# Samacheer Kalvi Class 7 Maths Term 2 Chapter 4 Solutions

## Samacheer Kalvi Class 7 Maths Term 2 Chapter 4 Geometry Solutions

Welcome to NCTB Solutions. Here with this post we are going to help 7th class students by providing Solutions for Samacheer Kalvi Class 7 Maths Term 2 chapter 4 Geometry. Here students can easily find all the solutions for Geometry Exercise 4.1, 4.2 and 4.3. Also here our Expert Maths Teacher’s solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 4 solutions. Here all the solutions are based on Tamil Nadu State Board latest syllabus.

#### Geometry Exercise 4.1 Solutions :

(1) Can 30°, 60° and 90° be the angles of a triangle

Solution :

Given angles 30°, 60° and 90°.

Sum of the angles

= 30° + 60° + 90°

= 180°

∴ The given angles can form a triangle.

(2) Can you draw a triangle with 25°, 65° and 80° as angles

Solution :

Given, angles 25°, 65° and 80°.

Sum of the angles

= 25° + 65° + 80°

= 170° ≠ 180°

The given angles can not form a triangle.

(3) In each of the following triangles, find the value of x.

Solution :

Figure – (i)

In ΔEFG, ∠E = 80° and ∠F = 55°

Let ∠G = x

We know, ∠E + ∠F + ∠G = 180° …. (Angel sum property)

80° + 55° + x= 180°

135° + x = 180°

x = 180° – 135°

x = 45°

Figure – (ii)

In ΔMNO , ∠N = 96° and ∠O = 22°

Let ∠M = x

We know, ∠M + ∠N + ∠O = 180° …. (Angel sum property)

x + 96° + 22° = 180°

x + 118° = 180°

x = 180° – 118°

x = 62°

Figure – (iii)

In ΔXYZ , ∠X = 29° and ∠Y = 90°

Let ∠Z = (2x+1)

We know, ∠X + ∠Y + ∠Z = 180° …. (Angel sum property)

29° + 90° + 2x +1 = 180°

120° + 2x = 180°

2x = 180° – 120°

2x = 60

x = 60/2

x = 30

Figure – (iv)

In ΔJKL, ∠K = 112° , ∠L = 3x and ∠J = x

We know, ∠K + ∠L + ∠J = 180° …. (Angel sum property)

112° + 3x + x = 180

4x = 180 – 112

4x = 68

x = 68/4

x = 17

Figure – (v)

In ΔRST , ∠R = 72° , ∠S = 3x and

SR = RT = 4.5 cm

Thus, ∠S = ∠T = 3x … [Angles opposite to equal sides]

We know, ∠R + ∠S + ∠T = 180° …. (Angel sum property)

72 + 3x + 3x = 180

6x = 180 – 72

6x = 108

x = 108/6

x = 18

Figure – (vi)

In Δ XYZ, ∠X = 3x , ∠Y = 2x and ∠Z = 4x

We know, ∠X + ∠Y +∠Z = 180°

3x + 2x + 4x = 180

9x = 180

x = 180/9

x = 20

Figure – (vii)

In ΔTUV, ∠T = x-4 , ∠U = 90 and ∠V = 3x-2

We know, ∠T + ∠U + ∠V = 180

x-4 + 90 + 3x – 2 = 180

4x + 84 = 180

4x = 180 – 84

4x = 96

x = 96/4

x = 24

Figure – (viii)

In ΔONP, ∠O = 3x – 10, ∠N = x+31 and ∠P = 2x-3

We know, ∠O + ∠N + ∠P = 180

3x – 10 + x + 31 + 2x -3 = 180

6x + 18 = 180

6x = 180 – 18

6x = 162

x = 162/6

x = 27

(4) Two line segments AD and BC intersect at O. Joining AB and DC we get two triangles, AOB and DOC as shown in the figure. Find the A and B

Solution :

From fig, In ΔDOC, ∠ D = 70° and ∠C = 30°

We know, ∠DOC + ∠ODC + ∠OCD = 180°

∠DOC + 70° + 30° = 180°

∠DOC + 100° = 180°

∠DOC = 180° – 100° = 80°

Since, ∠DOC and ∠BOA are vertical opposite angel and it’s equal.

