NCERT Solutions for Class 9 Math Chapter 7.1 Triangles

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EXERCISE 7.2 EXERCISE 7.3 EXERCISE 7.4 EXERCISE 7.5*

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1.    In quadrilateral ABCD, AC = AD and AB bisects ∠A (see Fig. 7.16). Show `\▵ABC≅▵ABD\ `. what can you say about BC and BD?

solution :

   Given,

   AC = AD and AB bisect ∠A.

   Find,

   `\▵ABC≅▵ABD\ `

   BC and BD

   In the Fig,

   AB bisect ∠A[ Given ]

   `\=>∠BAC=∠BAD\ `

   Now,

   In `\▵ABC\ ` and `\▵ABD\ `,

   AC = AD[ Given ]

   ∠CAB = ∠DAB[ Given ]

   AB = AB[ Common ]

   So,

   `\▵ABC≅▵ABD\ `[ SAS rule ]

   Then,

   BC = BD[ CPCT ]

   I say about BC and BD is equal.

2.    ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see Fig. 7.17). Prove that
(i)  `\▵ABD≅▵BAC\ `
(ii)  BD = AC
(iii)  ∠ABD = ∠BAC.

solution :

   Given,

   AD = BC and ∠DAB = ∠BAC,

   Find,

   (i)  `\▵ABD≅▵BAC\ `
   (ii)  BD = AC
   (iii)  ∠ABD = ∠BAC.

   Now,

   In `\▵ABD\ ` and `\▵BAC\ `,

   AD = BC[ Given ]

   ∠DAB = ∠CBA[ Given ]

   AB = AB[ Common ]

   So,

   (i) `\▵ABD≅▵BAC\ `[ SAS rule ]

   (ii) BD = AC[ CPCT ]

   (iii) ∠ABD = ∠BAC[ CPCT ]

3.    AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.

solution :

   Given,

   AD = BC and ∠CBO = ∠DAO.

   Find,

   CD bisect AB,

   In the Fig,

   BC and AD are perpendicular AB.

   We say,

   BC `\||\ ` AD

   CD is a transversal line.

   ∠BCD = ∠ADC`\-----(I)\ `[ Alternate interior angles ]

   Now,

   In `\▵CBO\ ` and `\▵DAO\ `,

   ∠BCD = ∠ADO[ From (I) ]

   BC = AD[ Given ]

   ∠CBO = ∠DAO[ Given ]

   so,

   `\▵CBO≅▵DAO\ `[ ASA rule ]

   Then,

   BO = AO[ CPCT ]

   CO = DO[ CPCT ]

   Now,

   I say that CD bisects AB.

4.    `\l\ ` and `\m\ ` are two parrel lines intersected by another pair of parallel lines `\p\ ` and `\q\ ` (see Fig. 7.19). Show that `\▵ABC≅▵CDA\ `.

solution :

   Given,

   l `\||\ ` m and p `\||\ ` q.

   Find,

   `\▵ABC≅▵CDA\ `

   In the Fig,

   l `\||\ ` m and AC is a transversal line.

   ∠DAC = ∠BCA`\-----(I)\ `[ Alternate interior angles ]

   ∠BAC = ∠DCA`\-----(II)\ `[ Alternate interior angles ]

   Now,

   In `\▵ABC\ ` and `\▵CDA\ `,

   ∠BAC = ∠DCA[ From (I) ]

   AC = AC[ Common ]

   ∠BCA = ∠DAC[ From (II) ]

   So,

   `\▵ABC≅▵CDA\ `[ ASA rule ]

5.    Line `\l\ ` is the bisector of an angle ∠A and B is any point on `\l\ `. BP and BQ are perpendiculars from B to the arms of ∠A (see Fig. 7.20). Show that:
(i)  `\▵APB≅▵AQB\ `
(ii)  BP = BQ or B is equidistant from the arms of ∠A.

solution :

   Given,

   l bisect ∠A and BP and BQ are perpendiculars from B to the arms of ∠A.

   Find,

   (i) `\▵APB≅▵AQB\ `

   (ii) BP = BQ or B is equidistant from the arms of ∠A.

