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)

Q13) In each figure, given below, ABCD is a square and ∆ BEC is an equilateral triangle.

Find, in each case : (i) ∠ABE(ii) ∠BAE

Solution:

We know that the sides of a square are equal and each angle is of 90°

Three sides of an equilateral triangle are equal and each angle is of 60°. Therefore,

In fig. (i), ABCD is a square and ∆ BEC is

an equilateral triangle.

(i) ∠ABE = ∠ABC + ∠CBE

= 90° + 60° = 150°

(ii) But in ∆ ABE

∠ABE + ∠BEA + ∠BAE = 180°

(Angles of a triangle)

⇒ 150° + ∠BAE + ∠BAE = 180°

(∵ AB = BE)

150° + 2∠BAE = 180°

⇒ 2∠BAE = 180° - 150° = 30°

∴ ∠BAE = 30°/2 = 15°

In figure (ii),

∵ ABCD is a square and ∆ BEC is an

Equilateral triangle,

(i) ∴ ∠ABE = ∠ABC - ∠CBE

= 90° - 60° = 30°

(ii) In ∆ ABE, ∠ABE + ∠AEB + ∠BAE = 180°

(Angles of a triangle)

= 30° + ∠BAE + ∠BAE = 180° (∵ AB = BE)

⇒ 30° + 2∠BAE = 180°

2∠BAE = 180° – 30° = 150°

⇒ ∠BAE = 150°/2 = 75°

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)

Q13) In each figure, given below, ABCD is a square and ∆ BEC is an equilateral triangle.

Find, in each case : (i) ∠ABE(ii) ∠BAE

Solution:

We know that the sides of a square are equal and each angle is of 90°

Three sides of an equilateral triangle are equal and each angle is of 60°. Therefore,

In fig. (i), ABCD is a square and ∆ BEC is

an equilateral triangle.

(i) ∠ABE = ∠ABC + ∠CBE

= 90° + 60° = 150°

(ii) But in ∆ ABE

∠ABE + ∠BEA + ∠BAE = 180°

(Angles of a triangle)

⇒ 150° + ∠BAE + ∠BAE = 180°

(∵ AB = BE)

150° + 2∠BAE = 180°

⇒ 2∠BAE = 180° - 150° = 30°

∴ ∠BAE = 30°/2 = 15°

In figure (ii),

∵ ABCD is a square and ∆ BEC is an

Equilateral triangle,

(i) ∴ ∠ABE = ∠ABC - ∠CBE

= 90° - 60° = 30°

(ii) In ∆ ABE, ∠ABE + ∠AEB + ∠BAE = 180°

(Angles of a triangle)

= 30° + ∠BAE + ∠BAE = 180° (∵ AB = BE)

⇒ 30° + 2∠BAE = 180°

2∠BAE = 180° – 30° = 150°

⇒ ∠BAE = 150°/2 = 75°

Still have questions? Our expert teachers can help you out

Book a free class nowLido

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