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GSEB Solutions for Class 9 Mathematics – Triangle

GSEB Solutions for Class 9 Mathematics – Triangle (English Medium)

Exercise 9:

Solution 1:

In ∆ABC,
m∠A + m∠B + m∠C = 180°
∴100° + m∠C = 180° … (∵ m∠A + m∠B = 100°)
∴ m∠C = 180° – 100°
∴ m∠C = 80°
Given, m∠B + m∠C = 130°
∴ m∠B + 80° = 130°
∴ m∠B = 130° – 80°
∴ m∠B = 50°
Now, m∠A + m∠B = 100°
∴ m∠A + 50° = 100°∴ m∠A = 100° – 50°
∴ m∠A = 50°
In ∆ABC, m∠A = 50°, m∠B = 50° and m∠C = 80°.

Solution 2:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9-2

Solution 3:

The length of each side of a triangle is greater than the positive difference of the lengths of the other two sides.
In ∆ABC,
AC < AB + BC and AC > |AB – BC|
∴|AB – BC| < AC < AB + BC
∴|10 – 18| < AC < 10 + 18
∴|-8| < AC < 28
∴8 < AC < 28

Solution 4:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9- 4

Solution 5:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9- 5

Solution 6:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9- 6

Solution 7:

The length of each side of a triangle is greater than the positive difference of the lengths of the other two sides.
In ΔABC, AB < BC + AC and AB > |BC – AC|
∴|BC – AC| < AB < BC + AC
∴|5 – 12| < AB < 5 + 12
∴|-7| < AB < 17
∴7 < AB < 17

Solution 8:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9- 8

Solution 9:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9- 9

Solution 10:

The length of each side of a triangle is greater than the positive difference of the lengths of the other two sides.
In DABC, AC < AB + BC and AC > |AB – BC|
∴|AB – BC| < AC < AB + BC
∴|8 – 5| < AC < 8 + 5
∴|3| < AB < 13
∴3 < AB < 13

Solution 11(1):

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9- 11(1)

Solution 11(2):

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9- 11(2)

Solution 11(3):

b. 75°
∠ACD is an exterior angle of ∆ ABC.
∴∠ ACD and ∠ ACB is linear pair of angle.
∴m∠ ACD + m∠ ACB = 180°
∴ 105° + m∠ ACB = 180°
∴ m∠ ACB = 180° – 105°
∴ m∠ ACB = 75°

Solution 11(4):

c. C
For the corresponding BAC ↔ YXZ between ∆ ABC and ∆ XYZ, the angle ∠C corresponds to ∠Z.

Solution 11(5):

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9- 11(5)

Solution 11(6):

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9- 11(6)

Solution 11(7):

a. 10
In ∆ ABC,
Given,
BC = 3, AC = 4
∠A ≅ ∠C
∴BC = AB
The perimeter of ∆ABC = AB + BC + AC = 3 + 3 + 4 = 10

Solution 11(8):

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9- 11(8)

Solution 11(9):

c. AAA
AAA – condition is not possible for the congruence of two triangles.

Solution 11(10):

d. AC < AB + BC
For ∆ABC, AC < AB + BC is true (if it is not a right triangle)

Solution 11(11):

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9- 11(11)

Solution 11(12):

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9- 11(12)

Solution 11(13):

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9- 11(13)

Solution 11(14):

a. scalene
In ∆ ABC,
m∠A + m∠B + m∠C = 180°
x + 2x + y = 180°
∴3x + y = 180° …(1)
∴2x – y = 40° … (2)
Add (1) and (2).
∴5x = 220°
∴x = 44°
Substitute in(1) we get,
3(44) + y = 180°
∴y = 48°

Solution 11(15):

a. 30°
Let m∠ A = k, m∠ B = 2k, m∠ C = 3k
In ∆ ABC
m∠ A : m∠ B : m∠ C = 1 : 2 : 3
∴m∠ A + m∠ B + m∠ C = 180°
∴k + 2k + 3k = 180°
∴6k = 180°
∴k = 30°
m∠A = 30°, m∠B = 60°, m∠C = 90°
The measure of the smallest angle is 30°.

Solution 11(16):

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9- 11(16)

Solution 11(17):

c. 60°
In ∆ ABC,
m∠A + m∠B + m∠C = 180°
∴ 120° + m∠C = 180°
∴ m∠C = 180° – 120° = 60°

Exercise 9.1:

Solution 1:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9.1-1
In ∆ABC,
∠ACD is an exterior angle and m∠ACD = 120.
The measure of one of the interior opposite angle is 40.
Let m∠A = 40.
We know that, measure of any exterior angle of a triangle is equal to the sum of the measures of its two interior opposite angles.
m∠ACD = m∠A + m∠B
∴ 120 = 40 + m∠B
∴ m∠B = 120 – 40= 80
In DABC,
The sum of the measures of all three angles of a triangle is 180.
m∠A + m∠B + m∠C = 180
∴ 40 + 80 + m∠C = 180
∴ m∠C = 180 – 40 – 80 = 60
Thus, the measures of the remaining angles of ∆ABC are 80 and 60.

Solution 2:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9.1-2

Solution 3:

Let in ∆PQR,
m∠P : m∠Q : m∠R = 2 : 3 : 5
Then,
m∠P = 2k, then m∠Q = 3k and m∠R= 5k.
In ∆PQR,
The sum of the measures of all three angles of triangle is 180.
m∠P : m∠Q : m∠R = 180
∴ 2k + 3k + 5k = 180
∴ 10k = 180
∴ k = 18
Substitute k = 18,we get
m∠P = 2k = 2 × 18 = 36
m∠Q = 3k = 3 × 18 = 54
m∠R = 5k = 5 × 18 = 90
Thus, the measures of the angles of ∆ABC are 36, 54 and 90.

