RD Sharma Solutions for Class 10thContents (Click on the Chapter Name to download the solutions for the desired Chapter) Chapter 1 : Real Numbers Chapter 2 : Polynomials Chapter 3 : Pair of Linear Equations in Two Variables Chapter 4 : Triangles Chapter 5 : Trigonometric Ratios Chapter 6 : Trigonometric Identities Chapter 7 : Statistics Chapter 8 : Quadratic Equations Chapter 9 : Arithmetic Progression Chapter 10 : Circles Chapter 11 : Constructions Chapter 12 : Some Applications of Trigonometry Chapter 13 : Probability Chapter 14 : Coordinate Geometry Chapter 15 : Areas related to Circles Chapter 16 : Surface Areas and Volumes Check out other Solutions by Vedantu : NCERT Solutions for Class 10 Previous Year Question Paper Solutions for Class 10 Need to solve doubts in a LIVE Session with our expert teachers? Book a FREE Trial Now! Class X Chapter 4 â Triangles Maths Exercise 4.1 1. (i) All circles are .......... (congruent, similar). (ii) All squares are .......... (similar, congruent). (iii) All .......... triangles are similar (isosceles, equilaterals): (iv) Two triangles are similar, if their corresponding angles are .......... (proportional, equal) (v) Two triangles are similar, if their corresponding sides are .......... (proportional, equal) (vi) Two polygons of the same number of sides are similar, if (a) their corresponding angles are and (b) their corresponding sides are .......... (equal, proportional). Sol: (i) All circles are similar (ii) All squares are similar (iii)All equilateral triangles are similar (iv) Two triangles are similar, if their corresponding angles are equal (v) Two triangles are similar, if their corresponding sides are proportional (vi) Two polygons of the same number of sides are similar, if (a) their corresponding angles are equal and (b) their corresponding sides are proportional. 2. Write the truth value (T/F) of each of the following statements: (i) Any two similar figures are congruent. (ii) Any two congruent figures are similar. (iii) Two polygons are similar, if their corresponding sides are proportional. (iv) Two polygons are similar if their corresponding angles are proportional. (v) Two triangles are similar if their corresponding sides are proportional. (vi) Two triangles are similar if their corresponding angles are proportional. Sol: (i) False (ii) True (iii)False (iv) False (v) True (vi) True Exercise 4.2 1. In âABC, D and E are points on the sides AB and AC respectively such that DE || BC (i) If AD = 6 cm, DB = 9 cm and AE = 8 cm, find AC. đŽđ· 3 (ii) If đ·đ” = 4 and AC = 15 cm, find AE đŽđ· 2 (iii) If đ·đ” = 3 and AC = 18 cm, find AE Printed from Vedantu.com. Book a free session with a top LIVE online tutor now. Class X Chapter 4 â Triangles Maths (iv) If AD = 4, AE = 8, DB = x â 4, and EC = 3x â 19, find x. (v) If AD = 8cm, AB = 12 cm and AE = 12 cm, find CE. (vi) If AD = 4 cm, DB = 4.5 cm and AE = 8 cm, find AC. (vii) If AD = 2 cm, AB = 6 cm and AC = 9 cm, find AE. đŽđ· 4 (viii) If đ”đ· = 5 đđđ đžđ¶ = 2.5 đđ, đđđđ đŽđž (ix) If AD = x, DB = x â 2, AE = x + 2 and EC = x â 1, find the value of x. (x) If AD = 8x â 7, DB = 5x â 3, AE = 4x â 3 and EC = (3x â 1), find the value of x. (xi) If AD = 4x â 3, AE = 8x â 7, BD = 3x â 1 and CE = 5x â 3, find the volume of x. (xii) If AD = 2.5 cm, BD = 3.0 cm and AE = 3.75 cm, find the length of AC. Sol: (i) We have, DE || BC Therefore, by basic proportionally theorem, đŽđ· đŽđž We have đ·đ” = đžđ¶ 6 8 â 9 = đžđ¶ 2 8 â 3 = đžđ¶ 8Ă3 â EC = 2 â EC = 12 cm â Now, AC = AE + EC = 8 + 12 = 20 cm ⎠AC = 20 cm (ii) Printed from Vedantu.com. Book a free session with a top LIVE online tutor now. đŽđ· 2 = 3 đđđ đ·đž | | đ”đ¶ đ·đ” đâđđđđđđđ. đŽđ· 3 = 4 đđđ đ·đž | |đ”đ¶ đ·đ” đâđđđđđđđ.Class X Chapter 4 â Triangles Maths We have.43 cm ⎠AE = 6. đŽđ· đžđ¶ = đŽđž đ·đ” 3 đžđ¶ â 2 = đŽđž Adding 1 on both sides. đđŠ đđđ đđ đđđđđđđĄđđđđđđđĄđŠ đĄâđđđđđ.com. đđŠ đđđ đđ đđđđđđđĄđđđđđđđĄđŠ đĄâđđđđđ. AE + EC = AC 60 â AE + = 15 7 60 â AE = 15 â 7 105â60 = 7 45 = 7 = 6. đ€đ âđđŁđ đŽđ· đŽđž = đ·đ” đžđ¶ Adding 1 on both sides. we get Printed from Vedantu. . we get đŽđ· đŽđž + 1 = đžđ¶ + 1 đ·đ” 3 đŽđž+đžđ¶ +1= 4 đžđ¶ 3+4 đŽđ¶ â = đžđ¶ [â” AE + EC = AC] 4 7 15 â 4 = đžđ¶ 15Ă4 â EC = 7 60 â EC = 7 Now. Book a free session with a top LIVE online tutor now. đ€đ âđđŁđ.43 cm (iii) We have. Printed from Vedantu. DE || BC Therefore. by basic proportionality theorem.Class X Chapter 4 â Triangles Maths 3 đžđ¶ + 1 = đŽđž + 1 2 3+2 đžđ¶+đŽđž â = 2 đŽđž 5 đŽđ¶ â 2 = đŽđž [â” AE + EC = AC] 5 18 â 2 = đŽđž [â” AC = 18] 18Ă2 â đŽđž = 5 36 â đŽđž = = 7. we have. . Book a free session with a top LIVE online tutor now.2 đđ 5 (iv) We have.com. đŽđ· đŽđž = đ·đ” đžđ¶ 4 8 = 3đ„â19 đ„â4 â 4(3x â 19) = 8(x â 4) â 12x â 76 = 8x â 32 â 12x â 8x = â32 + 76 â 4x = 44 44 âđ„= = 11đđ 4 ⎠x = 11 cm (v) We have. we have. . DE || BC Therefore. DE || BC Therefore. by basic proportionality theorem. Book a free session with a top LIVE online tutor now. AC = AE + EC =8+9 = 17 cm ⎠AC = 17 cm (vii) Printed from Vedantu.5 = đžđ¶ 8Ă4.com.5 â đžđ¶ = 4 â EC = 9cm Now.Class X Chapter 4 â Triangles Maths AD = 8cm. by basic proportionality theorem. AB = 12 cm ⎠BD = AB â AD = 12 â 8 â BD = 4 cm And. đŽđ· đŽđž = đ·đ” đžđ¶ 4 8 â 4. we have. đŽđ· đŽđž = đ”đ· đ¶đž 8 12 â 4 = đ¶đž 12Ă4 12 â CE = = 8 2 â CE = 6cm ⎠CE = 6cm (vi) We have. DE || BC Therefore.Class X Chapter 4 â Triangles Maths We have. we get 4 đžđ¶ + 1 = đŽđž + 1 2 4+2 đžđ¶+đŽđž â = 2 đŽđž 6 đŽđ¶ â 2 = đŽđž [â” EC + AE = AC] 6 9 â 2 = đŽđž [â” AC = 9cm] 9Ă2 đŽđž = 6 â đŽđž = 3đđ (viii) We have. AB = 6 cm ⎠DB = AB â AD =6â2 â DB = 4 cm And. we get. đ·đ” đžđ¶ = đŽđž đŽđ· 4 đžđ¶ = đŽđž 2 Adding 1 on both sides. Book a free session with a top LIVE online tutor now. DE || BC Printed from Vedantu. we have.com. đŽđ· đŽđž = đ·đ” đžđ¶ Taking reciprocal on both sides. . by basic proportionality theorem. AD = 2 cm. 5 â AE = 5 â AE = 2cm (ix) We have. Book a free session with a top LIVE online tutor now. by basic proportionality theorem.com. . We have. đŽđ· đŽđž = đ·đ” đžđ¶ đ„ đ„+2 â đ„â2 = đ„â1 â đ„(đ„ â 1) = (đ„ + 2)(đ„ â 2) â đ„ 2 â đ„ = đ„ 2 â (2)2 [â” (a â b) (a + b) = đ2 â đ 2 ] â âđ„ = â4 â x = 4 cm ⎠x = 4 cm (x) Printed from Vedantu. đŽđ· đŽđž = đ”đ· đžđ¶ 4 đŽđž â 5 = 2.Class X Chapter 4 â Triangles Maths Therefore. by basic proportionality theorem.5 4Ă2. We have. DE || BC Therefore. DE || BC Therefore. DE || BC Therefore. Book a free session with a top LIVE online tutor now. We have. đŽđ· đŽđž = đ·đ” đžđ¶ 8đ„â7 4đ„â3 â 5đ„â3 = 3đ„â1 â (8đ„ â 7)(3đ„ â 1) = (4đ„ â 3)(5đ„ â 3) â 24đ„ 2 â 8đ„ â 21đ„ + 7 = 20đ„ 2 â 12đ„ â 15đ„ + 9 â 24đ„ 2 â 20đ„ 2 â 29đ„ + 27đ„ + 7 â 9 = 0 â 4đ„ 2 â 2đ„ â 2 = 0 â 2[2đ„ 2 â đ„ â 1] = 0 â 2đ„ 2 â đ„ â 1 = 0 â 2đ„ 2 â 2đ„ + 1đ„ â 1 = 0 â 2đ„(đ„ â 1) + 1(đ„ â 1) = 0 â (2x + 1) (x â 1) = 0 â 2x + 1 = 0 or x â 1 = 0 1 â x = â 2 đđ đ„ = 1 1 đ„ = â 2 đđ đđđĄ đđđ đ đđđđ âŽx=1 (xi) We have. by basic proportionality theorem. we have.Class X Chapter 4 â Triangles Maths We have. đŽđ· đŽđž = đ·đ” đžđ¶ 4đ„â3 8đ„â7 â 3đ„â1 = 5đ„â3 â (4đ„ â 3)(5đ„ â 3) = (8đ„ â 7)(3đ„ â 1) â 4đ„(5đ„ â 3) â 3(5đ„ â 3) = 8đ„(3đ„ â 1) â 7(3đ„ â 1) â 20đ„ 2 â 12đ„ â 15đ„ + 9 = 24đ„ 2 â 8đ„ â 21đ„ + 7 â 4đ„ 2 â 2đ„ â 2 = 0 Printed from Vedantu.com. . by basic proportionality theorem. 7 cm.8 cm and AE = 2.75 â 3. D and E are points on the sides AB and AC respectively.75 + 4. BD = 9.com. AC = AE + EC = 3. BD = 4.5 = 8. we have.6cm. AC = 4.8 cm.5 250 15Ă3 â EC = 10 45 = 10 = 4. AE = 3. Sol: Printed from Vedantu.5 cm. (iv) AD = 5.8 cm.3 cm and EC = 5.4cm. (iii) AB = 10. AD = 8cm. AE = 12 cm and AC = l8cm.5 cm.2 cm and AE = 1.5 cm.5 3. AC= 7. AD = 1.8 cm. . Book a free session with a top LIVE online tutor now.Class X Chapter 4 â Triangles Maths â 2(2đ„ 2 â đ„ â 1)= 0 â 2đ„ 2 â đ„ â 1 = 0 â 2đ„ 2 â 2đ„ + 1đ„ â 1 = 0 â 2đ„(đ„ â 1) + 1(đ„ â 1) = 0 â (2x + 1) (x â 1) = 0 â 2x + 1 = 0 or x â 1 = 0 1 â x = â 2 or x = 1 1 x = â 2 is not possible âŽx=1 (xii) We have.25 ⎠AC = 8.25 cm 2.5 đđ Now. DE || BC Therefore. In a âABC. For each of the following cases show that DE || BC: (i) AB = 2cm.0 = đžđ¶ 3. đŽđ· đŽđž = đ·đ” đžđ¶ 2.75Ă3 375Ă3 â đžđ¶ = = 2. by basic proportionality theorem. (ii) AB = 5. 