Solution

CLASS IX WORKSHEET TRIANGLES

Class 09 - Mathematics

1.
(b) 50°
Explanation:
Given: △ABC, AB = AC and ∠ B = 50°.

As AB = AC
So it is an isosceles triangle.
So ∠ B = ∠ C
∠ B = 50° (given)

⇒ ∠ C = 50°

2.
(d) Isosceles but not necessarily congruent
Explanation:
In △DEF, DE = DF. So, △DEF is an isosceles.

∴ ∠ F = ∠E
Also, ∠ F = ∠ P and ∠ E = ∠ Q
From (i) and (ii), we get ∠ P = ∠ Q
Now, in △PQR, ∠ P = ∠ Q ⇒ RQ = PR
So, △PQR is an isosceles.
Hence, △DEF and △PQR are isosceles but not necessarily congruent.

3.
(c) PR
Explanation:
Since, by corresponding part of congruent triangle ED of △EFD is equal to the PR of △PQR.

4.
(c) SSA
Explanation:
The criteria for congruence of triangles are SSS(Side-Side-Side), SAS (Side-Angle-Side), ASA (Angle-Side- Angle) and RHS
(Right angle- Hypotenuse-Side)

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5.
(d) 15°
Explanation:
∠ PRS = (90° + 60°) = 150°
RQ = PR ... (i) [Sides of equilateral △]
and RQ = RS ... (ii) [Sides of square]
∴ RP = RS [from (i) and (ii)]

⇒ ∠ RPS = ∠ RSP = x

In △PSR, ∠ PRS + ∠ SPR + ∠ RSP = 180°

⇒ 150° + x + x = 180° => 2x = 30° => x = 15°

6.
(b) BC = PQ
Explanation:
According to the condition given in the question,
If △ABC ≅△P QR and △ABC is not congruent to △RP Q
Then, clearly BC ≠ PQ
∴ It is false

7.
(b) SAS
Explanation:
According to SAS criterion, if the corresponding sides and their included angles are equal, then the triangles are congruent.
Here, in △AOC and △XYZ, AO = XY, and AC = XZ are the corresponding sides and ∠A = ∠X are included angles, Hence,
△AOC ≅△XY Z , by SAS.

8.
∘
(d) 72 1

Explanation:

∠ C = 180° - ∠ ACD = 180° - 115° = 65°

In △ABC
∠ A + ∠ B + ∠ C = 180°
⇒ ∠ A = 180 - 30° - 65°

⇒ ∠ A = 85°

Now in △ALC
∠ ALC + ∠ LAC + ∠ C = 180°

⇒ ∠ ALC = 180° - ∠ LAC - ∠ C

= 180 ∘
−
∠A

2
− ∠C
∘

= 180 ∘
−
85

2
∘
− 65
∘

= 145

2
∘
= 72 1

9. (a) 40°
Explanation:

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 ∠ABC = 180° - 126° = 54°
∠ACB = 180° - 94° = 86°
Now, in △ABC
∠ BAC + ∠ ABC + ∠ ACB = 180^{\circ}
⇒ ∠ BAC = 180° - ∠ ABC - ∠ ACB
= 180° - 54° - 86°
⇒ ∠ BAC = 40°

10.
(b) 4 cm
Explanation:
In a triangle, if two of its angles are equal then the sides opposite to equal angles are also equal.
In △PQR, ∠ R = ∠ P
⇒ QR (side opposite to ∠ P) = PQ (side opposite to ∠ R)

Given that, QR = 4 cm
⇒ PQ = 4 cm

11.
(d) A is false but R is true.
Explanation:
A is false but R is true.

12. (a) Both A and R are true and R is the correct explanation of A.
Explanation:
In ΔACB and ΔADB
AC = AD [Given]
∠ CAB = ∠ DAB [AB bisects ∠ A]

AB = AB [Common]
ΔACB ≅ ΔADB [By SAS congruence rule]

13.
(c) A is true but R is false.
Explanation:
A is true but R is false.

14. (a) Both A and R are true and R is the correct explanation of A.
Explanation:

In ΔABC and ΔPQR

∠A = ∠P , ∠B = ∠Q, ∠C = ∠R

But sides are different

There is no rule of congruence as AAA.
△ABC ≆ △P QR

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15.
(c) A is true but R is false.
Explanation:
A is true but R is false.

