A semi-circle is inscribed in an isosceles right-angled triangle as sh

Question

https://gmatclub.com/forum/download/file.php?id=62324 A semi-circle is inscribed in an isosceles right-angled triangle as shown. What is the radius of the semi-circle? A) 2*(\sqrt{2} - 1) B) 1 C) \sqrt{3} - 1 D) 2 E) 2/3 Screenshot_3.png

chetan2u · Answer

Since AC is tangent to the semi circle, OD is perpendicular to AC, where O is the center and D, the point where line AC and semi circle intersect.

Take  	riangle AOD 
 \angle AOD = \angle DAO = 45 
This  	riangle AOD  is an isosceles right angled triangle.
OD=AD=r, and AO =  r\sqrt{2}

AB=BO+AO=r+r\sqrt{2}=r(1+\sqrt{2})=2

r=\frac{2}{(1+\sqrt{2})}

r=\frac{2*(\sqrt{2}-1)}{(1+\sqrt{2})(\sqrt{2}-1)}=\frac{2*(\sqrt{2}-1)}{2-1}=2*(\sqrt{2}-1)

IanStewart · Answer

I don't think the right answer is among the choices (the answer is twice as big as answer choice A).

If you draw the center of the circle, and call the radius r, then that divides the vertical line into two parts. The bottom part is the radius r, and the entire vertical length is 2, so the top part is 2-r.

Now we can connect the center of the circle to two points:

- we can draw a radius sloping upwards from the center to the point where the circle meets the hypotenuse (I'll call that point "P"). The hypotenuse is tangent to the circle, so that will create a right angle
- we can draw a very long line from the center to the distant bottom-right corner of the triangle

If you draw this picture, you'll see we've created two long and thin right triangles. The one at the bottom of the picture has one vertical side of length r, and a long horizontal side of length 2. This triangle shares a hypotenuse with the right triangle above it, which also has a side of length r. If these two right triangles have two equal sides (they have the same hypotenuse, and have another side of length r) their third side must also be equal, so the third side of the top triangle is 2.

Now, looking at the original 45-45-90 triangle, its hypotenuse was 2√2 (since in a 45-45-90, the hypotenuse is √2 times a side). We just determined the length of the section from our new point P down to the bottom right corner was 2. So the remaining shorter part at the top is 2√2 - 2.

Now we have a small right triangle at the top of our picture, where the hypotenuse is 2-r, and the two short sides are r and 2√2 - 2. Using Pythagoras we find

(2 - r)^2 = r^2 + (2√2 - 2)^2
4 - 4r + r^2 = r^2 + 8 - 8√2 + 4
4r = 8 - 8√2
r = 2 - 2√2

If stuck on a problem like this, an estimate might get down to one, two or three possible answers. The diameter of the semicircle is clearly less than 2, the length of the side of the triangle. So the radius must be less than 1, but it's not too much less than 1. So answer choices like '1' and '2' here are absolutely impossible, and answers like √2 - 1 is too small to be plausible.

jn0r · Answer

Thank you everyone for noticing the mistake in the answer choices. I've edited it

KetakiUpadhye · Answer

Hi Ian.... Can you please draw a diagram to elaborate your point.Thank you

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A semi-circle is inscribed in an isosceles right-angled triangle as sh

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