Kindly help me for this triangle problem

Sep 2009
135
1
The problem is attached. It is triangle and its peak is having value 1

the sides are at a and -a. The new point is on a/4. from which i have drawn perpendicular line

i want to know what is the value of that point on y-axis scale.
book says it is 0.733
but how???
 

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May 2010
27
5
The problem is attached. It is triangle and its peak is having value 1

the sides are at a and -a. The new point is on a/4. from which i have drawn perpendicular line

i want to know what is the value of that point on y-axis scale.
book says it is 0.733
but how???
Since I would consider this a typical proportional problem, I wonder that too?
Are you sure you don't have any more information on a?

If a can be any number, triangle height y can be 0<y<1 ?
 

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MHF Helper
Aug 2008
12,883
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The problem is attached. It is triangle and its peak is having value 1

the sides are at a and -a. The new point is on a/4. from which i have drawn perpendicular line

i want to know what is the value of that point on y-axis scale.
book says it is 0.733
but how???
The two known points on that line segment are

\(\displaystyle (a, 0)\) and \(\displaystyle (0, 1)\).

The gradient is

\(\displaystyle m = \frac{1 - 0}{0 - a}\)

\(\displaystyle = \frac{1}{-a}\)

\(\displaystyle = -\frac{1}{a}\).


The \(\displaystyle y\) intercept is \(\displaystyle 1\), so \(\displaystyle c = 1\).


Therefore the line segment has equation

\(\displaystyle y = -\frac{1}{a}\,x + 1\)

\(\displaystyle y = 1 - \frac{x}{a}\).


So if \(\displaystyle x = \frac{a}{4}\)

\(\displaystyle y = 1 - \frac{\frac{a}{4}}{a}\)

\(\displaystyle = 1 - \frac{1}{4}\)

\(\displaystyle = \frac{3}{4}\).
 
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May 2010
27
5
one can do that?
isn't that like saying a=y=1 ?

so all you get is a mirror?
y = 1-x ?

I'm asking since I really don't know.

//====edit
aren't you supposed to plug that back to the equation? so that:
3/4 = -(1/a)x + 1

//====edit 2
wow, that was really stupid of me :) of course 3/4
 
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