Thus, ∠BOA = 80°

Now, In ΔAOB, ∠BOA = 80°, ∠BAO= 3x and ∠OBA = 2x

We know, ∠BOA + ∠BAO + ∠OBA = 180°

80 + 3x + 2x = 180°

5x = 180 – 80

5x = 100

x = 100/5

x = 20

Therefore, ∠BAO= 3x = 3 × 20 = 60°

∠OBA = 2x = 2 × 20 = 40°

(6) If the three angles of a triangle are in the ratio 3 : 5 : 4 , then find them.

Solution :

Given, the three angles of a triangle are in the ratio 3 : 5 : 4

Let this angles are 3x , 5x and 4x.

Now, 3x + 5x + 4x = 180° ….( angel sum property )

12x = 180°

x = 180/12

x = 15

Thus, 3x = 3×15 = 45°

5x = 5 × 15 = 75°

4x = 4 × 15 = 60°

Hence, the angles are 45° , 75° and 60°

(7) In ∆RST, S is 10° greater than R and T is 5° less than S, find the three angles of the triangle.

Solution :

In ΔRST, ∠S is 10° greater than ∠R and ∠T is 5° less than ∠S.

Let ∠R be x.

Thus, ∠S = x + 10 and ∠T = x+10 – 5 = x + 5

Therefore,

∠R + ∠S + ∠T = 180°

x + x + 10 + x + 5 = 180°

3x + 15 = 180°

3x = 180 -15 = 165

x = 165/3

x = 55

Therefore, ∠R = x = 55°

∠S = x + 10 = 55 + 10 = 65°

∠T = x + 5 = 55 + 5 = 60°

(8) In ∆ABC, if B is 3 times A and C is 2 times A, then find the angles

Solution :

Given, In ΔABC, if ∠B is 3 times ∠A and ∠C is 2 times ∠A

Let ∠A = x, thus ∠B = 3x and ∠C = 2x .

Now, ∠A + ∠B + ∠C = 180°

x + 3x + 2x = 180°

6x = 180°

x = 180/6

x = 30°

∠A = x = 30°

∠B = 3x = 3 × 30 = 90°

∠C = 2x = 2 × 30 = 60°

(9) In ∆XYZ, if X : Z is 5 : 4 and Y=72°. find X and Z

Solution :

Given, In ΔXYZ, if ∠X: ∠Z is 5 : 4 and ∠Y= 72°

Let ∠X = 5x and ∠Z = 4x

Now, ∠X + ∠Y + ∠Z = 180°

5x + 72 + 4x = 180

9x + 72 = 180

9x = 180 – 72

9x = 108

x = 108/9

x = 12

Therefore,

∠X = 5x = 5 × 12 = 60°

∠Z = 4x = 4 × 12 = 48°

(10) In a right angled triangle ABC, B is right angle, A is x + 1 and C is 2x + 5. find A and C

Solution :

Given, In a right angled triangle ABC, ∠B is right angle

Thus, ∠B = 90, ∠ A = x+ 1 and ∠C = 2x +5

Now, ∠ A + ∠B + ∠C = 180

x + 1 + 90 + 2x + 5 = 180

3x + 96 = 180

3x = 180 – 96

3x = 84

x = 84/3

x = 28

Therefore,

∠ A = x+ 1 = 28 + 1 = 29°

∠C = 2x + 5 = 2*28 + 5 = 56 + 5 = 61°

(11) In a right angled triangle MNO, N = 90°, MO is extended to P. if NOP = 128°, find the other two angles of ∆MNO.

Solution :

Given, In a right angled triangle MNO, ∠N = 90°, MO is extended to P.

