R S AGGARWAL AND V AGGARWAL Solutions for Class 9 Maths Chapter 8 - Triangles

Page / Exercise

Chapter 8 - Triangles Exercise MCQ

Question 1

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles

Solution 1

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles

Question 2

In a ΔABC, if A - B = 42° and B - C = 21° then B = ?

(a) 32° 

(b) 63° 

(c) 53° 

(d) 95° 

Solution 2

Correct option: (c)

A - B = 42° 

A = B + 42° 

B - C = 21° 

C = B - 21° 

In ΔABC,

A + B + C = 180° 

B + 42° + B + B - 21° = 180° 

3B = 159

B = 53° 

Question 3

In a ΔABC, side BC is produced to D. If ABC = 50° and ACD = 110° then A = ?

 

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles 

(a) 160° 

(b) 60° 

(c) 80° 

(d) 30° 

Solution 3

Correct option: (b)

ACD = B + A (Exterior angle property)

110° = 50° + A

A = 60° 

Question 4

Side BC of ΔABC has been produced to D on left hand side and to E on right hand side such that ABD = 125° and ACE = 130°. Then A = ?

  1. 50°
  2. 55°
  3. 65°
  4. 75°

 

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles

Solution 4

Correct option: (d)

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles

Question 5

In the given figure, sides CB and BA of ΔABC have been produced to D and E respectively such that ABD = 110° and CAE = 135°. Than ACB =?

  1. 65°
  2. 45°
  3. 55°
  4. 35°

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles

Solution 5

Correct option: (a)

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles

Question 6

The sides BC, CA and AB of ΔABC have been produced to D,E and F respectively.  ∠BAE + ∠CBF + ACD =?

  1. 240°
  2. 300°
  3. 320°
  4. 360°

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles

Solution 6

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles

Question 7

In the given figure, EAD BCD. Ray FAC cuts ray EAD at a point A such that EAF = 30°. Also, in ΔBAC, BAC = x° and ABC = (x + 10)°. Then, the value of x is

 

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles 

(a) 20

(b) 25

(c) 30

(d) 35

Solution 7

Correct option: (b)

EAF = CAD (vertically opposite angles)

CAD = 30° 

In ΔABD, by angle sum property

A + B + D = 180° 

(x + 30)° + (x + 10)° + 90° = 180° 

2x + 130° = 180° 

2x = 50° 

x = 25° 

Question 8

In the given figure, two rays BD and CE intersect at a point A. The side BC of ΔABC have been produced on both sides to points F and G respectively. If ABF = x°, ACG = y° and DAE = z° then z = ?

 

 

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles 

(a) x + y - 180

(b) x + y + 180

(c) 180 - (x + y)

(d) x + y + 360° 

Solution 8

Correct option: (a)

ABF + ABC = 180° (linear pair)

x + ABC = 180° 

ABC = 180° - x

 

ACG + ACB = 180° (linear pair)

y + ACB = 180° 

ACB = 180° - y

 

In ΔABC, by angle sum property

ABC + ACB + BAC = 180° 

(180° - x) + (180° - y) + BAC = 180° 

BAC - x - y + 180° = 0

BAC = x + y - 180° 

 

Now, EAD = BAC (vertically opposite angles)

z = x + y - 180°  

Question 9

In the given figure, lines AB and CD intersect at a point O. The sides CA and OB have been produced to E and F respectively. such that OAE = x° and DBF = y°.

 

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles 

 

If OCA = 80°, COA = 40° and BDO = 70° then x° + y° = ?

(a) 190° 

(b) 230° 

(c) 210° 

(d) 270° 

Solution 9

Correct option: (b)

In ΔOAC, by angle sum property

OCA + COA + CAO = 180° 

80° + 40° + CAO = 180° 

CAO = 60° 

 

CAO + OAE = 180° (linear pair)

60° + x = 180° 

x = 120° 

 

COA = BOD (vertically opposite angles)

BOD = 40° 

 

In ΔOBD, by angle sum property

OBD + BOD + ODB = 180° 

OBD + 40° + 70° = 180° 

OBD = 70° 

 

OBD + DBF = 180° (linear pair)

70° + y = 180° 

y = 110° 

 

x + y = 120° + 110° = 230° 

Question 10

In a ΔABC it is given that A:B:C = 3:2:1 and ∠ACD = 90o. If it is produced to E, Then ECD =?

  1. 60°
  2. 50°
  3. 40°
  4. 25°

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles

Solution 10

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles

Question 11

In the given figure , BO and CO are the bisectors of B and C respectively. If A = 50°, then BOC= ?