16. Given : l || m and p || q

To prove : DABC ≅ DCDA
Proof : l || m and p || q . . . . [Given]
In DABC and DCDA
∠ BAC = ∠ DCA . . . . . [Alternate interior angles as AB || DC]

Similarly, ∠ ACB = ∠ CAD . . . [Alternate interior angles as BC || DA]

AC = AC . . . [Common]
DABC ≅ DCDA [By ASA congruency]
17. In ΔAM C and ΔBM D, we have
∠1 = ∠3 [Alt. ∠s because l||m]

∠2 = ∠4 [Vert. opp. ∠s ]

AM = BM [Given]
∴ ΔAM C ≅ΔBM D [By ASS congruence rule]
∴ C M = DM [CPCT]
Hence, M is also the mid-point of CD.
18. In the given △ABC
∠A, ∠ B and ∠ C satisfy the relation B - A = C - B
We need to fine the measure of ∠ B

As
B-A=C-B
B+B=C+A
2B = C + A
2B - A = C ...(i)
Now, using the angle sum property of the triangle, we get,
A + B + C = 180o
2B - A + A + B = 180o (Using i)
3B = 180o
= 60o
∘
180
B= 3

Therefore, ∠ B = 60o
19. In △ABE and △ACF,
∠ A = ∠ A [Common]

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 ∠AEB = ∠ AFC = [90°]
AB = AC [Given]
∴ △ABE ≅ △ACF [By ASA congruency]

BE = CF [By C.P.C.T.]
∴

So Altitudes are equal.

20. BC || AD (Given)
Therefore, ∠ CBO = ∠ DAO (Alternate interior angles)
and ∠ BCO = ∠ ADO (Alternate interior angles)
Also, BC = DA (Given)
So, △ BOC ≅△ AOD (ASA)
Therefore, OB = OA and OC = OD, i.e., O is the mid-point of both AB and CD.
21.
In △ACP, we have
∘
∠P AC + ∠AC P + ∠C P A = 180

∘ ∘ ∘
⇒ 45 + 100 + ∠C P A = 180
∘ ∘ ∘
⇒ ∠C P A = (180 − 145 ) = 35

∴ ∠DP B = ∠C P A = 35
∘
[vertically opp. △]
In △P BD we have
∠DP B + ∠P BD + ∠BDP = 180
∘
[sum of the angles of a triangle]
∘ ∘ ∘
⇒ 35 + 65 + ∠BDP = 180
∘ ∘
⇒ 100 + ∠BDP = 180

∘ ∘ ∘
⇒ ∠BDP = (180 − 100 ) = 80

Hence, ∠C P A = 35 , ∠DP B = 35 and ∠BDP

∘ ∘
= 80
∘
.
22. Given: An equilateral triangle ABC.
To Prove : ∠ A = ∠ B = ∠ C = 60 o
Proof : ABC is an equilateral triangle.
∴ AB = BC = CA . . . .(1)

As AB = BC
∠ A = ∠ C . . . [∠ s opposite to equal side of a △] . . . (2)

As BC = CA
∴ ∠ A = ∠ B . . . [∠ s opposite to equal side of a △] . . . (3)
∠ A = ∠ B = ∠ C . . .[From (2) and (3)] . . . (4)

In △ABC,
∠ A + ∠ B + ∠ C = 180o . . . [Sum of three angles of a triangle] . . . (5)
Let ∠ A = xo, Then ∠ B = ∠ C = xo . . .[From (4)]
xo + xo + xo = 180o
∴
3xo = 180o
∴ xo = 60o
∠ A = ∠ B = ∠ C = 60o

23.

Produce AD to E such that AD = DE. Join EC.