Solution 4:

Figure 1:
M ∠BAC + m ∠BAE = 180 (linear pair)
∴ m ∠BAC + 110 = 180
∴ m ∠BAC = 180 – 110 = 70
M ∠ACB + m ∠ACD = 180 (linear pair)
∴ m ∠ACB + 105 = 180
∴ m ∠ACB = 180 – 105
∴ m ∠ACB = 75
In DABC,
The sum of the measures of all three angles of triangle is 180.
∴ m ∠BAC + m ∠ABC + m ∠ACB = 180
∴ 70 + x + 75 = 180
∴ x = 180 – 70 – 75 = 35
Figure 2 :
We know that, measure of any exterior angle of a triangle is equal to the sum of the measure of its two interior opposite angles.
m∠ABE = m∠A + m∠ACB
∴ 105 = m∠A + m∠ACB ………….. (1)
m∠ACD = m∠A + m∠ABC
∴ 100 = m∠A + m∠ABC ………….. (2)
Add (1) and (2), we get
105 + 100 = m∠A + m∠ACB + m∠A + m∠ABC
∴ 205 = m∠A + (m∠ACB + m∠A + m∠ABC)
[The sum of the measures of all three angles of triangle is 180. ∴ m∠ACB + m∠A + m∠ABC = 180]
∴ 205 = m∠A + 180
∴ m∠A = 205 – 180
∴ m∠A = 25|
∴ x = 25

Solution 5:

Let, m∠B = x
m∠A – m∠B = 70
∴ m∠A – x = 70
∴ m∠A = x + 70
m∠B – m∠C = 40
∴ x – m∠C = 40
∴ x – 40 = m∠C
∴ m∠C = x – 40
In ∆ABC,
m∠A + m∠B + m∠C = 180 (The sum of the measures of all three angles of triangle is 180)
∴x + 70 + x + x – 40 = 180
∴ 3x + 30 = 180
∴ 3x = 180 – 30
∴ 3x = 150
∴ x = 50
m∠A = x + 70 = 50 + 70 = 120
m∠B = x = 50
m∠C = x – 4 = 50 – 40 = 10
Thus, m∠A = 120 and m∠C = 10.

Solution 6:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9.1-6

Solution 7:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9.1-7

Solution 8:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9.1-8

We know that, measure of any exterior angle of a triangle is equal to the sum of the measure of its two interior opposite angles.
Given, m∠ABE is an exterior angle of DABC.
∴m∠ABE = m∠BAC + m∠ACB  …[1]
∠CAD is an exterior angle of ∆ABC.
∴m∠CAD = m∠ABC + m∠ACB …[2]
Add equation [1] and [2], we get
m∠ABE + m∠CAD = m∠BAC + m∠ACB + m∠ABC + m∠ACB
∴ 100 + 125 = (m∠BAC + m∠ACB + m∠ABC) + m∠ACB
∴ 225 = 180 + m∠ACB (The sum of the measures of all three angles of triangle is 180.)
∴ m∠ACB = 225 – 180 = 45

Exercise 9.2:

Solution 1:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9.2-1

Solution 2:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9.2-2

Solution 3:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9.2-3

Solution 4:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9.2-4

Exercise 9.3:

Solution 1:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9.3-1

Solution 2:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9.3-2

Solution 3:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9.3-3

Solution 4:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9.3-4

Solution 5:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9.3-5

Solution 6:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9.3-6

Solution 7:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9.3-7

Solution 8:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9.3-8

Solution 9:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9.3-9

Solution 10:

Let the measure of two congruent angles of the triangles are x and x.
The measure of the third angle is x + 60.
Sum of measure of all three angles of triangle is 180°.
∴x + x + (x + 60) = 180°
∴3x + 60 = 180
∴3x = 180 – 60
∴3x = 120
∴x = 40
Then, x + 60 = 40 + 60 = 100
The measures of the two congruent angles of the triangle are 40° each and the measure of the third angle is 100°.

Exercise 9.4:

Solution 1:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9.4-1

Solution 2:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9.4-2

Solution 3:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9.4-3

Solution 4:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9.4-4

Solution 5:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9.4-5

Solution 6:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9.4-6

Given: P is in the interior of DABC such that PA = PB = PC.
To prove:  m∠A = m∠ABP + m∠ACP.
Proof : In ∆ABP,
PA = PB (Given)
∴ m∠ABP = m∠BAP …….. (1)
In ∆APC, PA = PC (data)
∴ m∠ACP = m∠PAC …….. (2)
Adding  (1) and (2), we get
m∠ABP + m∠ACP = m∠BAP + m∠PAC
Now, P is in the interior of ∆ABC.
m∠BAP + m∠PAC = m∠BAC
m∠ABP + m∠ACP = m∠BAC
⇒ m∠ABP + m∠ACP = m∠A
m∠A = m∠ABP + m∠ACP.

Solution 7:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9.4-7

Exercise 9.5:

Solution 1:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9.5-1

Solution 2:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9.5-2

Solution 3:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9.5-3

Solution 4:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9.5-4

Solution 5:

GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9.5-5

Solution 6:
GSEB Solutions for Class 9 Mathematics - Triangle-Ex-9.5-6

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