| Last visit was: 20 Sep 2026, 18:31 |
It is currently 20 Sep 2026, 18:31 |
|
|
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.
31 Aug 2020, 10:02
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
49% (01:35) correct
51%
(01:53)
wrong
based on 98
sessions
History
Date
Time
Result
Not Attempted Yet
In an isosceles triangle ABC, AD is a median of length \(5\sqrt{2}\). What is the perimeter of the triangle ABC?
1. AD=BD=CD
2. Triangle ABC is right-angled at A.
1. AD=BD=CD
2. Triangle ABC is right-angled at A.
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
yashikaaggarwal
Senior Moderator - Masters Forum
Joined: 19 Jan 2020
Last visit: 10 Aug 2026
Posts: 3,094
Own Kudos:
Given Kudos: 1,510
Location: India
Schools: Cranfield (A) Warwick MiM "23 (M)
GPA: 4
WE:Analyst (Internet and New Media)
31 Aug 2020, 11:05
Kudos
Bookmarks
Given: In an isosceles triangle ABC, AD is a median of length 5√2.
To find: What is the perimeter of the triangle ABC?
1. AD=BD=CD
Angle DAC= Angle DCA = Angle DBA = Angle DAB = 45
Angle A = 90
Area of triangle = 1/2*Height*base
Area of triangle = 1/2*5√2*2*5√2.
Area of triangle = 50
(Sufficient)
2. Triangle ABC is right-angled at A.
AD is median
therefore Angle DBA = Angle DAB = 45
the remaining angles by triangle sum property will be Angle DAC= Angle DCA = 45
Area of triangle = 1/2*Height*base
Area of triangle = 1/2*5√2*2*5√2.
Area of triangle = 50
(Sufficient)
Answer is D
To find: What is the perimeter of the triangle ABC?
1. AD=BD=CD
Angle DAC= Angle DCA = Angle DBA = Angle DAB = 45
Angle A = 90
Area of triangle = 1/2*Height*base
Area of triangle = 1/2*5√2*2*5√2.
Area of triangle = 50
(Sufficient)
2. Triangle ABC is right-angled at A.
AD is median
therefore Angle DBA = Angle DAB = 45
the remaining angles by triangle sum property will be Angle DAC= Angle DCA = 45
Area of triangle = 1/2*Height*base
Area of triangle = 1/2*5√2*2*5√2.
Area of triangle = 50
(Sufficient)
Answer is D
Attachments
Triangle.png [ 39.37 KiB | Viewed 2311 times ]
01 Sep 2020, 07:24
Kudos
Bookmarks
Given that ABC is an Isosceles triangle and AD is a median of length of \(5\sqrt{2}\)
To find the perimeter of triangle ABC
Statement 1: AD = BD = CD
Let \(\angle{DAC} = x\)
Then \(\angle{DAC} = \angle{DAB} = \angle{ABD} = \angle{ACD} = x\)
In \(\triangle{DAC}: \) x + x + x + x = 180 (Sum of angles of a triangle = \(180^o\))
\(4x = 180^o\), or \(x = 45^o\)
The triangle is an isosceles right angled triangle, right angled at A and hypotenuse BC = \(5\sqrt{2}+ 5\sqrt{2} = 10\sqrt{2}\)
The equal sides AC = AB = \(\frac{1}{\sqrt{2}} * Hypotenuse = 10\)
Therefore the perimeter = \(10 + 10 + 10\sqrt{2} = 20 + 10\sqrt{2}\)
Statement 1 is sufficient.. The answer Options could be A or D
Statement 2: Triangle ABC is right angled at A
Since it is right angled at A, again we have an isosceles right angled triangle, with \(\angle A = 90^o and \angleB = \angle C = 45^o\)
Therefore, the median from A is a perpendicular bisector and an angular bisector. (Property of an isosceles triangle for the median to the unequal side)
\(\triangle{ABD}\) is an isosceles right angled triangle, with AD = BD = \(5\sqrt{2}\)
\(\triangle{ADC}\) is an isosceles right angled triangle, with AD = DC = \(5\sqrt{2}\)
\(\triangle{ABC}\) is an isosceles right angled triangle, with AB = AC and hypotenuse = \(5\sqrt{2} + 5\sqrt{2} = 10\sqrt{2} \)
The equal sides AC = AB = \(\frac{1}{\sqrt{2}} * Hypotenuse = 10\)
Therefore the perimeter = \(10 + 10 + 10\sqrt{2} = 20 + 10\sqrt{2}\)
Statement 2 is sufficient..
Option D
Arun Kumar
To find the perimeter of triangle ABC
Statement 1: AD = BD = CD
Let \(\angle{DAC} = x\)
Then \(\angle{DAC} = \angle{DAB} = \angle{ABD} = \angle{ACD} = x\)
In \(\triangle{DAC}: \) x + x + x + x = 180 (Sum of angles of a triangle = \(180^o\))
\(4x = 180^o\), or \(x = 45^o\)
The triangle is an isosceles right angled triangle, right angled at A and hypotenuse BC = \(5\sqrt{2}+ 5\sqrt{2} = 10\sqrt{2}\)
The equal sides AC = AB = \(\frac{1}{\sqrt{2}} * Hypotenuse = 10\)
Therefore the perimeter = \(10 + 10 + 10\sqrt{2} = 20 + 10\sqrt{2}\)
Statement 1 is sufficient.. The answer Options could be A or D
Statement 2: Triangle ABC is right angled at A
Since it is right angled at A, again we have an isosceles right angled triangle, with \(\angle A = 90^o and \angleB = \angle C = 45^o\)
Therefore, the median from A is a perpendicular bisector and an angular bisector. (Property of an isosceles triangle for the median to the unequal side)
\(\triangle{ABD}\) is an isosceles right angled triangle, with AD = BD = \(5\sqrt{2}\)
\(\triangle{ADC}\) is an isosceles right angled triangle, with AD = DC = \(5\sqrt{2}\)
\(\triangle{ABC}\) is an isosceles right angled triangle, with AB = AC and hypotenuse = \(5\sqrt{2} + 5\sqrt{2} = 10\sqrt{2} \)
The equal sides AC = AB = \(\frac{1}{\sqrt{2}} * Hypotenuse = 10\)
Therefore the perimeter = \(10 + 10 + 10\sqrt{2} = 20 + 10\sqrt{2}\)
Statement 2 is sufficient..
Option D
Arun Kumar
Attachments
DS .jpg [ 444.01 KiB | Viewed 2155 times ]
