Experts Please Help on this Kaplan DS!

[srini1988]

Find the question in the image above.[/img]

[knight247]

(1) Circumference=18Pi
Simple formula substitution
2(Pi)(r)=18Pi..........{Circumference of circle=2(Pi)(r)}
r=18/2=9 and Hence Diameter=18. Which proves that l(AB) coincides with Diameter and also proves that triangle ABC is inscribed inside a semi circle. A triangle inscribed inside semi circle is always a right angled triangle. But this information is INSUFFICIENT to find area

(2)ABC=30. Not enough to find area. Hence INSUFFICIENT

Combining 1 and 2 we get. ABC is a right angled triangle with hypotenuse=18 and one 30 degree angle. Meaning the third angle has to be 60 degrees. Using 30-60-90 theorem we get AC=9,BC=9.√3. Area=0.5*9*9√3. Hence, C

BTW in a 30-60-90 triangle....If side opposite 30 is X, then Hypotenuse is 2x and Side opposite 60 is x√3

[tpr-becky]

although the angle C looks like a right angle in the diagram that is not stated anywhere in the problem.

area of a triangle is 1/2(b)(h). in a right triangle the base and height are legs of the triangle but we don't know whether this is a right triangle.

statment 1 tells us the radius of the triangle. if C=18ii then the radius is 9 which confirms that AD is a diameter. This means that C is a right angle. But we need two sides to solve the sides of a right triangle thus statement 1 is insufficient and we are left with BCE.

Statement 2 tells us that one of the angles is 30 but now we don't know whether C is a right angle. So we are left with CE.

Together we know that the triangle is a right triangle, and that it is a 30-60-90. with 30-60-90 triangles you only need one side to solve for all three sides. we have one side so we can solve for each side and then put it in the area formula.

answer is c.