Last visit was: 14 Jul 2024, 07:43 It is currently 14 Jul 2024, 07:43
Toolkit
GMAT 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

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

# Triangle ABC is inscribed in a semicircle. What is the area of the sh

SORT BY:
Tags:
Show Tags
Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 94342
Own Kudos [?]: 640711 [3]
Given Kudos: 85011
Senior Manager
Joined: 07 Jul 2012
Posts: 281
Own Kudos [?]: 325 [0]
Given Kudos: 71
Location: India
Concentration: Finance, Accounting
GPA: 3.5
Senior SC Moderator
Joined: 22 May 2016
Posts: 5327
Own Kudos [?]: 35754 [2]
Given Kudos: 9464
Target Test Prep Representative
Joined: 14 Oct 2015
Status:Founder & CEO
Affiliations: Target Test Prep
Posts: 19130
Own Kudos [?]: 22627 [0]
Given Kudos: 286
Location: United States (CA)
Re: Triangle ABC is inscribed in a semicircle. What is the area of the sh [#permalink]
Bunuel wrote:

Triangle ABC is inscribed in a semicircle. What is the area of the shaded region above?

(A) $$2\pi-2$$

(B) $$2\pi - 4$$

(C) $$4\pi - 4$$

(D) $$8\pi - 4$$

(E) $$8\pi - 8$$

Attachment:
2018-02-14_0952.png

First of all, we should observe that the angle ABC is an angle subtended by a diameter, therefore it is a right angle.

We see that we have an isosceles right triangle (45-45-90) in which AC = hypotenuse of triangle = diameter of the semicircle. Recall that the ratio of a leg of a 45-45-90 triangle to its hypotenuse is x : x√2.

Thus, AC = 2√2 x √2 = 4, so we know the circle’s diameter is 4 and its radius is 2.

The area of the semicircle is 1/2 x π x 2^2 = 2π.

The area of the triangle is (1/2)*(2√2)^2 = 8/2 = 4.

Thus, the area of the shaded region is 2π - 4.