right triangles

This topic has expert replies
Source: — Problem Solving |

User avatar
Master | Next Rank: 500 Posts
Posts: 385
Joined: Mon Apr 16, 2012 8:40 am
Location: Pune, India
Thanked: 186 times
Followed by:29 members

by aneesh.kg » Mon Jun 11, 2012 10:46 am
Q1:
Area of shaded region = 1/3*(Area of lmn)
Area of lmn = (3)^0.5/4 *(side)^2
Since side = 6 units,
Area of lmn = 9*(3)^0.3
Area of shaded region = (1/3)*(9)*(3)^0.5 = 3*(3)^0.5

[spoiler](C)[/spoiler] is correct.
Aneesh Bangia
GMAT Math Coach
[email protected]

GMATPad:
Facebook Page: https://www.facebook.com/GMATPad

User avatar
Master | Next Rank: 500 Posts
Posts: 385
Joined: Mon Apr 16, 2012 8:40 am
Location: Pune, India
Thanked: 186 times
Followed by:29 members

by aneesh.kg » Mon Jun 11, 2012 10:49 am
Q2:
Ratio of the sides given
= (27/36)^0.5
= (3)^0.5 / 2

In a right angled triangle which has angles of 90, 60 and 30 (in degrees), the ratio
hypotenuse:second longest side:shortest side = 2:3^(0.5):1

Since the ratio of the given sides is (3)^0.5 /2, the angle opposite to the side measuring (27)^0.5 must be 60 degrees.

[spoiler](C)[/spoiler] is correct.
Aneesh Bangia
GMAT Math Coach
[email protected]

GMATPad:
Facebook Page: https://www.facebook.com/GMATPad

User avatar
Master | Next Rank: 500 Posts
Posts: 385
Joined: Mon Apr 16, 2012 8:40 am
Location: Pune, India
Thanked: 186 times
Followed by:29 members

by aneesh.kg » Mon Jun 11, 2012 10:52 am
Q3:
AC = 4 in
If side of the square is 'a',
AC = (2)^0.5 times a
(2)^0.5 * a = 4
a = 2(2)^0.5

Area of square = a^2 = 8 square inches

[spoiler](C)[/spoiler] is correct.
Aneesh Bangia
GMAT Math Coach
[email protected]

GMATPad:
Facebook Page: https://www.facebook.com/GMATPad

User avatar
GMAT Instructor
Posts: 768
Joined: Wed Dec 28, 2011 4:18 pm
Location: Berkeley, CA
Thanked: 387 times
Followed by:140 members

by Mike@Magoosh » Mon Jun 11, 2012 10:53 am
Dear phoenix9801

Hi, there. I'm happy to help with these. Incidentally, I like the 9801 = 99^2 in your screenname.

For the first, it is awfully helpful to know that the area of an equilateral triangle of side s is
A = sqrt(3)*(s^2)/4
See this blog for an explanation:
https://magoosh.com/gmat/2012/gmat-math- ... emorizing/

Given that, area of the the whole equilateral is sqrt(3)*(6^2)/4 = 9sqrt(3), so one third of that is 3sqrt(3), answer C.

For the second, we know the hypotenuse is sqrt(36) = 6, and we know the long leg is sqrt(27) = 3*sqrt(3). With the Pythagorean Theorem, we can figure out, the square of the short leg is 9, so the length of the short leg is 3. This is a right triangle in which the short leg is half the length of the hypotenuse --- that has to be a 30-60-90 triangle. See this blog
https://magoosh.com/gmat/2012/the-gmats- ... triangles/
So the angle opposite the long leg in a 30-60-90 triangle is 60 degrees, answer = C.

For the third, we need the other special triangle, also discussed at
https://magoosh.com/gmat/2012/the-gmats- ... triangles/
We need the 45-45-90 triangle. Here's a quick approach. Let the side of the square = x. Then, consider right triangle ADC --- by the Pythagorean Theorem,
x^2 + x^2 = 4^2
2*x^2 = 16
x^2 = 8

Well, we need do nothing else. The side of the square, squared, is 8, which means the area = 8. Answer choice = C.

Does all that make sense?

Mike :-)
Magoosh GMAT Instructor
https://gmat.magoosh.com/