  1. 130°
  2. 100°
  3. 115°
  4. 120°

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles

Solution 11

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles

Question 12

In the given figure, side BC of ΔABC has been produced to a point D. If A = 3y°, B = x°, C = 5y° and ACD = 7y°. Then, the value of x is

 

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles 

(a) 60

(b) 50

(c) 45

(d) 35

Solution 12

Correct option: (a)

ACB + ACD = 180° (linear pair)

5y + 7y = 180° 

12y = 180° 

y = 15° 

 

Now, ACD = ABC + BAC (Exterior angle property)

7y = x + 3y

7(15°) = x + 3(15°)

105° = x + 45° 

x = 60° 

Chapter 8 - Triangles Exercise Ex. 8

Question 1

In ABC, if B = 76o and C = 48o, find A.

Solution 1

Since, sum of the angles of a triangle is 180o

A + B + C = 180o

A + 76o + 48o = 180o

A = 180o - 124o = 56o

A = 56o

Question 2

The angles of a triangle are in the ratio 2:3:4. Find the angles.

Solution 2

Let the measures of the angles of a triangle are (2x)o, (3x)o and (4x)o.

Then, 2x + 3x + 4x = 180         [sum of the angles of a triangle is 180o ]     

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles 9x = 180

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles The measures of the required angles are:

2x = (2 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles 20)o = 40o

3x = (3 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles 20)o = 60o

4x = (4 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles 20)o = 80o

Question 3

In ABC, if 3A = 4B = 6C, calculate A, B and C.

Solution 3

Let 3A = 4B = 6C = x (say)

Then, 3A = x

A =

4B = x

and 6C = x

C =

As A + B + C = 180o

A =

B =

C =

Question 4

In ABC, if A + B = 108o and B + C = 130o, find A, B and C.

Solution 4

A + B = 108o [Given]

But as A, B and C are the angles of a triangle,

A + B + C = 180o

108o + C = 180o

C = 180o - 108o = 72o

Also, B + C = 130o [Given]

B + 72o = 130o

B = 130o - 72o = 58o

Now as, A + B = 108o

A + 58o = 108o

A = 108o - 58o = 50o

A = 50o, B = 58o and C = 72o.

Question 5

In ABC, A + B = 125o and A + C = 113o. Find A, B and C.

Solution 5

Since. A , B and C are the angles of a triangle .

So, A + B + C = 180o

Now, A + B = 125o [Given]

125o + C = 180o

C = 180o - 125o = 55o

Also, A + C = 113o [Given]

A + 55o = 113o

A = 113o - 55o = 58o

Now as A + B = 125o

58o + B = 125o

B = 125o - 58o = 67o

A = 58o, B = 67o and C = 55o.

Question 6

In R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesPQR, if R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesP - R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesQ = 42o and R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesQ - R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesR = 21o, find R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesP, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesQ and R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesR.

Solution 6

Since, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesP, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesQ and R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesR are the angles of a triangle.

So,R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesP + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesQ + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesR = 180o(i)

Now,R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesP - R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesQ = 42o[Given]

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesP = 42o + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesQ(ii)

andR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesQ - R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesR = 21o[Given]

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesR = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesQ - 21o(iii)

Substituting the value of R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesP and R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesR from (ii) and (iii) in (i), we get,

42o + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesQ + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesQ + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesQ - 21o = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles3R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesQ + 21o = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles3R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesQ = 180o - 21o = 159o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesQ = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesP = 42o + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesQ

= 42o + 53o = 95o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesR = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesQ - 21o

= 53o - 21o = 32o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesP = 95o, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesQ = 53o and R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesR = 32o.

Question 7

The sum of two angles of a triangle is 116o and their difference is 24o. Find the measure of each angle of the triangle.

Solution 7

Given that the sum of the angles A and B of a R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC is 116o, i.e., R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB = 116o.

Since, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesC = 180o

So, 116o + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesC = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesC = 180o - 116o = 64o

Also, it is given that:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA - R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB = 24o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA = 24o + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB

Putting, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA = 24o + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB in R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB = 116o, we get,

24o + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB = 116o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles 2R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB + 24o = 116o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles 2R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB = 116o - 24o = 92o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles

Therefore, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA = 24o + 46o = 70o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA = 70o, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB = 46o and R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesC = 64o.

Question 8

Two angles of a triangle are equal and the third angle is greater than each one of them by 18o. Find the angles.

