GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Question 8: Triangle ABC is inscribed in a semicircle center
[#permalink]
19 May 2016, 16:19
19 May 2016, 16:19
00:00
Question Stats:
71% (02:33) correct
28% (02:28) wrong based on 7 sessions
Hide Show timer Statistics
Attachment:
semicircle.jpg [ 11.36 KiB | Viewed 2737 times ]
Triangle ABC is inscribed in a semicircle centered at D. What is the area of Triangle ABC?
A. \(\frac{12}{\sqrt{3}}\)
B. \(6 \sqrt{3}\)
C. \(12\)
D. \(12 \sqrt{3}\)
E. \(18 \sqrt{3}\)
Re: Question 8: Triangle ABC is inscribed in a semicircle center
[#permalink]
23 May 2016, 06:01
23 May 2016, 06:01
Expert Reply
Solution
We need to find the radius of the semi circle. Let it be R. Now we have triangle. ADB which is an isosceles triangle. With angles 120, 30 and 30.
Now we know if we drop a perpendicular from D on AB we will bisect the line AB. Let perpendicular from D to AB intersect at X
So we know that \(r *cos(30) = AX = \frac{6}{2}=3\).
So \(r= \frac{3}{cos 30}=2\sqrt{3}\)
Now triangle BDC is an equilateral triangle where its sides are equal to \(r = 2\sqrt{3}= BC\)
Finally we have the triangle ABC which is a right triangle with angle B as 90. So Area of triangle ABC.
\(=\frac{1}{2}*AB*BC = \frac{1}{2}*6*2\sqrt{3}= 6\sqrt{3}\)
Please follow the posting rules for the question and post the question in text format as possible. It is very important because otherwise people cannot search for the question if it is in image format.
Regards
We need to find the radius of the semi circle. Let it be R. Now we have triangle. ADB which is an isosceles triangle. With angles 120, 30 and 30.
Now we know if we drop a perpendicular from D on AB we will bisect the line AB. Let perpendicular from D to AB intersect at X
So we know that \(r *cos(30) = AX = \frac{6}{2}=3\).
So \(r= \frac{3}{cos 30}=2\sqrt{3}\)
Now triangle BDC is an equilateral triangle where its sides are equal to \(r = 2\sqrt{3}= BC\)
Finally we have the triangle ABC which is a right triangle with angle B as 90. So Area of triangle ABC.
\(=\frac{1}{2}*AB*BC = \frac{1}{2}*6*2\sqrt{3}= 6\sqrt{3}\)
Please follow the posting rules for the question and post the question in text format as possible. It is very important because otherwise people cannot search for the question if it is in image format.
Regards
Re: Question 8: Triangle ABC is inscribed in a semicircle center
[#permalink]
23 May 2016, 20:42
23 May 2016, 20:42
1
As ABC is inscribed in a semi circle, ABC is a right angle triangle. Angle ADB is 120, any perpendicular line from D to line AB bisects AB and let us assume the point of bisection is E. Now AED is 30, 60, 90 triangle. Given that AE is 3, AD will be 2*root 3. As triangle BDC is an equilateral, BC is also 2*root 3.
So the area of ABC is 1/2*6*2*root 3= 6*root 3, Hence answer is B
So the area of ABC is 1/2*6*2*root 3= 6*root 3, Hence answer is B
