQR and hence QR RS PR (iii) PSQ ~ QSR and hence QS PS SR (iv) PQ QR QS Sol: Given: PQR is right angled triangle at Q and QS PR To prove: (i) PSQ ~ PQR and hence PQ PS PR (ii) QSR ~ PQR and hence QR RS PR (iii) PSQ ~ QSR and hence QS PS SR (iv) PQ QR QS Proof: (i) In two triangles PSQ and PQR, we have: QPS QPR (Common) PSQ PQR (Each is right angle) PQS PRQ (Third angle) PSQ ~ PQR (AAA Similarity) Triangle PSQ and PQR are similar and so their corresponding sides must be proportional. PS SQ PQ PS PQ PQ PS PR This proves (i). PQ SR PR PQ PR (ii) Again consider two triangles QSR and PQR, we have QRS PRQ (Common) QSR PQR (Each is right angle) SQR QPR (Third angle) Triangle are similar and their corresponding sides must be proportional. QS SR QR i.e., QSR ~ PQR PR QR PR

9 Foundation for Mathematics 1.9 SR QR QR QR PR RS PR This proves (ii) (iii) In two triangles PSQ and QSR, we have QSP QSR [ , ] [ , ] PSQ ~ QSR (AAA criterion of similarity) Their corresponding sides must be proportional. PS SQ PQ QS SR QR PS QS QS SR QS PS SR QS is the mean proportional of PS and SR (iv) From (i), we have: PQ PS PR (ii), we have: QR RS PR (iii), we have: QS PS SR Consider PQ QR PS PR RS PR PR PS SR SR PS 1 PS SR 1 PR PQ QR PR PS SR PR PS SR PR PS SR PQ QR QS 1 1 PS SR QS (from (iii)) PLANCESS CONCEPTS If a perpendicular is drawn from the vertex containing the right angle of a right triangle to the hypotenuse then: (a) The triangle on each side of the perpendicular are similar to each other and also similar to the original triangle. (b) The square of the perpendicular is equal to the product of the length of two parts into which the hypotenuse is divided by the perpendicular. NOTE: This could also be proved from the properties of chords in a circle. Shivam Agarwal Gold Medalist, INPhO

10 1.10 Similarity 4. Results on Area of Similar Triangles Theorem-3: the areas of two similar triangles are proportional to the squares on their corresponding sides. Given: ABC ~ DEF To prove: Areaof ABC AB BC AC Areaof DEF DE EF DF Construction: Draw AL BC and DM EF. Proof: STATEMENT 1 BC AL Areaof ABC 1. Areaof DEF 1 EF DM Areaof ABC BC AL Areaof DEF EF DM. In ALB and DME, we have REASON (Area of Base Height) Figure 1.8 (i) ALB DME (each equal to 90 ) (ii) ABL DEM ( ABC ~ DEF B E) ALB ~ DME (AA-axiom) AL AB DM DE (Corresponding sides of similar s are Proportional.) 3. ABC ~ DEF (given.) AB BC AC DE EF DF (Corresponding sides of similar s are) proportional. 4. AL BC DM EF (From and 3.) AL BC 5. Substituting in 1, we get : DM EF Areaof ABC BC Areaof DEF EF 6. Combining 3 and 5, we get: Areaof ABC AB BC AC Areaof DEF DE EF DF

11 Foundation for Mathematics 1.11 Corollary-1: The areas of two similar triangles are proportional to the squares on their corresponding altitude. Given: ABC ~ DEF, AL BC and DM EF. To prove: Areaof ABC AL Areaof DEF DM Proof: STATEMENT REASON Figure BC AL Areaof ABC Areaof DEF 1 EF DM (Area of 1 Base Height) Areaof ABC BC AL Areaof DEF EF DM. In ALB and DME, we have (i) ALB DME (each equal to 90 ) (ii) ABL DEM ( ABC ~ DEF B E) ALB ~ DME (AA-axiom) AL AB DM DE (Corresponding sides of similar s are Proportional.) 3. ABC ~ DEF (given.) AB BC AC DE EF DF (Corresponding sides of similar s are proportional.) 4. BC AL EF DM (From and 3.) 5. Substituting in 1, we get : Areaof ABC AL. (Hence, proved.) Areaof DEF DM Corollary-: The areas of two similar triangles are proportional to the squares on their corresponding medians. Given: ABC ~ DEF and AP, DQ are their medians. To prove: Proof: Areaof ABC AP Areaof DEF DQ STATEMENT REASON 1. ABC ~ DEF (Given) Figure 1.10

12 1.1 Similarity Areaof ABC AB Areaof DEF DE. ABC ~ DEF ( Areas of two similar s are proportional to the squares on their corresponding sides.) AB BC BP BP (Corresponding sides of similar s are proportional.) DE EF EQ EQ AB BP and A D DE EQ (From and the fact the ABC ~ DEF) APB ~ DQE (By SAS-similarity axiom) BP AP EQ DQ AB AP (From and 3.) DE DQ AB DE AP DQ Areaof ABC AP Areaof DEF DQ Hence, proved. (From 1 and 3.) Corollary-3: The areas of two similar triangles are proportional to the squares on their corresponding angle bisector segments. Given: ABC ~ DEF and AX, DY are the bisectors of A and D respectively. To prove: Proof: 1. STATEMENT Areaof ABC AX Areaof DEF DY Areaof ABC AB Areaof DEF DE REASON Figure 1.11 ( Areas of two similar s are proportional to the squares of the corresponding sides.). ABC ~ DEF (Given) A D 1 A 1 D BAX EDY BAX 1 A and EDY 1 D 3. In ABX and DEY, we have (Given) BAX EDY (From.)