Thus, ∠NOP is external angle of ΔMNO

We know,

Exterior angle = sum of two interior opposite angles

∠NOP = ∠N + ∠M

128 = 90 + ∠M

∠M = 128 – 90

∠M = 38°

Now, ∠N + ∠M + ∠O = 180° …. (Angle sum property)

90° + 38° + ∠O = 180°

∠O = 180° – 128°

∠O = 52°

Hence, the other two angles of ΔMNO is 52° and 38°.

(12) Find the value of x in each of the given triangles.

Solution :

Figure – (i)

From fig, ∠ACB and ∠ LCB are linear pair angle

Thus, ∠ACB + ∠ LCB = 180°

∠ACB + 135° = 180° …. (Given, ∠LCB = 135° )

∠ACB = 180° – 135°

∠ACB = 45°

Since, ∠x is external angle of ΔABC

Thus, Exterior angle = sum of two interior opposite angles

x = ∠ACB + ∠ABC

x = 45° + 65°

x = 110°

Figure – (ii)

From fig, ∠XAZ and ∠BAC are opposite angel and it’s equal.

Thus, ∠BAC = ∠XAZ = 8x+7 … (1)

Since, ∠ACY is external angle of ΔABC

Thus, Exterior angle = sum of two interior opposite angles

∠ACY = ∠BAC + ∠ABC

120 = 8x+7 + 3x – 8

120 = 11x – 1

11x = 120 + 1

11x = 121

x = 121/11

x = 11°

(13) In ∆LMN, MN is extended to O. if MLN = 100 – x, LMN = 2x and LNO = 6x – 5, find the value of x

Solution :

Given, In ΔLMN, MN is extended O.

Thus, ∠LNO is external angle of ΔLMN.

We know, Exterior angle = sum of two interior opposite angles

∠LNO = ∠LMN + ∠MLN

6x – 5 = 2x + 100 – x

6x – 2x + x = 100 + 5

5x = 105

x = 105/5

x = 21°

So, value of x is 21°

(14) Using the given figure find the value of x.

Solution :

From fig, ∠BEC is external angle of ΔCDE .

We know, Exterior angle = sum of two interior opposite angles

∠BEC = ∠ECD + ∠EDC

x = 50 + 60

x = 110°

∴ The value of x is 110°.

(15) Using the diagram find the value of x.

Solution :

Since in a diagram all sides of the triangle are equal thus this triangle is equilateral triangle.

We know angles of equilateral triangle is 60.

Also, x be a external angle of the triangle.

We know, Exterior angle = sum of two interior opposite angles

x = 60 + 60

x = 120°

Objective Type Question Solutions :

(16) The angles of a triangle are in the ratio 2 : 3 : 4. Then the angles are

(i) 20, 30, 40

(ii) 40, 60, 80

(iii) 80, 20, 80

(iv) 10, 15, 20

Solution :

Correct Option → (ii)

The angles of a triangle are in the ratio 2:3:4. then the angles are 40, 60, 80.

(17) One of the angles of a triangle is 65°. if the difference of the other two angles is 45°, then the two angles are

(i) 85°, 40°

(ii)70°, 25°

(iii) 80°, 35°

(iv) 80°, 135°

Solution :

Correct Option → (iii)

One of the angles of a triangle is 65°. if the difference of the other two angles is 45°, then the two angles are 80°, 35°.

(18) In the given figure, AB is parallel to CD. Then the value of b is

(i) 112°

(ii) 68°

(iii) 102°

(iv) 62°

Solution :

Correct Option → (ii)

In the given figure, AB is parallel to CD. Then the value of b is 68°.

(19) In the given figure, which of the following statement is true

(i) x + y + z = 180°

(ii) x + y + z = a + b + c

(iii) x + y + z = 2(a + b + c)

(iv) x + y + z = 3(a + b + c)

Solution :

Correct Option → (iii)

x + y + z = 2(a + b + c) – This statement is True.

