Class 9 Exam  >  Class 9 Notes  >  Short Answer Type Questions: Triangles

Class 9 Maths Chapter 6 Question Answers - Triangles

Question 1. The angles of a triangle are in the ratio 2: 3: 4. Find the angles.
Solution:
Let the angles be 2x, 3x and 4x.
We know that the sum of the angles is 180°.
∴ 2x + 3x + 4x = 180°
⇒ 9x = 180°
⇒ x=(180°/9)= 20°
∴ 2x = 2 x 20° = 40° 3x = 3 x 20° = 60° 4x = 4 x 20° = 80°
∴ The required angles are 40°, 60° and 80°.
 

Question 2. In a ΔABC, if ∠ A + ∠ B = 110° and ∠ B + ∠ C = 132°, then find ∠ A, ∠ B and ∠ C.
Solution:
We have ∠ A + ∠ B = 110°            ...(1)
∠ B + ∠ C = 132°                ...(2)
Also ∠ A + ∠ B + ∠ C = 180°             ...(3)
∴ From (1) and (3), we have 110° + ∠ C = 180°
⇒ ∠ C = 180° - 110° = 70°

Now, from (2), we have ∠ B + 70° = 132°
⇒ ∠ B = 132° - 70° = 62° From (1),
∠ A + 62° = 110°
⇒ ∠ A = 110° - 62° = 48°
Thus, the required angles are ∠ A = 48°,
∠B = 62° and ∠ C = 70°


Question 3. Prove that ΔABC is isosceles if and only if altitude AD bisects BC.
Solution: 
In ΔABC, the altitude AD bisects BC. i.e. BD = DC Now, in ΔABD and ΔACD, we have

Class 9 Maths Chapter 6 Question Answers - Triangles

BD = DC             [Given]
AD = AD             [Common]
∠ADB = ∠ADC
            [each = 90°, ∵ AD is an altitude.]
∴ ΔACD ≌ ΔABD
∴ Their corresponding parts are equal.
⇒ AB = AC
⇒ ΔABC is an isosceles triangle.
 

Question 4. In the adjoining figure,
ΔOAD ≌ ΔOBC.
Find ∠ A and ∠ B.
Solution:
Since, ΔOAD ≌ ΔOBC

Class 9 Maths Chapter 6 Question Answers - Triangles

∴ ∠ A= ∠ B,
Since, ∠ A + ∠ D + ∠ AOD = 180°
⇒ ∠ A + 50° + 40° = 180°
⇒ ∠ A = 180° - 50° - 40°
= 90°
∴ ∠ B= ∠ A = 90°
Thus, ∠ A = 90° and ∠ B = 90°.
 

Question 5. In the figure, AD ⊥ BC and AB = AC. Find ∠ B.
Solution: 
∵ AD ⊥ BC and ∠ C = 55°             [Given]
∴ In right ΔADB and right ΔADC,

Class 9 Maths Chapter 6 Question Answers - Triangles

AD = AD             [Common]
Hypt. AB = Hypt. AC             [Given]
⇒ ΔABD ≌ ΔACD             [RHS criteria]
⇒ ∠ B= ∠ C
⇒ ∠ B = 55°


Question 6. In the adjoining figure, AB = BC = AC, then find the measure of ∠ A.
Solution:
∵ AB = BC = AC
∴ ΔABC is an equilateral triangle.
⇒ ∠ A= ∠ B = ∠ C
∴ ∠ A + ∠ B + ∠ C = 180°
⇒ ∠ A + ∠ A + ∠ A = 180°
⇒ 3∠ A = 180°
⇒ ∠ A =(180°/3)= 60°

Class 9 Maths Chapter 6 Question Answers - Triangles

Thus, the required measure of ∠A is 60°.


