BC=BD=DC=AD. If AB= 10, what is the length of AC?

You can do it using just the Pythagorean theorem too.
Notice that you have an equilateral triangle BCD. If its side is a, its altitude BE (shown by Bunuel in the diagram) will be \(\sqrt{3}a/2\).
Also the base AE will be a + a/2 (Since altitude of equilateral triangle bisects the base)

So \(10^2 = (3a/2)^2 + (\sqrt{3}a/2)^2\)
\(100 = 12a^2/4\)
\(a = 10\sqrt{3}/3\)

Length of AC \(= 2a = 20\sqrt{3}/3\)

From BC=BD=DC, we know angle DBC= 60 and Angle BDA=120.
And AD=AB so angle ABD=30.
Therefor triangle ABC is right angle trangle with angle B=90 and Angle A=30.

now Cos(A)= AB/AC=10/AC=Sqrt3/2

hence AC=20/srt3 or (20*srt3)/3

You are right that triangle ABC is right angled at B.
Also, AC = 2*BC, AB = 10

\(BC^2 + 10^2 = AB^2 = (2BC)^2\)
\(10^2 = 3(BC)^2\)
\(BC = 10\sqrt{3}/3\)

But the question asks you the length of AC.

\(AC = 2*BC = 20\sqrt{3}/3\)

Another way; since BD=BC=DC, triangle BDC equilateral triangle hence angle BCA = 60 degrees. it mens that other angles angle ABC + angle BAC = 180-60 = 120. Which means that AB = < AC. Which means AC >=10. Let BC = X ; Therefore AC= 2X.
Using pythagoreus theorem AC^2 = AB^2 + BC^2 = 100 +x
4x^2 = 100 + x^2
3x^2 = 100
x = 10[square_root]3/3
AC = 2X = 20 [square_root]3/3

Note: figure not drawn to scale

BC=BD=DC=AD. If AB= 10, what is the length of AC?

If BC=BD=DC then we know triangle DBC is an equilateral triangle. Furthermore, we know that ADC all lie on a line together which means angle ADB = 180-60 = 120. Because AD = DB we know that this triangle is isosceles and that the two other angle measures in this triangle are 30 each. Looking at both triangles together, we see that ABC is a 30:60:90 triangle. Knowing this, and one side length (the length opposite 60) we can solve for BC. Because BC = DC = AD we can find the length of AC (which is AD+DC)

The ratio of the sides in a 30:60:90 is (x/2) : (√3/2 x) : x
√3/2 x = 10
x = 20/√3
In a 30:60:90, the hypotenuse is twice the length of the shortest side. The shortest side is equal to x/2 or (20/√3)/2. Because the hypotenuse is twice that length, it is simply equal to 20/√3




Finally, to cancel out the root on the bottom multiply by (√3/√3) = 20√3/3

I would prefer nave method without a need to dissect the diagram. But you gain important insights about equilateral triangles from Karishma's solution.

Posted from my mobile device

I have a doubt here,

If BC=BD=DC=AD, Shouldn't the angle BAD be the same as the angels of these sides ? So, if the triangle BDC is equilateral, and all such angles are 60, shouldn't angle BAD also be 60?

This BAD angle is BAD :p ...

please provide some feedback . Thanks

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