Selina solutions

Selina solutions

Selina solutions

Grade 7

CHAPTERS

3. Fractions (Including Problems)

4. Decimal Fractions (Decimals)

5. Exponents (including Laws of Exponents)

6. Ratio and Proportion (Including Sharing in a Ratio)

7. Unitary Method (including Time and Work)

11. Fundamental Concepts (Including Fundamental Operations)

12. Simple Linear Equations (Including Word Problems)

13. Set Concepts (Some Simple Divisions by Vedic Method)

14. Lines and Angles (Including Construction of angles)

17. Symmetry (Including Reflection and Rotation)

18. Recognition of Solids (Representing 3-D in 2-D)

Q12) Find the unknown marked angles in the given figures :

Solution:

We know that in a triangle, if one side of it is produced, then

Exterior angle = sum of its interior opposite angles

(i) In Fig. (i), 110° = x° + 30°

⇒ x° = 110° - 30° = 80°

(ii) In Fig. (ii), 120° = y° + 60°

⇒ y° = 120° - 60° = 60°

(iii) In Fig.(iii), 122° = k° + 35°

⇒ k° = 122° - 35° = 87°

(iv) In Fig. (iv), 135° = a° + 73°

⇒ a° = 135° - 73° = 62°

(v) In Fig. (v), 125° = a + c ….(i)

And 140° = a + b …..(ii)

Adding, we get

a + c + a + b = 125° + 140°

⇒ a + a + b + c = 265°

But a + b + c = 180°

(Sum of angles of triangles)

∴ a + 180° = 265°

⇒ a = 265° - 180° = 85°

But a + b = 140°

⇒ 85° + b = 140°

⇒ b = 140° - 85° = 55°

And a + c = 125° ⇒ 85° + c = 125°

⇒ c = 125° - 85° = 40°

Hence, a = 85°, b = 55° = 40°

(vi) In Fig. (vi)

112° + x° = 180° (Linear pair)

⇒ x = 180° - 112° = 68°

And 112° = y + 63°

⇒ y = 112° - 63° = 49°

Hence x = 68°, y = 49°

(vii) In Fig. (vii),

120° = a + a ⇒ 2a = 120°

⇒ a = 120°/2 = 60°

∴ a = 60°

(viii) In Fig. (viii),

140° + a = 180° (Linear pair)

⇒ a = 180° - 140° = 40°

Now, 4m = 2m = a ⇒ 4m – 2m = a

⇒ 2m = 40° ⇒ m = 40°/2 = 20°

Hence, m = 20°

(ix) In Fig. (ix),

105° = b + b ⇒ 2b = 105°

⇒ b = 105°/2 = 52.5°

But a + 105° = 180° (Linear pair)

⇒ a = 180° - 105° = 75°

Hence a = 75° , b = 52.5°

Still have questions? Our expert teachers can help you out

Book a free class nowFractions (Including Problems)

Exponents (including Laws of Exponents)

Ratio and Proportion (Including Sharing in a Ratio)

Unitary Method (including Time and Work)

Fundamental Concepts (Including Fundamental Operations)

Simple Linear Equations (Including Word Problems)

Set Concepts (Some Simple Divisions by Vedic Method)

Lines and Angles (Including Construction of angles)

Symmetry (Including Reflection and Rotation)

Recognition of Solids (Representing 3-D in 2-D)

Q12) Find the unknown marked angles in the given figures :

Solution:

We know that in a triangle, if one side of it is produced, then

Exterior angle = sum of its interior opposite angles

(i) In Fig. (i), 110° = x° + 30°

⇒ x° = 110° - 30° = 80°

(ii) In Fig. (ii), 120° = y° + 60°

⇒ y° = 120° - 60° = 60°

(iii) In Fig.(iii), 122° = k° + 35°

⇒ k° = 122° - 35° = 87°

(iv) In Fig. (iv), 135° = a° + 73°

⇒ a° = 135° - 73° = 62°

(v) In Fig. (v), 125° = a + c ….(i)

And 140° = a + b …..(ii)

Adding, we get

a + c + a + b = 125° + 140°

⇒ a + a + b + c = 265°

But a + b + c = 180°

(Sum of angles of triangles)

∴ a + 180° = 265°

⇒ a = 265° - 180° = 85°

But a + b = 140°

⇒ 85° + b = 140°

⇒ b = 140° - 85° = 55°

And a + c = 125° ⇒ 85° + c = 125°

⇒ c = 125° - 85° = 40°

Hence, a = 85°, b = 55° = 40°

(vi) In Fig. (vi)

112° + x° = 180° (Linear pair)

⇒ x = 180° - 112° = 68°

And 112° = y + 63°

⇒ y = 112° - 63° = 49°

Hence x = 68°, y = 49°

(vii) In Fig. (vii),

120° = a + a ⇒ 2a = 120°

⇒ a = 120°/2 = 60°

∴ a = 60°

(viii) In Fig. (viii),

140° + a = 180° (Linear pair)

⇒ a = 180° - 140° = 40°

Now, 4m = 2m = a ⇒ 4m – 2m = a

⇒ 2m = 40° ⇒ m = 40°/2 = 20°

Hence, m = 20°

(ix) In Fig. (ix),

105° = b + b ⇒ 2b = 105°

⇒ b = 105°/2 = 52.5°

But a + 105° = 180° (Linear pair)

⇒ a = 180° - 105° = 75°

Hence a = 75° , b = 52.5°

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