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 BD = DC (given)
AD = DE (by construction)
∠ADB = ∠EDC

By SAS(Side-Angle-Side)
∴ ΔABD ≅ΔEC D

∴ AB = EC and ∠1 = ∠3
Also ∠1 = ∠2 [∵ AD bisects ∠A]

∴ ∠2 = ∠3

But, we know that the sides opposite to equal angles of a triangle are equal.
EC = AC.
so, AB = AC
Hence, ABC is isoceles
24. In △ABC and △ACE, we have
∠ADB = ∠AEC = 90 [Given]
∘

∠BAD = ∠C AE [Common angle]

and, BD = CE [Given]
So, by ASA(Angle-Side-Angle) congruence criterion, we obtain
ΔABD ≅ΔAC E

⇒ AB = AC [∵ Corresponding parts of congruent triangles are equal]

Hence, △ABC is isosceles.

25.

Given: ABC is an isosceles triangle with AB = AC.

To Prove : ∠ B = ∠ C
Construction: Draw AP ⊥ BC
Proof: In right triangle APB and right triangle APC.
AB = AC . . . . [given]
AP = AP . . . .[Common]
∴ △APB ≅ △APC . . . [RHS rule]

∴ ∠ ABP = ∠ ACP . . .[c.p.c.t.]

∴ ∠B = ∠C

26. Bisector of the angles B and C of an isosceles triangle ABC with AB = AC intersect each other at O. BO is produced to a point
M.Figure is drawn below.

In ΔABC , we have
AC = AB. [ Given ]
∴ ∠ABC = ∠AC B [∵ Angles opposite to equal sides of a triangle are equal]
Dividing both sides by 2, we get:-
1 1
⇒ ∠ABC = ∠AC B
2 2

⇒ ∠OBC = ∠OC B ∵ [ BO and CO are bisectors of ∠B and ∠C ]

⇒ ∠1 = ∠2 (from figure)......(1)

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 In ΔOBC , Exterior angle ∠M OC = ∠1 + ∠2
[∵ Exterior angle of a triangle is equal to the sum of interior opposite angles]
⇒ ∠M OC = 2∠1 [Since ∠1 = ∠2 ] .............. (2)
1
Also ,from figure, ∠1 = 2
∠ABC .........(3)∵ Angles opposite to equal sides of a triangle are equal]
From (2) & (3):-
∠M OC = 2 ×
1

2
∠ABC ∵ Angles opposite to equal sides of a triangle are equal]
⇒ ∠M OC =
1

2
∠ABC .
Hence Proved.
27. i. In △APD and △BQC
AD = BC (given)
AP = CQ (opposite sides of rectangle)
∠APD = ∠ BQC = 90o
By RHS criteria △APD ≅ △CQB
ii. △APD ≅ △CQB
Corresponding part of congruent triangle
side PD = side BQ
iii. In △ABC and △CDA
AB = CD (given)
BC = AD (given)
AC = AC (common)
By SSS criteria △ABC ≅ △CDA
OR
In △APD
∠ APD + ∠ PAD + ∠ ADP = 180o
⇒ 90o + (180o - 110o) + ∠ ADP = 180o (angle sum property of △)
⇒ ∠ ADP = m = 180o - 90o - 70o = 20o
∠ ADP = m = 20o
28. i. △ADE and △CFE
DE = EF (By construction)
∠ AED = ∠ CEF (Vertically opposite angles)

AE = EC(By construction)
By SAS criteria △ADE ≅△CFE
ii. △ADE ≅ △CFE
Corresponding part of congruent triangle are equal
∠ EFC = ∠ EDA

alternate interior angles are equal

⇒ AD ∥ FC
⇒ CF ∥ AB
iii. △ADE ≅ △CFE
Corresponding part of congruent triangle are equal.
CF = AD
We know that D is mid point AB
⇒ AD = BD

⇒ CF = BD

OR
BC
DE = 2
{line drawn from mid points of 2 sides of △ is parallel and half of third side}
DE ∥ BC and DF ∥ BC
DF = DE + EF
⇒ DF = 2DE(BE = EF)
⇒ DF = BC

29. In △ABC,
BP bisects ∠ ABC

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 ∴ ∠ ABP = ∠ PBC = 1

2
∠ B = (2∠ C) = ∠ C . . . . (1)
1

In PBC,
∴ ∠ PBC = ∠ PCB (= ∠C)

∴ PB = PC . . . [Sides opposite to equal angles] . . .(2)

In △APB and △DPC,

AB = CD . . . [Given]
PB = PC . . .[From (2)]
∠ ABP = ∠ DCP (= ∠ C)