13 Foundation for Mathematics B E ABX ~ DEY AB AX AB DE DY DE Areaof ABC AX Areaof DEF DY Hence, proved. AX DY ( ABC ~ DEF) (By AA similarity axiom) (from 1 and 3) Illustrations 11: It is given that PQR ~ ABC, Area ( ABC) 36 cm and area ( PQR) 5 cm. If QR 6 cm, find the length of BC. Sol: We know that the areas of similar triangles are proportional to the squares of their corresponding sides. Areaof ( ABC) BC Areaof ( PQR) QR Let BC x cm. Then, Hence BC 7. cm 36 x x 5 36 x x Illustrations 1: A and B are points on the sides PQ and PR respectively of PQR such that AB QR and divides PQR into two parts, equal in area. Find AQ : PQ. Sol: Area ( PAB) Area (trap. AQRB) [Given] Area ( PAB) [Area ( PQR) Area ( PAB)] Area ( PAB) Area ( PQR) Areaof ( PAB) 1 Areaof ( PQR)... (i) Now, in PAB and PQR, we have APB QPR [Common P] PAB PQR [corresponding s are equal] PAB ~ PQR. We known that the areas of similar s are proportional to the squares of their corresponding sides. Area of ( PAB) PA PA 1 Area of ( PQR) PQ PQ [Using (i)] PQ PA PQ (PQ AQ)

14 1.14 Similarity AQ ( 1) PQ AQ ( 1). PQ PB : AB ( 1) : Illustration 13: If the areas of two similar triangles are equal, prove that they are congruent. Sol: Let PQR ~ STU and area ( PQR) area ( STU). Since the ratio of the areas of two similar s is equal to the ratio of the squares on their corresponding sides, we have Areaof ( PQR) PQ PR QR Areaof ( STU) ST SU TU PQ PR QR 1 [ Area ( ABC) Area ( DEF)] ST SU TU PQ ST, PR SU and QR TU Q ST, PR SU and QR TU PQR STU [By SSS congruence] Theorem-4 [Pythagoras Theorem]: In a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides. Given: A ABC in which B 90. To prove: AC AB + BC. Construction: From B, Draw BD AC. Proof: STATEMENT REASON 1. In ADB and ABC, we have: BAD CAB A (Common) ADB ABC (Each 90 ) ADB ~ ABC (By AA axiom of similarity) AD AB (Corr. sides of similar s are proportional) AB AC AB AD AC..(i). In CDB and CBA, we have : CDB CBA (Each 90 ) Figure 1.1

15 Foundation for Mathematics 1.15 BCD ACB C CDB ~ CBA (Common) (By AA axiom of similarity) DC BC (Corr. sides of similar s are proportional) BC AC BC DC AC..(ii) 3. Adding (i) and (ii), we get AB + BC AD AC + DC AC (AD + DC) AC AC AD + DC AC Hence, AB + BC AC. Theorem 5: [Converse of pythagoras Theorem]: In a triangle if Figure 1.13 the square of one side is equal to the sum of the squares of the other two sides, then the triangle is right angled. Given: A ABC in which AB + BC AC To prove: B 90 Construction: Draw a DEF in which DE AB, EF BC and E 90 Proof: STATEMENT REASON 1. In DEF, we have: E 90 DE + EF DF (By Pythagoras Theorem) AB + BC DF ( DE AB and EF BC) AC DF ( AB + BC AC (Given)) AC DF. In ABC and DEF, we have : AB DE (By construction) BC EF (By construction) AC DF (Proved above) ABC DEF (By SSS congruence) B E (C.P.C.T.C.) Hence, B 90 E 90

16 1.16 Similarity Illustration 14: If PQR is an equilateral triangle of side a, prove that its altitude Sol: PQR is an equilateral triangle. We are given that PQ QR RP a. PS is the altitude, i.e., PS QR. Now, in right angled triangles PQS and PRS, we have PQ PR [Given] and PS PS [Common side] PQS PRS [By R.H.S. congruence] 3 a. QS RS QS SR 1 QR a From right triangle PQS, PQ PS + QS a PS a + a 3 PS a a PS a Illustration 15: In the given figure, R 90. S and T are any points on PR and RQ respectively. Prove that: PT + QS PQ + ST. [NCERT] Sol: In PRT, R 90 PT PR + RT... (i) In SRQ, R 90. QS SR + RQ... (ii) Adding (i) and (ii), we get: PT + QS (PR + RT ) + (SR + RQ ) PQ + ST [By Pythagoras Theorem] Hence, PT + QS PQ + ST. Illustration 16: In a trapezium PQRS, PQ SR and SR PQ. TU drawn parallel to PQ cuts PS in U and QR in T such that QT 3. Diagonal SQ intersects TU at V. Prove that 7TU 10PQ. TR 4 Sol: In SUV and SPQ, 1 USV PSQ [Common] [Corresponding s PQ UV] SUV ~ SPQ [By AA rule of similarity] SU UV...(i) SP PQ Again in trapezium PQRS, TU PQ SR PU QT SU TR PU 3 QT 3 (given) SU 4 TR 4

17 Foundation for Mathematics 1.17 PU SU 4 PS 7 SU 4 PU SU 7 SU 4 SU 4.(ii) PS 7 From (i) and (ii), we get, UV 4 i.e. UV 4 PQ...(iii) PQ 7 7 In QTV and QRS, we have QTV QRS VQT SQR [Corresponding angle TV QR] [Common] QTV ~ QRS [By AA rule of similarity] QT TV QR SR 3 TV 7 RS 3 3 TV RS (PQ) TV PQ...(iv) 7 Adding (iii) and (iv), we get UV TV PQ PQ PQ Hence, proved. 10 TU PQ i.e., 7TU 10PQ. 7 Illustration 17: Prove that in any triangle, the sum of the squares of any two sides is equal to twice the square of half of the third side together with twice the square of the median which bisects the third side. (Apollonius Theorem) Sol: Given: A PQR in which PS is a median. To prove: PQ + PR PS 1 + QR or PQ + PR (PS + QS ) Construction: Draw PT QR. Proof: PS is median QS SR Now, PQ + PR (PT + QT ) + (PT + RT ) AT + QT + RT PQ + PR [PS ST ] + QT + TR PQ + PR PS ST + (QS + ST) + (SR ST) PS ST + (QS + ST) + (QS ST) PQ + PR (PS + QS ) PS 1 + QR Hence, Proved.

