An equilateral triangle that has an area of 9√3 and is inscribed in a circle. What is the area of the circle?
a) 6∏ b) 9∏ c) 12∏ d) 9∏√3 e) 18∏√3
answer c
equilateral triangle inscribed in circle.
This topic has expert replies
-
- Senior | Next Rank: 100 Posts
- Posts: 48
- Joined: Fri Aug 08, 2008 4:44 am
- Thanked: 4 times
- GMAT Score:750
the area of the triangle is
a^2*sqrt(3)/4=9*sqrt(3) => a =6, where a is the side of the triangle.
the radius of the circle is 2/3*a*sqrt(3)/2=a*sqrt(3)/3
square of the circle = a^2*(pi)/3=36*(pi)/3=12*pi
a^2*sqrt(3)/4=9*sqrt(3) => a =6, where a is the side of the triangle.
the radius of the circle is 2/3*a*sqrt(3)/2=a*sqrt(3)/3
square of the circle = a^2*(pi)/3=36*(pi)/3=12*pi
-
- Legendary Member
- Posts: 829
- Joined: Mon Jul 07, 2008 10:09 pm
- Location: INDIA
- Thanked: 84 times
- Followed by:3 members
Hi pepeprepa.. this is how i do it.. i know its a bit difficult to remember the formulae but we have no options...
area of equilateral triangle is sqrt3/4*a^2
sqrt3/4*a^2 =9 rt 3
hence a^2 = 36, a =6
now we need to find the height
height of equilateral triangle is sqroot 3/2 * a
which is rt3/2*6 = 3sqrt3
now u shud know a rule ...
The centroid (i.e the centre of circumcircle divides this) height in 2 : 1 ratio
thus radius would be
2/3* 3 sqrt 3 = 2sqrt 3.
now area of a circle is pi* r^2
thus its pi*(2sqrt3)^2 = 12pi.
i hope its clear now..
do let me know if u have any doubts...
area of equilateral triangle is sqrt3/4*a^2
sqrt3/4*a^2 =9 rt 3
hence a^2 = 36, a =6
now we need to find the height
height of equilateral triangle is sqroot 3/2 * a
which is rt3/2*6 = 3sqrt3
now u shud know a rule ...
The centroid (i.e the centre of circumcircle divides this) height in 2 : 1 ratio
thus radius would be
2/3* 3 sqrt 3 = 2sqrt 3.
now area of a circle is pi* r^2
thus its pi*(2sqrt3)^2 = 12pi.
i hope its clear now..
do let me know if u have any doubts...
-
- Legendary Member
- Posts: 661
- Joined: Tue Jul 08, 2008 12:58 pm
- Location: France
- Thanked: 48 times
Thanks a lot for the knowledge sudhir.
I think you could have skipped the calculation of height.
Because the radius of a circumcircle with equilateral triangle inscribed is also
(one side)/sqrt(3)
So we have two formulas to find this radius with equilateral triangles and this one is more direct.
I think you could have skipped the calculation of height.
Because the radius of a circumcircle with equilateral triangle inscribed is also
(one side)/sqrt(3)
So we have two formulas to find this radius with equilateral triangles and this one is more direct.
Here is another way to do this if you don't know centeriod formula
Once you side of triangle as 6. Draw a perpendicular from cebter of circle to one side of triangle, this will bisect the side of triangle.
Make triangle 90-60-30
get radius as 6/squat3
Area as 12 pie
Once you side of triangle as 6. Draw a perpendicular from cebter of circle to one side of triangle, this will bisect the side of triangle.
Make triangle 90-60-30
get radius as 6/squat3
Area as 12 pie
-
- Master | Next Rank: 500 Posts
- Posts: 324
- Joined: Thu Dec 24, 2009 6:29 am
- Thanked: 17 times
- Followed by:1 members
Awesome stuff! Thank yousudhir3127 wrote:Hi pepeprepa.. this is how i do it.. i know its a bit difficult to remember the formulae but we have no options...
area of equilateral triangle is sqrt3/4*a^2
sqrt3/4*a^2 =9 rt 3
hence a^2 = 36, a =6
now we need to find the height
height of equilateral triangle is sqroot 3/2 * a
which is rt3/2*6 = 3sqrt3
now u shud know a rule ...