   In the Fig,

   l bisect ∠A[ Given ]

   ∠BAQ = ∠BAP`\-----(I)\ `

   Now,

   In `\▵APB\ ` and `\▵AQB\ `,

   ∠P = ∠Q[ 90° Given ]

   ∠BAP = ∠BAQ[ From (I) ]

   AB = AB[ Common ]

   So,

   (i) `\▵APB≅▵AQB\ `[ AAS rule ]

   Now,

   (ii) BP = BQ[ CPCT ]

   So,

   It can be said the point B is equidistant from the arms of A.

6.    In Fig. 7.21, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

solution :

   Given,

   AC = AE, AB = AD and ∠BAD = ∠EAC.

   Find,

   BC = DE

   In the Fig,

   ∠BAD = ∠EAC[ Given ]

   Add ∠DAC both side,

   ∠BAD + ∠DAC = ∠EAC + ∠DAC

   ⇒ ∠BAC = ∠EAD`\-----(I)\ `

   Now,

   In `\▵ABC\ ` and `\▵ADE\ `,

   AB = AD[ Given ]

   ∠BAC = ∠EAD[ From (I) ]

   AC = AE[ Given ]

   So,

   `\▵ABC≅▵ADE\ `[ SAS rule ]

   Now,

   BC = DE[ CPCT ]

7.    AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (see Fig. 7.22). Show that
(i)  `\▵DAP≅▵EBP\ `
(ii)  AD = BE

solution :

   Given,

   AP = BP, ∠BAD = ∠ABE and ∠EPA = ∠DPB.

   Find,

   (i) `\▵DAP≅▵EBP\ `

   (ii) AD = BE

   In the Fig,

   ∠EPA = ∠DPB[ Given ]

   Add ∠EPD both side.

   ∠EPA + ∠EPB = ∠DPB + ∠EPD

   ⇒ ∠APD = ∠EPB`\-----(I)\ `

   Now,

   In `\ ▵DAP\ ` and `\▵EBP\ `,

   ∠DAP = ∠EBP[ Given ]

   AP = BP[ Given ]

   ∠APD = ∠EPB[ From (I) ]

   So,

   `\▵DAP≅▵FPB\ `[ ASA rule ]

   Now,

   AD = BE[ CPCT ]

8.    In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. point D is joint B (see Fig. 7.23). Show that:
(i)  `\▵AMC≅▵BMD\ `
(ii)  ∠DBC is a right angle.
(iii)  `\▵DBC≅▵ACB\ `
(iv)  CM =`\frac{1}{2}\ `AB

solution :

   Given,

   `\▵ABC\ ` is right angled triangle, ∠C = 90°, AM = BM and DM = CM.

   Find,

   (i)  `\▵AMC≅▵BMD\ `

   (ii)  ∠DBC is a right angle.

   (iii)  `\▵DBC≅▵ACB\ `

   (iv)  CM =`\frac{1}{2}\ `AB

   In `\▵AMC\ ` and `\▵BMD\ `,

   AM = BM[ Given ]

   ∠AMC = ∠BMD[ Vertically opposit angles ]

   CM = DM[ Given ]

   So,

   (i) `\▵AMC≅▵BMD\ `[ SAS rule ]

   ∠ACM = ∠BDC[ CPCT ]

   AC = BD`\-----(I)\ `[ CPCT ]

   Now,

   ∠ACD = ∠BDC[ Alternate interior angles ]

   So,

   We say,

   AC `\||\ ` DB

   In the Fig.

   ∠DBC + ∠ACB = 180°[ Co-interior angles ]

   `\=>∠DBC + 90° = 180°\ `[ Given ∠C = 90° ]

   `\=> ∠DBC=180°-90°\ `

   `\=>∠DBC=90°\ `

   (ii) ∠DBC is a right angle.

   Now,

   In `\▵DBC\ ` and `\▵ACB\ `,

   BC = BC[ Common ]

   ∠DBC = ∠ACB[ right angles ]

   DB = AC[ From (I) ]

   So,

   (iii) `\▵DBC≅▵ACB\ `[ SAS rule ]

   DC = AB[ CPCT ]

   Now,

   DM = CM[ Given ]

   So,

   DC = AB

   ⇒ DM + CM = AB

   ⇒ CM + CM = AB

   ⇒ 2 CM = AB

   (iv) `\CM=\frac{1}{2}AB\ `

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EXERCISE 7.2 EXERCISE 7.3 EXERCISE 7.4 EXERCISE 7.5*