18 1.18 Similarity SUMMARY Congruent Figures: Two geometrical Figures are said to be congruent, provided they must have same shape and same size. Similar Figures: Two Figures are said to be similar, if they have the same shape. Similar Figures may differ in size. Two congruent Figures are always similar, but two similar Figures need not be congruent. Thales Theorem or Basic Proportionality Theorem: If a line is drawn parallel to one side of a triangle intersecting the other two sides, then the other two sides are divided in the same ratio. Converse of Basic Proportionality Theorem: If a line divides any two sides of a triangle proportionally, the line is parallel to the third side. Axioms of Similarity of Triangles AA (Angle-Angle) Axiom of Similarity: If two triangles have two pairs of corresponding angles equal, then the triangles are similar. SAS (Side-Angle-Side) Axiom of Similarity: If two triangles have a pair of corresponding angles equal and the sides including them proportional, then the triangles are similar. SSS (Side-Side-Side) Axiom of Similarity: If two triangles have three pairs of corresponding sides proportional, then the triangles are similar. Area of Similar Triangles The areas of two similar triangles are proportional to the squares on their corresponding sides. Pythagoras Theorem: In a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

19 Foundation for Mathematics 1.19 SOLVED EXAMPLES Exapmple1: Two isosceles triangles have equal vertical angles and their areas are in the ratio 16:5. Find the ratio of their corresponding heights. Sol: Let PQR and STU be the given triangles in which PQ PR, ST SU, P S and Areaof ( PQR) 16 Areaof( STU) 5 Draw PA QR and SB TU Now, PQ ST 1 and PR SU 1 [ PQ PR and ST SU] PQ PR. ST SU In PQR and STU, we have, PQ PR And P S ST SU PQR ~ STU [By SAS similarity axiom] But, the ratio of the areas of two similar s is the same as the ratio of the squares of their corresponding heights. Area( PQR) PA 16 PA Area( STU) SB 5 SB PA 4 PA : SB4 : 5, SB 5 i.e. The ratio of their corresponding heights 4:5. Example : A point O in the interior of a rectangle PQRS is joined with each of the vertices P, Q, R and S. Prove that: OP + OR OQ + OS. Sol: Through O, draw TOU PQ. Then, PQUT is a rectangle. In right triangles OTP and OUR, we have: OP OT + PT ; OR OU + RU OP + OR OT + OU + PT + RU Again, in right triangles OUQ and OTS, we have : OQ OU + QU ; OS OT + ST OQ + OS OU + OT + QU + ST... (i)

20 1.0 Similarity OT + OU + PT + RU...(ii) [ QU PT and ST RU] From (i) and (ii), we get OP + OR OQ + OS. Example3: In a PQR, S and T are points on the sides PQ and PR respectively such that ST QR. If PS 8x 6, PT 16x 14, QS 6x and RT 10x 6, find the value of x. [CBSE - 006] Sol: In PQR, we have ST QR PS PT SQ TR [By Basic Proportionality Theorem] 8x 6 16x 14 6x 10x 6 80x - 60x - 48x x - 84x 3x x - 108x x - 116x x - 8x x - x (x + 1) (x - 1) 0 x 1 or x - 1 So, the required value of x is 1. [x - 1 is neglected as length cannot be negative]. Example 4: In the given figure, PQR is right-angled at R. Let QR p, RP q, PQ r and RS a, where RS PQ. Prove that: (i) ra pq (ii) a p q Sol: (i) Area of PQR 1 PQ RS 1 ra. Also, area of PQR 1 QR PR 1 pq. 1 ra 1 pq. ra pq (ii) ra pq a pq r a pq 1 r p q [ r p + q ] r a pq pq a p q Example 5: In a PQR, obtuse angled at Q, if PS is perpendicular to RQ produced to S, prove that: PR PQ + QR + (QR QS)

21 Foundation for Mathematics 1.1 Sol: In PSQ, S 90. PS + SQ PQ [By Pythagoras Theorem] In PSR, S 90 PR PS + SR [By Pythagoras Theorem] PS + (SQ + QR) PS + SQ + QR + (QR QS ) PQ + QR + (QR QS) [Using (i)] Hence, PR PQ + QR + (QR QS).... (i) Example 6: QPR 90 o, PS is its bisector. If ST PR, prove that ST (PQ + PR) PQ PR. PQ QS Sol: It is given that PS is the bisector of P of PQR. PR SR PQ 1 QS 1 PR SR [Adding 1 on both sides] PQ PR QS SR PR SR PQ PR QR...(i) PR SR In s RST and RQP, we have SRT QRP [Common] STR QPR [Each equal to 90 0 ] So, by AA-criterion of similarity RST ~ RQP RS ST PQ QR...(ii) RQ QP ST SR From (i) and (ii), we have PQ PR PQ PR ST ST (PQ + PR) PQ PR. Example 7: In the given figure, AX, BY and CZ are each perpendicular to XZ. Prove that p r q Sol: In AXZ, we have YB XA YB ZY [ ZYB ~ ZXA] XA ZX

22 1. Similarity q ZY (i) p ZX In XZC, we have YB ZC YB XY [ XYB ~ XZC] ZC XZ q XY (ii) r XZ Adding (i) and (ii), we get q q ZY XY p r XZ XZ q q XY YZ q q xz p r XZ p r xz q q p r p r q Hence, Proved. Example 8: In the given figure, PQ RS. Find the value of x. Sol: Since the diagonals of a trapezium divide each other proportionally. PO QO OR OS 6x 38 x 8 x x x - 16x - 1x x - 76x x - 19x x - 11x - 8x (x - 8) (x - 11) 0 x 8 or x 11. Example 9: In the given figure, QR PQ, PT PQ and ST PR. Prove that ST QR PS PQ. Sol: In PQR and TSP, We have PQR PST [Each equal to 90 o ] PRQ TPS [Alternate angles] By AA Similarity PQR ~ TSP QR PS ST QR PS PQ. PQ ST

