Geometry questions needed help with

**nathan02079** · May 27th 2009, 05:19 PM

I was given 10 bonus questions today, and I managed to solve all but three. I am completely stuck on these three: If you could get me started with hints, I would really appreciated it. Thank you.

1. The three heights of a triangle are given and they are h1, h2, and h3. Find the area of the triangle. (Though I think it's supposed to be 'length')

If it's like this...

Then I think the area is just simply...
${H_3*B}/2$
Though I think I'm not comprehending the question correctly

2. Given that $r=10cm$ , the area of the triangle is $600cm^2$ . The line segment AB is tangent to the circle at point P. Find the coordinates of point A, B, and P.

3. It is given that the perimeter of the rectangle ABCD is four times greater than the radius of the circle. Find the ratio of the side lengths of the rectangle. AB/AD=?

Thank you!

N

**earboth** · May 27th 2009, 10:20 PM

Originally Posted by nathan02079

...

2. Given that $r=10cm$ , the area of the triangle is $600cm^2$ . The line segment AB is tangent to the circle at point P. Find the coordinates of point A, B, and P.

...

1. The triangle is a right triangle.
2. The given circle is the incircle of the triangle.
3. For convenience I've labeled the side OA = a, OB = b and AB = c.
4. Divide the triangle into three triangles with the radius of the circle as its' height. (see attachment)

5. Now you know:

$a^2+b^2=c^2~~~\text{Pythagorean theorem}$

$\dfrac12 \cdot a \cdot b = 600~~~\text{triangle is a half rectangle}$

$\dfrac12 \cdot a \cdot r+\dfrac12 \cdot b \cdot r+\dfrac12 \cdot c \cdot r= 600 ~~~\text{sum of the areas of the 3 partial triangles}$

6. Thus you have to solve for (a,b,c) the system of equations:

$\left|\begin{array}{rcl}a+b+c&=&120 \\ a^2+b^2&=&c^2 \\ a \cdot b&=&1200\end{array}\right.$

I've got $(a,b,c)= (30, 40, 50)$

7. I'll leave the rest for you.

**earboth** · May 28th 2009, 12:50 AM

Originally Posted by nathan02079

...

3. It is given that the perimeter of the rectangle ABCD is four times greater than the radius of the circle. Find the ratio of the side lengths of the rectangle. AB/AD=?

...

1. for convenience I've labeled the side AB = b and the side AC = a.
2. From the text of the question you know:

$2a+2b=4r~\implies~a+b=2r$

3. The upper left triangle is a right triangle. Use the pythagorean theorem:

$\left(\dfrac b2\right)^2+(a-r)^2=r^2~\implies~a^2-2ar+\dfrac{b^2}4=0$

4. Solve this system of equations for (a, b). I've got $a=\dfrac25 r~\wedge~b=\dfrac85 r$

5. AD is the diagonal of the rectangle:

$AD=\sqrt{\left(\dfrac25 r \right)^2 + \left( \dfrac85 r \right)^2 } = \dfrac15 r \sqrt{68}$

6. I'll leave the rest for you.

**nathan02079** · June 2nd 2009, 09:04 PM

Everything's clear, except that I don't get how you got this: would you please explain? Thank you.

Originally Posted by earboth

$a+b+c=120$

**Amer** · June 2nd 2009, 11:06 PM

Originally Posted by earboth

$\dfrac12 \cdot a \cdot r+\dfrac12 \cdot b \cdot r+\dfrac12 \cdot c \cdot r= 600 ~~~\text{sum of the areas of the 3 partial triangles}$

6. Thus you have to solve for (a,b,c) the system of equations:

$\left|\begin{array}{rcl}a+b+c&=&120 \\ a^2+b^2&=&c^2 \\ a \cdot b&=&1200\end{array}\right.$

I've got $(a,b,c)= (30, 40, 50)$

.

after permission of earboth I will explain it

he get it from $\dfrac12 \cdot a \cdot r+\dfrac12 \cdot b \cdot r+\dfrac12 \cdot c \cdot r= 600 ~~~\text{sum of the areas of the 3 partial triangles}$

multiply it with 2 then divide it with r since r=10
you will have a+b+c=1200/10

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