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## Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 4 Geometry Ex 4.1

Question 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 form a triangle.

Question 2.

Can you draw a triangle with 25°, 65° and 80° as angles?

Solution:

Given angle 25°, 65° and 80°.

Sum of the angles = 25° + 65° + 80° = 170° ≠ 180

∴ We cannot draw a triangle with these measures.

Question 3.

In each of the following triangles, find the value of x.

Solution:

(i) Let ∠G = x

By angle sum property we know that,

∠E + ∠F + ∠G = 180°

80° + 55° + x = 180°

135° + x = 180°

x = 45°

(ii) Let ∠M = x

By angle sum property of triangles we have

∠M + ∠M + ∠O = 180°

x + 96° + 22° = 180°

x + 118° = 180°

X = 180° – 118° = 620

(iii) Let ∠Z = (2x + 1)° and ∠Y = 90°

By the sum property of triangles we have

∠x + ∠y + ∠z = 180°

29° + 90° + (2x + 1)° = 180°

119° + (2x + 1)° = 180°

(2x + 1)° = 180° – 119°

2x + 1° = 61°

2x = 61° – 1°

2x = 60°

x = \(\frac{60^{\circ}}{2}\)

x = 30°

(iv) Let ∠J = x and ∠L – 3x.

By angle sum property of triangles we have

∠J + ∠K + ∠L = 180°

x + 112° + 3x = 180°

4x = 180° – 112°

x = 68°

x = \(\frac{68^{\circ}}{4}\)

x = 17°

(v) Let ∠S = 3x°

Given \(\overline{\mathrm{RS}}\) = Given \(\overline{\mathrm{RT}}\) = 4.5 cm

Given ∠S = ∠T = 3x° [∵ Angles opposite to equal sides are equal]

By angle sum property of a triangle we have,

∠R + ∠S + ∠T = 180°

72° + 3x + 3x = 180°

72° + 6x = 180°

x = \(\frac{108^{\circ}}{6}\)

x = 18°

(vi) Given ∠X = 3x; ∠Y = 2x; ∠Z = ∠4x

By angle sum property of a triangle we have

∠X + ∠Y + ∠Z = 180°

3x + 2x + 4x = 180°

∴ 9x = 180°

x = \(\frac{180^{\circ}}{9}\) = 20°

(vii) Given ∠T = (x – 4)°

∠U = 90°

∠V = (3x – 2)°

By angle sum property of a triang we have

∠T + ∠U + ∠V = 180°

(x – 4)° + 90° + (3x – 2)° = 180°

x – 4° + 90° + 3x – 2° = 180°

x + 3x + 90° – 4° – 2° = 180°

4x + 84° = 180°

4x = 180° – 84°

4x = 96°

x = \(\frac{96^{\circ}}{4}\) = 24°

x = 24°

(viii) Given ∠N = (x + 31)°

∠O = (3x – 10)°

∠P = (2x – 3)°

By angle sum property of a triangle we have

∠N + ∠O + ∠P = O

(x + 31)° + (3x – 10)° + (2x – 3)° = 180°

x + 31°+ 3x – 10° + 2x – 3° = 180°

x + 3x + 2x + 31° – 10° – 3° = 180°

6x + 18° = 180°

6x = 180° + 18°

6x = 162°

x = \(\frac{162^{\circ}}{6}\) = 27°

x = 27°

Question 4.

Two line segments \(\overline{A D}\) and \(\overline{B C}\) intersect at O. Joining \(\overline{A B}\) and \(\overline{D C}\) we get two triangles, ∆AOB and ∆DOC as shown in the figure. Find the ∠A and ∠B.

Solution:

In ∆AOB and ∆DOC,

∠AOB = ∠DOC [∵ Vertically opposite angles are equal]

Let ∠AOB = ∠DOC = y

By angle sum property of a triangle we have

∠A + ∠B + ∠AOB = ∠D + ∠C + ∠DOC = 180°

3x + 2x + y = 70° + 30° + y = 180°

5x + y = 100° + y = 180°

Here 5x + y = 100° + y

5x = 100° + y – y

5x = 100°

x = \(\frac{100^{\circ}}{5}\) = 20°

∠A = 3x = 3 × 20 = 60°

∠B = 2x = 2 × 20 = 40°

∠A = 60°

∠B = 40°

Question 5.

Observe the figure and find the value of

∠A + ∠N + ∠G + ∠L + ∠E + ∠S.

Solution:

In the figure we have two triangles namely ∆AGE and ∆NLS.

By angle sum property of triangles,

Sum of angles of ∆AGE = ∠A + ∠G + ∠E = 180° …(1)

Also sum of angles of ∆NLS = ∠N + ∠L + ∠S = 180° … (2)

(1) + (2) ∠A + ∠G + ∠E + ∠N + ∠L + ∠S = 180° + 180°

i.e., ∠A + ∠N + ∠G + ∠L + ∠E + ∠S = 360°

Question 6.

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

Solution:

Given three angles of the triangles are in the ratio 3 : 5 : 4.

Let the three angle be 3x, 5x and 4x.

By angle sum property of a triangle, we have

3x + 5x + 4x = 180°

12x = 180°

x = \(\frac{180^{\circ}}{12}\)

x = 15°

∴ The angle are 3x = 3 × 15° = 45°

5x = 5 × 15° = 75°

4x = 4 × 15° = 60°

Three angles of the triangle are 45°, 75°, 60°

Question 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 ∆RST. Let ∠R = x.

Then given S is ∠10° greater than ∠R

∴ ∠S = x + 10°

Also given ∠T is 5° less then ∠S.

So ∠T = ∠S – 5° = (x + 10)° – 5° = x + 10° – 5°

By angle sum property of triangles, sum of three angles = 180°.