23 Foundation for Mathematics 1.3 Hence Proved. Example 10: In an equilateral triangle PQR, the side QR is trisected at S. Prove that 9 PS 7PQ. [CBSE - 005] Sol: PQR be can equilateral triangle and S be point on QR such that QS 1 QR (Given) Draw PT QR, Join PS. 3 QT TR.. (Altitude drown from any vertex of an equilateral triangle bisects the opposite side) So, QT TR QR In PQR PQ PT + TQ...(i) PS PT + TS...(ii) From (i) and (ii) PQ PS TS + TQ PQ PS QR QR QR QR QR QR ( QS + ST ST ST ) 3 6 QR QR QR PQ PS ( TQ ) 36 4 PQ PQ PQ PS ( PQ QR) PQ PQ 9PQ 8PQ PS PS 7PQ 9PS Example 1: In figure, prove that PQR and STU are similar. If TU : ST 5 : 7, then find SU. Sol: In PQR, PQ 5 cm, QR 8 cm, PR 7 cm o o o o o and P 180 (0 30 ) 130 and in STU, T 130 (Given) PQ 5 TU 5 and.(1) PR 7 ST 7 Now for PQR and STU, o PQ TU P T 130 and [by eq. PR ST (1).] By SAS rule of congruency

24 1.4 Similarity PQR STU Q U 30 and S R o PQ QR QR TU 8 5 and SU 8cm TU SU PQ 5 Example 13: QY and RX are medians of PQR, right angled at P: Prove that 4(QY RX ) 5QR. Sol: In QPY,QY PY PQ (using Pythagoras theorem) (1) and in RPX, RX PX PR (using Pythagoras theorem) () Adding (1) and () and then multiplying by 4, we get 4(QY RX ) 4(PY PQ PX PR ) 4 PY PX (PQ PR ) 4(PY PX QR ) `[ PQR is a right triangle] 4(XY QR ) [ YPX is a right triangle] 4XY 4QR (A line joining mid-points of two sides is parallel to the third side and is equal to half of the base) QR 4QR 5QR (half of it, XY QR/) Example 14: X and Y are points on side PQ and PR respectively of PQR 5cm and YR 10cm, show that QR 3 XY.. If PX 3cm, XQ 6cm, PY Sol: In PQR, X and Y are points on PQ and PR.It is given that, PX 3cm, XQ 6cm, PY 5cm and YR 10cm. Now, PX 3 1 PY 5 1 PX PY ; ; XQ 6 YR 10 XQ YR Hence XY QR In PXY and PQR X Q [Corresponding angles] Y R Corresponding angles] P P [Common angle] PXY PQR [AAA Similarity] PX PY XY PQ PR QR 3 5 XY 9 15 QR

25 Foundation for Mathematics 1.5 [PQ3+69cm. PR5+1015cm] 1 XY QR 3XY 3 QR Example 15: In PQR,PQ PR and QR 6 cm. S is a point on side PR such that PS 5 cm and RS 4 cm. show that QRS PRQ and hence find QS. Sol: Consider PQR and QRS It is given that PQ PR, QR 6 cm, PS 5 cm and RS 4 cm. QR 6 6 RS 4 Then, and PR RQ 6 3 QR RS PR RQ Also, QRS PRQ (common) QRS PRQ (SAS similarly) QS RS QS ( PQ PR) PQ RQ 3 PR 3 QS PR (5 4) 9 6cm Example 16: L and M are the midpoints of the sides ZX and ZY respectively of a XYZ in which Z is a right angle. Prove that (i) 4XM 4XZ YZ and (ii) 4(XM YL ) 5XY o Sol: Given: Z 90, P is the midpoint of XZ, M is the midpoint of YZ. Proof: XM XZ ZM (Pythagoras s theorem) 1 1 XZ YZ XZ YZ 4 4XM 4XZ YZ (1) Similarly, 4YL 4YZ XZ () Adding (1) and (). 4XM 4YL (4XZ YZ ) (4YZ XZ ) 5XZ 5YZ 5(XZ YZ ) 4(XM YL ) 5XY (Pythagoras s theorem)

26 1.6 Similarity Example 17: LMN is right-angled at L. OPQR is square as shown in the figure. Prove that OP MO PN. Sol: Given: LMN is right-angled at. OPQR is a square To prove OP MO PN. Proof: In LRQ and OMR o RLQ MOR 90 LRQ OMR (corrsp. Angles) LRQ OMR (i) (AA similarlity) o In LRQ and PQN, RLQ NPQ 90 and, LQR QNP LRQ PQN (ii) (AA similarity) From (i) and (ii), OMR PQN OM OR, But PQOROP (sides of a square) PQ PN OM OP OP OMPN OP PN Example 18: In the given figure, D and E trisect the side BC of a right triangle ABC. Prove that 8AE 3AC 5AD Sol: D and E trisect the side BC. Let BD DE EC p units Let AB q units In ABD, AD AB BD q p (By pythagoras theorem)...(1) In ABE, AE AB BE (By Pythagoras theorem) q (p) q 4p...() In right ABC, AC AB BC (By Pythagoras theorem q (3p) q 9p (3) R.H.S. 3AC 5AD 3(q 9p ) 5(q p ) [From (1) and (3)] 3q 7p 5q 5p 8q 3p 8(q 4p ) 8AE L.H.S. [from ()] Thus, 8AE 3AC 5AD

