whats the solution?
Not able to get the answer as D.

This question is solved using similar triangles. I was not convinced with the solution so hoping some can help both of us.

You definitely missed something at the picture

With the given information i'm arriving at C.

No, I haven't.

AC will be the diameter, this can be deduced from question itself.

triangle ABC will be right angled at B, triangle inscribed in a semicircle.

We are going to start using similar triangle concept: We have 3 triangles in this figure, ABC,BDC and ABC right angled at D and B.

It would be much easier if I could attach a handwritten note... typing solution to geometry them becomes too difficult.

Actually official answer is correct, which is D. I have made a hand written attachment for how statement I is correct, similarly we can prove for statement II. please excuse my handwriting

OA is the correct answer which is D, I have made an handwritten note on how to solve for statement I, similarly we can solve statement II. So D is the answer. Thanks guys for making me do it, initially even I thought C was the answer...

PLS IGNORE THIS POST...

You are correct, infact I made a mistake in the final step of solution and everything cancels out after the substitution for AB^2+BC^2=AC^2.

So that step take me no where... I understood the mistake after I attached it in the forum.

And your method was much simpler to prove the case, I totally missed the big picture...

I have attached where I made the mistake...if any one has a solution, please share. I FAILED

The solution is C

Knowing just a part of the diameter is not enough to determine its length, in the picture there is a clear example.
In both cases \(x=2\) but in the second one \(x=\frac{diameter}{2}\) so diameter is 4, but if I move the height of the triangle I cannot determine its diameter.

Basically we need to "lock" one of the other aleatory factors. The remaining part is perfect so \(diameter=x+y=2+7=9\)

Does this make sense?

Makes sense... thanks Zarrolou and others...

Surely this cannot be a 700 level questions if the answer is simply 2+7 = 9!
There has got to be something to do this that we are missing?

Guys I found the following problem in GoGmat:


BC is a diameter. What is the length of the arc BC?
(1) \(x = 2\)
(2) \(y = 12.5\)



May be you want to solve this problem? At least it is more interesting.

This is a "stolen" question from OG13 plus it is incomplete.

OG question:


Discussed here: if-arc-pqr-above-is-a-semicircle-what-is-the-length-of-144057.html

This topic is locked.
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