∠R + ∠S + ∠T = 180°

x + x + 10° + x + 5° = 180°

3x + 15° = 180°

3x = 180° – 15°

x = \(\frac{165^{\circ}}{3}\) = 55°

∠R = x = 55°

∠S = x + 10° = 55° + 10° = 65°

∠T = x + 5° = 55° + 5° = 60°

∴ ∠R = 55°

∠S = 65°

∠T = 60°

Question 8.

In ∆ABC , if ∠B is 3 times ∠A and ∠C is 2 times ∠A, then find the angles.

Solution:

In ABC, Let ∠A = x,

then ∠B = 3 times ∠A = 3x

∠C = 2 times ∠A = 2x

By angle sum property of a triangles,

Sum of three angles of ∆ABC =180°.

∠A + ∠B + ∠C = 180

x + 3x + 2x = 180°

x (1 + 3 + 2) = 180°

6x = 180°

x = \(\frac{180^{\circ}}{6}\) = 30°

∠A = x = 30°

∠B = 3x = 3 × 30° = 90°

∠C = 2x = 2 × 30° = 60°

∴ ∠A = 30°

∠B = 90°

∠C = 60°

Question 9.

In ∆XYZ, if ∠X : ∠Z is 5 : 4 and ∠Y = 72°. Find ∠X and ∠Z.

Solution:

Given in ∆XYZ, ∠X : ∠Z = 5 : 4

Let ∠X = 5x; and ∠Z = 4x given ∠Y = 72°

By the angle sum property of triangles sum of three angles of a triangles is 180°.

∠X + ∠Y + ∠Z = 180°

5x + 72 + 4x = 180°

5x + 4x = 180° – 72°

9x = 108°

x = \(\frac{108^{\circ}}{9}\) = 12°

∠X = 5x = 5 × 12° = 60°

∠Z = 4x = 4 × 12° = 48°

∴ ∠X = 60°

∠Z = 48°

Question 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 ∆ABC ∠B = 90°

∠A = x + 1

∠B = 2x + 5

By angle sum property of triangles

Sum of three angles of ∆ABC = 180°

∠A + ∠B + ∠C = 180°

(x + 1) + 90° + (2x + 5) = 180°

x + 2x + 1° + 90° + 5° = 180°

3x + 96° = 180°

3x = 180° – 96° = 84°

x = \(\frac{84^{\circ}}{3}\) = 28°

∠A = x + 1 = 28 + 1 = 29

∠C = 2x + 5 = 2 (28) + 5 = 56 + 5 = 61

∴ ∠A = 29°

∠C = 61°

Question 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 ∠N = 90°

MO is extended to P, the exterior angle ∠NOP = 128°

Exterior angle is equal to the sum of interior opposite angles.

∴ ∠M + ∠N = 128°

∠M + 90° = 128°

∠M = 128° – 90°

∠M = 38°

By angle sum property of triangles,

∴ ∠M + ∠N + ∠O = 180°

38° + 90° + ∠O = 180°

∠O = 180° – 128°

∠O = 52°

∴ ∠M = 38° and ∠O = 52°

Question 12.

Find the value of x in each of the given triangles.

Solution:

(i) In ∆ABC, given B = 65°,

AC is extended to L, the exterior angle at C, ∠BCL = 135°

Exterior angle is equal to the sum of opposite interior angles.

∠A + ∠B = ∠BCL

∠A + 65° = 135°

∠A = 135° – 65°

∴ ∠A = 70°

x + ∠A = 180° [∵ linear pair]

x + 70° = 180° [∵ ∠A = 70°]

x = 180° – 70°

∴ x = 110°

(ii) In ∆ABC, given B = 3x – 8°

∠XAZ = ∠BAC [∵ vertically opposite angles]

8x + 7 + ∠BAC

i.e., In ∆ABC, ∠A = 8x + 7

Exterior angle ∠XCY = 120°

Exterior angle is equal to the sum of the interior opposite angles.

∠A + ∠B = 120°

8x + 7 + 3x – 8 = 120°

8x + 3x = 120° + 8 – 7

11x = 121°

x = \(\frac{121^{\circ}}{11}\) = 11°

Question 13.

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

Solution:

Exterior angle is equal to the sum of the opposite interior angles.

∠LNO = ∠MLN + ∠LMN

6x – 5 = 100° – x + 2x

6x – 5 + x – 2x = 100°

6x + x – 2x = 100° + 5°

5x = 105°

x = \(\frac{105^{\circ}}{5}\) = 21°

x = 21°

Question 14.

Using the given figure find the value of x.

Solution:

In ∆EDC, side DE is extended to B, to form the exterior angle ∠CEB = x.

We know that the exterior angle is equal to the sum of the opposite interior angles

∠CEB = ∠CDE + ∠ECD

x = 50° + 60°

x = 110°

Question 15.

Using the diagram find the value of x.

Solution:

Given triangle is an equilateral triangle as the three sides are equal. For an equilateral triangle all three angles are equal and is equal to 60° Also exterior angle is equal to sum of opposite interior angles.

x = 60° + 60°.

x = 120°

Objective Type Questions

Question 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

Answer:

(ii) 40, 60, 80

Question 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°

Answer:

(iii) 80°,35°

Question 18.

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

(i) 112°

(ii) 68°

(iii) 102°

(iv) 62° A

Answer:

(ii) 68°

Question 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)

Ans :

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

Question 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°

Answer:

(iii) 35°

Question 21.

In a ∆ABC, AB = AC. The value of x is _____.

(i) 80°

(ii) 100°

(iii) 130°

(iv) 120°

Answer:

(iii) 130°

Question 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°

Answer:

(ii) 65°, 80°