27 Foundation for Mathematics 1.7 Example 19: If a be the area of a right triangle and m one of the sides containing the right angle, prove am that the length of the altitude on the hypotenuse is 4 m 4a Sol: Let ABC be a right triangle right-angled at B such that BCm and a Area of ABC perpendicular to AC. We have, a area of ABC a AB (1) m Now, in s AXB and ABC, we have AXB ABC [each equal to 90 o ] And BAX BAC [Common] AB XB So, by AA-criterion of similarity, we have AXB ABC AC BC By Pythagoras theorem in ABC, we have, 4a AB BC AC m AC m 4 4 4a m 4a m AC m m a XB a From (1) and () we have XB m AC m AC 4 am 4a m XB AC 4 4a m m draw BX

28 1.8 Similarity EXERCISE 1 For School Examinations Fill in the Blanks Directions: Complete the following statements with an appropriate word/term to be filled in the blank space(s). Q.1. Every point on the circumference of circles are equidistant from its. Q.. Two polygons of the same number of sides are similar, if their corresponding angles are and their corresponding sides are in the same. Q.3. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the ratio. Q.4. If line divides any two sides of a triangle in the same ratio, then the line is parallel to the side. Q.5. Two polygons of the same number of sides are similar, if all the corresponding angles are. Q.6. Q.7. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO CO.ABCD is a BO DO A line drawn through the mid-point of one side of a triangle parallel to another side bisects the side. Q.8. Line joining the mid-points of any two sides of a triangle is to the third side. True / False Directions: Read the following statements and write your answer as true or false. Q.9. All the congruent Figures are similar but the converse is not true. True False Q.10. If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are in the same ratio (proportional). Then the triangles are similar. True False Q.11. In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. True False Q.1. If in a triangle, square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle. True False

29 Foundation for Mathematics 1.9 Q.13. Diagonals AC and BD of a trapezium ABCD with AB DC intersect each other at the point OA OB O,. OC OD True False Q.14. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. ABE is similar to CFB True False Match the Following Columns Directions: Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in column I have to be matched with statements (p, q, r, s) in Column II Q.15. In figure, the line segment XY is parallel to the side AC of ABC and it divides the triangle into two parts of equal areas, then match the column. Column I (A) AB : XB (p) Column II :1 (B) ( ABC) : ar( XBY) (q) : 1 (C) AX : AB (r) ( 1) : (D) X: A (s) 1 : 1 Very Short Answer Questions Directions: Give answer in one word or one sentence. Q.16. A clever outdoorsman whose eye-level is meters above the ground, wishes to find the height of a tree. He places a mirror horizontally on the ground 0 meters from the tree, and finds that if he stands at a point C which is 4 meters from the mirror B, he can see the reflection of the top of the tree. How high is the tree? Q.17. A ladder is placed against a wall such that its foot is at a distance of.5 m from the wall and its top reaches a window 6 m above the ground. Find the length of the ladder. Q.18. From the adjoining figure, prove that QR BA QB RA.

30 1.30 Similarity Q.19. Prove that PQ PR, if PQR is an isosceles triangle right angled at R. Q.0. The areas of two similar triangles ABC and PQR are in the ratio 9 : 16, If BC 4.5 cm., find the length of QR. Q.1. The areas of two similar triangles are 81 cm and 49 cm. If the altitude of the bigger is 4.5 cm. find the corresponding altitude of smaller triangle. Q.. In a triangle PQR, PS QR, If PS QS.SR, prove that PQR is right angle triangle. Short Answer Questions Directions: Give answer in two to three sentences. Q.3. If in an isosceles triangle a is the length of the base and b is the length of one of the equal side, then prove that its area is a 4b a 4 Q.4. In figure, AB AD. AE and AF are angle bisectors of BAC and DAC. Prove that BD EF. Q.5. In figure, T is a point on side RQ produced of an isosceles triangle PQR with PQ PR. If PS QR and TU PR prove that PQS TRU Q.6. A vertical stick 1m long casts a shadow 8 m long on the ground. At the same time a tower casts the shadow 40m long on the ground. Find the height of the tower. Q.7. ABC is an isosceles triangle with AC BC. If AB AC, prove that ABC is a right triangle. Q.8. In the given figure, S is a point on the side QR of PQR such that PSR QPR Prove that RP RQ RS RP

31 Foundation for Mathematics 1.31 Q.9. In figure PQ ST and QS TU. Prove that SR RUPR. Q.30. Let PQR STU and their area be respectively 49 cm and 100 cm. If TU 0 cm, find QR. Q.31. In the given figure PQR and SQR are two triangles on the same base QR. If PS intersects QR at ar( PQR) PO O. prove that. ar( SQR) SO Long Answer Questions Directions: Give answer in four to five sentences. Q3. Any point O, inside PQR, is joined to its vertices, From a point S on PO, ST is drawn so that ST PQ and TU QR as shown in figure. Prove that SU PR.

32 1.3 Similarity Q33. Prove that the sum of the squares of the sides of a rhombus is equal to sum of the squares of its diagonals. Q34. In adjoining figure, BE BC and 1. Prove that ABD EBC AC BD Q35. In adjoining figure if PQT PRS, prove that PAT PQR, Q36. In below given figure, DE BC and AD : DB 5:4. Find the ratio of areas of DEF and CFB. Q37. In the picture, ABCD is a parallelogram. AD is parallel to ZX and AZ XY equals /3. Then find ZB BD Q38. Through the mid-point X of the side RS of a parallelogram PQRS, the line QX is drawn intersecting PR at Y and PS produced at T. Prove that TY QY.