(20) An exterior angle of a triangle is 70° and two interior opposite angles are equal. then measure of each of these angle will be

(i) 110°

(ii) 120°

(iii) 35°

(iv) 60°

Solution :

Correct Option → (iii)

An exterior angle of a triangle is 70° and two interior opposite angles are equal. then measure of each of these angle will be 35°.

(21) In a ΔABC, AB = AC. the value of x is _____

(i) 80°

(ii) 100°

(iii) 130°

(iv) 120°

Solution :

Correct Option → (iii)

In a ΔABC, AB = AC. the value of x is 130°.

(22) If an exterior angle of a triangle is 115° and one of the interior opposite angles is 35°, then the other two angles of the triangle are

(i) 45°, 60°

(ii) 65°, 80°

(iii) 65°, 70°

(iv) 115°, 60°

Solution :

Correct Option → (ii)

Then the other two angles of the triangle are 65°, 80°

#### Geometry Exercise 4.2 Solutions :

(1) Given that ∆ABC ∆DEF (i) List all the corresponding congruent sides (ii) List all the corresponding congruent angles

Solution :

(i) Corresponding sides : AB, DE; BC, EF; AC, DF

(ii) Corresponding angles : ∠ABC, ∠DEF; ∠BCA, ∠EFD; ∠CAB, ∠FDE

(2) If the given two triangles are congruent, then identify all the corresponding sides and also write the congruent angles.

Solution :

Figure – (i)

Corresponding sides are,

PQ = LN,

PR = LM and RQ = MN

Congruent angles,

∠RPQ = ∠NLM,

∠PQR = ∠LNM and ∠PRQ = ∠LMN

Figure – (ii)

Corresponding sides are,

QR = LM , RP = LN and PQ = MN

Congruent angles,

∠ PQR = ∠LMN, ∠QRP = ∠MLN and ∠RPQ = ∠LNM

(3) If the given triangles ∆ABC and ∆EFG are congruent, determine whether the given pair of sides and angles are corresponding sides or corresponding angles or not

(i) Aand G

(ii) B and E

(iii) B and G

(iv) AC and GF

(v) BA and FG

(vi) EF and BC

Solution :

(i) ∠A and ∠G = Not corresponding angles

(ii) ∠B and ∠E = Not corresponding angles

(iii) ∠B and ∠G = corresponding angles

(iv) AC and GF = Not corresponding sides

(v) BA and FG = corresponding sides

(vi) EF and BC = Not corresponding sides

(6) For each pair of triangles state the criterion that can be used to determine the congruency?

Solution :

Figure → (i) By SSS

Figure → (ii) By ASA

Figure → (iii) By RHS

Figure → (iv) By ASA

Figure → (v) By ASA

Figure → (vi) By SAS

Objective Type Question Solutions :

(8) If two plane figures are congruent then they have

(i) same size

(ii) same shape

(iii) same angle

(iv) same shape and same size

Solution :

Correct Option → (iv)

If two plane figures are congruent then they have same shape and same size.

(9) Which of the following methods are used to check the congruence of plane figures?

(i) translation method

(ii) superposition method

(iii) substitution method

(iv) transposition method

Solution :

Correct Option → (ii)

Superposition method is used to check the congruence of plane figures.

(10) Which of the following rule is not sufficient to verify the congruency of two triangles.

(i) SSS rule

(ii) SAS rule

(iii) SSA rule

(iv) ASA rule

Solution :

Correct Option → (iii)

SSA rule is not sufficient to verify the congruency of two triangles.

(11) Two students drew a line segment each. What is the condition for them to be congruent?

(i) They should be drawn with a scale.

(ii) They should be drawn on the same sheet of paper.

(iii) They should have different lengths.

(iv) They should have the same length.

Solution :

Correct Option → (iv)

The condition for them to be congruent is they should have the same length.

