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Practice Exam 1 - GMAT Prep - Geometry

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lucas211 Senior | Next Rank: 100 Posts Default Avatar
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Practice Exam 1 - GMAT Prep - Geometry

Post Wed May 25, 2016 4:19 pm
Hello BTG

Would appreciate a little help on the following question:

Thanks Smile

ttp://postimg.org/image/rj0iwsrqj/" target="_blank">

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OptimusPrep Master | Next Rank: 500 Posts
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Post Thu May 26, 2016 6:53 pm
lucas211 wrote:
Hello BTG

Would appreciate a little help on the following question:

Thanks Smile

ttp://postimg.org/image/rj0iwsrqj/" target="_blank">
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

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Top Reply
Post Wed May 25, 2016 4:39 pm
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)
= √3/2
= B

RELATED RESOURCE
See this video to learn all of the circle properties that the GMAT tests: https://www.gmatprepnow.com/module/gmat-geometry/video/880

Cheers,
Brent

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