2 â 1. DE divides sides AB and AC of âABC in the same ratio.com.4 â DB = 4. (ii) We have. ⎠DB = AB â AD = 12 â 8 â DB = 4 cm And.8 cm ⎠DB = AB â AD = 5. by the converse of basic proportionality theorem.2 cm And. AD = 12 cm and AC = 18 cm.2 cm and AE = 1. .4 cm.8 â EC = 5. EC = AC â AE = 7. Therefore. AD = 1.4 cm Printed from Vedantu.6 cm. EC = AC â AE = 18 â 12 â EC = 6 cm đŽđ· 8 2 Now. Book a free session with a top LIVE online tutor now. DE || BC We have. AC = 7.6 â 1. đžđ¶ = =1 [â” EC = 6 cm] 6 đŽđ· đŽđž â đ·đ” = đžđ¶ Thus. AB = 5.Class X Chapter 4 â Triangles Maths AB = 12 cm. đ·đ” = 4 = 1 [â” DB = 4 cm] đŽđž 12 2 And. 8cm ⎠AD = AB â DB = 10.5 â AD = 6. DE divides sides AB and AC of âABC in the same ratio.3 cm] đŽđž 2.4 cm] Thus.Class X Chapter 4 â Triangles Maths đŽđ· 1.5cm.8 â EC = 2 cm đŽđ· 6. BD = 4. EC = AC â AE = 4.2 = 3 [â” DB = 4. đ·đ” = 4. Printed from Vedantu.8 â 2. Therefore.8 â 4. AC = 4.5 = 5 [â” AD = 6.4 1 Now. (iv) We have. Book a free session with a top LIVE online tutor now. by the converse of basic proportionality theorem. DE divides sides AB and AC of âABC in the same ratio. We have. AB = 10.8 28 75 And. đžđ¶ = 5.8 1 And. Therefore. đžđ¶ = = 20 = [â” EC = 2 cm] 2 5 Thus. đ·đ” = 4.4 = 3 [â” EC = 5. by the converse of basic proportionality theorem.com.2 cm] đŽđž 1.3 7 Now. .3 cm And.8 cm and AE = 2.8cm. (iii) We have. AD = 5. QC = 3 cm and BC = 6 cm. In a âABC.6 cm Now. Therefore. đžđ¶ = 5. AE = 3. by BPT We have. đŽđ đŽđ = đđ¶ đđ” 2.4 + 3. If AP = 2. such that PQ || BC.5 = 55 đŽđž 3 â đžđ¶ = 5 Thus DE divides sides AB and AC of âABC in the same ratio. Sol: We have || BC Therefore. find AB and PQ.6 = 6cm Now.5 = 95 đŽđ· 3 â đ”đ· = 5 đŽđž 3.3 cm and EC = 5. by the converse of basic proportionality theorem. In âAPQ and âABC â A = â A [common] â APQ = â ABC [â” PQ || BC â Corresponding angles are equal] â âAPQ ~ âABC [By AA criteria] đŽđ” đ”đ¶ â đŽđ = đđ [corresponding sides of similar triangles are proportional] 6Ă2.4 â PQ = 6 â PQ = 2. AB = AP + PB = 2.5 cm đŽđ· 5.4 3Ă24 3Ă6 18 â PB = = = = 2 2 5 5 â PB = 3. AQ = 2 cm. BD = 9.3 33 And.com.5 cm. P and Q are points on sides AB and AC respectively.4 cm Printed from Vedantu. Book a free session with a top LIVE online tutor now. We have DE || BC 3.4 2 =3 đđ” 3Ă2.7 57 Now đ”đ· = 9.7 cm. .Class X Chapter 4 â Triangles Maths DE || BC We have.4 cm. 2 = 4. AE = 3. .0 đđ 2 â AC = 8 cm ⎠CE = AC â AE = 8 â 3.4 cm 4.6 Ă 5 = 8.2 cm. D and E are points on AB and AC respectively such that DE || BC.0 cm â AB = 6 cm ⎠BD = 6 cm BD = AB â AD = 6 â 2.4 = 3.4cm.4Ă5 â AB = 2 â AB = 1.8 cm Printed from Vedantu. AB = 6 cm and PO = 2.6 cm Now.2 â = 5 2 3. DE = 2cm and BC = 5 cm.6 cm â DB = 3. BD = 3. find BD and CE.8 cm Hence. In a âABC. In âADE and âABC â A = â A [common] â ADE = â ABC [â” DE || BC â Corresponding angles are equal] â âADE ~ âABC [By AA criteria] đŽđ” đŽđ· â đ”đ¶ = đ·đž [corresponding sides of similar triangles are proportional] 2.Class X Chapter 4 â Triangles Maths Hence. DE || BC Now. Sol: We have. đŽđ¶ đŽđž = đ·đž [â” Corresponding sides of similar triangles are equal] đ”đ¶ đŽđ¶ 3.2Ă5 â đŽđ¶ = = 1.6 cm and CE = 4.2 Ă 5 = 6. Book a free session with a top LIVE online tutor now.com. If AD = 2. 6 36 3 And. . PQ is not parallel to EF 6.9 cm. For each of the following cases. In below Fig. PE = 3cm.3 13 Now. QM = 4. đđ = 4. DQ = 3. Book a free session with a top LIVE online tutor now. Sol: We have. đđ = 4. state if PQ || EF. đđč = 2.5 = 9 đđ 4 8 Also.5 cm.5 cm.com. đđž = = = 10 3 1 đ·đ 3.5 = 9 đđ đđ Hence. PN = 4 cm and NR = 4. QM = 4. đđ = đđ By converse of proportionality theorem MN || QR Printed from Vedantu. M and N are points on the sides PQ and PR respectively of a âPQR.5 cm đđ 4 8 Hence.Class X Chapter 4 â Triangles Maths 5.4 = 24 = 2 đ·đ đ·đ â đđž â đđč So. state whether MN || QR (i) PM = 4cm..4 cm đ·đ 3. PM = 4cm. DP = 3.5 cm Sol: (i) We have.6 cm and QF = 2.9 1. PN = 4 cm and NR = 4. com. Book a free session with a top LIVE online tutor now. Therefore. OB. In three line segments OA. Sol: We have. LN divides sides OA and OC of âOAC in the same ratio. M.Class X Chapter 4 â Triangles Maths 7. Show that LN ||AC. đđ¶ = đđ” âŠ(ii) đ¶đđđđđđđđ đđđąđđĄđđđ (đ)đđđ đđđąđđĄđđđ (đđ). N nor of A. đđ đđ = đŽđż đđ¶ Thus. Prove that âABC is isosceles. We have. and OC. by the converse of basic proportionality theorem. đ€đ đđđĄ. C are collinear. đđż đđ = đđ” âŠ(i) đŽđż đđ đđ đđđ. by basic proportionality theorem. Sol: Printed from Vedantu. N respectively are so chosen that LM || AB and MN || BC but neither of L. M. . LN || AC 8. points L. we have. If D and E are points on sides AB and AC respectively of a âABC such that DE || BC and BD = CE. B. LM || AB and MN || BC Therefore. find BD and DC.5 cm. (viii) If AB = 10cm. find AC. (ii) If BD = 2cm. AC = 6 cm and BC = 12 cm. find DC.6 cm.8 cm. DE || BC Therefore. đŽđ· đŽđž = đžđ¶ đ·đ” đŽđ· đŽđž â đ·đ” = đ·đ” [â”BD = CE] â AD = AE Adding DB on both sides â AD + DB = AE + DB â AD + DB = AE + EC [⎠BD = CE] â AB = AC â â ABC is isosceles Exercise 4. find AB (vi) If AB = 5. the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.2 Printed from Vedantu. (v) If AC = 4. we have. find AC. (i) If BD = 2.2 cm. AC = 4. BC = 6cm and BD = 3. (vii) If AD = 5. AB = 5cm and AC = 4. by BPT. AC = 6cm and DC = 3cm. DC = 6 cm and 10 cm. (iv) If AB = lo cm.2 cm and DC = 2.com.2cm.2 cm. đ”đ· đŽđ” ⎠đ·đ¶ = đŽđ¶ 2.5 5 â đ·đ¶ = 4. AD is the bisector of â A. (iii) If AB = 3. â BAD = â CAD We know that.Class X Chapter 4 â Triangles Maths We have.3 1. . find BD and DC. AC =14 cm and BC =6 cm. AB = 5cm and DC = 3cm. Sol: (i) We have. Book a free session with a top LIVE online tutor now. find BC. meeting side BC at D. In a âABC. find BD.5cm.6 cm. Class X Chapter 4 â Triangles Maths 2. đ”đ· đŽđ” ⎠đ·đ¶ = đŽđ¶ đ”đ· 3.2 3. the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. .com.5 cm (iii) In âABC. the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.1 đđ 100 ⎠DC = 2. We know that. đ”đ· đŽđ” ⎠= đ·đ¶ đŽđ¶ 2 5 â 3 = đŽđ¶ 5Ă3 15 â AC = = 2 2 â AC = 7. AD is the bisector of â A.5Ă4.1 cm (ii) We have.5Ă2 = 3 Printed from Vedantu.5 â 2. Book a free session with a top LIVE online tutor now.2 â đ·đ¶ = 5 25Ă42 5Ă42 210 = 5Ă100 = = 100 = 2. AD is the bisector of â A We know that.8 = 4. đ”đ· đŽđ” ⎠đ·đ¶ = đŽđ¶ đ„ 10 â 6âđ„ = 14 â 14x = 10(6 â x) â 24x = 60 60 5 â đ„ = 24 = 2 = 2.5 cm (v) We have.5đđ Since.33 đđ ⎠đ”đ· = 2. BC = 10 cm.Class X Chapter 4 â Triangles Maths 7 = 3 = 2.2 cm ⎠BD = BC â DC = 10 â 6 = 4 cm â BD = 4 cm Printed from Vedantu.5 = 3.5 cm Hence.3 đđ (iv) In âABC. Book a free session with a top LIVE online tutor now. the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. AD is the bisector of â A We know that. . DC = 6 â x = 6 â 2. DC = 6 cm and AC = 4.5cm.com. and DC = 3. BD = 2. 8 cm ⎠BC = 5. In âABC.8 cm (vi) We have.8đđ 6 2 â BD = 2. We know that. đ”đ· đŽđ” ⎠đ·đ¶ = đŽđ¶ 4 đŽđ” â 6 = 4.8 cm Since. BC = BD + DC = 2. . Printed from Vedantu.6 â = 3 6 5. AD is the bisector of â A.Class X Chapter 4 â Triangles Maths In âABC. We know that.8 cm (vii) We have. the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. Book a free session with a top LIVE online tutor now. the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.8 + 3 = 5.2 [â” BD = 4 cm] â AB = 2.6Ă3 5.6 â BD = = = 2. AD is the bisector of â A. đ”đ· đŽđ” ⎠đ·đ¶ = đŽđ¶ đ”đ· 5.com. 8 5.2 â đŽđ¶ = 2.5 = 4. We know that. the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the containing the angle.2 5.9 đđ (viii) In âABC. BD = 7. Book a free session with a top LIVE online tutor now. AE is the bisector of the exterior â CAD meeting BC produced in E. AD is the bisector of â A. If AB = 10cm.8 â đŽđ¶ = 3.6Ă7 = = 0.7 Ă 7 8 = 4.5 cm and DC = 12 â x = 12 â 7.5 đđ 16 ⎠BD = 7.2 = 6â3. AC = 6cm and BC = 12 cm.57. 4.2 [â” DC = BC â BD] đŽđ¶ 5.5 cm 2. đ”đ· đŽđ” ⎠đ·đ¶ = đŽđ¶ đ„ 10 â 12âđ§ = 6 â 6x = 10(12 â x) â 6x = 120 120 âđ„= = 7.Class X Chapter 4 â Triangles Maths In âABC. the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.6 3. .5 cm and DC = 4.6 3. find CE. We know that. Printed from Vedantu. đŽđ” đ”đ· ⎠đŽđ¶ = đ·đ¶ 5.6 Ă2.com. AD is the bisector of â A.5 cm Hence. In Fig. â đ¶ = 50°.58. Book a free session with a top LIVE online tutor now. âABC is a triangle such that đŽđ¶ = đ·đ¶ . if a line through one vertex of a triangle divides the opposite side in the ratio of the other two sides. đčđđđ â đ”đŽđ·.com. ⎠â 1 = â 2 In âABC â A + â B + â C = 180° â â A + 70 ° + 50° = 180° [â” â B = 70° and â C = 50°] â â A = 180° â 120° = 60° â â 1 + â 2 = 60° â â 1 + â 1 = 60° [â” â 1 = â 2] Printed from Vedantu. . 