NCERT solutions for class 10

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2. https://www.entrancei.com/ TRIANGLES 1. Similar Triangles: Two triangles are said to be similar if (a) their corresponding angles are equal or 2. (b) their corresponding sides are proportional. Basic Proportionality Theorem: In a triangle, a line drawn parallel to one side, to intersect the other two sides in distinct points, divides the two sides in the same ratio. A Given: A triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively M N E D AD= AE To prove: . DB EC C B Construction: Join BE and CD and draw DM ⊥AC and EN ⊥AB. 1 (  = 1 Proof: area of ADE = AD base height ) EN . 2 2 (Taking AD as base) 1 =2  [The area of ADE is denoted as ar (ADE)] ar ( ADE ) AD EN So, 1 =  ar ( BDE ) DB EN , Similarly, 2 1 1 =2  =  ar ( ADE ) AE DM ar ( DEC ) EC DM . and 2 (Taking AE as base) 1  AD EN ar ( ADE ) AD 2 = = Therefore, …(i) 1 ar( BDE ) DB  DB EN 2 1  AE DM ar ( ADE ) AE 2 = = and …(ii) 1 ar( DEC ) EC  EC DM 2 BDE = ... (iii) ar ( ) ar ( DEC ) [BDE and DEC are on the same base DE and between the same parallels BC and DE.] AD= AE Therefore, from (i), (ii) and (iii), we have: DB EC For Example: In the given figure (i) and (ii) DE || BC. Find EC in (i) and AD in (ii) https://www.entrancei.com/ncert-solutions-class-10-maths

3. https://www.entrancei.com/ A A 1.8cm 1.5cm 1cm D E E D 7.2cm 5.4 cm 3cm C C B B (ii) (i)  In ABC, DE || BC then by Basic Proportionality (BPT) Theorem, we have (i) AD= AE 5 . 1 1 3 =  = = or cm EC 2 DB EC 3 EC 5 . 1 In ABC, DE || BC then by B.P.T, we have AE DB (ii)  2 . 7  AD= AE DB 8 . 1 = = or AD = 2.4 cm EC EC 4 . 5 3. Converse of Basic Proportionality Theorem: If a line divides any two sides of a triangle in the same ratio, the line must be parallel to the third side. For Example: State whether MN II QR, given PQ = 15.2 cm, PR = 12.8 cm, PM = 5.7 cm, PN = 4.8 cm. P It has been given that 4.8 cm N PQ = 15.2 cm, PR = 12.8 cm, 5.7cm 12.8 cm PM = 5.7 cm and PN = 4.8 cm M MQ = PQ–PM = (15.2 – 5.7) cm = 9.5 cm and R 15.2 cm NR = PR – PN = (12.8 – 4.8) cm = 8 cm PM 7 . 5 3 PN 8 . 4 3 = = = = Now and Q NR 8 5 MQ 5 . 9 5 PM= PN  MQ NR Thus, in PQR, MN divides the sides PQ and PR in the same ratio. Therefore, by the converse of the Basic Proportionality Theorem, we have MN II QR. 4. Criteria forSimilarity of Triangles: (i) AAA similarity: If in two triangles, the corresponding angles are equal, then their corresponding sides are proportional (i.e., in the same ratio) and hence the triangles are similar. (ii) SSS similarity: If the corresponding sides of two triangles are proportional (i.e., in the same ratio), then, their corresponding angles are equal and hence the triangles are similar. (iii) SAS similarity: If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, the triangles are similar. For example: State the similarity criterions used and write the similarity relation in symbolic form. https://www.entrancei.com/ncert-solutions-class-10-maths

4. https://www.entrancei.com/ D A P P 6 0° 40° 5 cm 6 cm 3 cm 2.5 cm 60 80° 80° 40° R Q C R B 2 cm Q E F 4 cm (i) (ii) (i) In triangles ABC and PQR, we observe that  A = Q = 40°, B = P = 60° and C = R = 80° Therefore, by AAA-criterion of similarity BAC ~ PQR (ii) In triangle PQR and DEF, we observe that PQ QR PR 1 = = = DE EF DF 2 Therefore, by SSS-criterion of similarity, we have PQR ~ DEF 5. Pythagoras Theorem: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Given: A right triangle ABC, right angled at B B 2 2 2 = + AC AB BC To prove: Construction: Draw BD⊥AC Proof:In ADB and ABC, we have C A D ADB = ABC A = A ADB ~ ABC AD= [Each equal to 90°] [Common] [By AA similarity] AB  [Corresponding sides of similar triangles are AB AC proportional] 2  In BCD and ACB, we have CDB = CBA C = C By AA similarity criterion BCD ~ ACB DC AC =  …(i) AB AD AC [Each equal to 90°] [Common] BC=  BC 2  Adding equations (i) and (ii), we get =  …(ii) BC DC AC 2 2 + =  +  AB BC AD AC DC AC https://www.entrancei.com/ncert-solutions-class-10-maths

