Need some help with this one. It looks easy, but I solve and get another answer
- Attachments
-
- See attached problem
BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course
Redeem
Featured Course
Target Test Prep GMAT OnDemand
Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.
- 715+ score guarantee — highest in the industry (99th percentile)
- 52 chapters · 1,500+ lessons · 4,000+ practice questions
- 400 hours of video · 1,500+ instructor-led HD smartboard lessons
- 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
- 24/7 live support + weekly Zoom office hours with GMAT instructors
- TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
- 1,200+ flashcards + AI-powered study assistant & daily calendar
- OnDemand, LiveTeach & GMAT Bootcamp formats available
- Also: GRE, SAT Math & Executive Assessment courses
- MBA Admissions Consulting now available
- 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
- 200,000+ students served
- 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★
5.0
(559 reviews)
130-pt guarantee
$0 to start
then $127/mo
Powerprep : Co-oridnate Geom, Triangles
Source: Beat The GMAT — Problem Solving |
what i did is get the radius 1st...
coordinates (- sqrt 3, and 1) are given..i formed a triangle with x = sqrt 3 as the base, 1 as the height...Pythagorean theorem gives hypotenuse of 2
(sqrt 3)^2 + 1^2 = sqrt 4 = 2
so the radius is 2
a half circle means the arc is 180 degrees
form an imaginary triangle with radii 2..that triangle has a measure of 90 degrees at the (0,0)......so the other two triangles must = 90 degrees
look at the smaller triangle you formed with base sqrt 3 and height 1..the formula that stands out is 1-sqrt 3-2 which is a 30-60-90 triangle...
if the height is 1 then the angle opposite one is the shorter angle or 30 degrees
so..from the large imaginary triangle of 90 degrees and the small one of 30 degrees..the imaginary triangle in quadrant 1 must be 60 degrees...since the radius is 2...and we know one angle is 60 and one is 90..the other must be 30 or equation 1 - sqrt 3 - 2 again..
since we know the arc is 60 degrees...
since 60 degrees is the angle of the arc the height must be sqrt 3 (or the height) and the 90 degrees is the radius or 2..the shortest leg (in this case the x-coordinate) is 1 (the 30 degree line)
coordinates (- sqrt 3, and 1) are given..i formed a triangle with x = sqrt 3 as the base, 1 as the height...Pythagorean theorem gives hypotenuse of 2
(sqrt 3)^2 + 1^2 = sqrt 4 = 2
so the radius is 2
a half circle means the arc is 180 degrees
form an imaginary triangle with radii 2..that triangle has a measure of 90 degrees at the (0,0)......so the other two triangles must = 90 degrees
look at the smaller triangle you formed with base sqrt 3 and height 1..the formula that stands out is 1-sqrt 3-2 which is a 30-60-90 triangle...
if the height is 1 then the angle opposite one is the shorter angle or 30 degrees
so..from the large imaginary triangle of 90 degrees and the small one of 30 degrees..the imaginary triangle in quadrant 1 must be 60 degrees...since the radius is 2...and we know one angle is 60 and one is 90..the other must be 30 or equation 1 - sqrt 3 - 2 again..
since we know the arc is 60 degrees...
since 60 degrees is the angle of the arc the height must be sqrt 3 (or the height) and the 90 degrees is the radius or 2..the shortest leg (in this case the x-coordinate) is 1 (the 30 degree line)
Thanks, that helped.
The diagram as drawn made the figure look symmetrical wrt to the Y axis and that through me off.
The key was to guess given the sides as 1:sqrt3:2 as a 30-60-90 triangle on the left of the Y-axis.
Thanks, this was driving me nuts.
The diagram as drawn made the figure look symmetrical wrt to the Y axis and that through me off.
The key was to guess given the sides as 1:sqrt3:2 as a 30-60-90 triangle on the left of the Y-axis.
Thanks, this was driving me nuts.
