NCERT Solutions for Class 9 Math Chapter 7.2 Triangles

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EXERCISE 7.1 EXERCISE 7.3 EXERCISE 7.4 EXERCISE 7.5*

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1.    In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. join A to O. Show that:

(i) OB = OC

(ii) AO bisects ∠A

solution :

   Given,

   AB = AC and the bisectors of ∠B and ∠C intersect each other at O.

   Find,

   (i) OB = OC

   (ii) AO bisects ∠A

   In th Fig,

   ABC is a isosceles triangle with AB = AC[ Given ]

   ⇒ ∠B = ∠C[ equal sides of angle are equal ]

   `\=>\frac{1}{2}B=\frac{1}{2}C\ `

   `\=>∠OBC=∠OCB\ `[ Angle bisectors ]

   (i) OB = OC `\-----(I)\ `[ opposite side of the equal angle are equal ]

   Now,

   In `\▵AOB\ ` and `\▵AOC\ `,

   AB = AC[ Given ]

   AO = AO[ Common ]

   OB = OC[ From (I) ]

   So,

   `\▵AOB≅▵AOC\ `[ SSS rule ]

   ∠BAO = ∠CAO[ CPCT ]

   (ii) AO bisect ∠A.

2.    In `\▵ABC\ `, AD is the perpendicular bisector of BC (see Fig. 7.30). Show that `\▵ABC\ ` is an isosceles triangle in which AB = AC.

solution :

   Given,

   AD ⊥ BC and AD bisector of BC.

   Find,

   AB = AC

   In `\▵ADB\ ` and `\▵ADC\ `,

   AD = AD[ Common ]

   ∠ADB = ∠ADC[ 90° ∵ AD ⊥ BC ]

   BD = DC[ AD bisector of BC ]

   So,

   `\▵ADB≅▵ADC\ `[ SAS rule ]

   AB = AC[ CPCT ]

   Now,

   We say that `\▵ABC\ ` is isosceles triangle.

3.    ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig. 7.31). Show that these altitudes are equal.

solution :

   Given,

   AB = AC

   Find,

   BE = CF

   In `\▵AFC\ ` and `\▵AEB\ `,

   ∠AFC = ∠AEB[ 90° ]

   ∠A = ∠A[ Common ]

   AC = AB[ Given ]

   So,

   `\▵AFC≅▵AEB\ `[ AAS rule ]

   BE = CF[ CPCT ]

   These altitudes are equal.

4.    ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. 7.32). Show that
   (i) `\▵ABE≅▵ACF\ `
   (ii) AB = AC, i.e., ABC is an isosceles triangle.

solution :

   Given,

   BE = CF

   Find,

   (i) `\▵ABE≅▵ACF\ `

   (ii) AB = AC

   In `\▵ABE\ ` and `\▵ACF\ `,

   ∠A = ∠A[ Common ]

   ∠AEB = ∠AFC[ 90° ]

   BE = FC[ Given ]

   So,

   (i) `\▵ABE≅▵ACF\ `[ AAS rule ]

   (ii) AB = AC[ CPCT ]

   So,

   We say that `\▵ABC\ ` is an isosceles triangle.

5.    ABC and DBC are two isosceles triangles on the same base BC (see Fig. 7.33). Show that ∠ABD = ∠ACD.

solution :

   Given,

   AB = AC and BD = CD

   Find,

   ∠ABD = ∠ACD.

   In the Fig.,

   Let, Draw a line A to D.

   

   In `\▵ABD\ ` and `\▵ACD\ `,

   AD = AD[ Common ]

   AB = AC[ Given ]

   BD = CD[ Given ]

   So,

   `\▵ABD≅▵ACD\ `[ SSS rule ]

   ∠ABD = ∠ACD[ CPCT ]

6.    `\▵ABC\ ` is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. 7.34). Show that ∠BCD is a right angle.

solution :

   Given,

   AB = AC and AD =AB,

   Find,

   ∠BCD = 90°

   In `\▵ABC\ `,

   AB = AC[ Given ]

   So,

   ∠ABC = ∠ACB[ Angles opposite to equal sides ]

   Now,

   ∠ABC + ∠ACB + ∠CAB = 180°[ Angles sum property of a triangle ]

   ⇒ ∠ACB + ∠ACB + ∠CAB = 180°[ ∠ABC = ∠ACB ]

   ⇒ 2 ∠ACB + ∠CAB = 180°

   ⇒ ∠CAB = 180° − 2 ∠ACB`\-----(I)\ `

   Now,

   In `\▵ADC\ `,

   AC = AD[ Given ]

   So,

   ∠ACD = ∠ADC[ Angles opposite to equal sides ]

   Now,

   ∠ACD + ∠ADC + ∠DAC = 180°[ Angles sum property of a triangle ]

   ⇒ ∠ACD + ∠ACD + ∠DAC = 180°[ ∠ACD = ∠ADC ]

   ⇒ 2 ∠ACD + ∠DAC = 180°

   ⇒ ∠DAC = 180° − 2 ∠ACD`\-----(II)\ `

   In the Fig.,

   ∠CAB + ∠CAD = 180°`\-----(III)\ `[ Linear pair of angles ]

   Adding (I) and (II),

   We get,

   ⇒ ∠CAB + ∠CAD = (180° − 2 ∠ACB) + (180° − 2 ∠ACD)

   ⇒ 180° = 360° − 2 ∠ACB − 2 ∠ACD[ From (III) ]

   ⇒ 2 ∠ACB + 2 ∠ACD = 360° − 180°

   ⇒ 2 (∠ACB + ∠ ACD) = 180°

   ⇒ ∠ACB + ∠ACD = `\frac{180°}{2}\ `

   ⇒ ∠BCD = 90°[ ∠ACB + ∠ACD = ∠BCD ]

   Now,

   We say that ∠BCD is a right angle.

   

7.    ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.

solution :

   Given,

   ∠A = 90° and AB = AC,

   Find,

   ∠B and ∠C,

   In `\▵ABC\ `,

   AB = AC[ Given ]

   So,

   ∠ABC = ∠ACB[ Angles opposite to equal sides ]

   Now,

   ∠ABC + ∠ACB + ∠CAB = 180°[ Angles sum property of a triangle ]

   ⇒ ∠ACB + ∠ACB + ∠CAB = 180°[ ∠ABC = ∠ACB ]

   ⇒ 2 ∠ACB + 90° = 180°[ ∠CAB = 90° ]

   ⇒ 2 ∠ACB = 180° − 90°

   ⇒ 2 ∠ACB = 90°

   ⇒ ∠ACB = `\frac{90°}{2}\ `

   ⇒ ∠ACB = 45°

   ⇒ ∠ABC = 45°[ ∠ABC = ∠ACB ]

   So,

   we say,

   ∠B and ∠C are 45° respectively.

8.    Show that the angles of an equilateral triangle are 60° each.

solution :

   In the Fig.,

   AB = BC = AC[ Given ]

   So,

   ∠A = ∠B = ∠C[ Angles opposite to equal sides ]

   In `\▵\ `ABC,

   ∠A + ∠B + ∠C = 180°[ Angles sum property of a triangle ]

   ⇒ ∠A + ∠A + ∠A = 180°[ ∠A + ∠B + ∠C ]

   ⇒ 3∠A = 180°

   ⇒ ∠A = `\frac{180°}{3}\ `

   ⇒ ∠A = 60°

   So,

   We say ∠A, ∠B and ∠C are 60° respectively.

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EXERCISE 7.1 EXERCISE 7.3 EXERCISE 7.4 EXERCISE 7.5*