33 Foundation for Mathematics 1.33 EXERCISE For Competitive Examinations Multiple Choice Questions Directions: This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct. Q.1. If in an isosceles triangle, a is the length of the base and b the length of one of the equal sides, then its area is - a (a) 4b a (b) b 4b a 4 4 a b a b (c) a b (d) b a 4 4 Q.. PQR is a triangle. If S is a point in the plane of the triangle such that the perpendicular distance from S to the three sides of the triangle are all equal, then there exist(s)- (a) Just one such point as S (b) Three such point as S (c) Four such points as S (d) None of the above Q.3. PSR is a triangle right angled at S.D is the mid-point of SR. If the bisector of PSR and perpendicular bisector of SR meet at O, then triangle OSD is (a) Isosceles (b) Equilateral (c) Isosceles right angled (d) Acute-angled Q.4. If any two sides of a triangle are produced beyond its base and the exterior angles thus obtained are bisected, then these bisectors will include an angle equal to (a) Half the sum of the base angles (b) Sum of the base angles (c) Half the difference of the base angles (d) Difference of the base angles Q.5. If x is the length of the median of an equilateral triangle, then its area is Q.6. Q.7. (a) x (b) 3 x (c) 3 x 1 (d) x 3 In a triangle XYZ, points P, Q and R are the mid-points of the sides XY, YZ and ZX respectively. If the area of the triangle XYZ is 4 sq. then area of the triangle PQR equal to (a) 10 sq. units (b) 5 5 sq. units (c) 5 sq. units (d) 6 sq. units The area of a right angled triangle is 0 sq. cm. and its perimeter is 0 cm. The length of its hypotenuse is (a) 16 cm (b) 8 cm (c) 17cm (d) Data insufficient

34 1.34 Similarity Q.8. Q.9. If each side of triangle ABC is of length 4 and if AD is 1 cm and ED AB. What is area of region BCDE? (a) 8 3 cm (b) 4 3 cm (c) cm (d) cm ar(abc) 16 If ABC QRP,, AB 18 cm and BC 16 cm; then PR ar(pqr) 9 is equal 1 to (a) 10 cm (b) 1 cm (c) 0 cm (d) 8 cm 3 Q.10. It is given that ABC PQR with BC 1 ar( PRQ). Then is equal to QR 4 ar( BCA) (a) 9 (b) 16 (c) 1 (d) Q.11. In triangle ABC, D, E, F are points of trisection of BC, AC and AB respectively. Which of the following statement is not true? (a) Area EDC / 9 area ABC (b) Area FBD / 7 area AFDC (c) Area DEF / 9 area ABC (d) ( EDC DBF AFE) area DEF Q.1. A certain right angled triangle has its area numerically equal to its perimeter. The length of its each side is an even integer. What is the perimeter? (a) 4 units (b) 36 units (c) 3 units (d) 30 units More than One Correct Directions: This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d) out of which ONLY OR MORE may be correct. Q.13. Which among the following is/are not correct? (a) The ratio of the areas of two similar triangles is equal to the ratio of their corresponding sides. (b) The ratio of areas of two similar triangles are in the ratio of the corresponding altitudes. (c) The ratio of area of two similar triangles are in the ratio of the corresponding medians. (d) If the areas of two similar triangles are equal, then the triangles are congruent.

35 Foundation for Mathematics 1.35 Q.14. Which among the following is/are correct? (I) If the altitudes of two similar triangles are in the ratio : 1, then the ratio of their areas is 4 : 1. area( APQ) 1 (II) PQ BC and AP : PB1 :. Then, area( ABC) 4 (III) The areas of two similar triangles are respectively 9cm and 16cm. The ratio of their corresponding sides is 3 : 4. (a) I (b) II (c) III (d) None of these Passage Based Questions Directions: Study the given passage (s) and answer the following questions. Q.15. In figure, AD is a median of a triangle ABC and AM BC. BC i. AD BC.DM a) AC b) AB c) BC d) none of these BC ii. AD BC.DM a) AC b) AB c) BC d) none of these 1 iii. AD BC a) AC BC c) AC AB b) AB BC d) none of these Assertion and Reason Directions: Each of these questions contains an Assertion followed by Reason. Read them carefully and answer the questions on the basis of following options. You have to select the one that best describes the two statements. a) If both assertion and reason are correct and reason is the correct explanation of assertion. b) If both assertion and reason are correct, but reason is not the correct explanation of assertion. c) If assertion is correct but reason is incorrect. d) If assertion is incorrect but reason is correct. Q.16. Assertion: ABC is an isosceles, right triangle, right angled at C. Then AB AC Reason: In an isosceles triangle ABC if AC BC and AB AC, then o C 90

36 1.36 Similarity Multiple Matching Questions Directions: Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in column I have to be matched with statements (p, q, r, s) in Column II Q.17. Column-I Column-II (a) (p) 36 : 49 ABC is an isosceles right angled triangle. AB? (b) ABC DEF, such that AB 1. cm and DE 1.4 cm area( ABC)? area( DEF) (c) ABC APQ and area( APQ) 36 area( ABC) 49 BC? PQ (q) (r) AB AC AB AC +BC (d) If DE BC and AD 6 then, DB 7 AE EC? (s) 6/7

37 Foundation for Mathematics 1.37 Subjective Questions Directions: Answer the following questions. Q.18. In the given figure, ABC is a right triangle, right angled at B. AD and CE are the two medians drawn from A and C respectively. If AC 5 cm and 3 5 AD cm, find the length of CE. Q.19. Two poles of height a metres and b metres are p metres apart. Prove that the height of the point of intersection of the lines joining the top of each pole to the foot of the opposite pole is given ab by metres. a b Q.0. If two sides and a median bisecting the third side of a triangle are respectively proportional to the corresponding sides and the median of another triangle, then prove that the two triangles are similar.

