Geometry
This topic has expert replies
-
- Senior | Next Rank: 100 Posts
- Posts: 69
- Joined: Sat May 10, 2008 3:05 pm
- Thanked: 1 times
-
- Senior | Next Rank: 100 Posts
- Posts: 62
- Joined: Fri Dec 21, 2007 2:58 pm
- Location: Fremont , California
- Thanked: 3 times
Answer: C
A(S) = 2 A(s)
= 2 *( 1/2 * s * h) ( area = 1/2 * base * height)
A(S) = 1/2 * S * H
From above, we can conclude that if S = sqrt(2) * s and H = sqrt(2) *h
then the A(S) = 2 A(s)
There fore S = sqrt(2) * s
A(S) = 2 A(s)
= 2 *( 1/2 * s * h) ( area = 1/2 * base * height)
A(S) = 1/2 * S * H
From above, we can conclude that if S = sqrt(2) * s and H = sqrt(2) *h
then the A(S) = 2 A(s)
There fore S = sqrt(2) * s
the area is twice as large does not mean the length of the sides are twice as long.
what is given is the following:
similar triangles = proportions of sides on the two triangles are equal
i.e. b/B=s/S=any side of the smaller triangle/the same side of the larger triangle
area of big = 2x area of small => (1/2*s*b)*2 = 1/2*S*B
=>s*b=1/2*S*B
=>2*s*b/B=S
=>2*s*(s/S)=S
=>2s^2=S^2
=>S = SQRT(2)*s
what is given is the following:
similar triangles = proportions of sides on the two triangles are equal
i.e. b/B=s/S=any side of the smaller triangle/the same side of the larger triangle
area of big = 2x area of small => (1/2*s*b)*2 = 1/2*S*B
=>s*b=1/2*S*B
=>2*s*b/B=S
=>2*s*(s/S)=S
=>2s^2=S^2
=>S = SQRT(2)*s
I beat the GMAT! 760 (Q49/V44)
- Stuart@KaplanGMAT
- GMAT Instructor
- Posts: 3225
- Joined: Tue Jan 08, 2008 2:40 pm
- Location: Toronto
- Thanked: 1710 times
- Followed by:614 members
- GMAT Score:800
The key fact to remember to answer this question quickly is that area is a square function and length is a linear function.
We have similar triangles, which means that one is simply a bigger copy of the other; in other words, everything is proportional from one triangle to the next.
If we were told that the bigger triangle had LENGTHS twice the size of the smaller, then the area of the larger (since it's a square function) would simply be 2^2 = 4 times as big. If the bigger had lengths three times the size of the smaller, then the area of the larger would be 3^2 = 9 times as big.
However, we're not going from length to area, we're going from area to length... so instead of squaring, we need to square root.
So, if the area is twice as big, then each linear dimension will be sqrt2 as big: choose (sqrt2)s.
We have similar triangles, which means that one is simply a bigger copy of the other; in other words, everything is proportional from one triangle to the next.
If we were told that the bigger triangle had LENGTHS twice the size of the smaller, then the area of the larger (since it's a square function) would simply be 2^2 = 4 times as big. If the bigger had lengths three times the size of the smaller, then the area of the larger would be 3^2 = 9 times as big.
However, we're not going from length to area, we're going from area to length... so instead of squaring, we need to square root.
So, if the area is twice as big, then each linear dimension will be sqrt2 as big: choose (sqrt2)s.
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
- Stuart@KaplanGMAT
- GMAT Instructor
- Posts: 3225
- Joined: Tue Jan 08, 2008 2:40 pm
- Location: Toronto
- Thanked: 1710 times
- Followed by:614 members
- GMAT Score:800
First, it's side is NOT twice the side of the smaller triangle.. it's side is sqrt2 * the side of the smaller triangle.chidcguy wrote:Stuart,
Are you saying that the height of the larger triangle is also double the height of the smaller triangle, because its side is twice the side of the smaller triangle?
And yes, the height of the larger triangle is also sqrt2 * the height of the smaller triangle.
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
-
- Master | Next Rank: 500 Posts
- Posts: 438
- Joined: Mon Feb 12, 2007 9:44 am
- Thanked: 26 times
Thanks Stuart.
How can we say that the height of similar triangles are also proportional? AFAIK, all the material I have looked at (Kaplan,MGMAT) only pointed out that angles are equal and all the sides are in same proportion. If its mandatory that the heights of similar triangles are in same proportion as the sides are, why did every one miss this?
I looked for mathematical theorem or proof that says similar triangles have same height. I could not find any
How can we say that the height of similar triangles are also proportional? AFAIK, all the material I have looked at (Kaplan,MGMAT) only pointed out that angles are equal and all the sides are in same proportion. If its mandatory that the heights of similar triangles are in same proportion as the sides are, why did every one miss this?
I looked for mathematical theorem or proof that says similar triangles have same height. I could not find any
-
- Legendary Member
- Posts: 1169
- Joined: Sun Jul 06, 2008 2:34 am
- Thanked: 25 times
- Followed by:1 members
IMO this is how it goes...
First,as look at the problem,we get to know the triangles are similar triangles.
For similar triangles -- square of the ratio of the sides = Ratio of the area of the triangles.
(s/S)^2 = As/AS
Now it's been given in the problem that AS=2As
Substitute-- we get s^2/S^2=1/2
Cross multiple-- 2s^2 = S^2
Therefore S = sqrt2 s
First,as look at the problem,we get to know the triangles are similar triangles.
For similar triangles -- square of the ratio of the sides = Ratio of the area of the triangles.
(s/S)^2 = As/AS
Now it's been given in the problem that AS=2As
Substitute-- we get s^2/S^2=1/2
Cross multiple-- 2s^2 = S^2
Therefore S = sqrt2 s
-
- Master | Next Rank: 500 Posts
- Posts: 154
- Joined: Tue Aug 26, 2008 12:59 pm
- Location: Canada
- Thanked: 4 times
This one is easy to prove mathematically:chidcguy wrote:Thanks Stuart.
How can we say that the height of similar triangles are also proportional? AFAIK, all the material I have looked at (Kaplan,MGMAT) only pointed out that angles are equal and all the sides are in same proportion. If its mandatory that the heights of similar triangles are in same proportion as the sides are, why did every one miss this?
I looked for mathematical theorem or proof that says similar triangles have same height. I could not find any
Take triangle 1 and 2 to have angle x and 90degree angle
Then we know for smaller triangle
tan(x) = h div s
For bigger triangle
tan(x) = H div S
Equate the two equations:
h div s = H div S
Height to side ratios are proportional.
z
x 90
Clinton
Math Minor