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# In the figure, $$\bigtriangleup$$ ABC are in adjacent planes. AB = AC = 5cm, BC ...

### Question

In the figure, $$\bigtriangleup$$ ABC are in adjacent planes. AB = AC = 5cm, BC = 6cm and o then AE is equal to

### Options

A) 3$$\sqrt{2}$$

B) 2$$\sqrt{3}$$

C) $$\frac{\sqrt{3}}{2}$$

D) $$\frac{2}{\sqrt{3}}$$

### Explanation:

BC = 6 : DC = $$\frac{6}{2}$$ = 3cm

By construction < EDE = 180$$^{\circ}$$(90$$^{\circ}$$ + 60$$^{\circ}$$) = 180$$^{\circ}$$ - 150$$^{\circ}$$

= 30$$^{\circ}$$(angle on a strt. line)

= 25 - 9 = 6

From < AEC, let AS = x

$$\frac{x}{sin 60^o}$$ - $$\frac{4}{sin 90^o}$$

sin 90$$^{\circ}$$ = 1

sin 60$$^{\circ}$$ = $$\frac{\sqrt{3}}{2}$$

x = 4sin 60$$^{\circ}$$

x = 3 x $$\frac{\sqrt{3}}{2}$$

= 2$$\sqrt{3}$$

## Dicussion (1)

• BC = 6 : DC = $$\frac{6}{2}$$ = 3cm

By construction < EDE = 180$$^{\circ}$$(90$$^{\circ}$$ + 60$$^{\circ}$$) = 180$$^{\circ}$$ - 150$$^{\circ}$$

= 30$$^{\circ}$$(angle on a strt. line)

= 25 - 9 = 6

From < AEC, let AS = x

$$\frac{x}{sin 60^o}$$ - $$\frac{4}{sin 90^o}$$

sin 90$$^{\circ}$$ = 1

sin 60$$^{\circ}$$ = $$\frac{\sqrt{3}}{2}$$

x = 4sin 60$$^{\circ}$$

x = 3 x $$\frac{\sqrt{3}}{2}$$

= 2$$\sqrt{3}$$