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:
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:
Solution 5:
Solution 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:
Solution 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):
Solution 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):
Solution 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):
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):
Solution 11(12):
Solution 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):
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:
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:
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:
Solution 7:
Solution 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:
Solution 2:
Solution 3:
Solution 4:
Exercise 9.3:
Solution 1:
Solution 2:
Solution 3:
Solution 4:
Solution 5:
Solution 6:
Solution 7:
Solution 8:
Solution 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:
Solution 2:
Solution 3:
Solution 4:
Solution 5:
Solution 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:
Exercise 9.5:
Solution 1:
Solution 2:
Solution 3:
Solution 4:
Solution 5:
Solution 6:
