Last visit was: 02 Aug 2025, 16:18 It is currently 02 Aug 2025, 16:18

Close

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

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

Track
Your Progress

every week, we’ll send you an estimated GRE 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.

Close

Request Expert Reply

Confirm Cancel
avatar
Manager
Manager
Joined: 12 Jan 2016
Posts: 142
Own Kudos [?]: 202 [17]
Given Kudos: 0
Send PM
avatar
Intern
Intern
Joined: 02 Jul 2016
Posts: 25
Own Kudos [?]: 18 [1]
Given Kudos: 0
Send PM
avatar
Intern
Intern
Joined: 03 Jun 2016
Posts: 37
Own Kudos [?]: 148 [4]
Given Kudos: 0
Send PM
User avatar
Retired Moderator
Joined: 07 Jun 2014
Posts: 4815
Own Kudos [?]: 11678 [2]
Given Kudos: 0
GRE 1: Q167 V156
WE:Business Development (Energy and Utilities)
Send PM
Re: Area of triangle ADB..... [#permalink]
2
Expert Reply
Solving for x we see than ABC is a right triangle with 90 degree at A, 30 degrees at B and 60 at C.

Let side AC = 1 then side AB = \(\sqrt{3}\) and side BC= 2.

DC = 1 since ADC is an equilateral triangle




Area of ADC = \(\sqrt{3}\frac{1}{4}* (1)^2\)=\(\frac{1}{4}*\sqrt{3}\).


Area of big triangle ABC= \(\frac{1}{2}*1*\sqrt{3}\).

Area of triangle ABD = Area of big triangle ABC - Area of ADC = \(\frac{1}{2}*1*\sqrt{3}-\)\(\frac{1}{4}*\sqrt{3}\)=\(\frac{1}{4}*\sqrt{3}\).

Hence option C is the right answer.

PS: phoenixio's method is much better.
avatar
Intern
Intern
Joined: 16 Jul 2021
Posts: 17
Own Kudos [?]: 18 [0]
Given Kudos: 22
Send PM
Re: Area of triangle ADB..... [#permalink]
How is it possible for sides CD = BD when angle x corresponds to CD and angle 2x corresponds to BD? Kudos for explanation.
Retired Moderator
Joined: 02 Dec 2020
Posts: 1831
Own Kudos [?]: 2156 [1]
Given Kudos: 140
GRE 1: Q168 V157

GRE 2: Q167 V161
Send PM
Re: Area of triangle ADB..... [#permalink]
1
In triangle \(ADC\), when solved for \(x\), the triangle will be an equilateral one.

Hence, \(AD = AC = DC\)

Now, in triangle \(ABD\)

\(AD\) & \(BD\) are the sides opposite to the same angle \(x\), that's why they will be same.

So, \(AD = BD\)

Finally, \(AD = BD = AC = DC\)

rubytuesdays21 wrote:
How is it possible for sides CD = BD when angle x corresponds to CD and angle 2x corresponds to BD? Kudos for explanation.
avatar
Intern
Intern
Joined: 13 Jul 2025
Posts: 1
Own Kudos [?]: 0 [0]
Given Kudos: 0
Send PM
Re: Area of triangle ADB..... [#permalink]
phoenixio wrote:
In triangle ADB, angle DAB= angle DBA which means AD = BD
similarly in triangle ADC, angle DAC= angle ACD, which means AD=CD
therefore, BD=CD, and height is same for both triangles, hence area will be same for both ADB and ACD.
Answer is C



I understood Your point but isn't the area maximum when its a regular figure, and figure ACD is regular so shouldn't the area of ACD be bigger?
if possible can you tell where my logic is flawed? I understood your explanation but was stuck at this point
Prep Club for GRE Bot
Re: Area of triangle ADB..... [#permalink]
Moderators:
GRE Instructor
121 posts
GRE Forum Moderator
37 posts
Moderator
1141 posts
GRE Instructor
234 posts
Moderator
29 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne