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2 Geometry 3 SIMILARITY & CONGRUENCY Congruency: When two figures have same shape and size, then they are said to be congruent figure. The phenomena between these two figures is said to be congruency. CONDITIONS FOR TRIANGLES TO BE CONGRUENT When three sides are equal When two sides and their included angles are equal When two angles and one side is equal CONDITIONS FOR TRIANGLES TO BE CONGRUENT When three sides are equal AB=QR BC =PQ AC = PR ABC RQP. Reference: SSS When two sides and their included angles are equal If AB = PQ BC = QR, and B = Q Then ABC PQR. Reference: SAS
3 When two angles and one side is equal If B = Q A = P and AB = PQ Then ABC PQR. Reference: ASA Similiarity: Two figures are said to be similar when they have same shape. Example : Proto type of a building and the actual building are similar When we deal with two dimension figures we use similarity to prove the given figures are similar. Similarity can be proved by conditions given below CONDITIONS FOR TRIANGLES TO BE SIMILAR Three Angles are Equal If A = X B = Y and C = Z then ABC ~ XYZ. Reference: AAA WHEN TWO TRIANGLE ARE SIMILAR Three Angles are Equal If ABC ~ XYZ. Then, BC = AC = AB YZ XZ XY Reference: By similar Triangles
4 BASIC THEOREM OF PROPORTIONALITY Thales Theorem: If a straight line divides any two sides of a triangle in the same ratio, then the straight line is parallel to the third side of the triangle. If, AD = AE DB EC Then DE BC MID POINT THEOREM If the midpoints of two adjacent sides of a triangle are joined by a line segment, then this segment is parallel and half of the third side, If, AD = BD & AE = CE, Then DE BC and DE = 1 2 BC THEOREM The areas of similar triangles are proportional to the squares of corresponding sides, altitude, or median. Then, ar ( ABC) = AB2 = AC2 = BC2 = AM2 or ( DEF) DE 2 DF 2 EF 2 DN 2 SOME IMPORTANT CONCLUSION APB is a right triangle with P = 90. PN is perpendicular, drawn from P, to the hypotenuse AB. Then, (i) PN 2 = AN.NB (ii) AP 2 = AN.AB (iii) BP 2 = BN.BA
5 Ex: In Fig. AB and DE are perpendiculars to BC. If AB = 9 cm, DE = 3 cm and AC = 24 cm. Calculate AD. Sol: In similar ABC and CDE AB DE = AC DC 9 = 24 3 DC DC = 8 cm AD = AC DC = 24 8 = 16 cm Ex: In Fig. AQ and PB are perpendiculars to AB. If AQ = 14 cm, PB = 3.5 cm and AO = 6 cm. Calculate OP.? (Approx.) Sol: In similar OAQ and OBP, OA = OB AQ PB 6 = OB , OB = 1.5 cm In right triangle OBP, OP 2 = OB 2 + PB 2 OP 2 = = 14.5, OP = 3.8 cm Ex: The sides A, BC of a trapezium ABCD are parallel & the diagonals AC, BD meet at O. The area of the BOB = 5cm 2, and the ratio between OA : OC = 3 :5. Calculate the area of triangle AOD?
6 Sol: In similar AOD and COB Then, ar( AOD) ar( BOC) =AO2 BO 2 ar( AOD) 5 = 9 25 ar( AOD) = 9 = 1.8 cm2 5 Ex: In the given figure, ABC and CEF are two triangles where BA is parallel to CE, and AF: AC = 5: 8. Find AD, if CE = 6 cm Sol: In similar triangles ADF and CEF AD = AF CE CF AD = 5x 6 3x AD = 10 cm (By theorem) (AF = 5x, AC = 8x, CF= 3x) Ex: BE and CF be the two medians of a ABC and G be their intersection. Also let EF cut AG at O. Then find AO: OG? Sol: In similar AOE and ADC, OA AD = AE AC = 1 2 But AD AG = 3 2 OA AD x AD AG = OA = 3 AG 4 OA = 3(OA + OG) OA = 3 OG (E is the mid point of AC) OA : OG = 3 : 1
7 Ex: In PQR, S and T are points on side PR and PQ respectively such that PQR = PST. If PT = 5cm, PS = 3cm and TQ = 3cm, then find the length of SR? Sol: In PQR and PST, PQR = PST and P is common PQ = PR PS PT 8 3 = PR 5 PR = 8 X 5 3 = 40 40, then SR = =31 9 Ex: In ABC it is given that AB = 6cm, AC = 8cm AD is the angle bisector of BAC. Then find the ratio of BD and DC? Sol: In ABC, AD is angle bisector of BAC so BD = AB 6 DC AC 8 BD : AC = 3 : 4 Ex: In ABC D is point on BC in such that AD BC and E is point on AD in such that AE : ED = 5:1, If BAD = 30 0 and tan ( ACB) = 6 tan ( DBE) then find the ( ACB)? Sol. tan ( ACB) = 6 tan ( DBE) ( ACB) = 6 ( DBE) DBA = = 60 0 In ADB & EDB
8 DBA = AD 600 = 6 DBE ED DBE 1 DBE = 10 0 ACB = 6 x 10 = 60 0 Ex: A straight line parallel to BC of ABC intersects AB and AC at point P and Q given. AP = QC, PB =4 units and AQ = 9 units, then find the length of AP? Sol: In ABC, PQ BC AP AB = AQ AC AP+PB AP = AQ+QC AQ PB = QC AP AQ = AP AQ Then AP 2 =PB x AQ = 4 x 9 = 36 AP = 6 Ex: In the given figure BAD = CAD, AB= 4 cm., AC = 5.2 BD = 3 cm then find length of the BC? Sol: BAD = CAD AB AC = BD CD = 3 CD CD= 3.9 BC = = 6.9 cm
9 Ex: In ABC, D and E point onn AB and AC in such that AD = 1 3 AB and AE = 1 AC. If BC = 15cm then find the length of DE? 3 Sol: AD AC = AE AC = DE BC =1 3 DE 15 = 1 3 DE= 5cm Ex: In ABC, ab = 10 cm, BC = 8cm, CA = 6cm and M is mid point of BC is a point on AC in such that MN AB, then find the area of trapezium ABMN? Sol: In ABC MN AB So, CN = NM CA AB 3 = NM 6 10 NM = 5 cm area of ABC = 24 cm 2 area of CNM = 6 So area trapezium ABMN = 24 6 = 18cm 2
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