Question

ABCD is a square of side 4 cm. If E is a point in the interior of the square such that ΔCED is equilateral, then area of Δ ACE is
(a)2(√3-1) cm²
(b)4(√3-1) cm²
(c)6(√3-1) cm²
(d)8(√3-1 )cm²

Answer 1

Answer:

The area of ∆AEC is  4(√3 - 1) cm² .

Among the given options option (b) 4(√3 - 1) cm² is the correct answer.

Step-by-step explanation:

Given :

Side of a square, ABCD = 4 cm

∆CED is an equilateral triangle.

EC = CD = DE = 4 cm

∠ECD = 60°  

[Angle of an Equilateral triangle]

AC is a diagonal of a square ABCD.

Therefore, ∠ACD = 45°

∠ECA = ∠ECD -  ∠ACD  

∠ECA = 60° - 45°

∠ECA = 15°

In ∆ACE , draw perpendicular EM the base AC.

Now in ∆EMC ,  

sin 15° = P/H = EM/EC  

sin 15° = EM/4

(√3 -  1)/2√2 = EM/4

(√3 -  1) × √2 /(2√2 ×√2) = EM/4

[By rationalising the denominator]

√2(√3 - 1) /4 = EM/4

√2(√3 - 1) = EM

EM = √2(√3 - 1)

Diagonal Of A Square, AC = √2 a

AC = √2 × 4 = 4√2

Diagonal Of A Square = 4√2 cm

Now in ∆AEC,

Area of ∆AEC, A = ½ × AC × EM

A = ½ × 4√2 ×  √2(√3 - 1)

A = 2 × 2 (√3 - 1)

A = 4(√3 - 1) cm²

Area of ∆AEC =  4(√3 - 1) cm²

Hence, the area of ∆AEC is  4(√3 - 1) cm² .

 ★★ [sin(A - B) = sin A cos B - cos A sin B  

sin 15° = sin(45° - 30°)  

sin 15° = sin 45° cos 30°- cos 45° sin 30°  

sin 15° = (1/√2) (√3/2) -  (1/√2) (1/2)  

sin 15° = (√3 -  1)/2√2]

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

ABCD is a square of side 4 cm. If E is a point in the interior of the square such that ΔCED is equilateral, then area of Δ ACE is

(a)2(√3-1) cm²

(b)4(√3-1) cm²

(c)6(√3-1) cm²

(d)8(√3-1 )cm² ✔✔✔