A 30-60-90 triangle is drawn on the exterior of equilateral tria

Question

A 30^o-60^o-90^o triangle is drawn on the exterior of equilateral triangle ABC as shown in the figure below so that the hypotenuse of the right triangle forms one side of the equilateral triangle. If the length of CD is 2 , what is the length of AD? 6.10.png A. 3 B. 4 C. 5 D. √7 E. 2√7

AbbasAli12 · Answer

Cos 60 = CD/CB. From here you can get CB. Now CB = AB = AC. You can also find BD using Pythagoras theorem and similarly you can find the length of AD because triangle ADB is right angled triangle.

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sakshamchhabra · Answer

since this is a 30-60-90 triangle so we can figure out, BD and BC

triangle ABC is equilateral so, BC=AB,

Triangle ABD is a right triangle so we can find AD.

harish1986 · Answer

One way to solve it.

Extend a line parallel to  BD  from the point  A  and mark the other end as  F .  AF  will be parallel to  BD  with length as  2 \sqrt{3} .

Now it becomes a  RECTANGLE  with  EQUILATERAL TRIANGLE  in it with the base as length

AB will be  4 , which will the length of the rectangle and width as  2 \sqrt{3}

Diagonal of the rectangle will be  2 \sqrt{7}

Archit3110 · Answer

using 30;60;90 
we get side to 90 ; 4
so 
height of ∆ ; 2√3
∆ADB ; AD = (2√3)^2+(4^2 = √28 
=2√7
IMO E

MathRevolution · Answer

=>

Since the triangle  BCD  is a  30-60-90  triangle,  CD:BC:BD = 1:2: √3.

So,  BC = 4, AB = 4,  and  BD = 2 √3.

Angle  ABD  is a right angle since angle  ABC  is  60  degrees, and angle  CBD  is  30  degrees.

By Pythagoras’ theorem,  AD2 = AB2 + BD2 = 16 + 12 = 28 . Therefore,  AD = 2√7.

Therefore, E is the answer. 
Answer: E

A 30-60-90 triangle is drawn on the exterior of equilateral tria

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A 30-60-90 triangle is drawn on the exterior of equilateral tria

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