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.
The area of the right triangle below is 2√3. What is the length of the
[#permalink]
16 Feb 2022, 06:00
16 Feb 2022, 06:00
Expert Reply
1
Bookmarks
00:00
Question Stats:
68% (01:22) correct
31% (01:59) wrong based on 16 sessions
Hide Show timer Statistics
The area of the right triangle below is 2√3. What is the length of the altitude h?
GRE The area of the right triangle below.jpg [ 12.28 KiB | Viewed 2306 times ]
(A) √2
(B) √3
(C) 2
(D) 2√2
(E) 3
Attachment:
GRE The area of the right triangle below.jpg [ 12.28 KiB | Viewed 2306 times ]
(A) √2
(B) √3
(C) 2
(D) 2√2
(E) 3
The area of the right triangle below is 2√3. What is the length of the
[#permalink]
16 Feb 2022, 11:03
16 Feb 2022, 11:03
1
Lets find length of other side --> 1/2 * other side * 2 = 2\(\sqrt{3} ==> 2\sqrt{3}\)
Apply Pythagoras to bigger triangle \(2^2 + (2\sqrt{3})^2 = 4 +12 = \sqrt{16} ==> 4\)
So we can calculate the area of triangle again using the altitude ==> \(\frac{1}{2} * 4 * h = 2\sqrt{3} ==>> h =\sqrt{3}\)
Apply Pythagoras to bigger triangle \(2^2 + (2\sqrt{3})^2 = 4 +12 = \sqrt{16} ==> 4\)
So we can calculate the area of triangle again using the altitude ==> \(\frac{1}{2} * 4 * h = 2\sqrt{3} ==>> h =\sqrt{3}\)
Re: The area of the right triangle below is 23. What is the length of the
[#permalink]
04 Jun 2024, 19:24
04 Jun 2024, 19:24
AkkuJi wrote:
Lets find length of other side --> 1/2 * other side * 2 = 2\(\sqrt{3} ==> 2\sqrt{3}\)
Apply Pythagoras to bigger triangle \(2^2 + (2\sqrt{3})^2 = 4 +12 = \sqrt{16} ==> 4\)
So we can calculate the area of triangle again using the altitude ==> \(\frac{1}{2} * 4 * h = 2\sqrt{3} ==>> h =\sqrt{3}\)
Apply Pythagoras to bigger triangle \(2^2 + (2\sqrt{3})^2 = 4 +12 = \sqrt{16} ==> 4\)
So we can calculate the area of triangle again using the altitude ==> \(\frac{1}{2} * 4 * h = 2\sqrt{3} ==>> h =\sqrt{3}\)
Can you please explain it how base is taken as 4??
So we can calculate the area of triangle again using the altitude ==> \(\frac{1}{2} * 4 * h = 2\sqrt{3} ==>> h =\sqrt{3}\)[/quote]
