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.
In triangle ABC, AB = AC = 5 and BC = 8
[#permalink]
15 Jul 2020, 22:50
15 Jul 2020, 22:50
1
1
Bookmarks
00:00
Question Stats:
78% (01:02) correct
21% (01:05) wrong based on 19 sessions
Hide Show timer Statistics
In triangle ABC, AB = AC = 5 and BC = 8. What is the area of the triangle?
A. 10
B. 12
C. 15
D. 24
E. 30
A. 10
B. 12
C. 15
D. 24
E. 30
Moderator
Joined: 02 Jan 2020
Status:GRE Quant Tutor
Posts: 1141
Given Kudos: 9
Location: India
Concentration: General Management
Schools: XLRI Jamshedpur, India - Class of 2014
GMAT 1: 700 Q51 V31
GPA: 2.8
WE:Engineering (Computer Software)
Re: In triangle ABC, AB = AC = 5 and BC = 8
[#permalink]
16 Jul 2020, 04:29
16 Jul 2020, 04:29
1
Refer Attached Image
In triangle ABC, AB = AC = 5, that means its an isosceles triangle. This question is based on a property of isosceles triangle.
"In an Isosceles Triangle the perpendicular from the vertex to the non equal side is a perpendicular bisector to the non-equal side"
In the attached figure, AD which is a perpendicular draw from A to BC will touch BC at its mid-point, which is D. => BD = DC = 8/2 = 4
To find the Area of ABC we need to find the height of ABC which is AD.
Using Pythagoras theorem on ADB we get
\(AB^2\) =\( BD^2\) + \(AD^2\)
=> \(5^2\) =\( 4^2\) + \(AD^2\)
=> \(AD^2\) = 25 - 16 = 9 = \(3^2\)
=> AD = 3
So, Area of ABC = \(\frac{1}{2}\)BC * AD = \(\frac{1}{2}\) * 8 * 3 = 12
So, Answer will be B
Hope it helps!
In triangle ABC, AB = AC = 5, that means its an isosceles triangle. This question is based on a property of isosceles triangle.
"In an Isosceles Triangle the perpendicular from the vertex to the non equal side is a perpendicular bisector to the non-equal side"
In the attached figure, AD which is a perpendicular draw from A to BC will touch BC at its mid-point, which is D. => BD = DC = 8/2 = 4
To find the Area of ABC we need to find the height of ABC which is AD.
Using Pythagoras theorem on ADB we get
\(AB^2\) =\( BD^2\) + \(AD^2\)
=> \(5^2\) =\( 4^2\) + \(AD^2\)
=> \(AD^2\) = 25 - 16 = 9 = \(3^2\)
=> AD = 3
So, Area of ABC = \(\frac{1}{2}\)BC * AD = \(\frac{1}{2}\) * 8 * 3 = 12
So, Answer will be B
Hope it helps!
Attachments
Image.jpg [ 42.46 KiB | Viewed 2639 times ]
Re: In triangle ABC, AB = AC = 5 and BC = 8
[#permalink]
07 Jan 2025, 21:30
07 Jan 2025, 21:30
1
Going by Heron's formula, A= sqrt(s*(s-a)(s-b)(s-c)) where a, b, and c are sides and s is the semi perimeter;
Here s = (5+5+8)/2 = 9
Therefore, A= sqrt(9*4*4*1) = 12.
Here s = (5+5+8)/2 = 9
Therefore, A= sqrt(9*4*4*1) = 12.
