sushbis wrote:A triangle has coordinates (0,0),(0,4) and (6,0), if line y=c divided the
triangle equally, c=?

Ans: c = 2.

Answer is Y=1.....

Wat are the options..... it will be easy to solve looking at the options....

IMO: C=3

leonswati wrote:Wat are the options..... it will be easy to solve looking at the options....

+1, please post questions only with all 5 options!

is it 3*root2 ?

user123321

sushbis wrote:A triangle has coordinates (0,0),(0,4) and (6,0), if line y=c divided the
triangle equally, c=?

Okay,

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.

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)

shankar.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)

Your approach is correct, looks like approximation is the order of the day

4 - 2sqrt(2) = 1.172 ~ 1. Pity, the options weren't provided on this one.

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]