Ans: c = 2.sushbis wrote:A triangle has coordinates (0,0),(0,4) and (6,0), if line y=c divided the
triangle equally, c=?
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Geometry question
Source: Beat The GMAT — Problem Solving |
+1, please post questions only with all 5 options!leonswati wrote:Wat are the options..... it will be easy to solve looking at the options....
Okay,sushbis wrote:A triangle has coordinates (0,0),(0,4) and (6,0), if line y=c divided the
triangle equally, c=?
Let's have A(0,0) B(0,4) and C(6,0)
A(T ABC) = 1/2 * 4* 6 = 12
Since the goal is to divide the triangle into half such that each half has an equal area (Assumption) then each half should have an area of 6 units.
thus we reach the eqn: 1/2* (x intcpt) * (4 - y intcpt) = 6
for ease, x intcpt = p and y intcpt = q
--> p*(4-q) = 12
--> Factorising 12 we get factors as 1,2,3,4,6,12
now the limits are p<6 and q<4 since the line y=q is dividing the triangle.
Hence,
@p=4 4-q =3 --> q=1
@p=3 4-q=4 --> q=0 (discard)
@p=2 4-q=6 --> q=-2(discard)
@p=1 4-q=12 --> q=-8 (discard)
leaves us with q=1. Thus y=1 is the line which divides the T ABC into two halves equally.
Last edited by Proleefeek111 on Tue Oct 11, 2011 10:09 pm, edited 1 time in total.
For some reason there are so many answers here 
My attempt and a different answer as well.
You know the 2 triangles obtained are similar (AAA)
Their sides would be proportional.
The height and base are 4 & 6 and area = 12.
Since the sides are in the ratio 2x,3x.(Height/Base) (Area would be 1/2 * 2x *3x = 12)
x^2 = 4 ; x =2. (Hence we get corresponding sides as 4 &6)
Now area is to be half, i.e 6
So, we have 1/2 * 2x * 3x = 6
x^2 = 2 ; x = Sqrt(2)
Hence corresponding height(2x)= 2Sqrt(2) and base(3x) = 3Sqrt(2)
Difference in height would be 4 - 2Sqrt(2).
y = c = 4 - 2Sqrt(2)
My attempt and a different answer as well.
You know the 2 triangles obtained are similar (AAA)
Their sides would be proportional.
The height and base are 4 & 6 and area = 12.
Since the sides are in the ratio 2x,3x.(Height/Base) (Area would be 1/2 * 2x *3x = 12)
x^2 = 4 ; x =2. (Hence we get corresponding sides as 4 &6)
Now area is to be half, i.e 6
So, we have 1/2 * 2x * 3x = 6
x^2 = 2 ; x = Sqrt(2)
Hence corresponding height(2x)= 2Sqrt(2) and base(3x) = 3Sqrt(2)
Difference in height would be 4 - 2Sqrt(2).
y = c = 4 - 2Sqrt(2)
Your approach is correct, looks like approximation is the order of the dayshankar.ashwin wrote:For some reason there are so many answers here
My attempt and a different answer as well.
You know the 2 triangles obtained are similar (AAA)
Their sides would be proportional.
The height and base are 4 & 6 and area = 12.
Since the sides are in the ratio 2x,3x.(Height/Base) (Area would be 1/2 * 2x *3x = 12)
x^2 = 4 ; x =2. (Hence we get corresponding sides as 4 &6)
Now area is to be half, i.e 6
So, we have 1/2 * 2x * 3x = 6
x^2 = 2 ; x = Sqrt(2)
Hence corresponding height(2x)= 2Sqrt(2) and base(3x) = 3Sqrt(2)
Difference in height would be 4 - 2Sqrt(2).
y = c = 4 - 2Sqrt(2)
4 - 2sqrt(2) = 1.172 ~ 1. Pity, the options weren't provided on this one.
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sohrabkalra
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I am assuming that by equal Parts you mean parts of equal area !
So the co-ordinates
A(0,0)
B(0,6)
C(4,0)
Leave us with a right angled triangle ABC with
AB=6 , AC = 4 , Angle BAC = 90
Hence the area of triangle ABC = 1/2 * AB * BC
= 1/2*6*4
= 12
Now lets say Y=C cuts AC at D and BC at E
So DE will create a right angled triangle DCE whose area is 1/2* (Area of triangle of ABC)
Area of triangle DCE = 12
Area of triangle DCE = 1/2 * DC * DE
DC = AC - AD = 4 - C
And since trinagle DEF and triangle ABC can be proved similar by AAA property
DC/AC = DE/AB
Thus DE = (DC/AC)* AB = (3/2) * (4 - C)
Hence Area of triangle DCE = 1/2 * (4-C) * (3/2) * (4-C) = 12
Solving this equation will give you two solutions
4+2root2 and 4-2root2
since the former lies outside the triangle
Answer should be 4-2root2[/img]
So the co-ordinates
A(0,0)
B(0,6)
C(4,0)
Leave us with a right angled triangle ABC with
AB=6 , AC = 4 , Angle BAC = 90
Hence the area of triangle ABC = 1/2 * AB * BC
= 1/2*6*4
= 12
Now lets say Y=C cuts AC at D and BC at E
So DE will create a right angled triangle DCE whose area is 1/2* (Area of triangle of ABC)
Area of triangle DCE = 12
Area of triangle DCE = 1/2 * DC * DE
DC = AC - AD = 4 - C
And since trinagle DEF and triangle ABC can be proved similar by AAA property
DC/AC = DE/AB
Thus DE = (DC/AC)* AB = (3/2) * (4 - C)
Hence Area of triangle DCE = 1/2 * (4-C) * (3/2) * (4-C) = 12
Solving this equation will give you two solutions
4+2root2 and 4-2root2
since the former lies outside the triangle
Answer should be 4-2root2[/img]
