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# Short Answer Type Questions- Triangles Class 9 Notes | EduRev

## Class 9 Mathematics by Full Circle

Created by: Full Circle

## Class 9 : Short Answer Type Questions- Triangles Class 9 Notes | EduRev

The document Short Answer Type Questions- Triangles Class 9 Notes | EduRev is a part of the Class 9 Course Class 9 Mathematics by Full Circle.
All you need of Class 9 at this link: Class 9

Question 1. The angles of 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 180Â°.
âˆ´ 2x + 3x + 4x = 180Â°
â‡’ 9x = 180Â°
â‡’ x=(1800/9)= 20Â°
âˆ´ 2x = 2 x 20Â° = 40Â° 3x = 3 x 20Â° = 60Â° 4x = 4 x 20Â° = 80Â°
âˆ´ The required angles are 40Â°, 60Â° and 80Â°.

Question 2. In a Î”ABC, if âˆ  A + âˆ  B = 110Â° and âˆ  B + âˆ  C = 132Â°, then find âˆ  A, âˆ  B and âˆ  C.
Solution:
We have âˆ  A + âˆ  B = 110Â°            ...(1)
âˆ  B + âˆ  C = 132Â°                ...(2)
Also âˆ  A + âˆ  B + âˆ  C = 180Â°             ...(3)
âˆ´ From (1) and (3), we have 110Â° + âˆ  C = 180Â°
â‡’ âˆ  C = 180Â° - 110Â° = 70Â°

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

Question 3. Prove that Î”ABC is an isosceles if and only if altitude AD bisects BC.
Solution:
In Î”ABC, the altitude AD bisects BC. i.e. BD = DC Now, in Î”ABD and Î”ACD, we have

BD = DC             [Given]
[each = 90Â°, âˆµ AD is an altitude.]
âˆ´ Î”ACD â‰Œ Î”ABD
âˆ´ Their corresponding parts are equal.
â‡’ AB = AC
â‡’ Î”ABC is an isosceles triangle.

Question 4. In the adjoining figure,
Find âˆ  A and âˆ  B.
Solution:

âˆ´ âˆ  A= âˆ  B,
Since, âˆ  A + âˆ  D + âˆ  AOD = 180Â°
â‡’ âˆ  A + 50Â° + 40Â° = 180Â°
â‡’ âˆ  A = 180Â° - 50Â° - 40Â°
= 90Â°
âˆ´ âˆ  B= âˆ  A = 90Â°
Thus, âˆ  A = 90Â° and âˆ  B = 90Â°.

Question 5. In the figure, AD âŠ¥ BC and AB = AC. Find âˆ  B.
Solution:
âˆµ AD âŠ¥ BC and âˆ  C = 55Â°             [Given]

Hypt. AB = Hypt. AC             [Given]
â‡’ Î”ABD â‰Œ Î”ACD             [RHS criteria]
â‡’ âˆ  B= âˆ  C
â‡’ âˆ  B = 55Â°

Question 6. In the adjoining figure, AB = BC = AC, then find the measure of âˆ  A.
Solution:
âˆµ AB = BC = AC
âˆ´ Î”ABC is an equilateral triangle.
â‡’ âˆ  A= âˆ  B = âˆ  C
âˆ´ âˆ  A + âˆ  B + âˆ  C = 180Â°
â‡’ âˆ  A + âˆ  A + âˆ  A = 180Â°
â‡’ 3âˆ  A = 180Â°
â‡’ âˆ  A =(1800/3)= 60Â°

Thus, the required measure of âˆ A is 60Â°.

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

â‡’ Largest angle is 80Â°,
i.e. âˆ A = 80Â°
âˆ´ Side opposite of âˆ A is the longest.
â‡’ BC is the longest side of Î”ABC
Again âˆ C is the smallest angle
âˆ´ Side opposite of âˆ C is the smallest. i.e. AB is the smallest side of Î”ABC.

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 âˆ  BCP = 90Â°.
Solution:
âˆµ AB = AC             [Given]
âˆ´ The angles opposite to AB and AC are equal.

