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.

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**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=(180

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]

AD = AD [Common]

âˆ ADB = âˆ ADC

[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,**

Î”**OAD â‰Œ Î”OBC. Find âˆ A and âˆ B. Solution:** Since, Î”OAD â‰Œ Î”OBC

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]

âˆ´ In right Î”ADB and right Î”ADC,

AD = AD [Common]

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 =(180

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 =(180^{0}/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**

âˆ´ âˆ BAD = âˆ CAD

In Î”ABD,

Exterior âˆ ADC > âˆ BAD

â‡’ âˆ ADC > âˆ CAD

Now, in Î”ADC,

âˆ ADC > âˆ CAD

â‡’ AC > CD [Side opposite to greater angle in a D is greater]

Also, in Î”ABD, âˆ ADB > âˆ BAD [âˆµ Ext. âˆ ADB is greater than âˆ CAD and âˆ CAD = âˆ BAD]

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

Produce AD to E, such that AD = DE

Join CE.

In Î”ABD and Î”ECD

BD = DC [AD is a median]

AD = DE [Construction]

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) > AD + AD [âˆµ AD = DE by construction]

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)

Adding (1) and (2),

[(PQ + QS) + (PR + SR)] > 2 PS

â‡’ PQ + (QS + SR) + PR > 2 PS

â‡’ PQ + QR + RP > 2 PS

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