What is the area of the region encircled by y<=3 X>=2, and Y>=x-6?
pls explain ..
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
area on encircled region
This topic has expert replies
-
clock60
- Legendary Member
- Posts: 759
- Joined: Mon Apr 26, 2010 10:15 am
- Thanked: 85 times
- Followed by:3 members
hi mariah
i am slightly confused with the word encircled (which imply circle), to me we must find the area restricted with lines
it looks like right triangle, so if we find the legs of triangle we can find its area
it is easy to construct lines x=2, y=3, and y=x-6, and then to find coordinates of the vertex
x=2, and y=x-6 insert x=2 to the y=x-6,y=2-6=-4, so the first leg will equal=3+4=7
y=3, and y=x-6, here insert y=3 into the y=x-6, 3=x-6, x=9 thte second leg will equal 9-2=7 also 7
and now the area=1/2*7*7=49/2
but if we need to find the area if the circle inscribed into the given triangle we must first find radius, r= (leg1+leg2-hypotenuse)/2=(7+7-2*7^1/2)/2=7-7^1/2
S=pi*(7-7^1/2)^2
i am slightly confused with the word encircled (which imply circle), to me we must find the area restricted with lines
it looks like right triangle, so if we find the legs of triangle we can find its area
it is easy to construct lines x=2, y=3, and y=x-6, and then to find coordinates of the vertex
x=2, and y=x-6 insert x=2 to the y=x-6,y=2-6=-4, so the first leg will equal=3+4=7
y=3, and y=x-6, here insert y=3 into the y=x-6, 3=x-6, x=9 thte second leg will equal 9-2=7 also 7
and now the area=1/2*7*7=49/2
but if we need to find the area if the circle inscribed into the given triangle we must first find radius, r= (leg1+leg2-hypotenuse)/2=(7+7-2*7^1/2)/2=7-7^1/2
S=pi*(7-7^1/2)^2
-
force5
- Legendary Member
- Posts: 582
- Joined: Tue Mar 08, 2011 12:48 am
- Thanked: 61 times
- Followed by:6 members
- GMAT Score:740
hi Clock60 you are correct. the area covered should be 49/2 (i found the area using the same way). I an sure encircled would mean enclosed because there cannot be any circle in this question. i mean yes we can find the incircle however then the question would change.
-
atulmangal
- Legendary Member
- Posts: 1112
- Joined: Sat Feb 26, 2011 11:16 am
- Thanked: 77 times
- Followed by:49 members
Guys i have a completely different view, The figure formed is actually a square...and IMO the answer is to your surprise 49...moreover the area question is talking about is i think the area enclosed between the lines...as we get 4 lines after solving the inequalities
Explanation:-
As u already found the values...
X max = 9 and X min = 2 (X=9 and X=2 are 2 lines)
Y max = 3 and Y min = -4 (Y=3 and Y=(-4) are 2 lines)
so we get 4 lines
if u draw on X-Y plane, u get a square (shaded region enclosed b/w them) with
side || to X-axis of length = 9-2 = 7
side || to Y-axis of length = 3-(-4) = 7
Hence the enclosed area = 7*7 = 49
correct me if m wrong???
Explanation:-
As u already found the values...
X max = 9 and X min = 2 (X=9 and X=2 are 2 lines)
Y max = 3 and Y min = -4 (Y=3 and Y=(-4) are 2 lines)
so we get 4 lines
if u draw on X-Y plane, u get a square (shaded region enclosed b/w them) with
side || to X-axis of length = 9-2 = 7
side || to Y-axis of length = 3-(-4) = 7
Hence the enclosed area = 7*7 = 49
correct me if m wrong???
-
Tani
- Legendary Member
- Posts: 1255
- Joined: Fri Nov 07, 2008 2:08 pm
- Location: St. Louis
- Thanked: 312 times
- Followed by:90 members
Graph the three lines y=3 x=2 and y=x-6
the first is parallel to the x axis at y = 3, the second is parallel to the y axis at x = 2 and the third has a slope of 1 and a y intercept of -6. To see the relevant figure, shade everything above the line y = 3, everything to the left of the line x = 2 and everything below the line y=x-6. The unshaded area is then "encircled" by those three lines and is, indeed, a triangle.
the first is parallel to the x axis at y = 3, the second is parallel to the y axis at x = 2 and the third has a slope of 1 and a y intercept of -6. To see the relevant figure, shade everything above the line y = 3, everything to the left of the line x = 2 and everything below the line y=x-6. The unshaded area is then "encircled" by those three lines and is, indeed, a triangle.
Tani Wolff
