# Math Help - Geometry help needed!

1. ## Geometry help needed!

Hi all.

I need a help with a construction question in Geometry. I would really appreciate it if anyone could tell me how to go about it, cause I'm stumped.

Q: Construct a triangle ABC, having given l_ ACB = 90, the hypotenuse AB = 6.4 cm, and the sum of the remaining sides AC and BC as 8 cm.

Thanks!

2. Originally Posted by Insaeno
Hi all.

I need a help with a construction question in Geometry. I would really appreciate it if anyone could tell me how to go about it, cause I'm stumped.

Q: Construct a triangle ABC, having given l_ ACB = 90, the hypotenuse AB = 6.4 cm, and the sum of the remaining sides AC and BC as 8 cm.

Thanks!
Are you talking about a literal compass and straightedge construction or do you just want the lengths of the legs of the right triangle?

Call the remaining two legs x and y.

Then we know that
$x + y = 8$
and
$x^2 + y^2 = 6.4^2$ <-- Since we have a right triangle.

Solving the first for y gives:
$y = 8 - x$

Inserting this into the second equation gives:
$x^2 + (8 - x)^2 = 48.96$

$x^2 + 64 - 16x + x^2 = 48.96$

$2x^2 - 16x + 23.04 = 0$

$x^2 - 8x + 11.52 = 0$

$x = \frac{-(-8) \pm sqrt{(-8)^2 - 4(1)(11.52)}}{2 \cdot 1}$

$x = \frac{8 \pm sqrt{17.92}{2}$

$x \approx 1.8834~cm$ or $x \approx 6.1166~cm$

So let x = 1.8834 cm, then y = 6.1166 cm. (Or x = 6.1166 cm and y = 1.8834 cm. Whichever.)

-Dan

3. Hello, Insaeno!

This is not a simple problem . . .

Construct a triangle ABC, having given $\angle ACB = 90^o$, the hypotenuse $AB = 6.4$ cm,
and the sum of the remaining sides $AC$ and $BC$ is 8 cm.

Since right triangle $ACB$ can be inscribed in a semicircle,
we have this diagram:
Code:
                |
* * *   C
*     |    o*
*       | o   o *
*       o|      o *
o   |       o
*  o      |        o*
- A o - - - - + - - - - o B -
-3.2                 3.2

The circle has the equation: . $x^2 + y^2 \:=\:3.2^2$ . [1]

The locus of point $C$ where $AC + BC \,= \,8$ is an ellipse.
. . We have: . $a = 4$ and $c = 3.2$
Then: . $b^2 \:=\:a^2-c^2\:=\:4^2 - 3.2^2 \:=\:5.76$
. . The equation of the ellipse is: . $\frac{x^2}{16} + \frac{y^2}{5.76} \:=\:1$ . [2]

And solve the system of equations.

I got: . $x \,= \,\pm\sqrt{7},\;\;y \,= \,\pm1.8$

4. Are you talking about a literal compass and straightedge construction or do you just want the lengths of the legs of the right triangle?
I need to construct such a triangle, and thats the information given to me.
Thanks a lot though, now that the sides are discovered the triangle can be constructed very easily.

Soroban, I'm sorry but I didn't understand your post. I didn't understand how x can be root 7 while x + y = 8.

5. Originally Posted by Insaeno
I need to construct such a triangle, and thats the information given to me.
Thanks a lot though, now that the sides are discovered the triangle can be constructed very easily.

Soroban, I'm sorry but I didn't understand your post. I didn't understand how x can be root 7 while x + y = 8.
x and y are the coordinates of the point C.

Once you have x and y you can calculate the lengths of the sides, and
the lengths in this case are constructible, so a construction can be
concocted once you know what they are (though probably not worth
the effort).

RonL

6. Hello, Insaeno!

I didn't understand how x can be root 7 while x + y = 8.
Read the problem again . . .

$x + y \,=\,8$ is not part of the problem.
. . This locates a point where the sum of the coordinates equals 8.

They ask for a point whose distances from A and B has a sum of 8.

7. Oh, Alright, I understand now. Thanks a lot, you've helped tremendously!!