5. https://www.entrancei.com/ 2 = + =  = AC ( AD DC ) AC AC AC 2 2 2 = + Hence, AC AB BC . For example: A ladder is placed in such a way that its foot is at a distance of 5 m from a wall and its top reaches a window 12 m above the ground. Determine the length of the ladder. Let AB be the ladder, B be the window and BC be the wall B Then BC = 12 m, AC = 5m and ACB = 90° In right triangle ACB, we have 12 m 2 2 2 = + AB AC BC [By Pythagoras Theorem] 2 2 2 = + = + = AB ( 5 ) 12 ( ) ( 25 144 ) 169  C A = AB 13 m  Hence, the length of the ladder is 13 m. 5 m 6. Converse of Pythagoras Theorem: In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle. P A A 2 triangle . AC  = 90 ABC such that Given: AB 2 2 + = B BC  To prove: Construction: Draw a PQR right angled at Q such that PQ = AB and QR = BC. Proof: In right triangle PQR, we have 2 2 2 PR QR PQ = + C Q R B  [By Pythagoras Theorem] 2 2 2 + = [PQ = AB and QR = BC (by …(i) AB construction)] 2 BC AB + BC PR 2 2 = But From equation (i) and equation (ii), we get AC PR =  AC PR = Now, in ABC and PQR, we have AB = PQ and BC = QR and PR AC = ABC PQR  B = Q = 90° Hence, B = 90°. For example: [Given] …(ii) AC 2 2 …(iii) [By construction] [From (iii)] [By SSS congruency] [c.p.c.t.] The sides of some triangles are given below. Determine which of them right triangles are. (i) 7 cm, 24 cm, 25 cm (ii) 7 cm, 8 cm, 6 cm If the sum of the squares of the smaller sides of a triangle is equal to the square of the larger side, then the triangle is right angled. https://www.entrancei.com/ncert-solutions-class-10-maths

6. https://www.entrancei.com/ (7 cm)2 + (24 cm)2 = (49 + 576) cm2 = 625 cm2 and (25 cm)2 = 625 cm2 (i) Hence, the given triangle is a right triangle. (7 cm)2 + (6 cm)2 = (49 + 36) cm2 = 85 cm2 and (8 cm)2 = 64 cm2 7. (ii) Hence, the given triangle is not a right triangle. The ratio of the areas of two similar triangles is equal to the ratio of the squares of the corresponding sides. Given: ABC and PQR such that ABC ~ PQR. P prove: CA To A 2 2 2 ABC  ar ( ) AB BC = = = PQR  2 2 2 ar ( ) PQ QR RP B C D Construction:Draw AD ⊥ BC and PS⊥QR S Q R 1   BC AD ABC  1 ar ( ) 2 base  height] = = Proof: [ Area of triangle 1 PQR  2 ar ( )   QR PS 2 ABC  ar ( ) BC AD  =  …(i) PQR  ar ( ) QR PS In ADB and PSQ B = Q [ABC ~ PQR] ADB = PSQ [Both 90°]  ADB ~ PSQ [By AA similarity] AD= AB  …(ii) PS PQ [Corresponding sides of similar triangles are proportional] AB= BC [ABC ~ PQR] But PQ QR AD= BC  [Using (ii)] …(iii) PS QR From (i) and (iii), we have 2 ABC  ar ( ) BC BC BC =  = …(iv) PQR  2 ar ( ) QR QR QR Since ABC ~ PQR AB BC CA  = = …(v) PQ QR RP 2 2 2 ABC  ar ( ) AB BC CA = = = Hence, [From (iv) and (v)] PQR  2 2 2 ar ( ) PQ QR RP https://www.entrancei.com/ncert-solutions-class-10-maths

7. https://www.entrancei.com/ For Example: The areas of two similar triangles ABC and PQR are 64 cm2 and 36 cm2 respectively. If QR = 16.5 cm, find BC. Since the ratio of the areas of two similar triangles is equal to the ratio of the squares of the corresponding sides. 2 2 2 ABC  ar ( ) BC 64 cm BC   =  = PQR  2 2 2 ar ( ) QR 36 cm 16 ( 5 . cm )  8 16 5 . 8 cm BC 5 .   = = = BC cm 22 cm 6 6 cm 16 cm 8. Hence BC = 22 cm. The area of two similar triangles are in the ratio of the squares of the corresponding altitudes. 9. If two sides and a ‘median bisecting one of these sides’ of one triangle are respectively proportional to the two sides and the corresponding median of another triangle, then the triangles are similar. 10. 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 the triangles are similar. 11. If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then the triangles on both sides of the perpendicular are similar to the whole triangle and also to each other. 12. Ratio of the perimeters of two similar triangles is the same as the ratio of their corresponding sides. 13. The areas of two similar triangles are in the ratio of the squares of the corresponding altitudes. 14. The areas of two similar triangles are in the ratio of the squares of the corresponding medians. 15. The areas of two similar triangles are in the ratio of the squares of the corresponding angle bisector segments. https://www.entrancei.com/ncert-solutions-class-10-maths

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