Selina solutions

Selina solutions

Grade 7

Triangles | Exercise 15(B)

Question 13

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°

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Question 13

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°

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