A triangle is inscribed in a circle. A point is randomly chosen inside

Since we are looking at probability, even finding ratio of areas will be sufficient...

Statement I does not say anything about radius of circle or area of triangle..
Insufficient

Statement II states ABC is isosceles right angle triangle....
Remember the rule - any angle in the semicircle with two sides meeting the end of diameter is 90..
So here the Dia of circle is hypotenuse of triangle...
Since triangle is isosceles, sides of triangle and radius are related..
We can find the ratio of two areas..
Sufficient
. B

Otherwise..
Sides of isosceles right angle triangle = Dia/√2 = 2r/√2=√2r
Area of isosceles triangle =\(\frac{1}{2}*√2r*√2r=r^2..\)
Area of circle=\(π*r^2\)..
Probability = \(\frac{r^2}{π*r^2}=\frac{1}{π}\)
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B.

A) No information about triangle's type (scalene or isosceles) and radius or diameter of the circle. Not Sufficient.
B) Right isosceles triangle means diameter is the hypotenuse and is equal to 2r.
Area of triangle = 1/2*2r*r = r^2
Area of circle = 22/7*r^2
Therefore probability = r^2/(22/7*r^2) = 7/22 = Sufficient.

We don't need to do calculations to solve this one!

If you know that it is a right isosceles triangle, then the longest side must be the diameter and the relative orientation and relative length of other sides are known. Therefore, we would also know the ratio of areas of the triangle to the circle. Hence statement B alone is sufficient.

thank you all for the explanation.

Statement 2- The isosceles triangle can be at any point in the circle, and not necessarily cut through the diameter of the circle. In this case, this question is not solvable. Hence, answer is E

Appreciate if someone can advise where did I go wrong on the below thought.

Is there only a single size right-angle triangle that can fit in a circle?

Hi philipssonicare,

No. There are many right angled triangles that can fit inside a circle. But one side will ALWAYS be the diameter (As we know, angle in a semicircle is 90 degrees).

But Option (2) says right angled ISOSCELES triangle.
There will be only one such triangle, with hypotenuse as diameter.
Eg: Assume a circle with Center origin (0, 0) and radius 2 units and let opposite ends of diameter are A(2, 0) and B(-2, 0) lying on x-axis.
Third vertices will lie on y-axis at (0, 2) or (0, -2).

Hope I’m clear!

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Hi, since 2r is the diameter (base of the triangle), the height of the triangle would be r? is this true in all cases when the triangle is right-angled isosceles triangle inscribed in a circle?

Great C-Trap

Key is to remember that the probability is only a “ratio” value. We do not need to know the Actual Areas of the inscribed triangle and circumscribed circle. We just need to know the Areas RELATIVE to one another.

Statement 2 alone:

Any Right triangle inscribed within a Circle must have as its Hypotenuse the DIAMETER of the circle.

Right Isosceles Triangle area = (1/2) (X) (X) = (X^2) / 2

Diameter = Hypotenuse = X * sqrt(2)

To get the Radius we take (1/2) the diameter — and then to get the Area of the circumscribed circle we square the radius and multiply by (pi) ——> Area of circumscribed Circle =

(X * sqrt(2) / 2)^2 * (pi) =

(X^2 (pi) ) / 2 = Circle Area

Probability = (Area of inscribed triangle) / (Area of circumscribed circle) =

(X^2 / 2)
_________
(X^2 (pi) / 2)

The X^2 cancels out and we get:

(2) / (2 * pi) =

1 / pi

B: statement 2 is sufficient alone



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the desired probability = area of triangle / area of circle
= half (base*height) / pi*r^2
St1. Longest side of triangle ABC measures 10.
Longest side of a triangle inscribed in a circle has to be the diameter of the circle.
but infinite triangle's can be drawn with this condition so desired probability cannot be determined. Not Sufficient

St2. ABC is a right isosceles triangle.
since triangle is right angle and isosceles, it implies that the triangle has a side which is diameter of the circle and also its height (considering diameter as base) is radius of the circle. (we make use of Thales theorem - Inscribed angles subtended by a diameter are right). the area of triangle is r^2.
If we observe carefully, only one such triangle can exist.
Hence, the desired probability can be found from statement 2 alone. Sufficient
Answer B

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