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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## The rhombus (AFCE) is inscribed in a rectangle (ABCD).... tagged by: AAPL ##### This topic has 1 expert reply and 0 member replies ### Top Member ## The rhombus (AFCE) is inscribed in a rectangle (ABCD).... The rhombus (AFCE) is inscribed in a rectangle (ABCD). The length has a width of 20 yards and a length of 25 yards, what would be the total length (the perimeter) of a fence along the sides defined by AFCE? A. 80 yards. B. 82 yards. C. 84 yards. D. 85 yards. E. 90 yards. The OA is B. Please, can any expert assist me with this PS question? I don't have it clear and I appreciate if any explain it for me. Thanks. ### GMAT/MBA Expert GMAT Instructor Joined 04 Oct 2017 Posted: 551 messages Followed by: 11 members Upvotes: 180 Quote: The rhombus (AFCE) is inscribed in a rectangle (ABCD). The length has a width of 20 yards and a length of 25 yards, what would be the total length (the perimeter) of a fence along the sides defined by AFCE? A. 80 yards. B. 82 yards. C. 84 yards. D. 85 yards. E. 90 yards. The OA is B. Please, can any expert assist me with this PS question? I don't have it clear and I appreciate if any explain it for me. Thanks. Hi AAPL, Let's take a look at your question. Since AFCE is a rhombus, there it has all sides of equal length let's say 'x'. $$CD=25$$ $$AD=20$$ Let's suppose, $$DE\ =\ y$$ Therefore, we can write: $$DE\ +EC=\ 25$$ $$y+x=\ 25$$ $$x=\ 25-y...\left(i\right)$$ In the right triangle ADE, $$\left(AE\right)^2=\left(AD\right)^2+\left(DE\right)^2$$ Since AE is the side of the rhombus so it is x as we supposed in the beginning and DE is y, therefore, $$\left(x\right)^2=\left(20\right)^2+\left(y\right)^2$$ $$\left(x\right)^2=400+\left(y\right)^2$$ Pluggingin the value of x from eq(i), we get: $$\left(25-y\right)^2=400+\left(y\right)^2$$ $$625-50y+y^2=400+y^2$$ $$625-50y=400$$ $$50y=625-400$$ $$50y=225$$ $$y=\frac{225}{50}$$ $$y=4.5$$ We need to find the perimeter of the rhombus, for that purpose we will calculate the side length of the rhombus i.e. x. We know from equation (i), that: $$x=25-y$$ $$x=25-4.5$$ $$x=20.5$$ Perimeter of Rhombus = 4 * (Length of side) $$Perimeter\ =\ 4x$$ $$Perimeter\ =\ 82$$ Therefore, Option B is correct. Hope it helps. I am available if you'd like any follow up. _________________ GMAT Prep From The Economist We offer 70+ point score improvement money back guarantee. Our average student improves 98 points. Free 7-Day Test Prep with Economist GMAT Tutor - Receive free access to the top-rated GMAT prep course including a 1-on-1 strategy session, 2 full-length tests, and 5 ask-a-tutor messages. Get started now. • Magoosh Study with Magoosh GMAT prep Available with Beat the GMAT members only code • 5 Day FREE Trial Study Smarter, Not Harder Available with Beat the GMAT members only code • Award-winning private GMAT tutoring Register now and save up to$200

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