4. â đ” = 70°. In Fig. AD is the bisector of â A. Sol: We have. the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.Class X Chapter 4 â Triangles Maths Sol: In âABC. We know that. đ”đ· đŽđ” đ„ 10 ⎠đ·đ¶ = đŽđ¶ â 12âđ„ = 6 â 6(12 + x) = 10x â 72 + 6x = 10x â 4x â 72 72 âx= = 18 đđ 4 ⎠CE = 18 cm đŽđ” đ”đ· 3. then the line bisects the angle at the vertex. In âABC (Fig. (i) [Alternate angles] And. â 1 â â 4 âŠ. prove that đŽđ¶ = đ·đ¶ . in âACE.. . In âBCE. ⎠â 2 â â 3 âŠ. â 1 â â 2 [Given] From (i) and (ii). we get â 3 â â 4 Thus. CE || DA and AC cuts them. if â 1 = â 2. we have â 3 = â 4 â AE = AC ⊠(iii) [Sides opposite to equal angles are equal] Now.Class X Chapter 4 â Triangles Maths â 2â 1 = 60° â â 1 = 30° ⎠â BAD = 30° đŽđ” đ”đ· 4. Proof: since.59).com. 4. we have DA || CE Printed from Vedantu. Book a free session with a top LIVE online tutor now. Sol: Given: A âABC in which â 1 = â 2 đŽđ” đ”đ· To prove: đŽđ¶ = đ·đ¶ Construction: Draw CE || DA to meet BA produced in E.(ii) [Corresponding angles] But. determine AP.Class X Chapter 4 â Triangles Maths đ”đ· đ”đŽ â đ«đȘ = đšđŹ [Using basic proportionality theorem] đ”đ· đŽđ” â đ·đ¶ = đŽđ¶ [â” BA â AB and AE â AC from (iii)] đŽđ” đ”đ· Hence. D. E and F are the points on sides BC. BE bisects â B. đŽđč đŽđ¶ ⎠đčđ” = đ”đ¶ đŽđč 4 â 5âđŽđč = 8 [â” FB = AB â AF = 5 â AF] đŽđč 1 â 5âđŽđč = 2 â 2AF = 5 â AF â 2AF + AF = 5 â 3AF = 5 5 â AF = 3 cm Again. CE and BD. BC = 8 cm and CA = 4 cm. Sol: In âABC. If AB = 5 cm. . Book a free session with a top LIVE online tutor now. In âABC. the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. CF bisects â C. đŽđ¶ = đ·đ¶ 5. We know that. BE bisects â B and CF bisects â C.com. đŽđž đŽđ” ⎠đžđ¶ = đ”đ¶ 4âđ¶đž 5 â =8 [â” AE = AC â CE = 4 â CE] đ¶đž â 8(4 â đ¶đž) = 5 Ă đ¶đž â 32 â 8CE = 5CE â 32 = 13CE Printed from Vedantu. CA and AB respectively of âABC such that AD bisects â A. 6 2 And. đ”đ· 1.5 3 = 3.5 cm and CD = 3.5 cm and BC = 9cm Sol: Now. (v) AB = 5 cm.4 = 3 đŽđ” đ”đ· â đŽđ¶ = đ¶đ· â AD is the bisector of â A. 4.60. BD = 2. Now. In fig. AC = 24 cm.5 = 7 đ¶đ· đŽđ” 5 1 And. BD = 6 cm and BC = 24cm (iv) AB = 6 cm. BD = 1. . đŽđ” 8 1 Now. AC = 8cm. BD = l.Class X Chapter 4 â Triangles Maths 32 â CE = 13 cm Similarly. đŽđ” 4 2 =6=3 đŽđ¶ đ”đ· 1. đŽđ¶ = 10 = 2 đ”đ· đŽđ” â đ¶đ· â đŽđ¶ â AD is not the bisector of â A. AC = l2 cm. AC = 10cm.com.. Book a free session with a top LIVE online tutor now. đ”đ· đŽđ· = đ·đ¶ đŽđ¶ đ”đ· 5 â 8âđ”đ· = 4 [â” DC = BC â BD = 8 â BD] â 4BD = 40 â 5BD â 9BD = 40 40 â BD = cm 9 5 32 40 Hence.6 cm and CD = 2. AC = 6 cm. check whether AD is the bisector of â A of âABC in each of the following: (i) AB = 5cm. AF = 3 đđ. đ¶đ· = 2.5 cm (ii) AB = 4 cm.5 cm and CD= 2 cm. CE = 13 cm and BD = cm. đŽđ¶ = 24 = 3 Printed from Vedantu. BD = 1.4 cm (iii) AB = 8 cm. 9 6. com.5 = 6. đŽđ” 6 3 =8=4 đŽđ¶ đ”đ· 2.5 = 9â2. đ¶đ· = đ”đ¶âđ”đ· [â” CD = BC â BD] đ”đ· = 24â6 6 = 18 1 =3 đŽđ” đ”đ· ⎠đŽđ¶ = đ¶đ· ⎠đŽđ· đđ đĄâđ đđđ đđđĄđđ đđ â đŽ đđ âđŽđ”đ¶.. . Sol: Since diagonals of a trapezium divide each other proportionally. Exercise 4. (i) In below fig. đŽđ đ”đ ⎠đđ¶ = đđ· 4 đ„+1 â 4đ„â2 = 2đ„+4 â 4(2đ„ + 4) = (đ„ + 1)(4đ„ â 2) â 8x + 16 = x(4x â 2) +1(4x â 2) â 8x + 16 = 4x2 + 2x â 2 â 4đ„ 2 + 2đ„ â 8đ„ â 2 â 16 = 0 â 4đ„ 2 â 6đ„ â 18 = 0 â 2[2đ„ 2 â 3đ„ â 9] = 0 â 2đ„ 2 â 3đ„ â 9 = 0 Printed from Vedantu. Book a free session with a top LIVE online tutor now.4 1.5 And. If AB || CD.5 1 =3 đŽđ” đ”đ· ⎠đŽđ¶ â đ¶đ· ⎠AD is not the bisector of â A of âABC. find the value of x.Class X Chapter 4 â Triangles Maths đ”đ· đ”đ· And.5 2. = [â” CD = BC â BD] đ¶đ· đ”đ¶âđ”đ· 2. com. find x. If AB || CD. 3đ„â1 2đ„+1 â 5đ„â3 = 6đ„â5 â (3x â 1) (6x â 5) = (2x + 1) (5x â 3) â 3x (6x â 5) â 1(6x â 5) = 2x (5x â 3) + 1 (5x â 3) â 18đ„ 2 â 15đ„ â 6đ„ + 5 = 10đ„ 2 â 6đ„ + 5đ„ â 3 â 8đ„ 2 â 20đ„ + 8 = 0 â 4(2đ„ 2 â 5đ„ + 2) = 0 â 2đ„ 2 â 4đ„ â 1đ„ + 2 = 0 â 2đ„(đ„ â 2) â 1(đ„ â 2) = 0 â (2đ„ â 1)(đ„ â 2) = 0 â 2x â 1 = 0 or x â 2 = 0 1 â đ„ = 2 đđ đ„ = 2 1 đ„ = 2 is not possible.. because. OB = x â 4. AB || CD. If OA = 3x â 19. .Class X Chapter 4 â Triangles Maths â 2đ„(đ„ â 3) + 3(đ„ â 3) = 0 â (đ„ â 3)(2đ„ + 3) = 0 â đ„ â 3 = 0 or 2đ„ + 3 = 0 3 â đ„ = 3 or đ„ = â 2 3 â x = 3 or đ„ = â 2 3 3 1 đ„ = â 2 is not possible. find the value of x. because OB = x + 1 = â 2 + 1 = â 2 Length cannot be negative đŽđ đ”đ ⎠đđ¶ = đđ· (ii) In the below fig. OC = 5x â 3 1 = 5 (2) â 3 5â6 1 = = â2 2 (iii) In below fig.. Printed from Vedantu. Book a free session with a top LIVE online tutor now. OC = x â 3 and OD = 4. 5 â 2.8 cm. Book a free session with a top LIVE online tutor now. . PQ = 4 cm.5 1. đŽđ đ”đ ⎠đđ¶ = đđ· 3đ„â19 đ„â4 â = đ„â3 4 â 4(3đ„ â 19) = (đ„ â 4)(đ„ â 3) â 12x â 76 = x (x â 3) â4(x â 3) â 12đ„ â 76 = đ„ 2 â 3đ„ â 4đ„ + 12 â đ„ 2 â 7đ„ â 12đ„ + 12 + 76 = 0 â đ„ 2 â 19đ„ + 88 = 0 â đ„ 2 â 11đ§ â 8đ§ + 88 = 0 â đ„(đ„ â 11) â 8(đ„ â 11) = 0 â (đ„ â 11)(đ„ â 8) = 0 â đ„ â 11 = 0 or x â 8 = 0 â x = 11 or x = 8 Exercise 4.Class X Chapter 4 â Triangles Maths Since diagonals of a trapezium divide each other proportionally. 4. In fig. đŽđ = đđ = đŽđ [corresponding parts of similar â are proportional] đŽđ¶ 8 6.5 â 2. âACB ~ âAPQ.5 cm and AP = 2. Sol: Given âACB ~ âAPQ đŽđ¶ đ”đ¶ đŽđ” Then.8 đđđ đŽđ = 6. If BC = 8 cm.com.8 = 4 = đŽđ đŽđ¶ 8 8 6.136. find CA and AQ. BA = 6.8 = 4 đđđ 4 â đŽđ 8 4 â AC = 4 Ă 2.5 Ă 8 Printed from Vedantu. 5 đ 8 3. âABC ~ âPQR [By AA similarity] đŽđ” đ”đ¶ ⎠đđ = đđ 10đđ â đđ â = 3000 8 đđ 10 ââ= Ă 3000 = 3750 đđ = 37. Find the length of PB. Sol: We have.25 cm 2. AB || QR. Sol: Length of stick = 10 cm Length of shadow of stick = 8 cm Length of shadow of tower = h cm In âABC and âPQR â B = â Q = 90° And.6 cm and AQ = 3.Class X Chapter 4 â Triangles Maths â AC = 5. â C = â R [Angular elevation of sun] Then. In Fig. 4. A vertical stick 10 cm long casts a shadow 8 cm long.137. Determine the height of the tower. At the same time a shadow 30 m long.com. Book a free session with a top LIVE online tutor now. . âPAB and âPQR â P = â P [common] â PAB = â PQR [corresponding angles] Printed from Vedantu. XY || BC. In fig. âAXY ~ âABC [By AA similarity] đŽđ đđ ⎠đŽđ” = đ”đ¶ [Corresponding parts of similar â are proportional] 1 đđ â4= 6 6 â XY = 4 = 1.5đđ 5. . In a right angled triangle with sides a and b and hypotenuse c. Book a free session with a top LIVE online tutor now. XY || BC In âAXY and âABC â A = â A [common] â AXY = â ABC [corresponding angles] Then. the altitude drawn on the hypotenuse is x. Find the length of XY Sol: We have.com.138. 4.Class X Chapter 4 â Triangles Maths Then. âPAB ~ âPQR [By AA similarity] đđ” đŽđ” ⎠đđ = đđ [Corresponding parts of similar âare proportional] đđ” 3 â =9 6 3 â PB = 9 Ă 6 = 2 đđ 4. Prove that ab = cx. Sol: We have: â C = 90° and CD â„ AB In âACB and âCDB Printed from Vedantu. in âBCD] Compare equations (i) & (ii) â ABD = â C âŠ(iii) In âABD and âBCD â ABD = â C [From (iii)] â ADB = â BDC [Each 90°] Then. .7 cm. âACB ~ âCDB [By AA similarity] đŽđ¶ đŽđ” ⎠đ¶đ· = [Corresponding parts of similar â are proportional] đ¶đ” đ đ âđ„=đ â ab = cx 6. Sol: We have.Class X Chapter 4 â Triangles Maths â B = â B [common] â ACB = â CDB [Each 90°] Then. BD = 3. â C + â DBC â 90° âŠ(ii) [By angle sum prop.4 cm.com. find CD. 4.139.14. Book a free session with a top LIVE online tutor now. In Fig. If AB = 5.8 cm and CD = 5. â ABC = 90° and BD â„ AC. 4. âABD ~ âBCD [By AA similarity] đ”đ· đŽđ· ⎠đ¶đ· = đ”đ· [Corresponding parts of similar â are proportional] 8 4 â đ¶đ· = 8 8Ă8 â CD = = 16 đđ 4 7. â ABC = 90° and BD â„ AC Now. In Fig. If BD = 8 cm and AD = 4 cm. Printed from Vedantu. â ABC = 90° and BD â„ AC. â ABD + â DBC â 90° âŠ(i) [â” â ABC â 90°] And. find BC. 8 Ă 8.7 đ”đ¶ â 3. In Fig. âABC ~ âBDC [By AA similarity] đŽđ” đ”đ¶ ⎠đ”đ· = đ·đ¶ [Corresponding parts of similar â are proportional] 5.Class X Chapter 4 â Triangles Maths Sol: We have. 4. If AB = 6 cm. Sol: 1 We have. DE || BC. DE || BC such that AE = (1/4) AC.141. â ABC = 90° and BD â„ AC In âABC and âBDC â ABC = â BDC [Each 90°] â C = â C [Common] Then. AB = 6 cm and AE = 4 đŽđ¶ In âADE and âABC â A = â A [Common] â ADE = â ABC [Corresponding angles] Then.com.4 5.8 = 5. find AD.5 cm Printed from Vedantu. âADE ~ âABC [By AA similarity] đŽđ· đŽđž â đŽđ” = đŽđ¶ [Corresponding parts of similar â are proportional] 1 đŽđ· đŽđ¶ 1 4 â = [â” AE = 4 đŽđ¶ đđđŁđđ] 6 đŽđ¶ đŽđ· 1 â =4 6 6 â AD = 4 = 1.1 đđ 8.7 â BC = 3. . Book a free session with a top LIVE online tutor now. .Class X Chapter 4 â Triangles Maths 1 1 1 9. In below fig. Also.(i) In â AQB and âARC â QAB = â RAC [common] â ABQ = â ACR [Each 90°] Then. find the value of x. PA â„ AC.com. PA. QB â„ AC and RC â„ AC Let. Prove that âCAB ~ âCED. . In fig.(ii) Adding equations (i) & (ii) đŠ đŠ đ đ + đ§ = đ+đ + đ+đ đ„ 1 1 đ+đ â đŠ (đ„ + đ§) = đ+đ 1 1 â y(đ„ + đ§) = 1 1 1 1 âđ„+đ§ =đŠ 10. QB and RC are each perpendicular to AC..142. Printed from Vedantu. 