GRE Prep 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.
∠ ABC = 45 ° and ED = DF. The area of triangle ABC is 4
[#permalink]
03 Oct 2018, 22:51
03 Oct 2018, 22:51
1
Expert Reply
3
Bookmarks
00:00
Question Stats:
44% (02:12) correct
55% (01:54) wrong based on 95 sessions
Hide Show timer Statistics
Attachment:
triangles.jpg [ 8.03 KiB | Viewed 6632 times ]
∠ ABC = 45 ° and ED = DF. The area of triangle ABC is 4 times the area of triangle DEF.
Quantity A |
Quantity B |
\(\frac{AC}{EF}\) |
\(\sqrt{2}\) |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Re: ∠ ABC = 45 ° and ED = DF. The area of triangle ABC is 4
[#permalink]
04 Oct 2018, 04:47
04 Oct 2018, 04:47
1
1
Bookmarks
Carcass wrote:
Attachment:
triangles.jpg
∠ ABC = 45 ° and ED = DF. The area of triangle ABC is 4 times the area of triangle DEF.
Quantity A |
Quantity B |
\(\frac{AC}{EF}\) |
\(\sqrt{2}\) |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
In triangle ABC we have ∠ ABC = 45 ° , ∠ CAB = 90 °
Therefore ∠ ACB = 45 °
So the triangle is isosceles triangle
Similarly in triangle DEF, we have DF = DE and ∠ FDE = 90 °
Therefore ∠ DEF = ∠ DFE = 45 °
Now we can write as
\(\frac{1}{2} AC^2 = 4* \frac{1}{2} DF^2\)
or \(AC = 2*DF\)
In triangle DEF, since it is 90-45-45 the sides are distributed in
\(1 : 1 : \sqrt2\)
i.e. \(DF = DE = 1\) and \(EF = \sqrt2\)
Therefore AC = 2 * 1 = 2
Now \(QTY A =\frac{AC}{EF} = \frac{2}{{\sqrt2}}\) =\(2 *\frac{{\sqrt2}}{{\sqrt2 *\sqrt2}}\) = \(\sqrt2\)
Hence QTY A = QTY B
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
∠ ABC = 45 ° and ED = DF. The area of triangle ABC is 4
[#permalink]
30 Mar 2021, 07:47
30 Mar 2021, 07:47
1
Carcass wrote:
Attachment:
triangles.jpg
∠ ABC = 45 ° and ED = DF. The area of triangle ABC is 4 times the area of triangle DEF.
Quantity A |
Quantity B |
\(\frac{AC}{EF}\) |
\(\sqrt{2}\) |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Triangle ABC and DEF, both are Isosceles Right Triangles with ratio of their sides \(1 : 1 : \sqrt{2}\)
Also, both the triangles are Similar
i.e. \(\frac{AB}{ED} = \frac{AC}{DF} = \frac{BC}{EF} = \sqrt{\frac{Ar. (ABC)}{ Ar. (DEF)}
}\)
In Triangle ABC:
Let AB = AC = \(Y\)
So, BC = \(\sqrt{2}Y\)
Area = \(\frac{Y^2}{2}\)
In Triangle DEF:
Let ED = DF = \(X\)
So, EF = \(\sqrt{2}X\)
Area = \(\frac{X^2}{2}\)
Given, \(\frac{Y^2}{2} = 4(\frac{X^2}{2})\)
\(Y = 2X\)
Col. A: \(\frac{Y}{\sqrt{2}X}\)
Col. B: \(\sqrt{2}\)
Col. A: \(\frac{2X}{\sqrt{2}X}\)
Col. B: \(\sqrt{2}\)
Col. A: \(\sqrt{2}\)
Col. B: \(\sqrt{2}\)
Hence, option C