(12) In the given figure, AD = CD and AB = CB. Identify the other three pairs that are equal

(i) ADB = CDB, ABD = CBD, BD = BD

(ii) AD = AB, DC = CB, BD = BD

(iii) AB = CD, AD = BC, BD = BD

(iv) ADB = CDB, ABD = CBD, DAB = DBC

Solution :

Correct Option → (i)

∠ADB = ∠CDB, ∠ABD = ∠CBD, BD = BD

(13) In ∆ABC and ∆PQR, A=50°=P, PQ=AB, and PR=AC. by which property ∆ABC and ∆PQR are congruent

(i) SSS property

(ii) SAS property

(iii) ASA property

(iv) RHS property

Solution :

Correct Option → (ii)

By SAS property ∆ABC and ∆PQR are congruent.

#### Geometry Exercise 4.3 Solutions :

(1) In an isosceles triangle one angle is 76°. If the other two angles are equal find them.

Solution :

Given, in an isosceles triangle one angle is 76°

Since, the other two angles are equal

Thus, x + x + 76 = 180 …. (Angel sum property)

2x = 180 – 76

2x = 104

x = 104/2

x = 52

Hence, the other two angles are 52°, 52°.

(2) If two angles of a triangle are 46° each, how can you classify the triangle?

Solution :

Given, two angles of a triangle are 46° each.

Therefore this triangle is isosceles triangle.

(3) If one angle of a triangle is equal to the sum of the other two angles, find the type of the triangle.

Solution :

Let x, y and z be angles of triangles

According to question, x = y + z … (1)

We know,

x + y + z = 180

x + x = 180 (from, 1)

2x = 180

x = 180 /2

x = 90

Since, one of the angel is 90

Therefore, this triangle is right angle triangle.

(4) If the exterior angle of a triangle is 140° and its interior opposite angles are equal, find all the interior angles of the triangle.

Solution :

Let x, y and z be the interior angles of the triangle

Given, the exterior angle of a triangle is 140° and its interior opposite angles are equal

Supposed y and z are equal.

We know,

Exterior angle = sum of two interior opposite angles

140 = y + z

140 = y+y

2y = 140

y = 140/2 =70

Thus, z = y = 70

Now, x + y + z = 180 …( Angles sum property )

x + 70 + 70 = 180

x = 180 – 140

x = 40

Hence, angles of triangles are 40°, 70° and 70°.

(5) In ∆JKL, if J = 60° and K = 40°, then find the value of exterior angle formed by extending the side KL.

Solution :

Given, n ΔJKL, if ∠J = 60° and ∠K = 40°

Since, exterior angle formed by extending the side KL

We know,

Exterior angle = sum of two interior opposite angles

= ∠J + ∠K

= 60° + 40°

= 100°

Therefore, the value of exterior angle is 100°

(6) Find the value of ‘x’ in the given figure.

Solution :

From fig, ∠ABD and ∠CBD is liner pair

Thus, ∠ABD +∠CBD = 180°

128° +∠CBD = 180°

∠CBD = 180° – 128°

∠CBD = 52° …. (1)

Also, ∠BDE is external angle of ΔBCD

Thus, Exterior angle = sum of two interior opposite angles

∠BDE = ∠CBD + ∠BCD

x = 52° + 100°

x = 152°

Therefore, the value of ‘x’ is 152°

(7) If ∆MNO ∆DEF, M = 60° and E = 45° then find the value of O

Solution :

Given, ΔMNO is congruent to ΔDEF

Therefore, ∠M is congruent to ∠D,

∠N is congruent to ∠E and ∠O is congruent to ∠F

Thus, ∠D = ∠M = 60°,

∠N = ∠E = 45°

Now, ΔMNO, ∠M = 60°, ∠N = 45°

We know, ∠M + ∠N + ∠O = 180° … (Angle sum property)

60° + 45° + ∠O = 180°

∠O = 180° – 105°

∠O = 75°

Challenge Problem Solutions :

(11) In given figure BD = BC, find the value of x.

Solution :

From fig, In Δ BCD, BD = BC

Thus, ∠BDC = ∠BCD … (Opposite angel of equal sides) … (1)

Since, ∠BCX and ∠BCD are linear pair

Therefore, ∠BCX + ∠BCD = 180°

115° + ∠BCD = 180°

∠BCD = 180° – 115°

∠BCD = 65°

From 1, ∠BDC = 65°

Since, ∠BDC is external angle of ΔABD

Thus, Exterior angle = sum of two interior opposite angles

∠BDC = ∠CBA + ∠BAC

65° = x + 35°

x = 65° – 35°

x = 30°

Therefore, The value of x is 30°.

(12) In the given figure find the value of x.

Solution :

From fig ∠JMK is external angle of ΔLMN

We know, Exterior angle = sum of two interior opposite angles

∠JMK = ∠JNM + ∠MLN

∠JMK = 26° + 30°

∠JMK = 56°

Also ∠LJK is external angle of ΔJKM

We know, Exterior angle = sum of two interior opposite angles

∠LJK = ∠JMK + ∠JKM

x = 56° + 58°

x = 114°

(13) In the given figure find the values of x and y.

Solution :

From fig, ∠XAB is an external angle of ΔABC.

We know, Exterior angle = sum of two interior opposite angles

∠XAB = ∠ABC + ∠ACB

62° = 28° + x

x = 62° – 28°

x = 34°

Now, In ΔABC

∠ABC + ∠ACB + ∠ BAC = 180° … (Angle sum property)

28° + 34° + y = 180°

y = 180° – 62°

y = 118°

Hence, the value of x and y is 34° and 118° resp.

(14) In ΔDEF, F = 48°, E = 68° and bisector of D meets FE at G. find FGD

Solution :

Given, In ΔDEF, ∠F = 48°, ∠E = 68°

Now, ∠FDE + ∠DFE + ∠DEF = 180°

∠FDE + 48° + 68° = 180°

∠FDE + 116° = 180°

∠FDE = 180° – 116°

∠FDE = 64°

Since, bisector of ∠D meets FE at G.

Thus, ∠GDE = 1/2∠FDE = 1/2 *64 = 32°

From fig, ∠DGF is an external angle of ΔDGE.

We know, Exterior angle = sum of two interior opposite angles

∠DGF = ∠GDE +∠GDE

∠DGF = 32° + 68°

A∠DGF = 100°

(15) In the figure find the value of x.

Solution :

From fig, ∠UTP is an external angle of the ΔSTP.

Thus, ∠UTP = ∠TSP + ∠TPS

105° = 75° + ∠TPS

∠TPS = 105° – 75°

∠TPS = 30°

Also From fig,

∠TPS + ∠TPR + ∠RPQ = 180°

30° + 90° + ∠RPQ = 180°

∠RPQ = 180° – 120°

∠RPQ = 60° …. (1)

Since, ∠PRQ and ∠VRQ is a linear pair

Thus, ∠PRQ + ∠VRQ = 180°

∠PRQ + 145° = 180°

∠PRQ = 180° – 145 °

∠PRQ = 35° …. (2)

Now, ∠PQY is an external angle of ΔPRQ.

Thus, Exterior angle = sum of two interior opposite angles

∠PQY = ∠PRQ + ∠RPQ

x = 35° + 60° (from 1 and 2)

x = 95°

(16) From the given figure find the value of y

Solution :

From fig, ∠YCX and ∠ACB are opposite angel and it’s equal.

∠ACB = ∠YCX =48°

Now, In ΔABC

∠ACB + ∠CAB + ∠CBA = 180° … (Angles sum property)

48° + 57° + ∠CBA = 180°

∠CBA = 180° – 105°

∠CBA = 75°

Also from fig,

∠CBA + ∠CBE + ∠EBD = 180°

75° + 65° + ∠EBD = 180°

∠EBD = 180° – 140°

∠EBD = 40°

Since, ∠BDZ is an external angle of ΔBED.

Thus, Exterior angle = sum of two interior opposite angles

∠BDZ = ∠BED + ∠EBD

y = 40° + 97°

y = 137°

Next Chapter Solution :

Updated: July 28, 2023 — 3:58 pm