Solution 8

Let the two equal angles, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA and R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB, of the triangle be xo each.

We know,

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesC = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Trianglesxo + xo + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesC = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles2xo + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesC = 180o(i)

Also, it is given that,

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesC = xo + 18o(ii)

Substituting R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesC from (ii) in (i), we get,

2xo + xo + 18o = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles3xo = 180o - 18o = 162o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Trianglesx = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles

Thus, the required angles of the triangle are 54o, 54o and xo + 18o = 54o + 18o = 72o.

Question 9

Of the three angles of triangle, one is twice the smallest and another one is thrice the smallest. Find the angles.

Solution 9

Let C be the smallest angle of ABC.

Then, A = 2C and B = 3C

Also, A + B + C = 180o

2C + 3C + C = 180o

6C = 180o

C = 30o

So, A = 2C = 2 30o = 60o

B = 3C = 3 30o = 90o

The required angles of the triangle are 60o, 90o, 30o.

Question 10

In a right-angled triangle, one of the acute angles measures 53o. Find the measure of each angle of the triangle.

Solution 10

Let ABC be a right angled triangle and C = 90o

Since, A + B + C = 180o

A + B = 180o - C = 180o - 90o = 90o

Suppose A = 53o

Then, 53o + B = 90o

B = 90o - 53o = 37o

The required angles are 53o, 37o and 90o.

Question 11

In a right-angled triangle, one of the acute angles measures 53o. Find the measure of each angle of the triangle.

Solution 11

Let ABC be a right angled triangle and C = 90o

Since, A + B + C = 180o

A + B = 180o - C = 180o - 90o = 90o

Suppose A = 53o

Then, 53o + B = 90o

B = 90o - 53o = 37o

The required angles are 53o, 37o and 90o.

Question 12

If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.

Solution 12

Let ABC be a triangle.

So, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA < R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesC

Adding R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA to both sides of the inequality,

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA < R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesC

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA < 180o [Since R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesC = 180o]

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles 

Similarly, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB <R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesC

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB < 90o

and R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesC < R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesC < 90o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC is an acute angled triangle.

Question 13

If one angle of a triangle is greater than the sum of the other two, show that the triangle is obtuse angled.

Solution 13

Let ABC be a triangle and B > A + C

Since, A + B + C = 180o

A + C = 180o - B

Therefore, we get,

B > 180o - B

Adding B on both sides of the inequality, we get,

B + B > 180o - B + B

2B > 180o

B >

i.e., B > 90o which means B is an obtuse angle.

ABC is an obtuse angled triangle.

Question 14

In the given figure, side BC of ABC is produced to D. If ACD = 128o and ABC = 43o, find BAC and ACB.

Solution 14

Since ACB and ACD form a linear pair.

So, ACB + ACD = 180o

ACB + 128o = 180o

ACB = 180o - 128 = 52o

Also, ABC + ACB + BAC = 180o

43o + 52o + BAC = 180o

95o + BAC = 180o

BAC = 180o - 95o = 85o

ACB = 52o and BAC = 85o.

Question 15

In the given figure, the side BC of R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC has been produced on the right-hand side from B to D and  on the right-hand side from C and E. If R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesABD = 106o and R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesACE= 118o, find the measure of each angle of the triangle.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles

Solution 15

As R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesDBA and R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC form a linear pair.

So,R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesDBA + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles106o + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC = 180o - 106o = 74o

Also, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesACB and R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesACE form a linear pair.

So,R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesACB + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesACE = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesACB + 118o = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesACB = 180o - 118o = 62o

In R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC, we have,

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesACB + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesBAC = 180o

74o + 62o + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesBAC = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles136o + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesBAC = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesBAC = 180o - 136o = 44o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesIn triangle ABC, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA = 44o, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB = 74o and R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesC = 62o

Question 16

Calculate the value of x in each of the following figures.

(i)

(ii)

(iii)

Given: AB || CD

(vi)

Solution 16

(i) EAB + BAC = 180o [Linear pair angles]

110o + BAC = 180o

BAC = 180o - 110o = 70o

Again, BCA + ACD = 180o [Linear pair angles]

BCA + 120o = 180o

BCA = 180o - 120o = 60o

Now, in ABC,

ABC + BAC + ACB = 180o

xo + 70o + 60o = 180o

x + 130o = 180o

x = 180o - 130o = 50o

x = 50

(ii)

In ABC,

A + B + C = 180o

30o + 40o + C = 180o

70o + C = 180o

C = 180o - 70o = 110o

Now BCA + ACD = 180o [Linear pair]

110o + ACD = 180o

ACD = 180o - 110o = 70o

In ECD,

ECD + CDE + CED = 180o

70o + 50o + CED = 180o

120o + CED = 180o

CED = 180o - 120o = 60o

Since AED and CED from a linear pair

So, AED + CED = 180o

xo + 60o = 180o

xo = 180o - 60o = 120o

x = 120

(iii)

EAF = BAC [Vertically opposite angles]

BAC = 60o

In ABC, exterior ACD is equal to the sum of two opposite interior angles.

