Hello BTG
Would appreciate a little help on the following question:
Thanks
Practice Exam 1 - GMAT Prep - Geometry
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- Brent@GMATPrepNow
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Important: AC is the DIAMETER of the circle.
Since ∠ABC holds (or contains) the DIAMETER, ∠ABC = 90 degrees.
So, we have a RIGHT TRIANGLE.
Side AC is the HYPOTENUSE and it has length 2
Side BC is a leg and it has length 1
To find the missing side, we'll us the Pythagorean Theorem:
1² + (side AB)² = 2²
1 + (side AB)² = 4
So, (side AB)² = 3
So, side AB = √3
Since we have a RIGHT TRIANGLE, the area = (1/2)(base)(height)
If we let side AB be the base, then side BC is the height.
So, the area = (1/2)(√3)(1)
= [spoiler]√3/2[/spoiler]
= B
RELATED RESOURCE
See this video to learn all of the circle properties that the GMAT tests: https://www.gmatprepnow.com/module/gmat ... /video/880
Cheers,
Brent
Since ∠ABC holds (or contains) the DIAMETER, ∠ABC = 90 degrees.
So, we have a RIGHT TRIANGLE.
Side AC is the HYPOTENUSE and it has length 2
Side BC is a leg and it has length 1
To find the missing side, we'll us the Pythagorean Theorem:
1² + (side AB)² = 2²
1 + (side AB)² = 4
So, (side AB)² = 3
So, side AB = √3
Since we have a RIGHT TRIANGLE, the area = (1/2)(base)(height)
If we let side AB be the base, then side BC is the height.
So, the area = (1/2)(√3)(1)
= [spoiler]√3/2[/spoiler]
= B
RELATED RESOURCE
See this video to learn all of the circle properties that the GMAT tests: https://www.gmatprepnow.com/module/gmat ... /video/880
Cheers,
Brent
- OptimusPrep
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Always remember: A circle inscribed in a semicircle with always be a right angle triangle.
In this case,
∠ABC = 90
Hence in the triangle ABC, we have base = BC = 1
Hypotenuse = CA = 2
Height = AB^2 = 2^2 + 1^2
Hence AB = √3
Therefore, area = 1/2*√3*1 = √3/2
Correct option: B