The centroid (i.e the centre of circumcircle divides this) height in 2 : 1 ratio
thus radius would be
2/3* 3 sqrt 3 = 2sqrt 3.
now area of a circle is pi* r^2
thus its pi*(2sqrt3)^2 = 12pi.
i hope its clear now..
do let me know if u have any doubts...
-
- Senior | Next Rank: 100 Posts
- Posts: 40
- Joined: Tue Mar 29, 2011 8:08 pm
- Thanked: 1 times
- Followed by:1 members
This is obviously the easiest way to find. The only formula you have to know is area of equiateral triangle (squareroot3/4 *(side^2)) which gives 6 as side of triangle. Then use 30-60-90 rules (x, x*squareroot3, 2x) to find radius which is 6/squareroot3. Anyway, ya I like this way.arorag wrote:Here is another way to do this if you don't know centeriod formula
Once you side of triangle as 6. Draw a perpendicular from cebter of circle to one side of triangle, this will bisect the side of triangle.
Make triangle 90-60-30
get radius as 6/squat3
Area as 12 pie
- MartyMurray
- Legendary Member
- Posts: 2131
- Joined: Mon Feb 03, 2014 9:26 am
- Location: https://martymurraycoaching.com/
- Thanked: 955 times
- Followed by:140 members
- GMAT Score:800
- GMATinsight
- Legendary Member
- Posts: 1100
- Joined: Sat May 10, 2014 11:34 pm
- Location: New Delhi, India
- Thanked: 205 times
- Followed by:24 members
Truly Basic methods are the best... However for those who are Quant Freaks...
There is an expression to calculate the radius of the circle encircling any Triangle
This is called Circumradious, R = (a x b x c)/ (4 x Area of Triangle)
where a, b and c are sides of Triangle
Area od Equilateral Triangle = (√3/4)Side^2 = 9√3 ==> Side^2 = 36 ==> Side = 6
Here R = (6x6x6)/(4x9√3) = 6/√3 = 2√3
Area of Circle = � R^2 = � (2√3)^2 = 12�
Answer: Option C
There is an expression to calculate the radius of the circle encircling any Triangle
This is called Circumradious, R = (a x b x c)/ (4 x Area of Triangle)
where a, b and c are sides of Triangle
Area od Equilateral Triangle = (√3/4)Side^2 = 9√3 ==> Side^2 = 36 ==> Side = 6
Here R = (6x6x6)/(4x9√3) = 6/√3 = 2√3
Area of Circle = � R^2 = � (2√3)^2 = 12�
Answer: Option C
"GMATinsight"Bhoopendra Singh & Sushma Jha
Most Comprehensive and Affordable Video Course 2000+ CONCEPT Videos and Video Solutions
Whatsapp/Mobile: +91-9999687183 l [email protected]
Contact for One-on-One FREE ONLINE DEMO Class Call/e-mail
Most Efficient and affordable One-On-One Private tutoring fee - US$40-50 per hour
Most Comprehensive and Affordable Video Course 2000+ CONCEPT Videos and Video Solutions
Whatsapp/Mobile: +91-9999687183 l [email protected]
Contact for One-on-One FREE ONLINE DEMO Class Call/e-mail
Most Efficient and affordable One-On-One Private tutoring fee - US$40-50 per hour
- GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
I posted an explanation here:lutp44 wrote:Can someone please explain to me how we get the radius? In a 30-60-90 triangle, the side corresponding to 60 degree angle is 3. I don't understand how the radius is found to be 6sqrt3.
https://www.beatthegmat.com/gmat-set-7-q8-t282169.html
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3