Distance formula vs similar triangles

[fayan77]

Hello, took a year off school, now shaking the rust off. so according to the book using similar triangles d=9.6 I understand how they got the answer, but i used distance formula from the point to the origin and got 9.8 I checked 9.6 and it checks out with the numbers but Idk why using the distance formula yields a different answer. am I doing something wrong?

 

[scottdave]

To use the distance formula, you need the coordinates of that point. How did you find the coordinates? What were the steps that you used?

 

[fayan77]

the book shows point A (4,9) sorry for the picture quality, my phone sucks.

 

[mfb]

I moved your thread to your homework section.
The picture quality is really problematic.

(4,9) are not the coordinates of any interesting point in the sketch.

9.6 is correct.

 

[scottdave]

(4,9) is a point on that line between (0,12) and (16,0), but it is not the point of intersection {it does not satisfy the (4/3) slope from the origin}. You can find the intersection point by finding where the two lines meet.
It is not hard, just two straight lines to find the (x,y) coordinate. The actual intersection point is (5.76,7.68) which you can now use the distance formula and get the same answer.

 

[mfb]

I wouldn't solve this via similar triangles, by the way. The area of a triangle can be calculated as length*height/2. You can use the vertical and horizontal sides as length and height to find the area. Then you can use the long diagonal and the unknown length (also a length/height pair) to find the unknown length.

 

[fayan77]

yeah that is what i did i took the slope and line then derived an orthogonal line passing through origin and found the point. That point on the book is confusing tho. thanks.

 

[mfb]

yeah that is what i did i took the slope and line then derived an orthogonal line passing through origin and found the point.

That's not what I meant.

12=3*4 and 16=4*4, and you should recognize 3,4,5 as right triangle. Therefore, the hypothenuse has a length of 5*4=20.

The area of the triangle is 12*16/2.

If we call the unknown length h, we get 12*16/2 = 20*h/2.
The unknown length is then h=12*16/20, and that is something you can calculate easily.

 

[Ray Vickson]

I wouldn't solve this via similar triangles, by the way. The area of a triangle can be calculated as length*height/2. You can use the vertical and horizontal sides as length and height to find the area. Then you can use the long diagonal and the unknown length (also a length/height pair) to find the unknown length.

Wouldn't similar triangles be the easiest possible way? Solving $16/20 = d/12$ is about as simple as it gets.

Of course, one would first need to recognize that the large triangle has hypotenuse 20, as you have indicated.

 

[mfb]

Well, it leads to the same formula, but you have to think which side corresponds to what.