In the diagram, what is the value of x?

a) 1+ sqrt(2)
b) 1+ sqrt(3)
c) 2sqrt(2)
d) sqrt(2) + sqrt(3)
e) 2sqrt(3)

How can you say A is the mid-point of DE? It is not the midpoint.

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My dad once said to me: Son, nothing succeeds like success.

let the triangle be ABC where AB = AC = x and BC = sqrt(2)

angle A of triangle ABC is 30 degrees. = > angle B = angle C = 75 (as its an isosceles triangle).

draw a perpendicular BD from B to AC . this forms two right angle triangles ABD and BCD.

ABD forms 30-60-90 triangle ,

where AB = x, AD = x * sqrt(3)/2 and BD = x/2 -------------equation 1


now consider BCD triangle,

where BD = x/2 ( from equation 1)
BC = 2
CD = x-(x*sqrt(3)/2)

using Pythagorean theorem we have
BC^2 = CD^2+BD^2

Solving this will give x^2 = 2/(2-sqrt(3))

= 2(2+sqrt(3))

= 4 + 2 sqrt(3)
= (sqrt(3)^2) +1+ 2 * sqrt(3)

= (1+sqrt(3)^2)

=> x = 1+sqrt(3)

Answer is B.

Please advise where i am wrong with this

Sin 30 = Opp side/Hyp side.

1/2 = sq(2)/x

x = sq(2)*2

I guess this cannot be solved without the use of Trigonometry!!

Spidy001, Thanks for the explanation!!
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Cheers,
Varun

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Used my Trigonometry skills

The two isoscales base angles are 75 degrees each.
Cos (45+30) = Cos (45) Cos (30) - Sin(45) Sin(30)
Cos(75) = (\sqrt{3} - 1)/2 = 1/\sqrt{2}x

Hence, x = \sqrt{3}+1

Answer B.

Cheers!
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i see,the best way to solve this is using cosine law or sin law.

2=2x^2-2x^2cos30
1=x^2(1-cos30)
x=1+sqrt3

or

x=sqrt2*sin75/sin30 =1+sqrt3

\sqrt{2}/sin30 = \frac{x}{sin75}

sin30 = 1/2
sin75 = (\sqrt{6}+\sqrt{2})/4

solve for x
x = 1+\sqrt{3}
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