Last visit was: 16 Nov 2024, 10:11 It is currently 16 Nov 2024, 10:11

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
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Own Kudos [?]: 3221 [3]
Given Kudos: 172
Location: India
WE:Education (Education)
Send PM
Manager
Manager
Joined: 26 Mar 2021
Posts: 82
Own Kudos [?]: 162 [1]
Given Kudos: 6
GMAT 1: 660 Q49 V31
GMAT 2: 660 Q48 V33
Send PM
Intern
Intern
Joined: 23 May 2020
Posts: 17
Own Kudos [?]: 2 [0]
Given Kudos: 8
Send PM
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Own Kudos [?]: 3221 [2]
Given Kudos: 172
Location: India
WE:Education (Education)
Send PM
Re: In the figure above, what is the area of triangle ABC? [#permalink]
2
wewake06298 wrote:
Can you explain this 3/root(2) how it came please?


wewake06298

In 45-45-90 Right triangle, the sides are in ratio \(x : x : \sqrt{2}x\)

i.e. opposite to 45 degree angle, side would be \(x\) whereas opposite to angle 90, side would be \(\sqrt{2}x\)

In the question,
\(BC = 3 = \sqrt{2}x\)

\(x = \frac{3}{\sqrt{2}}\)

Therefore, AB = AC = \(x = \frac{3}{\sqrt{2}}\)
Retired Moderator
Joined: 02 Dec 2020
Posts: 1833
Own Kudos [?]: 2146 [1]
Given Kudos: 140
GRE 1: Q168 V157

GRE 2: Q167 V161
Send PM
Re: In the figure above, what is the area of triangle ABC? [#permalink]
1
Hi wewake06298

So in Triangle ABC, Angle ABC = angle ACB = 45°, and angle BAC = 90°

Now this is 45-45-90 triangle. so sides are in ratio of 1:1:\(\sqrt{2}\)
WHY?

in triangle ABC , sin C = \(\frac{AB}{BC}\) & cos C = \(\frac{AC}{BC}\)
Now we know that angle C = 45° so sin C = sin 45° = \(\frac{1}{\sqrt{2}}\)

\(\frac{1}{\sqrt{2}}\) = \(\frac{AB}{BC}\) = \(\frac{AC}{BC}\)

AB = AC = \(\sqrt{2}\)BC
so they are in proportion of 1:1:\(\sqrt{2}\)

And if one side given is BC = 3 so other will be \(\frac{3}{\sqrt{2}}\)

Ask if the doubt still remains.
Also please check this : GRE MATH book Special Triangle theory prepared by Carcass

Regards,
R

wewake06298 wrote:
Can you explain this 3/root(2) how it came please?
Manager
Manager
Joined: 26 Mar 2021
Posts: 82
Own Kudos [?]: 162 [1]
Given Kudos: 6
GMAT 1: 660 Q49 V31
GMAT 2: 660 Q48 V33
Send PM
Re: In the figure above, what is the area of triangle ABC? [#permalink]
1
wewake06298 wrote:
Can you explain this 3/root(2) how it came please?


Notice triangle ABC is 45-45-90 triangle. The sides are always in fixed ratio \(1 : 1 : \sqrt{2}\).
I assume you are not able to understand how to use this fixed ratio in order to find the length of sides for the triangle ABC.

However, this is not very difficult to understand. You may write the fixed ratio as \(k : k : \sqrt{2}*k\).
Sides opposite to 45° are equal, and the length is k here. The side opposite to 90° is \(\sqrt{2}*k\).

Now BC is opposite to 90°, so \(BC = \sqrt{2}*k\)
We already know BC = 3.
Equating both equations for BC,
\(\sqrt{2}*k = 3\)
\(k = \frac{3}{\sqrt{2}}\)

Therefore, the sides opposite to 45°, AC and AB is \(\frac{3}{\sqrt{2}}\).
Prep Club for GRE Bot
Re: In the figure above, what is the area of triangle ABC? [#permalink]
Moderators:
GRE Instructor
78 posts
GRE Forum Moderator
37 posts
Moderator
1111 posts
GRE Instructor
234 posts

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