38 1.38 Similarity SOLUTIONS EXERCISE 1 For School Examinations Fill in the Blanks 1. Centre. Equal, ratio 3. Same 4. Third 5. Equal 6. Trapezium 7. Third 8. Parallel True / False 9. True 10. True 11. True 1. True 13. True 14. True Match the Following Columns 15. (A) (p); (B) (q); (C) (r); (D) (s) Very Short Answer Questions 16. The height of the tree is 10 meters. 17. Thus, length of the ladder is 6.5 m. 18. In PQR:.(1) o QR PQ PR [ P 90 ] o In PAB: AB PA PB [ P 90 ] () QR AB PQ PR PA PB [Adding (1) and ()] (PQ PB ) (PR PA ) [By grouping] QB RA QR BA QB RA 19. Proof: PQR is an isosceles right angled triangle. PR QR Using Pythagoras theorem, we have PQ PR QR PQ PR PR..(1) [ PR QR (Given] PQ PR

39 Foundation for Mathematics Proof: We know that if two triangles are similar then their areas are proportional to the squares of the corresponding sides, ar( ABC) BC 9 (4.5) ar( PQR) 16 QR QR 3QR QR 6cm. 1. Let ABC and DEF be two given triangles such that ABC DEF. Draw AL BC,DM EF,AL 4.5 cm. area( ABC) AL We know that, area( DEF) DM DM 81 (4.5) (4.5) DM 81 (4.5) (7) 31.5 DM 3.5cm PQ PS QS and PR PS SR PQ PR PS QS SR (QSSR) [ PS QS.RS] QR o PQR is rt triangle, QPR 90 (By converse of Pythagoras theorem) Q.E.D. Short Answer Questions 3. Let ABC be an isosceles triangle, where base AB a and equal sides AC BC b. Let CD be the perpendicular on AB, so, Altitude, CD height of the ABC is given by 1 a AD DB AB 1 h AC AD h 4b a a Area of the ABC base altitude a 4b a 4b a 4 4. AB AD and AE, AF are angle bisector of BAC and DAC respectively. Now in ABC,AE is angle bisector of BAC AB BE AC EC.(1) AD DF AB DF Similarly, ( AB AD) AC FC AC FC.() Comparing eq. (1) and (), we get BE DF EC FC EF BD in BCD.

40 1.40 Similarity 5. Given: In PQR, PQ PR PS QR and TU PR Proof: Since, PQ PR PQR PRQ Now in PQS and TRU o PSQ TUR 90 and PQR PRQ PQS URT PQS ~ TRU 6. In figure, AB represents the stick and BC is its shadow. Therefore AB 1 m and BC 8m. Again PQ is tower and QR is its shadow. Therefore QR 40 m Now, ABC PQR PQ AB PQ 1 PQ 60m QR BC In ACB AB AC and AC BC Now in ABC, AB AC AC AC AB AC BC (BC AC) o ACB 90 and ABC is a right triangle (Converse of Pythagoras theorem) 8. In PQR and PSR, 9. In QPR PSR [Given] PRQ PRS [Common] QPR PSR [AA similarity] RP RQ RS RP PQR, ST PQ RS RT PR QR.(1) [Converse of basic proportional theorem]

41 Foundation for Mathematics 1.41 Again in RSQ, TU QS RU RT RS.() RQ [Converse of Basic proportionality theorem] From (1) and (), we get RS RU RS RUPR PR RS ar( PQR) QR 49 QR ar( STU) TU 100 TU 10 (0)(7) QR 7 14cm Draw PX QR, SY QR In PXO and SYO, o (each 90 ) 3...(vertically opposite s) PXO SYO (By A.A. rule of similarity) PO PX.. (i) SO SY 1 (QR)(PX) area( PQR) PX PO Now, area( SQR) 1 SY SO (QR)(SY) Hence, area( PQR) PO area( SQR) SO.from (i) Long Answer Questions 3. Given: In PQR,O is any point inside it. PO, OQ, OR are joined. S is a point on OP. ST PQ meeting OQ at T and TU QR meeting OR at U. To prove : SU PR Proof : In OPQ, ST PQ OS OT PS TQ [Basic proportionality theorem].(1) Again in OQR, TU QR OT OU TQ RU [Basic proportionality theorem].()

42 1.4 Similarity 33. From (1) and (), we get, OS OU SU PR PS RU [ In a triangle if a line divides the two sides proportionally then it is parallel to the third side] 1 1 OR PO PR and OS QO QS..(1) [Diagonals of a rhombus bisect each other] We know that diagonals of a rhombus bisect each other at right angles. In right angled POQ, OP OQ PQ..() In right angled QOR OQ OR QR..(3) In right angled ROS OR OS RS.. (4) In right angled SOP OS OP PS.(5) Adding () to (5), we get (OP OQ OR OS ) PQ QR RS SP PR QS PR QS PQ QR RS SP [From (1)] PR QS PQ QR RS SP PQ QR RS SP PR QS BE BC 34. We have (Given) AC BD BE AC BC BD (i) Now 1 (Given) AB AC (Sides opposite the equal angles) BE AB BC BD [From (i) and (ii)] Now in triangles, ABD and EBC We have AB BE from (iii) BD BC and B B ( 1in each) ABD EBC (SAS similarity) (ii)

43 Foundation for Mathematics We are given that PQT PRS, Therefore, PQ PR (CPCT) And PT PS (CPCT) PQ PR PS PT [From (I) and (II)] (I).(II) PQ PS i.e., PR PT (III) Now in PST, P (i.e., SPT) is included between sides PS and PT and in PQR, P (i.e., QPR) is included between sides PQ and PR and SPT QPR (Common angles) PQ PS Further PR PT [From (III)] PST PQR (SAS similarity) 36. In ABC, we have DE BC ADE ABC and ACD ACB [Corresponding angles] Thus, in triangles ADE and ABC, we have A A [Common] ADE ABC and AED ACB ADE ABC [By AAA similarity] AD DE AB BC AD 5 We have, DB 4 DB 4 AD 5 DB (By adding 1 both sides) AD 5 DB AD 9 AB 9 AD 5 AD 5 AD 5 AB 9 DE 5 ( ADE ABC) BC 9 In DFE and CFB, we have 1 3 [Alternate interior angles] 4 [Vertically opposite angles] Therefore, by AA-similarity criterion, we have DFE CFB Area( DFE) DE 5 5 Area ( CFB) BC 9 81