So, ACD = BAC + ABC

115o = 60o + xo

xo = 115o - 60o = 55o

x = 55

(iv)

Since AB || CD and AD is a transversal.

So, BAD = ADC

ADC = 60o

In ECD, we have,

E + C + D = 180o

xo + 45o + 60o = 180o

xo + 105o = 180o

xo = 180o - 105o = 75o

x = 75

(v)

In AEF,

Exterior BED = EAF + EFA

100o = 40o + EFA

EFA = 100o - 40o = 60o

Also, CFD = EFA [Vertically Opposite angles]

CFD = 60o

Now in FCD,

Exterior BCF = CFD + CDF

90o = 60o + xo

xo = 90o - 60o = 30o

x = 30

(vi)

In ABE, we have,

A + B + E = 180o

75o + 65o + E = 180o

140o + E = 180o

E = 180o - 140o = 40o

Now, CED = AEB [Vertically opposite angles]

CED = 40o

Now, in CED, we have,

C + E + D = 180o

110o + 40o + xo = 180o

150o + xo = 180o

xo = 180o - 150o = 30o

x = 30

Question 17

In the figure given alongside, AB CD, EF BC, BAC = 60° and DHF = 50°. Find GCH and AGH.

 

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles 

Solution 17

AB CD and AC is the transversal.

BAC = ACD = 60° (alternate angles)

i.e. BAC = GCH = 60° 

 

Now, DHF = CHG = 50° (vertically opposite angles)

 

In ΔGCH, by angle sum property,

GCH + CHG + CGH = 180° 

60° + 50° + CGH = 180° 

CGH = 70° 

 

Now, CGH + AGH = 180° (linear pair)

70° + AGH = 180° 

AGH = 110° 

Question 18

Calculate the value of x in the given figure.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles

Solution 18

Produce CD to cut AB at E.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles

Now, in R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesBDE, we have,

Exterior R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesCDB = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesCEB + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesDBE

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles xo = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesCEB + 45o     .....(i)

In R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles AEC, we have,

Exterior R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesCEB = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesCAB + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesACE

= 55o + 30o = 85o

Putting R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesCEB = 85o in (i), we get,

xo = 85o + 45o = 130o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles x = 130

Question 19

In the given figure, AD divides R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesBAC in the ratio 1: 3 and AD = DB. Determine the value of x.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles

Solution 19

The angle R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesBAC is divided by AD in the ratio 1 : 3.

Let R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesBAD and R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesDAC be y and 3y, respectively.

As BAE is a straight line,

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesBAC + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesCAE = 180o        [linear pair]

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesBAD + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesDAC + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles CAE = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles y + 3y + 108o = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles 4y = 180o - 108o = 72o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles

Now, in R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC,

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesBCA + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesBAC = 180o

y + x + 4y = 180o

[Since, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesBAD (given AD = DB) and R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesBAC = y + 3y = 4y]

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles 5y + x = 180

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles 5 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles 18 + x = 180

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles 90 + x = 180

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles x = 180 - 90 = 90

Question 20

If the sides of a triangle are produced in order, prove that the sum of the exterior angles so fomed is equal to four right angles.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles

Solution 20

Given : A R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC in which BC, CA and AB are produced to D, E and F respectively.

To prove : Exterior R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesDCA + Exterior R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesBAE + Exterior R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesFBD = 360o

Proof : Exterior R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesDCA = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB(i)

Exterior R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesFAE = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesC(ii)

Exterior R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesFBD = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesC(iii)

Adding (i), (ii) and (iii), we get,

Ext. R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesDCA + Ext. R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesFAE + Ext. R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesFBD

= R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesC + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesC

= 2R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA +2R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB + 2R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesC

= 2 (R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesC)

= 2 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles180o

[Since, in triangle the sum of all three angle is 180o]

= 360o

Hence, proved.