Question 7. In ΔABC, if ∠ A = 80°, ∠ B = 70°, then identify the longest and the shortest sides of the triangle.
Solution: 
Since, ∠ A = 80° and ∠ B = 70°
∠ C = 180° - (∠ A + ∠ B)
= 180° - (80° + 70°)
= 180° - 150° = 30°

Class 9 Maths Chapter 6 Question Answers - Triangles

⇒ The largest angle is 80°,
i.e. ∠A = 80°
∴ The side opposite of ∠A is the longest.
⇒ BC is the longest side of ΔABC
Again ∠C is the smallest angle
∴ The side opposite of ∠C is the smallest. i.e. AB is the smallest side of ΔABC.


Question 8. In the figure, ABC is a triangle in which AB = AC. The side BA is produced to P such that AB = AP. Prove that ∠ BCP = 90°.
Solution:
∵ AB = AC             [Given]
∴ The angles opposite to AB and AC are equal.

Class 9 Maths Chapter 6 Question Answers - Triangles

⇒ ∠ ABC = ∠ ACB             ...(1)
Also, AC = AP             [Given]
⇒ ∠ APC = ∠ ACP             ...(2)
Adding (1) and (2), we get
∠ ABC + ∠ APC = ∠ ACB + ∠ ACP
⇒ ∠ ABC + ∠ APC = ∠ BCP
⇒ ∠ PBC + ∠ BPC = ∠ BCP               [∵ ∠ ABC = ∠ PBC and ∠ APC = ∠ BPC]
⇒ ∠ PBC + ∠ BPC + ∠ BCP
= ∠ BCP + ∠ BCP            [Adding ∠ BCP to both sides]
⇒ 180° = 2∠ BCP
⇒ ∠ BCP =(1800/2)= 90°
Thus, ∠ BCP = 90°.


Question 9. In the adjoining figure, O is the centre of the circle and AB is a diameter. If AC is any chord, then show that ∠ A = (1/2)∠COB.
Solution: 
∵ O is the centre of the circle.
∴ OA = OC [Radii of the same circle]
Now, in ΔOAC,
∠ OAC = ∠ OCA             [Angles opposite to equal sides are equal.]
∠ OAC + ∠ OCB = ∠ COB
[∵ Exterior angle is equal to the sum of interior opposite angles.]

Class 9 Maths Chapter 6 Question Answers - Triangles

⇒ ∠ OAC + ∠ OAC = ∠ COB
⇒ 2∠ OAC = ∠ COB
⇒ ∠ OAC = (1/2)∠COB
Thus, ∠ A= (1/2)∠COB


Question 10. In the figure, ABC is a triangle such that ∠ B = 40° and ∠ C = 50°. The bisector of ∠ A meets BC in X. Write AX, BX and CX in ascending order.
Solution: 
∵ ∠ B = 40° and ∠ C = 50°             [Given]

Class 9 Maths Chapter 6 Question Answers - Triangles

∴ ∠ A = 180° - (∠ B + ∠ C)
⇒ ∠ A = 180°- (40° + 50°)
⇒ ∠ A = 180° - 90° = 90°
Since AX bisects ∠BAC.
∴ ∠ BAX = 45° = ∠ CAX
Now, ∠ BAX > ∠ ABX             [∵ 45° > 40°]
⇒ BX > AX             ...(1)
Also ∠ ACX > ∠ CAX
⇒ AX > CX             ...(2)
From (1) and (2), we have
BX > AX > CX
Thus, the required ascending order of AX, BX and CX is Class 9 Maths Chapter 6 Question Answers - Triangles


Question 11. In the given figure, AD is the bisector of ∠ A of  ΔABC, where D lies on BC. Show that AB > BD and AC > CD.
Solution: In ΔABC, ∵ AD is the bisector of ∠A

Class 9 Maths Chapter 6 Question Answers - Triangles

∴ ∠ BAD = ∠ CAD
In ΔABD,
Exterior ∠ADC > ∠BAD
⇒ ∠ADC > ∠CAD
Now, in ΔADC,
∠ ADC > ∠ CAD
⇒ AC > CD             [Side opposite to greater angle in a D is greater]
Also, in ΔABD, ∠ ADB > ∠ BAD             [∵ Ext. ∠ ADB is greater than ∠ CAD and ∠ CAD = ∠ BAD]
⇒ AB > BD [Side opposite to greater angle in a D is greater]