44 1.44 Similarity Hence, ratio of areas of DEF and CFB is 5: Given that, AZ AZ 1 1 ZB 3 ZB 3 AZ ZB 5 BA 5 BZ 3 ZB 3 BZ 3 BA 5 Now, XZ AD. BXZ and BDA are similar, so BX 3 (i) BD 5 In XYZ and BYC XYZ BYC [opposite angles] ZXY YBC [XZ BC and BX meets them] XZY YCB [XZ BC and CZ meets them] Triangle XYZ and BYC are similar, then XY ZX 3 YB BC 5 XY 3 YB 5 As YB 5 XY 3 YB 5 YB XY 8 BX XY 3 XY 3 XY 3 XY 3...(ii) BX 8 XY BX XY From (i) and (ii), BX BD BD In s QXR and TXS, we have QXR TXS [Vertically opposite angles] XR XS [ X is the mid-point of RS] XRQ XST [Alternate angles] So, by AAS-congruence criterion, we have QXR TXS QR TS [ Corresponding parts of congruent triangles are equal] In s PTY and RQY, we have PYT RYQ [Vertically opposite angles] TPY QRY [Alternate angles]

45 Foundation for Mathematics 1.45 So, by AA-criterion of similarity, we have PTY RQY PT TY PY QR QY RY Taking first two terms, we get TY PT PS ST QR ST QR QY QR QR QR QR TY QY. EXERCISE For Competitive Examinations Multiple Choice Questions 1. (a). (a) A point which is equidistant from the sides of a triangle is the incentre and such a point is one and only one in the plane of the triangle. 3. (c) 4. (a) Half the sum of the base angles. 5. (c) 6. (d) As P, Q and R are the midpoints of XY, YZ and ZX respectively, XYZ is divided into 4 4 triangles of equal areas. Therefore, area ( PQR) 6 sq. units (b) 8. Area of ABC will be 3 (4) 4 ADE is right angle triangle where AD 1 cm, so we will get DE 3 cm and AE cm So area of ADE will be cm So area of BCDE / 3.5 3cm 9. (b) 10. (b) Since, ABC PQR ar( PRQ) QR 16 QR 4 16 ar( BCA) BC 1 BC (d) In triangle ACD,DB DC o o DBC DCB 60 In ABC, AB AC o

46 1.46 Similarity o o ABC ACB 75 o or o o o EBD (a) Also, in o DEB, BDE 10 o o o o BED 180 (10 15 ) 45 More than One Correct Answers 13. (a, b, c) 14. (a, c) Passage Based Questions 15. i. (a) AD is the median, so D is the mid point of BC. 1 So, BD DC BC..(1) In right angled AMC, AC AM MC..() In right angled AMD, AM AD MD..(3) Putting AM from (3) in (), We get AC AD MD MC AD MD (MD DC) BC BC AD DM. BC AC AD BC.DM ii. (b) In right angled ABM, AB AM BM From AMD, AM AD MD So, AB AD MD BM AD MD (BD MD) AD MD BD BD.MD MD BC AB AD BC.DM proved. iii. (c) From the solution of above two questions BC AC AD BC.DM (i)

47 Foundation for Mathematics AB AD BC.DM BC.(ii) 4 And Adding results of (i) and (ii) we get. BC AC AB AD BC.DM BC AD BC.DM 1 AC AB AD (BC) Assertion and Reason 16. (a) In right angled ABC, AB AC BC (By pythagorus Theorem) AC AC [ BC AC] AC AB AC Assertion (A) is true. Again since AB AC AC AC ( AC BC given) AC BC o C 90 (by converse of Pythagoras theorem) Reason is true Multiple Matching Questions 17. (A) q, r (B) p (C) s (D) s (A) AB AC BC Since, ABC is an isosceles right angled triangle. AC BC Now, AB AC AC AC (B) area( ABC) (AB) (1.) 1.44 (DE) (1.4) area( DEF)

48 1.48 Similarity (C) (D) area( APQ) (BC) 36 (PQ) area( ABC) 49 BC 6 PQ 7 DE BC AD AE 6 DB EC 7 Subjective Questions 18. In right ABD, AD AB BD 1 AD AB BC 1 AD AB BC.(1) 4 1 Similarly, in right EBC, EC BC AB.() Adding (1) and (), we get AD EC AB BC EC (5) EC EC 0cm CE Let AB and CD be two poles of heights a metres and b metres respectively. Poles are p metres apart AC p metres. Let lines AD and BC meet at O such that OL AC and OL h metres. Let CL x and LA y. Then, x + y p. In ABC and LOC, we have CAB CLO [Each is 90 o ] BCA OCL [Common] CAB CLO [AA criterion of similarity] CA AB CL LO p a ph x.(1) x h a

49 Foundation for Mathematics 1.49 In ALO and ACD, we have ALO ACD [Each equal to 90 o ] DAC OAL [Common] LAO CAD [By AA-criterion of similarity] AL OL y h AC DC p b ph y [ AC x y p].() b ph ph From (1) and (), we have x y a b 1 1 p ph [ x y p] a b a b ab 1 h h metres ab a b 0. Produce AL to P such that LP AL. Join CP and produce DM to Q such that MQ DM join FQ. In ALB and ALC BL CL [ AL is the median] AL PL (By const.) ALB CLP (vertically opposite angles) ALB CLP (By SAS) AB PC Now, in DME and MFQ EM MF ( DM is the median) DM MQ (By Const.) DME FMQ (vertically opposite angles) DME FQM (By SAS) DE QF AB AC AL Now, (Given) DE DF DM PC AC AL QF DF DM.(i)..(ii)

50 1.50 Similarity PC AC AP QF DF DQ [AL AP and DM DQ] In ABC and DEF, APC DQF (By AA) 1 Similarly A D And AB AC (Given) DE DF ABC DEF (By SAS) Q.E.D.

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