4. Book a free session with a top LIVE online tutor now. âCQB ~ âCPA [By AA similarity] đđ” đ¶đ” ⎠đđŽ = đ¶đŽ [Corresponding parts of similar â are proportional] đŠ đ â đ„ = đ+đ âŠ. Prove that đ„ + đ§ + đŠ Sol: We have. â A = â CED. âAQB ~ âARC [By AA similarity] đđ” đŽđ” ⎠đ đ¶ = [Corresponding parts of similar â are proportional] đŽđ¶ đŠ đ â đ§ = đ+đ âŠ. AB = a and BC = b In âCQB and âCPA â QCB = â PCA [Common] â QBC = â PAC [Each 90°] Then. If AL â„ BC and DM â„ EF. ratio of perimeter of triangles = ratio of corresponding sides 25 đŽđ” â 12 = đđ 25 9 â 15 = đđ 15Ă9 â PQ = = 5. it is being given that: AB = 5 cm.4 cm. If one side of first triangle is 9 cm. find AL: DM. âABC ~ âPQR Then.4 đđ 25 12. Book a free session with a top LIVE online tutor now. âCAB ~ âCED [By AA similarity] đ¶đŽ đŽđ” ⎠đ¶đž = đžđ· [Corresponding parts of similar â are proportional] 15 9 â 10 = đ„ 10Ă9 âđ„= = 6 đđ 15 11.com.2 cm. Sol: Printed from Vedantu. BC = 4 cm and CA = 4.Class X Chapter 4 â Triangles Maths Sol: We have. The perimeters of two similar triangles are 25 cm and 15 cm respectively. In âABC and âDEF. EF = 8 cm and FD = 8. â A = â CED In âCAB and âCED â C = â C [Common] â A = â CED [Given] Then. âABC ~ âPQR Perimeter of â ABC = 25 cm Perimeter of â PQR = 15 cm AB = 9 cm PQ = ? Since. DE=10cm. what is the corresponding side of the other triangle? Sol: Assume ABC and PQR to be 2 triangles We have. . đ·đž = đžđč = đ·đž = 2 Then. Sol: We have. AE = 6 cm and CE = 9 cm. đŽđ· 8 2 = 12 = 3 đ·đ” đŽđž 6 2 And. In âABL ~ âDEM â B = â E [â ABC ~ âDEF] â ALB = â DME [Each 90°] Then. DB = 12 cm. âABC ~ âDEF [By SSS similarity] Now. đžđ¶ = 9 = 3 đŽđ· đŽđž Since. Prove that BC = 5/2 DE.com. by converse of basic proportionality theorem DE || BC Printed from Vedantu. âABL ~ âDEM [By AA similarity] đŽđ” đŽđż ⎠đ·đž = đ·đ [Corresponding parts of similar â are proportional] 5 đŽđż â 10 = đ·đ 1 đŽđż â 2 = đ·đ 13. D and E are the points on the sides AB and AC respectively of a âABC such that: AD = 8 cm. . đ·đ” = đžđ¶ Then. Book a free session with a top LIVE online tutor now.Class X Chapter 4 â Triangles Maths đŽđ” đ”đ¶ đŽđ¶ 1 Since. (i) In âCDF and âCBX [By AA similarity] đ¶đ· đ·đč ⎠đ¶đ” = đ”đ [Corresponding parts of similar â are proportional] 1 đ·đč â 2 = đ”đž+đžđ [BD = DC given] â BE + EX = 2DF â BE + EX = 4EX â BE = 4EX â EX [By using (i)] â BE = 4EX â EX Printed from Vedantu.Class X Chapter 4 â Triangles Maths In âADE and âABC â A = â A [Common] â ADE = â B [Corresponding angles] Then. draw DF || BX Proof: In âEAX and âADF â EAX = â ADF [Common] â AXE = â DAF [Corresponding angles] Then. Prove that BE : EX = 3 : 1 Sol: Given: In âABC. âADE ~ âABC [By AA similarity] đŽđ· đ·đž ⎠đŽđ” = [Corresponding parts of similar â are proportional] đ”đ¶ 8 đ·đž â 20 = đ”đ¶ 2 đ·đž â5= đ”đ¶ 5 â BC = 2 DE 14. To prove: BE : EX = 3 : 1 Const: Through D. âAEX ~ âADF [By AA similarity] đžđ đŽđž ⎠đ·đč = đŽđ· [Corresponding parts of similar â are proportional] đžđ đŽđž â đ·đč = 2đŽđž [AE = ED given] â DF = 2EX âŠ. .com. AD is bisected at the point E and BE produced cuts AC at the point X. D is the mid-point of BC and E is the mid-point of AD. D is the mid-point of side BC of a âABC. Book a free session with a top LIVE online tutor now. đđ· = đ”đ¶ â AB Ă BC = QD Ă BP 16. ABCD is a parallelogram and APQ is a straight line meeting BC at P and DC produced at Q. Sol: Given: ABCD is a parallelogram To prove: BP Ă DQ = AB Ă BC Proof: In âABP and âQDA â B = â D [Opposite angles of parallelogram] â BAP = â AQD [Alternate interior angles] Then. DA = BC [Opposite sides of parallelogram] đŽđ” đ”đ Then. AL and CM are the perpendiculars from the vertices A and C to BC and AB respectively. If AL and CM intersect at O. âABP ~ âQDA [By AA similarity] đŽđ” đ”đ ⎠đđ· = đ·đŽ [Corresponding parts of similar â are proportional] But. In âABC. Book a free session with a top LIVE online tutor now. prove that: (i) â OMA and âOLC đđŽ đđ (ii) = đđ¶ đđż Sol: Printed from Vedantu.com. Prove that the rectangle obtained by BP and DQ is equal to the AB and BC.Class X Chapter 4 â Triangles Maths đ”đž 3 â đžđ = 1 15. . EF = 10 cm. Book a free session with a top LIVE online tutor now.com.(ii) In âACD and âAEF â CAD = â EAF [common] â ACD = â AEF [corresponding angles] Then. Sol: We have AB || CD || EF.(i) [By AA similarity] đžđ¶ đ¶đ· ⎠đžđŽ = đŽđ” [Corresponding parts of similar â are proportional] đžđ¶ đ„ â đžđŽ = 6 âŠ. âOMA ~ âOLC [By AA similarity] đđŽ đđ ⎠đđ¶ = [Corresponding parts of similar â are proportional] đđż 17. BD = 4 cm and DE = y cm In âECD and âEAB â CED = â AEB [common] â ECD = â EAB [corresponding angles] Then.75 cm 8 đ·đ¶ đžđ· From (i) đŽđ” = đ”đž Printed from Vedantu. EF = 10 cm. âECD ~ âEAB âŠ. If AB = 6 cm. If AB = 6 cm. âACD ~ âAEF [By AA similarity] đŽđ¶ đ¶đ· ⎠đŽđž = đžđč đŽđ¶ đ„ â = âŠ(iii) đŽđž 10 Add equations (iii) & (ii) đžđ¶ đŽđ¶ đ„ đ„ ⎠đžđŽ + đŽđž = 6 + 10 đŽđž 5đ„+3đ„ â đŽđž = 30 8đ„ â 1 = 30 30 âđ„= = 3. BD = 4 cm and DE = y cm. . CD = x cm. AL â„ BC and CM â„ AB In â OMA and âOLC â MOA = â LOC [Vertically opposite angles] â AMO = â CLO [Each 90°] Then. CD = x cm.Class X Chapter 4 â Triangles Maths We have. calculate the values of x and y. In Fig below we have AB || CD || EF. Q.Class X Chapter 4 â Triangles Maths 3. if AB â„ BC. DC â„ BC and DE â„ AC. below. by mid-point theorem 1 QR || AD and QR = 2 đŽđ· âŠ(ii) Compare (i) and (ii) PS || QR and PS = QR Since one pair of opposite sides is equal as well as parallel then PQRS is a parallelogram . S be the mid-points of AB.com. If P. In âABC. show that PQRS is a rhombus. by mid-point theorem 1 PQ || BC and PQ = 2 BC âŠ(iv) And. CD and BD respectively. Q.25 = 6. AC.25y = 15 15 â đŠ = 2.. .. In Fig. PQRS is a parallelogram with PS = PQ then PQRS is a rhombus 19.67 đđ 18. In âBAD.(iii) Now.75y + 15 â 2. ABCD is a quadrilateral in which AD = BC. R. AC. Printed from Vedantu. by mid-point theorem 1 PS || AD and PS = 2 AD âŠ(i) In âCAD.75 đŠ â = đŠ+4 6 â 6y = 3. Prove that â CED ~ ABC. AD = BC âŠ(v) [given] Compare equations (i) (iv) and (v) PS = PQ âŠ(vi) From (iii) and (vi) Since. show that PQRS is a rhombus. R and S are the mid-points of sides AB. Book a free session with a top LIVE online tutor now. Sol: AD = BC and P. CD and BD respectively. CA = CB and AP Ă BQ = AC2 To prove: âAPC ~ âBCQ Proof: AP Ă BQ = AC2 [Given] â AP Ă BQ = AC Ă AC â AP Ă BQ = AC Ă BC [AC = BC given] đŽđ đŽđ¶ â đ”đ¶ = đ”đ âŠ(i) Since.com. In an isosceles âABC.Class X Chapter 4 â Triangles Maths Sol: Given: AB â„ BC. â CAB + â CAP = 180° âŠ(iii) [Linear pair of angles] And. Prove that âAPC ~ âBCQ. â CAB = â CBA âŠ(ii) [Opposite angles to equal sides] Now. CA = CB [Given] Then. DC â„ BC and DE â„ AC To prove: âCED ~ âABC Proof: â BAC + â BCA = 90° âŠ(i) [By angle sum property] And. . the base AB is produced both the ways to P and Q such that AP Ă BQ = AC2. â CBA + â CBQ = 180° âŠ(iv) [Linear pair of angles] Printed from Vedantu. â BCA + â ECD = 90° âŠ(ii) [DC â„ BC given] Compare equation (i) and (ii) â BAC = â ECD âŠ(iii) In âCED and âABC â CED = â ABC [Each 90°] â ECD = â BAC [From (iii)] Then. Sol: Given: In âABC. Book a free session with a top LIVE online tutor now. âCED ~ âABC [By AA similarity] 20. âđŽđđ¶~âđ”đ¶đ [By SAS similarity] 21.6 đ„+4.8 đ„ 1 â đ„+4.8m Let length of shadow (AC) = x cm In âABC and âAPQ â ACB = â AQP [Each 90°] â BAC = â PAQ [Common] Then.8 â 4x â x = 4.8 âx= = 1. Height of girl = 90 cm = 0.8 4.6 m above the ground. .8 = 4 â 4x = x + 4. âAB ~ âAPQ [By AA similarity] đŽđ¶ đ”đ¶ ⎠đŽđ = đđ [Corresponding parts of similar â are proportional] đ„ 0. A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.9 â = 3.9 m Height of lamp-post = 3.2m/sec. find the length of her shadow after 4 seconds.com. Sol: We have.2 Ă 4 = 4.6 m Speed of girl = 1.8 â 3x = 4.Class X Chapter 4 â Triangles Maths Compare equation (ii) (iii) & (iv) â CAP = â CBQ âŠ(v) In âAPC and âBCQ â CAP = â CBQ [From (v)] đŽđ đŽđ¶ = đ”đ [From (i)] đ”đ¶ đâđđ. If the lamp is 3.2 m/sec ⎠Distance moved by girl (CQ) = Speed Ă Time = 1.6 đ 3 Printed from Vedantu. Book a free session with a top LIVE online tutor now. â BAC = â MAP In âABC and âAMP â B = â M [Each 90°] â BAC = â MAP [Given] Then. âABC ~ âAMP [By AA similarity] Printed from Vedantu.Class X Chapter 4 â Triangles Maths ⎠Length of shadow = 1. Prove that (i) âABC ~ âAMP đ¶đŽ đ”đ¶ (ii) = đđ đđŽ Sol: We have. If âABC and âAMP are two right triangles. . âAOB ~ âCOD [By AA similarity] đđŽ đđ” ⎠đđ¶ = đđ· [Corresponding parts of similar â are proportional] 23. Sol: We have.com. Using similarity criterion for two triangles. right angled at B and M respectively such that â MAP = â BAC. ABCD is a trapezium with AB || DC In âAOB and âCOD â AOB = â COD [Vertically opposite angles] â OAB = â OCD [Alternate interior angles] Then. â B = â M = 90° And. show that đđ¶ = đđ·. Book a free session with a top LIVE online tutor now. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the đđŽ đđ” point O.6m 22. Prove that âABC ~ âADE and Hence find the lengths of AE and DE. Sol: In âACB. In below Fig.com. đŽđ” đ”đž = đ·đč đ¶đ· đŽđ” 28 = 6 4 6 AB = 28 Ă 4 = 42đ So.. Sol: Let AB be a tower CD be a stick. Book a free session with a top LIVE online tutor now. â DCF = â BAE And â DFC = â BEA â CDF = â ABE (tower and stick are vertical to ground) Therefore â ABE ~ âCDF (By AA similarity) So. . CD = 6m Shadow of AB is BE = 28m Shadow of CD is DF = 4m At same time light rays from sun will fall on tower and stick at same angle.Class X Chapter 4 â Triangles Maths đ¶đŽ đ”đ¶ ⎠đđŽ = đđ [Corresponding parts of similar â are proportional] 24. Find the height of the tower. by Pythagoras theorem đŽđ” 2 = đŽđ¶ 2 + đ”đ¶ 2 â đŽđ” 2 = (5)2 + (12)2 Printed from Vedantu. âABC is right angled at C and DE â„ AB. A vertical stick of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. So. height of tower will be 42 metres. 25. (iii)If AC = 19cm and DF = 8 cm.2 cm.6 1. Book a free session with a top LIVE online tutor now. âABC ~ âDEF Area(âABC) = 9 cm2 Area (âDEF) = 64 cm2 Printed from Vedantu.Class X Chapter 4 â Triangles Maths â đŽđ” 2 = 25 + 144 = 169 â AB = â169 = 13 đđ In âAED and âACB â A = â A [Common] â AED = â ACB [Each 90°] Then. âAED ~ âACB [By AA similarity] đŽđž đ·đž đŽđ· ⎠đŽđ¶ = = đŽđ” [Corresponding parts of similar â are proportional] đ¶đ” đŽđž đ·đž 3 â = = 13 5 12 đŽđž 3 đ·đž 3 â = 13 and 12 = 13 5 15 36 â AE = 13 cm and DE = 13 cm Exercise 4. Area (âđ·đžđč) = 25 đđ2 And BC = 2.3)2 â 25 = đžđč2 4 2. Sol: (i) We have.com. đŽđđđ (â đ·đžđč) = đžđč2 [By area of similar triangle theorem] 16 (2. find AB.4 cm. area (âđ·đžđč) = 64 đđ2 and DE = 6.3 cm. find the ratio of the areas of âABC and âDEF.3 â5= [By taking square root] đžđč 11. âABC ~ âDEF đŽđđđ (âđŽđ”đ¶) đ”đ¶ 2 Then.5 â EF = = 2.3 cm Since.2 cm and DE = 1.875 đđ 4 (ii) We have. (ii) If area (âđŽđ”đ¶) = 9đđ2 . find EF. (v) If AB = 1. area (âđ·đžđč) = 64 đđ2 and DE = 5. find the ratio of the area of two triangles. . (iv) If area (âđŽđ”đ¶) = 36đđ2 . find AB. âABC ~âDEF Area (âđŽđ”đ¶) = 16 đđ2 . area (âđ·đžđč) = 25 đđ2 and BC = 2. Triangles ABC and DEF are similar (i) If area (âđŽđ”đ¶) = 16đđ2 .1 cm. âABC ~ âDEF AC = 19 cm and DF = 8 cm By area of similar triangle theorem đŽđđđ (âđŽđ”đ¶) đŽđ¶ 2 (19)2 361 = đ·đč2 = = đŽđđđ (âđ·đžđč) 82 64 (iv) We have.Class X Chapter 4 â Triangles Maths And DE = 5.2)2 = (1.2 cm And. Area (âABC) = 36 đđ2 Area (âđ·đžđč) = 64 đđ2 DE = 6.com. âABC ~ âDEF đŽđđđ (âđŽđ”đ¶) đŽđ”2 Then.4)2 Printed from Vedantu. âABC ~ âDEF AB = 1. âABC ~ âDEF By area of similar triangle theorem đŽđđđ( âđŽđ”đ¶) đŽđ”2 = đ·đž2 đŽđđđ (âđ·đžđč) 36 đŽđ”2 â 64 = (6. âABC ~ âDEF AC = 19 cm and DF = 8 cm By area of similar triangle theorem đŽđđđ (â đŽđ”đ¶) đŽđ¶ 2 (19)2 361 = đ·đč2 = = đŽđđđ (âđ·đžđč) 82 64 We have.65 đđ 8 (v) We have.9125 đđ 8 (iii) We have.1)2 3 đŽđ” â 8 = 5.1 [By taking square root] 3Ă5.2 â AB = = 4. . đŽđđđ (âđ·đžđč) = đ·đž2 [By area of similar triangle theorem] 9 đŽđ”2 â 64 = (5.2)2 [By taking square root] 6Ă6.1 cm Since.4 cm By area of similar triangle theorem đŽđđđ (â đŽđ”đ¶) đŽđ”2 = đ·đž2 đŽđđđ (â đ·đžđč) (1.1 â AB = = 1. Book a free session with a top LIVE online tutor now.2 cm and DF = 1. 96 36 = 49 2.5 Ă 10 5 â AC = 5. find CA and AQ. BA = 6.8 đđđ đŽđ = 6. đŽđ = đđ = đŽđ [Corresponding parts of similar â are proportional] đŽđ¶ 10 6. The areas of two similar triangles are 81 cm2 and 49 cm2 respectively. Also.8 = = đŽđ 5 đŽđ¶ 10 10 6.6 cm and AQ = 3. Book a free session with a top LIVE online tutor now. What is the ratio of their corresponding medians? Sol: Printed from Vedantu. In fig. find the area (âđŽđ¶đ”): đđđđ (âđŽđđ) Sol: We have. If BC = 10 cm.Class X Chapter 4 â Triangles Maths 1. below âACB ~ âAPQ. âACB ~ âAPQ đŽđ¶ đ¶đ” đŽđ” Then.8 cm.com. PQ = 5 cm.25 cm By area of similar triangle theorem đŽđđđ (âđŽđ¶đ”) đ”đ¶ 2 = đđ2 đŽđđđ (âđŽđđ) (10)2 = (5)2 100 = 25 4 =1 3. .5 â 2.5 â 2. Find the ratio of their corresponding heights.44 = 1.5 cm and AP = 2.8 = đđđ = đŽđ 5 5 10 5 â AC = Ă 2. Area (âPQR) = 49 cm2 And AD and PS are the altitudes By area of similar triangle theorem đŽđđđ (âđŽđ”đ¶) đŽđ”2 = đđ2 đŽđđđ (âđđđ ) 81 đŽđ”2 â 49 = đđ2 9 đŽđ” â = âŠ. âABC ~ âPQR Area(âABC) = 169 cm2 Area(âPQR) = 121 cm2 And AB = 26 cm Printed from Vedantu. The areas of two similar triangles are 169 cm2 and 121 cm2 respectively.com. ratio of altitudes = Ratio of medians = 9 : 7 4. find the longest side of the smaller triangle. âABD ~ âPQS [By AA similarity] đŽđ” đŽđ· ⎠đđ = âŠ(ii) [Corresponding parts of similar â are proportional] đđ Compare (1) and (2) đŽđ· 9 =7 đđ 9 ⎠Ratio of altitudes = 7 Since. Hence.(i) [Taking square root] 7 đđ đŒđ âđŽđ”đ· đđđ âđđđ â B = â Q [âABC ~ âPQR] â ADB = â PSQ [Each 90°] Then. Sol: We have. âABC ~ âPQR Area (âABC) = 81 cm2. the ratio of the area of two similar triangles is equal to the ratio of the squares of the squares of their corresponding altitudes and is also equal to the squares of their corresponding medians. . If the longest side of the larger triangle is 26 cm.Class X Chapter 4 â Triangles Maths We have. Book a free session with a top LIVE online tutor now. âđŽđ”đ¶~âđđđ [By SAS similarity] đŽđđđ (âđŽđ”đ¶) đŽđ”2 ⎠= đđ2 âŠ. Book a free session with a top LIVE online tutor now. đŽđ¶ = 1 and đđ = 1 đŽđ” đđ ⎠đŽđ¶ = đđ đŽđ” đŽđ¶ â đđ = đđ âŠ(ii) In âABC and âPQR â A = â P [Given] đŽđ” đŽđ¶ = đđ [From (2)] đđ đâđđ. đŽđđđ (â đđđ ) = 25 âŠ(i) đŽđ· To find: đđ Proof: Since.(iv) Printed from Vedantu.(iii) [By area of similar triangle theorem] đŽđđđ (â đđđ ) đ¶đđđđđđ đđđąđđĄđđđ (đ)đđđ (đđđ) đŽđ”2 36 = đđ 2 25 đŽđ” 6 â đđ = 5 âŠ. AB = AC and PQ = PR đŽđ” đđ Then. Find the ratio of their corresponding heights. Sol: Given: AB = AC. PQ = PQ and â A = â P And. .Class X Chapter 4 â Triangles Maths By area of similar triangle theorem đŽđđđ (âđŽđ”đ¶) đŽđ”2 = đŽđđđ (â đđđ ) đđ 2 169 (26) 2 â 121 = đđ 2 13 26 â 11 = đđ [Taking square root] 11 â PQ = 13 Ă 26 = 22 đđ 5.com. Two isosceles triangles have equal vertical angles and their areas are in the ratio 36 : 25. AD and PS are altitudes đŽđđđ (âđŽđ”đ¶) 36 And. . Sol: We have.4 cm And AD and PS are the altitudes To find: PS Proof: Since.(ii) [Corresponding parts of similar â are proportional] đđ Compare (i) and (ii) Printed from Vedantu.(i) đŒđ âđŽđ”đ· đđđ âđđđ â B = â Q [â ABC ~ âPQR] â ADB ~ â PSQ [Each 90°] Then.4 cm. Book a free session with a top LIVE online tutor now. by area of similar triangle theorem đŽđđđ (âđŽđ”đ¶) đŽđ”2 = đđ2 đŽđđđ (âđđđ ) 25 đŽđ”2 â = 36 đđ 2 5 đŽđ” â 6 = đđ âŠ. If the altitude of the first triangle is 2. âABC ~ âPQR Area (âABC) = 25 cm2 Area (âPQR) = 36 cm2 AD = 2.com. âABD ~ âPQS [By AA similarity] đŽđ” đŽđ· ⎠đđ = đđ 6 đŽđ· â5= [From (iv)] đđ 6.Class X Chapter 4 â Triangles Maths In âABD and âPQS â B = â Q [âABC ~ âPQR] â ADB = â PSQ [Each 90°] Then. âABC ~ âPQR Then. find the corresponding altitude of the other. The areas of two similar triangles are 25 cm2 and 36 cm2 respectively. âABD ~ âPQS [By AA similarity] đŽđ” đŽđ· ⎠đđ = âŠ. ANâ„ BC.4Ă6 â đđ = = 2.4 5 â =6 đđ 2. The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively. PS = 9 cm By area of similar triangle theorem đŽđđđ (âđŽđ”đ¶) đŽđ”2 = đđ2 âŠ(i) đŽđđđ (âđđđ ) đŒđ âđŽđ”đ· đđđ âđđđ â B = â Q [âABC ~ âPQR] â ADB = â PSQ [Each 90°] Then.com. âABD ~ âPQS [By AA similarity] đŽđ” đŽđ· ⎠đđ = [Corresponding parts of similar â are proportional] đđ đŽđ” 6 â đđ = 9 đŽđ” 2 â đđ = 3 âŠ(ii) Compare equations (i) and (ii) đŽđđđ(âđŽđ”đ¶) 2 2 4 = (3) = 9 đŽđđđ (âđđđ ) 8. Sol: Printed from Vedantu. ABC is a triangle in which â A =90°. Find the ratio of the areas of âANC and âABC. âABC ~ âPQR AD = 6 cm And.Class X Chapter 4 â Triangles Maths đŽđ· 5 =6 đđ 2. Find the ratio of their areas. Sol: We have. Book a free session with a top LIVE online tutor now.88 đđ 5 7. BC = 12 cm and AC = 5cm. . (ii) If DE = 4cm. DE = 4 cm. . Book a free session with a top LIVE online tutor now. (iii)If DE : BC = 3 : 5. âANC ~ âBAC [By AA similarity] By area of similarity triangle theorem đŽđđđ (â đŽđđ¶) đŽđ¶ 2 = đ”đ¶ 2 đŽđđđ (âđ”đŽđ¶) 52 = 122 25 = 144 9. 