â‡’ âˆ  ABC = âˆ  ACB             ...(1)
Also, AC = AP             [Given]
â‡’ âˆ  APC = âˆ  ACP             ...(2)
Adding (1) and (2), we get
âˆ  ABC + âˆ  APC = âˆ  ACB + âˆ  ACP
â‡’ âˆ  ABC + âˆ  APC = âˆ  BCP
â‡’ âˆ  PBC + âˆ  BPC = âˆ  BCP               [âˆµ âˆ  ABC = âˆ  PBC and âˆ  APC = âˆ  BPC]
â‡’ âˆ  PBC + âˆ  BPC + âˆ  BCP
= âˆ  BCP + âˆ  BCP            [Adding âˆ  BCP to both sides]
â‡’ 180Â° = 2âˆ  BCP
â‡’ âˆ  BCP =(1800/2)= 90Â°
Thus, âˆ  BCP = 90Â°.

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 âˆ  A = (1/2)âˆ COB.
Solution:
âˆµ O is the centre of the circle.
âˆ´ OA = OC [Radii of the same circle]
Now, in Î”OAC,
âˆ  OAC = âˆ  OCA             [Angles opposite to equal sides are equal.]
âˆ  OAC + âˆ  OCB = âˆ  COB
[âˆµ Exterior angle is equal to sum of interior opposite angles.]

â‡’ âˆ  OAC + âˆ  OAC = âˆ  COB
â‡’ 2âˆ  OAC = âˆ  COB
â‡’ âˆ  OAC = (1/2)âˆ COB
Thus, âˆ  A= (1/2)âˆ COB

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

âˆ´ âˆ  A = 180Â° - (âˆ  B + âˆ  C)
â‡’ âˆ  A = 180Â°- (40Â° + 50Â°)
â‡’ âˆ  A = 180Â° - 90Â° = 90Â°
Since AX bisects âˆ BAC.
âˆ´ âˆ  BAX = 45Â° = âˆ  CAX
Now, âˆ  BAX > âˆ  ABX             [âˆµ 45Â° > 40Â°]
â‡’ BX > AX             ...(1)
Also âˆ  ACX > âˆ  CAX
â‡’ AX > CX             ...(2)
From (1) and (2), we have
BX > AX > CX
Thus, the required ascending order of AX, BX and CX is

Question 11. In the given figure, AD is the bisector of âˆ  A of  Î”ABC, where D lies on BC. Show that AB > BD and AC > CD.
Solution: In Î”ABC, âˆµ AD is bisector of âˆ  A

In Î”ABD,
â‡’ AC > CD             [Side opposite to greater angle in a D is greater]
â‡’ 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 external angle adjacent to âˆ  ABC is equal to âˆ  BOC.
Solution:
In Î”ABC, AB = AC
â‡’ âˆ B = âˆ  C             [âˆ´ Angles opposite to equal sides are equal]

â‡’ (1/2)âˆ B = (1/2)âˆ C
â‡’ âˆ  OBC = âˆ  OCB Now, in Î”BOC,
âˆ  BOC = 180Â° - [âˆ  OBC + âˆ  OCB]
= 180Â° - (1/2) [âˆ B + âˆ C]
= 180Â° - (1/2) [âˆ  B + âˆ  B] [âˆµ âˆ  B = âˆ  C]
= 180Â° - âˆ  B ... (1)
âˆµ âˆ  ABD + âˆ  B = 180Â°
âˆ´ âˆ ABD = 180Â° - âˆ  B ... (2)
From (1) and (2), we have âˆ  BOC = âˆ  ABD
But âˆ ABD is external angle adjacent to âˆ ABC
Thus, [external angle adjacent to âˆ ABC] = âˆ 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 Î”ABC, âˆµ AD is a median

âˆ´ BD = DC
Join CE.
In Î”ABD and Î”ECD

BD = DC             [AD is a median]
âˆ  1 = âˆ  2             [Vertically opp. angles]
â‡’ Î”ABD â‰Œ Î”ECD [SAS]
âˆ´ AB = CE             ... (1)
In Î”AEC, (AC + CE) > AE             [âˆµ In a D sum of any two sides is greater than the third side]
â‡’ (AC + AB) > AE   [From (1)]
â‡’ (AC + AB) > AD + DE
â‡’ AC + AB > 2 AD
â‡’ [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 Î”PQR. Prove that PQ + QR + RP > 2 PS.

Solution: âˆµ In a D, the sum of any two sides is greater than the third side.
âˆ´ In Î”PQS, (PQ + QS) > PS             ... (1)
Similarly, in Î”PRS, (PR + SR) > PS             ... (2)
[(PQ + QS) + (PR + SR)] > 2 PS
â‡’ PQ + (QS + SR) + PR > 2 PS
â‡’ PQ + QR + RP > 2 PS

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