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## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 4 Geometry Ex 4.1

Question 1.

In the figure, AB is parallel to CD, find x

Solution:

(i) From the figure

∠1 = 140° (∴ corresponding angles are equal)

∠2 = 40° (∴ ∠1 + ∠2= 180°)

∠3 = 30° (∵ ∠3 + 150= 180°)

∠4 = 110° (∵ ∠2 + ∠3 + ∠4 = 180°)

∴ ∠x = 70° (∵ ∠4 + ∠x = 180°)

(ii) From the figure

∠1 = 48°

∠3 = 108° (∠1 +24° + ∠3 = 180°)

∠4 = 108° (If two lines are intersect, then the vertically the opposite angles are equal)

∠5 = 72° (∵ ∠3 + ∠5 = 180°)

∴ ∠3 + ∠4 + ∠5 = 108° + 108° + 72°

x = 288°

(iii) From the figure

∠D = 53° ( ∵ ∠B and ∠D are alternate interior angles)

Sum of the three angles of a triangle is 180°

∠x° = 180°- (38°+ 53°)

= 180°- 91° = 89°

Question 2.

The angles of a triangle are in the ratio 1 : 2 : 3, find the measure of each angle of the triangle.

Solution:

Let the angles be x, 2x and 3x respectively.

Sum of the three angles of a triangle = 180°

The 3 angles of the triangle are 30°, 60°, 90°.

Question 3.

Consider the given pairs of triangles and say whether each pair is that of congruent triangles. If the triangles are congruent, say ‘how’; if they are not congruent say ‘why’ and also say if a small modification would make them congruent:

Solution:

(i) Consider ∆PQR and ∆ABC

Given, RQ = BC

PQ = AB

∆ABC is not congruent to ∆PQR

If PR = AC, then ∆ABC ≅ ∆PQR

(ii) Consider ∆ABD and ∆BCD for the triangles to be congruent.

Given, AB = DC

AD = BC and AB is common side.

∴ By SSS rule ∆ABD ≅ ∆BCD.

(iii) Consider ∆PXY and ∆PXz,

Given, XY = XZ

PY = PZ and PX is common

∴ By SSS rule ∆PXY ≅ ∆PXZ.

(iv) Consider ∆OAB and ∆ODC,

Given, OA = OC

∠ABO = ∠ODC and ∠AOB = ∠DOC (vertically opposite angles)

∴ By AAS rule, AOAB = AODC.

(v) Consider ∆AOB and ∆DOC,

Given, AO = OC

OB = OD

and ∠AOB = ∠DOC [vertically opposite angles]

∴ By SAS rule, ∆AOB = ∆DOC.

(vi) Consider ∆AMB and ∆AMC,

Given, AB = AC

∠AMB = ∠AMC = 90°

∴ AM is common.

∴ By RHS rule

∆AMB ≅ ∆AMC.

Question 4.

∆ABC and ∆DEF are two triangles in which AB = DF, ∠ACB = 70°, ∠ABC = 60°; ∠DEF = 70° and ∠EDF = 60°. Prove that the triangles are congruent.

Solution:

∴ By ASA rule ∆ABC ≅ ∆FDE

Question 5.

Find all the three angles of the ∆ABC

Solution:

Exterior angle = Sum of the two opposite interior angles.