∴ △APB = △DPC . . . . [By SAS property]

∴ ∠ BAP = CDP (= ∠ A) . . .[c.p.c.t.] . . . (3)

and AP = DP . . . [c.p.c.t.] . . . .(4)

In △APD,
As AP = DC . . . [From (4)]
∴ ∠ PDA = ∠ PAD =
A

∴ ∠ DPA = π − ( A

2
+
A

2
) = π − A . . . . (5)
From ∠ DPC
∠ DPC = π – (A + C)

∴ ∠ DPA = π – ∠ DPC = π – [π – (A + C)] = A + C . . .(6)

From (5) and (6)

π – A = A + C ∴ 2A + C = π . . . (7)

Again
A + B + C = π . . .[Sum of three angles of a triangle = π ]
∴ A + 2C + C = π . . .[As B = 2C]

∴ A + 3C = π . . . (8)

Multiplying (7) by 3, we get

6A + 3C = 3π
5A = 2π . . .[By subtracting (8) from (9)]
2π 2 ∘ ∘
∴ A= = × 180 = 72
5 5

∴ ∠ BAC = 72o
30. In △ ABC, ∠ B = 60, ∠ C = 80 and the bisectors of ∠ B and ∠ C meet at O.
We need to find the measure of ∠ BOC

Since, BO is the bisector of ∠ B

1
∠ OBC = 2
∠B

= 1

2
∘
(60 ) = 30 ∘

Similarly, CO is the bisector of ∠ C

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 1
∠OC B ∠C
2

= 1

2
(60 )
∘

= 40 ∘

Now, applying angle sum property of the triangle, in △BOC , we get,

∠ OCB + ∠ OBC+ ∠ BOC = 180
∘

∘ ∘ ∘
30 + 40 + ∠BOC = 180

∘ ∘
∠BOC + 70 = 180

BOC = 180 ∘
− 70
∘

= 110 ∘

Therefore, ∠ BOC = 110 ∘

31.

As AB = AC
∴ ∠ ACB = ∠ ABC . . . . [Angles opposite to equal sides of triangle]

∴ 2∠ 2 = 2∠ 1

As CE and BD are the bisectors of ∠ C and ∠ B respectively.

∠2 = ∠1

BP = PC . . . [Side opposite to equal angles of triangle] . . .(1)

In △BPE and △CPD,
BP = CP . . .[Proved as above]
∠ EBP = ∠ DCP . . . [Proved as above]

∠BPE = ∠ CPD . . .[Vertically opposite angles]

△BPE ≅△CPD. . .[By ASA property]

PE = PD . . .[c.p.c.t.]
PD = PE . . . . (2)
BP + PD = PC + PD . . . .[Adding (1) and (2)]
BD = CE
∴

32. The point which is equidistant from all the sides of a triangle is known as its incentre and it is also the point of intersection of the
angular bisectors of angles of the triangle. Hence we will proceed with finding the incentre of the given triangle.
Let ABC be a triangle.

Draw bisectors of ∠ B and ∠ C.

Let these angular bisectors intersect each other at point I.
Draw IK ⊥ BC
Also, draw IJ ⊥ AB and IL ⊥ AC.
Join BI & CI
In ∆ BIK and ∆ BIJ,
∠ IKB = ∠ IJB = 90° [By construction]

∠ IBK = ∠ IBJ

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[ ∵ BI is the bisector of ∠ B (By construction)]
BI = BI [Common side]
∴ ΔBIK ≅ΔBIJ [ASA congruency criterion of triangle]

∴ IK = IJ [C.P.C.T.] ……….(i)
Now, in ∆CIK and ∆CIL,
∠I K C = ∠I LC (each 90°)

∠I C K = ∠I C L [ since, CI is angular bisector of ∠C ]

CI = CI [ common side ]
Hence, ΔClK ≅ΔClL [ ASA congruency criterion of triangle]
∴ IK = IL [ C.P.C.T.] ……….(ii)

From eq (i) and (ii),

IK = IJ = IL
Hence, I is the point of intersection of angular bisectors of any two angles of ∆ ABC and is also equidistant from the sides of the
triangle.

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