Question 21

In the given figure, show that

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesC + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesD + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesE + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesF = 360o.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles

Solution 21

In R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesACE, we have,

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesC + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesE = 180o (i)

In R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesBDF, we have,

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesD + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesF = 180o (ii)

Adding both sides of (i) and (ii), we get,

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesC+R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesE + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesD + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesF = 180o + 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesC + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesD + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesE + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesF = 360o.

Question 22

In the given figure, AM BC and AN is the bisector of A. If ABC = 70° and ACB = 20°, find MAN.

 

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles 

Solution 22

In ΔABC, by angle sum property,

A + B + C = 180° 

A + 70° + 20° = 180° 

A = 90° 

 

In ΔABM, by angle sum property,

BAM + ABM + AMB = 180° 

BAM + 70° + 90° = 180° 

BAM = 20° 

 

Since AN is the bisector of A,

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles

 

Now, MAN + BAM = BAN

MAN + 20° = 45° 

MAN = 25° 

Question 23

In the given figure, BAD EF, AEF = 55° and ACB = 25°, find ABC.

 

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles 

Solution 23

BAD EF and EC is the transversal.

AEF = CAD (corresponding angles)

CAD = 55° 

 

Now, CAD + CAB = 180° (linear pair)

55° + CAB = 180° 

CAB = 125° 

 

In ΔABC, by angle sum property,

ABC + CAB + ACB = 180° 

ABC + 125° + 25° = 180° 

ABC = 30° 

Question 24

In the given figure, ABC is a triangle in which R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA : R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB : R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesC = 3 : 2 : 1 and AC R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles CD. Find the measure of R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles

Solution 24

In the given R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC, we have,

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA : R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB : R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesC = 3 : 2 : 1

Let R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA = 3x, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB = 2x, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesC = x. Then,

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesC = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles 3x + 2x + x = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles 6x = 180o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles x = 30o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA = 3x = 3 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles 30o = 90o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB = 2x = 2 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles 30o = 60o

and, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesC = x = 30o

Now, in ABC, we have,

Ext R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesACE = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesA + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesB = 90o + 60o = 150o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesACD + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesECD = 150o

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesECD = 150o - R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesACD 

 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesECD = 150o - 90o    [since R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles]

 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - TrianglesECD= 60o


Question 25

In the given figure, AB DE and BD FG such that ABC = 50° and FGH = 120°. Find the values of x and y.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles 

Solution 25

FGH + FGE = 180° (linear pair)

120° + y = 180° 

y = 60° 

 

AB DF and BD is the transversal.

ABC = CDE (alternate angles)

CDE = 50° 

 

BD FG and DF is the transversal.

EFG = CDE (alternate angles)

EFG = 50° 

 

In ΔEFG, by angle sum property,

FEG + FGE + EFG = 180° 

x + y + 50° = 180° 

x + 60° + 50° = 180° 

x = 70° 

Question 26

In the given figure, AB CD and EF is a transversal. If AEF = 65°, DFG = 30°, EGF = 90° and GEF = x°, Find the value of x.

 

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles 

Solution 26

AB CD and EF is the transversal.

AEF = EFD (alternate angles)

AEF = EFG + DFG

65° = EFG + 30° 

EFG = 35° 

 

In ΔGEF, by angle sum property,

GEF + EGF + EFG = 180° 

x + 90° + 35° = 180° 

x = 55° 

Question 27

In the given figure, AB CD, BAE = 65° and OEC = 20°. Find ECO.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles 

Solution 27

AB CD and AE is the transversal.

BAE = DOE (corresponding angles)

DOE = 65° 

 

Now, DOE + COE = 180° (linear pair)

65° + COE = 180° 

COE = 115° 

 

In ΔOCE, by angle sum property,

OEC + ECO + COE = 180° 

20° + ECO + 115° = 180° 

ECO = 45° 

Question 28

In the given figure, AB CD and EF is a transversal, cutting them at G and H respectively. If EGB = 35° and QP EF, find the measure of PQH.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles 

Solution 28

AB CD and EF is the transversal.

EGB = GHD (corresponding angles)

GHD = 35° 

 

Now, GHD = QHP (vertically opposite angles)

QHP = 35° 

 

In DQHP, by angle sum property,

PQH + QHP + QPH = 180° 

PQH + 35° + 90° = 180° 

PQH = 55° 

Question 29

In the given figure, AB CD and EF AB. If EG is the transversal such that GED = 130°, find EGF.

 

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Triangles 

Solution 29

AB CD and GE is the transversal.

EGF + GED = 180° (interior angles are supplementary)

EGF + 130° = 180° 

EGF = 50°