|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
11 Jun 2025, 09:40
Kudos
Bookmarks
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct
100% (02:37) wrong based on 1 sessions
History
Date
Time
Result
Not Attempted Yet
∠ ABC = 45° and ED = DF. The area of triangle ABC is 4 times the area of triangle DEF.
Quantity A
\(\frac{AC}{EF}\)
Quantity B
\(\sqrt{2}\)
Attachment:
GMAT-Club-Forum-ywitpwj4.png [ 19.67 KiB | Viewed 152 times ]
14 Jun 2025, 07:08
Kudos
Bookmarks
OE
Since ∠ ABC is 45°, ∠ BCA is 180° - 90° - 45° = 45°, so ABC is an isosceles right triangle. Because ED = DF, triangle DEF is also an isosceles right triangle. The formula for the area of a triangle is Area =\(\frac{1}{2}\) base × height. Since the triangles in this question are isosceles right triangles, the formula for the area of a triangle can be restated as \(\frac{1}{2}\) \(base^2\) by substituting the base in for the height, since they are the same value. Given that the area of triangle ABC is four times the area of triangle DEF, set up the equation\((\frac{1}{2})AC^2=4(\frac{1}{2})DF^2\) Even though there are two variables but only one equation, you can solve by Picking Numbers because only the ratio matters, not the actual values. Try 2 for the value of AC to get \(\frac{1}{2}2^2\)=\(2DF^2\), so \(DF^2=1\) and \(DF=1.\) Given the side ratios of an isosceles right triangle, \(1:1:\sqrt{2}\), side \(EF=\sqrt{2}\). So Quantity A becomes \(\frac{AC}{EF}=\frac{2}{\sqrt{2}}\) Multiply both the numerator and the denominator by \(\sqrt{2}\) to get \(\frac{(\sqrt{2})2}{(\sqrt{2})\sqrt{2}}\)=\( \frac{2\sqrt{2}}{2}\)=\(\sqrt{2}\). This is the same as Quantity B, so (C) is correct.
Answer: C
Since ∠ ABC is 45°, ∠ BCA is 180° - 90° - 45° = 45°, so ABC is an isosceles right triangle. Because ED = DF, triangle DEF is also an isosceles right triangle. The formula for the area of a triangle is Area =\(\frac{1}{2}\) base × height. Since the triangles in this question are isosceles right triangles, the formula for the area of a triangle can be restated as \(\frac{1}{2}\) \(base^2\) by substituting the base in for the height, since they are the same value. Given that the area of triangle ABC is four times the area of triangle DEF, set up the equation\((\frac{1}{2})AC^2=4(\frac{1}{2})DF^2\) Even though there are two variables but only one equation, you can solve by Picking Numbers because only the ratio matters, not the actual values. Try 2 for the value of AC to get \(\frac{1}{2}2^2\)=\(2DF^2\), so \(DF^2=1\) and \(DF=1.\) Given the side ratios of an isosceles right triangle, \(1:1:\sqrt{2}\), side \(EF=\sqrt{2}\). So Quantity A becomes \(\frac{AC}{EF}=\frac{2}{\sqrt{2}}\) Multiply both the numerator and the denominator by \(\sqrt{2}\) to get \(\frac{(\sqrt{2})2}{(\sqrt{2})\sqrt{2}}\)=\( \frac{2\sqrt{2}}{2}\)=\(\sqrt{2}\). This is the same as Quantity B, so (C) is correct.
Answer: C