4. Sol: We have. find the area of âABC. Calculate the ratio of the areas of âADE and the trapezium BCED. BC = 8 cm and Area (âADE) = 25 cm2.178.Class X Chapter 4 â Triangles Maths In âANC and âABC â C = â C [Common] â ANC = â BAC [Each 90°] Then. DE || BC. BC = 6 cm and Area (âADE) = 16 cm2. BC = 6 cm and area (âđŽđ·đž) = 16đđ2 In âADE and âABC â A = â A [Common] â ADE = â ABC [Corresponding angles] Then. DE || BC (i) If DE = 4 cm.com. In Fig. âADE ~ âABC [By AA similarity] ⎠By area of similar triangle theorem đŽđđđ (âđŽđ·đž) đ·đž 2 đŽđđđ (âđŽđ”đ¶) = đ”đ¶ 2 16 42 â đŽđđđ (âđŽđ”đ¶) = 62 Printed from Vedantu. find the area of âABC. đ·đž = 4 đđ.DECB) â 16 ar(âADE) = 9 ar (trap.đ·đžđ¶đ”) = 16 Printed from Vedantu. âADE ~ âABC [By AA similarity] By area of similar triangle theorem đŽđđđ (âđŽđ·đž) đ·đž 2 â đŽđđđ (âđŽđ”đ¶) = đ”đ¶ 2 25 42 = đŽđđđ (âđŽđ”đ¶) 82 25Ă64 â đŽđđđ (âđŽđ”đ¶) = = 100 đđ2 16 đ·đž 3 We have. DE = 4 cm. DE || BC. DECB) đđ (âđŽđ·đž) 9 â đđ (đĄđđđ. Book a free session with a top LIVE online tutor now. âADE ~ âABC [By AA similarity] By area of similar triangle theorem đŽđđđ (âđŽđ·đž) đ·đž 2 â đŽđđđ (âđŽđ”đ¶) = đ”đ¶ 2 đđ (âđŽđ·đž) 32 â đđ(âđŽđ·đž)+đđ(đĄđđđ.com. đ”đ¶ = 8 đđ đđđ đđđđ (âđŽđ·đž) = 25 đđ2 đŒđ âđŽđ·đž đđđ âđŽđ”đ¶ â A = â A [Common] â ADE = â ABC [Corresponding angles] Then. . DE || BC. BC = 8 cm and area (âADE) = 25cm2 In âADE and âABC â A = â A [Common] â ADE = â ABC [Corresponding angles] Then.(i) In âADE and âABC â A = â A [Common] â ADE = â B [Corresponding angles] Then. and đ”đ¶ = 5 âŠ. âđŽđ·đž~âđŽđ”đ¶ [By AA similarity] By area of similar triangle theorem đŽđđđ (âđŽđ·đž) đ·đž 2 = đ”đ¶ 2 đŽđđđ (â đŽđ”đ¶) 16 42 â = đŽđđđ (âđŽđ”đ¶ ) 62 16Ă36 â Area (âABC) = = 36 đđ2 16 We have.Class X Chapter 4 â Triangles Maths 16Ă36 â đŽđđđ (âđŽđ”đ¶) = = 36đđ2 16 đ€đ âđđŁđ. đ·đžđ¶đ”) â 25 ar (âADE â 9ar) (âADE) = 9ar (trap.đ·đžđ¶đ”) = 52 [From (i)] â 25ar (âADE) = 9đđ (âđŽđ·đž) + 9đđ (đĄđđđ. đ·đž | |đ”đ¶. 4.179. D and E as the mid-points of AB and AC So. đđđđ (âđŽđ”đ¶) đŽđ prove that đđđđ (âđ·đ”đ¶) = đ·đ Sol: 1 We know that area of a triangle = 2 Ă đ”đđ đ Ă âđđđâđĄ Since âABC and âDBC are one same base. . Find the ratio of the areas of âADE and âABC Sol: We have. In Fig. Book a free session with a top LIVE online tutor now. according to the mid-point theorem 1 DE || BC and DE = 2 đ”đ¶ âŠ(i) In âADE and âABC â A = â A [Common] â ADE = â B [Corresponding angles] Then.. In âABC.Class X Chapter 4 â Triangles Maths 10. âADE ~ âABC [By AA similarity] By area of similar triangle theorem đđ(âđŽđ·đž) đ·đž 2 = đ”đ¶ 2 đđ(âđŽđ”đ¶) 1 2 ( đ”đ¶) 2 = [From (i)] đ”đ¶ 2 1 đ”đ¶ 2 4 = đ”đ¶ 2 1 =4 11. If AD and BC intersect at O. Therefore ratio between their areas will be as ratio of their heights. âABC and âDBC are on the same base BC. D and E are the mid-points of AB and AC respectively. Let us draw two perpendiculars AP and DM on line BC.com. Printed from Vedantu. Class X Chapter 4 â Triangles Maths In âAPO and âDMO. The diagonals AC and BD intersect at O. â APO = â DMO (Each is 90°) â AOP = â DOM (vertically opposite angles) â OAP = â ODM (remaining angle) Therefore âAPO ~ âDMO (By AAA rule) đŽđ đŽđ Therefore đ·đ = đ·đ đđđđ (âđŽđ”đ¶) đŽđ Therefore đđđđ (âđ·đ”đ¶) = đ·đ 12. âAOB ~ âCOD [By AA similarity] (a) By area of similar triangle theorem đđ (âđŽđđ”) đđŽ2 62 36 9 = = = = đđ (âđ¶đđ·) đđ¶ 2 82 64 16 (b) Draw DP â„ AC 1 đđđđ (âđŽđđ·) ĂđŽđĂđ·đ ⎠đđđđ (âđ¶đđ·) = 21 Ăđ¶đĂđ·đ 2 đŽđ = đ¶đ 6 =8 3 =4 Printed from Vedantu. OC = 8 cm.com. Prove that: (i) âAOB and âCOD (ii) If OA = 6 cm. . Book a free session with a top LIVE online tutor now. ABCD is a trapezium in which AB || CD. AB || DC In âAOB and âCOD â AOB = â COD [Vertically opposite angles] â OAB = â OCD [Alternate interior angles] Then. Find: đđđđ (âđŽđđ”) (a) đđđđ (âđ¶đđ·) đđđđ (âđŽđđ·) (b) đđđđ (âđ¶đđ·) Sol: We have. If the altitude the bigger triangle is 5 cm. In ABC. BPQC) đđ (âđŽđđ) 1 â đđ(đĄđđđ. Q is a point in AC such that PQ || BC.com. âAPQ ~âABC [By AA similarity] By area of similar triangle theorem đđ (âđŽđđ) đŽđ 2 = đŽđ”2 đđ (âđŽđ”đ¶) đđ (âđŽđđ) 12 đŽđ 1 â đđ(âđŽđđ)+đđ(đĄđđđ. Sol: We have. đ”đđđ¶) â 8ar(APQ) = ar(trap. PQ || BC đŽđ 1 And đđ” = 2 In âAPQ and âABC â A = â A [Common] â APQ = â B [Corresponding angles] Then. Find the ratio of the areas of âAPQ and trapezium BPQC. Book a free session with a top LIVE online tutor now.đ”đđđ¶) = 8 14. P divides the side AB such that AP : PB = 1 : 2. The areas of two similar triangles are 100 cm2 and 49 cm2 respectively. find the corresponding altitude of the other.Class X Chapter 4 â Triangles Maths 13. . Sol: We have.đ”đđđ¶) = 32 [đđ” = 2] â 9ar (đŽđđ) = đđ(âđŽđđ) + đđ(đĄđđđ. đ”đđđ¶) â 9ar (đŽđđ) â đđ(âđŽđđ) + đđ(đĄđđđ. âABC ~ âPQR Printed from Vedantu. Book a free session with a top LIVE online tutor now. If the median of the first triangle is 12. âABD ~ âPQS [By AA similarity] đŽđ” đŽđ· ⎠đđ = âŠ(ii) [Corresponding parts of similar â are proportional] đđ Compare (i) and (ii) đŽđ· 10 = đđ 7 5 10 â đđ = 7 5Ă7 â PS = = 3.1 cm And AD and PS are the medians By area of similar triangle theorem đŽđđđ(âđŽđ”đ¶) đŽđ”2 = đđ2 đŽđđđ(âđđđ ) Printed from Vedantu. find the corresponding median of the other. Sol: We have.com. âABC ~ âPQR Area (âABC) = 121 cm2. Area (âPQR) = 49 cm2 AD = 5 cm And AD and PS are the altitudes By area of similar triangle theorem đŽđđđ (âđŽđ”đ¶) đŽđ”2 = đđ2 đŽđđđ (âđđđ ) 100 đŽđ”2 â = đđ2 49 10 đŽđ” â = đđ âŠ(i) 7 đŒđ âđŽđ”đ· đđđ âđđđ â B = â Q [âABC ~ âPQR] â ADB = â PSQ [Each 90°] Then.1 cm. The areas of two similar triangles are 121 cm2 and 64 cm2 respectively. .Class X Chapter 4 â Triangles Maths Area(âABC) = 100 cm2.5 đđ 10 15. Area (âPQR) = 64 cm2 AD = 12. 1 â PS = = 8. determine DE. Find đŽđ” Printed from Vedantu. Sol: We have.8 đđ đđ 16. âABC ~ âDEF such that AB = 5 cm.1 â PS = đđ 8 Ă12. âđŽđ”đ· ~ âđđđ [By SAS similarity] đŽđ” đŽđ· ⎠đđ = âŠ(iii) [Corresponding parts of similar â are proportional] đđ đ¶đđđđđđ (đ) đđđ (đđđ) 11 đŽđ· = 8 đđ 11 12. If âABC ~ âDEF such that AB = 5 cm. PQ is a line segment intersecting AB at P and AC at Q such that PQ || BC and đ”đ PQ divides âABC into two parts equal in area.com. Book a free session with a top LIVE online tutor now. đđ = đđ [Corresponding parts of similar â are proportional] đŽđ” 2đ”đ· â đđ = [AD and PS are medians] 2đđ đŽđ” đ”đ· â đđ = âŠ(ii) đđ In âABD and âPQS â B = â Q [âABC ~ âPQS] đŽđ” đ”đ· = [From (ii)] đđ đđ đâđđ.Class X Chapter 4 â Triangles Maths 121 đŽđ”2 â = đđ2 64 11 đŽđ” â = đđ âŠ(i) 8 Since. In âABC.5 đđ 2 17. . Area (âABC) = 20 cm2 and area(âDEF) = 45 cm2 By area of similar triangle theorem đŽđđđ(âABC) đŽđ”2 = đ·đž2 đŽđđđ(âDEF) 20 52 â 45 = đ·đž2 4 52 â 9 = đ·đž2 2 5 â 3 = đ·đž [Taking square root] 3Ă5 â đ·đž = = 7. âABC ~ âPQR đŽđ” đ”đ¶ Then. area (âABC) = 20 cm2 and area (âDEF) = 45 cm2.1 â = 8 đđ 8 Ă12. Book a free session with a top LIVE online tutor now.5 cm. If BC = 4.Class X Chapter 4 â Triangles Maths Sol: We have. Printed from Vedantu.com. âAPQ ~ âABC [By AA similarity] ⎠By area of similar triangle theorem đđ(âđŽđđ) đŽđ 2 = đŽđ”2 đđ(âđŽđ”đ¶) đđ(âđŽđđ) đŽđ 2 â đđ(âđŽđđ) = đŽđ”2 [By using (i)] 1 đŽđ 2 â 2 = đŽđ”2 1 đŽđ â = đŽđ”2 â2 1 đŽđ â = đŽđ” [Taking square root] â2 1 đŽđ”âđ”đ â = â2 đŽđ” 1 đŽđ” đ”đ â = đŽđ” â đŽđ” â2 1 đ”đ â = 1 â đŽđ” â2 đ”đ 1 = đŽđ” = 1 â â2 đ”đ â2â1 â đŽđ” = â2 18. PQCB) â ar(âAPQ) = ar(âABC) â ar(âAPQ) â 2ar(âAPQ) = ar(âABC) âŠ(i) In âAPQ and âABC â A = â A [common] â APQ = â B [corresponding angles] Then. . The areas of two similar triangles ABC and PQR are in the ratio 9:16. PQ || BC And ar(âAPQ) = ar(trap. find the length of QR. Sol: We have. prove that area of âAPQ is one. AP = 1 cm.Class X Chapter 4 â Triangles Maths âABC ~ âPQR đđđđ (âđŽđ”đ¶) đ”đ¶ 2 = đđđđ (âđđđ ) đđ 2 9 (4. AQ = 1.5 â QR = = 6đđ 3 19.com. PB = 3 cm. AQ = 1. Sol: We have. If AP = 1 cm.5 â 4 = đđ [Taking square root] 4Ă4. Find the ratio of areas of âABC and âBDE. If D is a point on the side AB of âABC such that AD : DB = 3. Book a free session with a top LIVE online tutor now. QC = 4.5 m In âAPQ and âABC â A = â A [Common] đŽđ đŽđ 1 = đŽđ¶ [Each equal to 4] đŽđ” đâđđ.5 m. ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q.2 and E is a Point on BC such that DE || AC. PB = 3 cm.5 cm and QC = 4.5)2 â 16 = đđ 2 3 4. .5 cm. Sol: Printed from Vedantu. âđŽđđ ~âđŽđ”đ¶ [By SAS similarity] đ”đŠ đđđđ đđ đ đđđđđđ đĄđđđđđđđ đĄâđđđđđ đđ(âđŽđđ) 12 = 42 đđ(âđŽđ”đ¶) đđ(âđŽđđ) 1 â đđ(âđŽđ”đ¶) = 16 Ă đđ(âđŽđ”đ¶) 20.sixteenth of the area of ABC. .com. If âABC and âBDE are equilateral triangles. đŽđ· 3 = đ·đ” 2 đ·đ” 2 â đŽđ· = 3 In âBDE and âBAC â B = â B [common] â BDE = â A [corresponding angles] Then. Book a free session with a top LIVE online tutor now. âABC and âBDE are equilateral triangles then both triangles are equiangular ⎠âABC ~ âBDE [By AAA similarity] By area of similar triangle theorem Printed from Vedantu. find the ratio of areas of âABC and âBDE. Sol: We have. where D is the mid-point of BC.Class X Chapter 4 â Triangles Maths We have. âBDE ~ âBAC [By AA similarity] By area of similar triangle theorem đđ(âđŽđ”đ¶) đŽđ”2 = đ”đ·2 đđ(âđ”đ·đž) 52 đŽđ· 3 = 22 [đ·đ” = 2] 25 = 4 21. com. âABC is an equilateral triangle Then. AB = BC = AC Let. another equilateral triangle ADE is constructed. . Book a free session with a top LIVE online tutor now. On AD as base. Prove that Area (âADE): Area (âABC) = 3: 4 Sol: We have. âđŽđ”đ¶ đđđ âđŽđ·đž đđđĄâ đđđ đđđąđđđđĄđđđđ đĄđđđđđđđđ đĄâđđ đĄâđđŠ đđđ đđđąđđđđđąđđđ ⎠âABC ~ âADE [By AA similarity] By area of similar triangle theorem đđ(âđŽđ·đž) đŽđ· 2 = đŽđ”2 đđ(âđŽđ”đ¶) 2 (â3đ„) = (2đ„)2 3đ„ 2 = 4đ„ 2 3 =4 Printed from Vedantu. AB = BC = AC = 2x Since. AD â„ BC then BD = DC = x In âADB. by Pythagoras theorem đŽđ” 2 = (2đ„)2 â (đ„)2 â đŽđ·2 = 4đ„ 2 â đ„ 2 = 3đ„ 2 â đŽđ· = â3đ„ đđ đđđđđ. AD is an altitude of an equilateral triangle ABC.Class X Chapter 4 â Triangles Maths đđ(âđŽđ”đ¶) đ”đ¶ 2 = đ”đ·2 đđ (âđ”đ·đž) 2(đ”đ·)2 = [D is the mid-point of BC] đ”đ· 2 4đ”đ· 2 = đ”đ· 2 4 =1 22. triangle is not a right triangle. đ2 + đ 2 = 81 + 256 = 337 â đ2 Then. by converse of Pythagoras theorem. given triangle is not a right triangle.com. by converse of Pythagoras theorem. a = 7 cm. Sol: We have.6 cm.8 cm and c = 4 cm (iv) a = 8 cm. a = 1. đ 2 = 100 đđđ đ 2 = 36 Since. b = 16 cm and c = 18 cm ⎠đ2 = 81. (i) a = 7 cm.Class X Chapter 4 â Triangles Maths Exercise 4. đ 2 = 576 đđđ đ 2 = 625 Since. Determine which of them right triangles are. determine whether the triangle is a right-angled triangle. and 6 cm long. b = 10 cm and c = 6 cm Sol: We have. by converse of Pythagoras theorem. b = 24 cm and c = 25 cm (ii) a = 9 cm. given triangle is a right triangle. Book a free session with a top LIVE online tutor now. We have. 4 cm. b = l6 cm and c = 18 cm (iii) a = 1. b = 3. đŽđ” 2 + đ”đ¶ 2 â đŽđ¶ 2 Then. a = 9 cm. Sides of triangle AB = 3 cm BC = 4 cm AC = 6 cm ⎠đŽđ” 2 = 32 = 9 đ”đ¶ 2 = 42 = 16 đŽđ¶ 2 = 62 = 36 Since. đ2 + đ 2 = 49 + 576 = 625 = đ2 Then. đ2 + đ 2 = 64 + 36 = 100 = đ 2 Printed from Vedantu. b = 24 cm and c = 25 cm ⎠đ2 = 49. b = 3. If the sides of a triangle are 3 cm.8 cm and C = 4 cm ⎠đ2 = 64.7 1. đ 2 = 256 đđđ đ 2 = 324 Since. . 2. We have.6 cm. The sides of certain triangles are given below. by Pythagoras theorem đŽđ2 = đŽđ” 2 + đ”đ2 â đŽđ2 = 82 + 152 â đŽđ2 = 64 + 225 = 289 â AO = â289 = 17đ ⎠He is 17m far from the starting point. given triangle is a right triangle. by converse of Pythagoras theorem. ⎠In âABO. A man goes 15 metres due west and then 8 metres due north. A ladder 17 m long reaches a window of a building 15 m above the ground. by Pythagoras theorem đŽđ” 2 + đ”đ¶ 2 = đŽđ¶ 2 â 152 + đ”đ¶ 2 = 172 â 225 + đ”đ¶ 2 = 172 â đ”đ¶ 3 = 289 â 225 â đ”đ¶ 2 = 64 Printed from Vedantu. Book a free session with a top LIVE online tutor now.Class X Chapter 4 â Triangles Maths Then. Find the distance of the foot of the ladder from the building. 3. . Sol: In âABC. How far is he from the starting point? Sol: Let the starting point of the man be O and final point be A. 4.com. If the distance between their feet is 12 m. by applying Pythagoras theorem đŽđ2 + đđ¶ 2 = đŽđ¶ 2 122 + 52 = đŽđ¶ 2 đŽđ¶ 2 = 144 + 25 = 169 AC = 13 Therefore distance between their tops = 13m. Two poles of heights 6 m and 11 m stand on a plane ground. BC = 14 cm. Calculate the altitude from A on BC. Therefore CP = 11 â 6 = 5 m From the figure we may observe that AP = 12m In triangle APC. Book a free session with a top LIVE online tutor now. find the distance between their tops.Class X Chapter 4 â Triangles Maths â đ”đ¶ = 8 đ ⎠đ·đđ đĄđđđđ đđ đĄâđ đđđđĄ đđ đĄâđ đđđđđđ đđđđ đđąđđđđđđ = 8 đ 5.com. Sol: Let CD and AB be the poles of height 11 and 6 m. In an isosceles triangle ABC. 6. . Sol: We have AB = AC = 25 cm and BC = 14 cm In âABD and âACD â ADB = â ADC [Each 90°] AB = AC [Each 25 cm] Printed from Vedantu. AB = AC = 25 cm. Class X Chapter 4 â Triangles Maths AD = AD [Common] Then, âABD â âACD [By RHS condition] ⎠BD = CD = 7 cm [By c.p.c.t] In âADB, by Pythagoras theorem đŽđ·2 + đ”đ·2 = đŽđ” 2 â đŽđ·2 + 72 = 252 â đŽđ·2 = 625 â 49 = 576 â đŽđ· = â576 = 24 đđ 7. The foot of a ladder is 6 m away from a wall and its top reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its tip reach? Sol: Let, length of ladder be AD = BE = l m In âACD, by Pythagoras theorem đŽđ·2 = đŽđ¶ 2 + đ¶đ·2 â đ 2 = 82 + 62 âŠ.(i) đŒđ âđ”đ¶đž, đđŠ đđŠđĄâđđđđđđ đĄâđđđđđ đ”đž 2 = đ”đ¶ 2 + đ¶đž 2 â đ 2 = đ”đ¶ 2 + 82 âŠ.(ii) đ¶đđđđđđ (đ)đđđ (đđ) đ”đ¶ 2 + 82 = 82 + 62 â đ”đ¶ 2 = 62 â đ”đ¶ = 6đ 8. Two poles of height 9 m and 14 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops. Sol: Printed from Vedantu.com. Book a free session with a top LIVE online tutor now. Class X Chapter 4 â Triangles Maths We have, AC = 14 m, DC = 12m and ED = BC = 9m Construction: Draw EB â„ AC ⎠AB = AC â BC = 14 â 9 = 5m And, EB = DC = 12 m In âABE, by Pythagoras theorem, đŽđž 2 = đŽđ” 2 + đ”đž 2 â đŽđž 2 = 52 + 122 â đŽđž 2 = 25 + 144 = 169 â đŽđž = â169 = 13 đ ⎠đ·đđ đĄđđđđ đđđĄđ€đđđ đĄâđđđ đĄđđđ = 13 đ 9. Using Pythagoras theorem determine the length of AD in terms of b and c shown in Fig. 4.219 Sol: We have, In âBAC, by Pythagoras theorem đ”đ¶ 2 = đŽđ” 2 + đŽđ¶ 2 â đ”đ¶ 2 = đ 2 + đ 2 â đ”đ¶ = âđ 2 + đ 2 âŠ(i) đŒđ âđŽđ”đ· đđđ âđ¶đ”đŽ â B = â B [Common] Printed from Vedantu.com. Book a free session with a top LIVE online tutor now. Class X Chapter 4 â Triangles Maths â ADB = â BAC [Each 90°] Then, âABD ~ âCBA [By AA similarity] đŽđ” đŽđ· ⎠đ¶đ” = [Corresponding parts of similar â are proportional] đ¶đŽ đ đŽđ· â âđ 2 = +đ2 đ đđ â AD = âđ 2 +đ 2 10. A triangle has sides 5 cm, 12 cm and 13 cm. Find the length to one decimal place, of the perpendicular from the opposite vertex to the side whose length is 13 cm. Sol: Let, AB = 5cm, BC = 12 cm and AC = 13 cm. Then, đŽđ¶ 2 = đŽđ” 2 + đ”đ¶ 2 . This proves that âABC is a right triangle, right angles at B. Let BD be the length of perpendicular from B on AC. 1 Now, Area âABC = 2 (đ”đ¶ Ă đ”đŽ) 1 = 2 (12 Ă 5) = 30 cm2 1 1 Also, Area of âABC = 2 đŽđ¶ Ă đ”đ· = 2 (13 Ă đ”đ·) â (13 Ă đ”đ·) = 30 Ă 2 60 â BD = 13 cm 11. ABCD is a square. F is the mid-point of AB. BE is one third of BC. If the area of âFBE = 108 cm2, find the length of AC. Sol: Printed from Vedantu.com. Book a free session with a top LIVE online tutor now. Book a free session with a top LIVE online tutor now. by Pythagoras theorem đŽđ·2 + đ”đ·2 = 132 â 25 + đ”đ·2 = 169 â đ”đ·2 = 169 â 2 = 144 â đ”đ· = â144 = 12 đđ đŒđ âđŽđ·đ” đđđ âđŽđ·đ¶ â ADB = â ADC [Each 90°] AB = AC [Each 13 cm] Printed from Vedantu.904 đđ 12. find BC. đđŠ đđŠđĄâđđđđđđ đĄâđđđđđ đŽđ¶ 2 = đŽđ” 2 + đ”đ¶ 2 â đŽđ¶ 2 = đ„ 2 + đ„ 2 â đŽđ¶ 2 = 2đ„ 2 â đŽđ¶ 2 = 2 Ă (36)2 â đŽđ¶ = 36â2 = 36 Ă 1. Sol: In âADB. ABCD is a square Then. BE is one third of BC đ„ Then. area of âFBE = 108 cm2 1 â 2 Ă đ”đž Ă đčđ” = 108 1 đ„ đ„ â 2 Ă 3 Ă 2 = 108 â đ„ 2 = 108 Ă 2 Ă 3 Ă 2 â đ„ 2 = 1296 â đ„ = â1296 = 36đđ đŒđ âđŽđ”đ¶. BE = 3 cm We have. In an isosceles triangle ABC. F is the mid-point of AB đ„ Then. .414 = 50.Class X Chapter 4 â Triangles Maths Since. if AB = AC = 13 cm and the altitude from A on BC is 5 cm. AF = đčđ” = 2 đđ Since.com. AB = BC = CD = DA = x cm Since. p. by Pythagoras theorem đŽđ·2 + đ”đ·2 = đŽđ” 2 â đŽđ·2 + (đ)2 = (2đ)2 â đŽđ·2 + đ2 = 4đ2 â đŽđ·2 = 4đ2 â đ2 = 3đ2 â đŽđ· = đâ3 1 (ii) Area of âABC = 2 Ă đ”đ¶ Ă đŽđ· 1 = 2 Ă 2đ Ă đâ3 = â3đ2 14. âABD â âACD [By RHS condition] ⎠BD = CD = a [By c.c. âADB â âADC [By RHS condition] ⎠BD = CD = 12 cm [By c. In a âABC. The lengths of the diagonals of a rhombus are 24 cm and 10 cm. Find each side of the rhombus.t] Hence.Class X Chapter 4 â Triangles Maths AD = AD [Common] Then. Sol: Printed from Vedantu. BC = 12 + 12 = 24 cm 13.p.com. . AB = BC = CA = 2a and AD â„ BC.c. Prove that (i) AD = aâ3 (ii) Area (âABC) = â3 a2 Sol: (i) In âABD and âACD â ADB = â ADC [Each 90°] AB = AC [Given] AD = AD [Common] Then. Book a free session with a top LIVE online tutor now.t] In âADB. com. Sol: We have. Each side of a rhombus is 10 cm.Class X Chapter 4 â Triangles Maths We have. ABCD is a rhombus with side 10 cm and diagonal BD = 16 cm We know that diagonals of a rhombus bisect each other at 90° ⎠BO = OD = 8 cm In âAOB. by Pythagoras theorem đŽđ” 2 = đŽđ2 + đ”đ2 â đŽđ” 2 = 52 + 122 â đŽđ” 2 = 25 + 144 = 169 â đŽđ” = â169 = 13 đđ 15. . đŽđ¶ = 6 + 6 = 12 đđ Printed from Vedantu. If one of its diagonals is 16 cm find the length of the other diagonal. ABCD is a rhombus with diagonals AC = 10 cm and BD = 24 cm We know that diagonal of a rhombus bisect each other at 90° ⎠AO = OC = 5 cm and BO = OD = 12 cm In âAOB. Book a free session with a top LIVE online tutor now. by pythagoras theorem đŽđ2 + đ”đ2 = đŽđ” 2 â đŽđ2 + 82 = 102 â đŽđ2 = 100 â 64 = 36 â đŽđ = â36 = 6 đđ [By above property] âđđđđ. In âABC. AD is a median. In an acute-angled triangle. .