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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) \(\frac{2}{\sqrt{3}}\)

The correct answer is B.

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)

From rt < triangle ADC, AD2 = 52 - 32

= 25 - 9 = 6

AD = 4

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\(\sqrt{3}\)


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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)

    From rt < triangle ADC, AD2 = 52 - 32

    = 25 - 9 = 6

    AD = 4

    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\(\sqrt{3}\)

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