 

Question 12. Bisectors of the angles B and C of an isosceles triangle ABC with AB = AC intersect each other at O. Show that the external angle adjacent to ∠ ABC is equal to ∠ BOC.
Solution:
In ΔABC, AB = AC
⇒ ∠B = ∠ C             [∴ Angles opposite to equal sides are equal]

Class 9 Maths Chapter 6 Question Answers - Triangles

⇒ (1/2)∠B = (1/2)∠C
⇒ ∠ OBC = ∠ OCB Now, in ΔBOC,
∠ BOC = 180° - [∠ OBC + ∠ OCB]
= 180° - (1/2) [∠B + ∠C]
= 180° - (1/2) [∠ B + ∠ B] [∵ ∠ B = ∠ C]
= 180° - ∠ B ... (1)
∵ ∠ ABD + ∠ B = 180°
∴ ∠ABD = 180° - ∠ B ... (2)
From (1) and (2), we have ∠ BOC = ∠ ABD
But ∠ABD is external angle adjacent to ∠ABC
Thus, [external angle adjacent to ∠ABC] = ∠BOC.


Question 13. Prove that any two sides of a D are together greater than twice the median drawn to the third side.  
 Solution
: In ΔABC, ∵ AD is a median

Class 9 Maths Chapter 6 Question Answers - Triangles

∴ BD = DC
Produce AD to E, such that AD = DE
Join CE.
In ΔABD and ΔECD

BD = DC             [AD is a median]
AD = DE             [Construction]
∠ 1 = ∠ 2             [Vertically opp. angles]
⇒ ΔABD ≌ ΔECD [SAS]
∴ AB = CE             ... (1)
In ΔAEC, (AC + CE) > AE             [∵ In a D sum of any two sides is greater than the third side]
⇒ (AC + AB) > AE   [From (1)]
⇒ (AC + AB) > AD + DE
⇒ (AC + AB) > AD + AD            [∵ AD = DE by construction]
⇒ AC + AB > 2 AD
⇒ [Sum of any two sides of a D] > [Twice the median to the third side]


Question 14. In the given figure, S is any point on the side QR of ΔPQR. Prove that PQ + QR + RP > 2 PS.

Class 9 Maths Chapter 6 Question Answers - Triangles

Solution: ∵ In a D, the sum of any two sides is greater than the third side.
∴ In ΔPQS, (PQ + QS) > PS             ... (1)
Similarly, in ΔPRS, (PR + SR) > PS             ... (2)
Adding (1) and (2),
[(PQ + QS) + (PR + SR)] > 2 PS
⇒ PQ + (QS + SR) + PR > 2 PS
⇒ PQ + QR + RP > 2 PS

The document Class 9 Maths Chapter 6 Question Answers - Triangles is a part of Class 9 category.
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FAQs on Class 9 Maths Chapter 6 Question Answers - Triangles

1. What are the different types of triangles?
Ans. There are three types of triangles based on their sides: equilateral triangles (all sides are equal), isosceles triangles (two sides are equal), and scalene triangles (no sides are equal).
2. How do you identify a right triangle?
Ans. A right triangle can be identified by one of its angles being a right angle (90 degrees). The sides opposite the right angle are called the legs, and the side opposite the right angle is called the hypotenuse.
3. How do you calculate the area of a triangle?
Ans. The area of a triangle can be calculated using the formula: Area = (base × height) / 2. The base is the length of the triangle's bottom side, and the height is the perpendicular distance from the base to the opposite vertex.
4. What is the sum of angles in a triangle?
Ans. The sum of angles in a triangle is always 180 degrees. This property is known as the Triangle Angle Sum Theorem.
5. How can you determine if three given side lengths form a triangle?
Ans. In order for three side lengths to form a triangle, the sum of the lengths of any two sides must be greater than the length of the third side. This is known as the Triangle Inequality Theorem.
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