(i) [BC = 2BD given] 4 Again. express a median in terms of its sides.com. by pythagoras theorem đŽđ” 2 = đŽđž 2 + đ”đž 2 â đŽđ” 2 = đŽđ·2 â đ·đž 2 + (đ”đ· â đ·đž)2 [By Pythagoras theorem] 2 2 2 2 2 â đŽđ” = đŽđ· â đ·đž + đ”đ· + đ·đž â 2đ”đ· Ă đ·đž â đŽđ” 2 = đŽđ·2 + đ”đ·2 â 2đ”đ· Ă đ·đž đ”đ¶ 2 â đŽđ” 2 = đŽđ·2 + â đ”đ¶ Ă đ·đž âŠ. Draw AE â„ BC In âAEB. Sol: We have. In âAEC. by pythagoras theorem đŽđ¶ 2 = đŽđž 2 + đžđ¶ 2 â đŽđ¶ 2 = đŽđ·2 â đ·đž 2 + (đ·đž + đ¶đ·)2 [By Pythagoras theorem] âđŽđ¶ 2 = đŽđ·2 + đ¶đ· 2 + 2đ¶đ· Ă đ·đž đ”đ¶ 2 â đŽđ¶ 2 = đŽđ·2 + + đ”đ¶ Ă đ·đž âŠ.Class X Chapter 4 â Triangles Maths 16. Book a free session with a top LIVE online tutor now.(ii) [BC = 2CD given] 4 đŽđđ đđđąđđĄđđđđ (đ) đđđ (đđ) đ”đ¶ 2 đŽđ” 2 + đŽđ¶ 2 = 2đŽđ· 2 + 2 â 2đŽđ” 2 + 2đŽđ¶ 2 = 4đŽđ·2 + đ”đ¶ 2 [Multiply by 2] â4đŽđ·2 = 2đŽđ” 2 + 2đŽđ¶ 2 â đ”đ¶ 2 2đŽđ”2 +2đŽđ¶ 2 âđ”đ¶ 2 â đŽđ·2 = 4 Printed from Vedantu. In right-angled triangle ABC in which â C = 90°. â C = 90° and D is the mid-point of BC In âACB. Sol: We have. if D is the mid-point of BC. by Pythagoras theorem đŽđ” 2 = đŽđ¶ 2 + đ”đ¶ 2 â đŽđ” 2 = đŽđ¶ 2 + (2đ¶đ·)2 [D is the mid-point of BC] Printed from Vedantu.Class X Chapter 4 â Triangles Maths 17. âABD â âACD [By RHS condition] 2 2 2 ⎠đŽđ· + đ”đ· = đŽđ” â đŽđ·2 + 62 = 122 â đŽđ·2 = 144 â 36 = 108 â AD = â108 = 10. Draw AE â„ BC In âABD and âACD â ADB = â ADC [Each 90°] AB = AC [Each 12 cm] AD = AD [Common] Then. âABC is an equilateral â with side 12 cm. Sol: We have. .39 cm 18.com. Book a free session with a top LIVE online tutor now. Calculate the height of an equilateral triangle each of whose sides measures 12 cm. prove that AB2 = 4 AD2 â3 AC2. D as the mid-point of BC (i) đŽđ¶ 2 = đŽđž 2 + đžđ¶ 2 đ 2 = đŽđž 2 + (đžđ· + đ·đ¶)2 [By pythagoras theorem] 2 2 2 đ = đŽđ· + đ·đ¶ + 2đ·đ¶ Ă đžđ· đ 2 đ đ 2 = đ2 + (2 ) + 2 (2 ) Ă đ„ [BC = 2CD given] đ2 â đ 2 = đ2 + + đđ„ âŠ(i) 4 (ii) In âAEB. ED = x. If BC = a. show that Printed from Vedantu. AB = c.com. by pythagoras theorem đŽđ” 2 = đŽđž 2 + đ”đž 2 â đ 2 = đŽđ· 2 â đžđ·2 + (đ”đ· â đžđ·)2 [By pythagoras theorem] â đ 2 = đ2 â đžđ·2 + đ”đ·2 + đžđ·2 â 2đ”đ· Ă đžđ· đ 2 đ â đ 2 = đ2 + (2) â 2 (2) Ă đ„ âŠ(ii) (iii) Add equations (i) and (ii) đ2 đ 2 + đ 2 = 2đ2 + 2 20. In Fig. In Fig.Class X Chapter 4 â Triangles Maths đŽđ” 2 = đŽđ¶ 2 + 4đ¶đ· 2 â đŽđ” 2 = đŽđ¶ 2 + 4(đŽđ·2 â đŽđ¶ 2 ) [In âACD.220. . D is the mid-point of side BC and AE â„ BC. AD = p and AE = h. Book a free session with a top LIVE online tutor now.221. 4. â B < 90° and segment AD â„ BC. 4. AC = b. by Pythagoras theorem] â đŽđ” 2 = đŽđ¶ 2 + 4đŽđ·2 â 4đŽđ¶ 2 â đŽđ” 2 = 4đŽđ·2 â 3đŽđ¶ 2 19. prove that: đ2 (i) đ 2 = đ2 + đđ„ + 4 2 2 đ2 (ii) đ = đ â đđ„ + 4 2 2 2 đ2 (iii) đ + đ = 2đ + 2 Sol: We have.. PB â„AC and QC â„ AB. by pythagoras theorem đŽđ¶ 2 = đŽđ·2 + đ·đ¶ 2 â đ 2 = â2 + (đ â đ„)2 â đ 2 = â2 + đ2 + đ„ 2 â 2đđ„ â đ 2 = đ2 + (â2 + đ„ 2 ) â 2đđ„ â đ 2 = đ2 + đ 2 â 2đđ„ by Pythagoras theorem 21. Prove that: (i) AB Ă AQ = AC Ă AP (ii) đ”đ¶ 2 = (đŽđ¶ Ă đ¶đ + đŽđ” Ă đ”đ) Sol: Then. âAPB ~ âAQC [By AA similarity] đŽđ đŽđ” ⎠= [Corresponding parts of similar â are proportional] đŽđ đŽđ¶ â đŽđ Ă đŽđ¶ = đŽđ Ă đŽđ” âŠ(i) (ii) In âBPC. In âABC. by pythagoras theorem đ”đ¶ 2 = đ”đ2 + đđ¶ 2 â đ”đ¶ 2 = đŽđ” 2 â đŽđ2 + (đŽđ + đŽđ¶)2 [By pythagoras theorem] â đ”đ¶ 2 = đŽđ” 2 + đŽđ¶ 2 + 2đŽđ Ă đŽđ¶ âŠ(ii) đŒđ âđ”đđ¶.com. đđŠ đđŠđĄâđđđđđđ đĄâđđđđđ. . đ”đ¶ 2 = đ¶đ 2 + đ”đ 2 â đ”đ¶ 2 = đŽđ¶ 2 â đŽđ 2 + (đŽđ” + đŽđ)2 [By pythagoras theorem] Printed from Vedantu. Book a free session with a top LIVE online tutor now.Class X Chapter 4 â Triangles Maths (i) đ 2 = â2 + đ2 + đ„ 2 â 2đđ„ (ii) đ 2 = đ2 + đ 2 â 2đđ„ Sol: In âADC. â A is obtuse. . prove that đ”đ¶ 2 = 4(đŽđ·2 â đŽđ¶ 2 ) Sol: To prove: đ”đ¶ 2 = 4[đŽđ·2 â đŽđ¶ 2 ] We have. prove that â ACD = 90°. â B = 90° and đŽđ·2 = đŽđ” 2 + đ”đ¶ 2 + đ¶đ·2 ⎠đŽđ·2 = đŽđ” 2 + đ”đ¶ 2 + đ¶đ·2 [Given] But đŽđ” 2 + đ”đ¶ 2 = đŽđ¶ 2 [By pythagoras theorem] Printed from Vedantu. In a right âABC right-angled at C. if D is the mid-point of BC. Book a free session with a top LIVE online tutor now. AD2 = AB2 + BC2 + CD2. â B = 90°. â C = 90° and D is the mid-point of BC. Sol: We have. LHS = đ”đ¶ 2 = (2đ¶đ·)2 [D is the mid-point of BC] 2 = 4CD = 4[đŽđ·2 â đŽđ¶ 2 ] [In âACD. In a quadrilateral ABCD. by pythagoras theorem] = đ đ»đ 23.com.Class X Chapter 4 â Triangles Maths â đ”đ¶ 2 = đŽđ¶ 2 â đŽđ 2 + đŽđ” 2 + đŽđ 2 + 2đŽđ” Ă đŽđ â đ”đ¶ 2 = đŽđ¶ 2 + đŽđ” 2 + 2đŽđ” Ă đŽđ âŠ(iii) đŽđđ đđđąđđĄđđđđ (đđ)& (đđđ) 2đ”đ¶ 2 = 2đŽđ¶ 2 + 2đŽđ” 2 + 2đŽđ Ă đŽđ¶ + 2đŽđ” Ă đŽđ â 2đ”đ¶ 2 = 2đŽđ¶ 2 + 2đŽđ” 2 + 2đŽđ Ă đŽđ¶ + 2đŽđ” Ă đŽđ â 2đ”đ¶ 2 = 2đŽđ¶[đŽđ¶ + đŽđ] + đŽđ”[đŽđ” + đŽđ] â 2đ”đ¶ 2 = 2đŽđ¶ Ă đđ¶ + 2đŽđ” Ă đ”đ â đ”đ¶ 2 = đŽđ¶ Ă đđ¶ + đŽđ” Ă đ”đ [Divide by 2] 22. Class X Chapter 4 â Triangles Maths Then. âABD. âABC is an equilateral â and AD â„ BC In âADB and âADC â ADB = â ADC [Each 90°] AB = AC [Given] AD = AD [Common] Then. Show that: (i) đŽđ” 2 = đ¶đ” Ă đ”đ· (ii) đŽđ¶ 2 = đ·đ¶ Ă đ”đ¶ (iii) đŽđ·2 = đ”đ· Ă đ¶đ· đŽđ”2 đ”đ· (iv) = đ·đ¶ đŽđ¶ 2 Sol: Printed from Vedantu.com. Sol: We have. In an equilateral âABC. . by Pythagoras theorem đŽđ” 2 = đŽđ·2 + đ”đ·2 â đ”đ¶ 2 = đŽđ·2 + đ”đ·2 [AB = BC given] 2 2 2 â [2đ”đ·] = đŽđ· + đ”đ· [From (i)] 2 2 2 â 4đ”đ· â đ”đ· = đŽđ· â 3đ”đ·2 = đŽđ·2 25. đŽđ·2 = đŽđ¶ 2 + đ¶đ·2 By converse of by pythagoras theorem â ACD = 90° 24. AD â„ BC. âADB â âADC [By RHS condition] đ”đ¶ ⎠BD = CD = âŠ(i) [corresponding parts of similar â are proportional] 2 In. prove that đŽđ·2 = 3đ”đ·2 . Book a free session with a top LIVE online tutor now. âABD is a right triangle right angled at A and AC â„ BD. âADB ~ âCAB (by AAA similarity) đŽđ” đ”đ· Therefore đ¶đ” = đŽđ” â đŽđ” 2 = đ¶đ” Ă đ”đ· (ii) Let â CAB = x In âCBA â CBA = 180° â 90° â đ„ â CBA = 90° â đ„ Similarly in âCAD â CAD = 90° â â đ¶đŽđ· = 90° â đ„ â CDA = 90° â â CAB = 90° â đ„ â CDA = 180° â90° â (90° â đ„) â CDA = x Now in âCBA and âCAD we may observe that â CBA = â CAD â CAB = â CDA â ACB = â DCA = 90° Therefore âCBA ~ âCAD (by AAA rule) đŽđ¶ đ”đ¶ Therefore đ·đ¶ = đŽđ¶ â đŽđ¶ 2 = đ·đ¶ Ă đ”đ¶ (iii) In âDCA & âDAB â DCA = â DAB (both are equal to 90°) â CDA = â ADB (common angle) â DAC = â DBA (remaining angle) âDCA ~ âDAB (AAA property) đ·đ¶ đ·đŽ Therefore đ·đŽ = đ·đ” âđŽđ·2 = đ”đ· Ă đ¶đ· (iv) From part (i) đŽđ” 2 = đ¶đ” Ă đ”đ· From part (ii) đŽđ¶ 2 = đ·đ¶ Ă đ”đ¶ đŽđ”2 đ¶đ”Ăđ”đ· Hence đŽđ¶ 2 = đ·đ¶Ăđ”đ¶ đŽđ”2 đ”đ· = đ·đ¶ đŽđ¶ 2 Hence proved Printed from Vedantu.com.Class X Chapter 4 â Triangles Maths (i) In âADB and âCAB â DAB = â ACB = 90° â ABD = â CBA (common angle) â ADB = â CAB (remaining angle) So. . Book a free session with a top LIVE online tutor now. An aeroplane leaves an airport and flies due north at a speed of 1000km/hr. Book a free session with a top LIVE online tutor now. Therefore by pythagoras theorem.Class X Chapter 4 â Triangles Maths 26. At the same time. đŽđ” 2 = đđ” 2 + đđŽ2 242 = 182 + đđŽ2 đđŽ2 = 576 â 324 OA = â252 = â6 Ă 6 Ă 7 = 6â7 Therefore distance from base = 6â7 đ 27. . distance travelled by the plane flying towards west in 1 2 âđđ 1 = 1200 Ă 1 2 = 1800 đđ Let these distances are represented by OA and OB respectively. Now applying Pythagoras theorem 1 Distance between these planes after 1 2 âđđ AB = âđđŽ2 + đđ” 2 = â(1500)2 + (1800)2 = â2250000 + 3240000 Printed from Vedantu. How far from the base of the pole should the stake be driven so that the wire will be taut? Sol: Let OB be the pole and AB be the wire. another aeroplane leaves the same airport and flies due west at a speed of 1200 km/hr. A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far apart will be the two planes after 1 hours? Sol: 1 Distance traveled by the plane flying towards north in 1 2 hrs 1 = 1000 Ă 1 2 = 1500 đđ 1 Similarly.com. Determine whether the triangle having sides (a â 1) cm. 2âđ cm and (a + 1) cm is a right- angled triangle. distance between these planes will be 300â61 km. Book a free session with a top LIVE online tutor now.com. đŽđ” 2 = (đ â 1)2 = đ2 + 1 â 2đ 2 đ”đ¶ 2 = (2âđ) = 4đ đ¶đŽ2 = (đ + 1)2 = đ2 + 1 + 2đ Hence đŽđ” 2 + đ”đ¶ 2 = đŽđ¶ 2 đđ â đŽđ”đ¶ đđ đđđâđĄ đđđđđđ â đđĄ đ”. Sol: Let ABC be the â with AB = (a â 1) cm BC = 2âđ cm. Printed from Vedantu. CA = (a + 1) cm Hence. .Class X Chapter 4 â Triangles Maths = â5490000 = â9 Ă 610000 = 300â61 1 So. after 1 2 âđđ 28.