# Math Help - mathematical model

1. ## mathematical model

Can anyone help me with this ? Thank you in advance. I really need the answers because I will report this. Sorry to be of bother, I really have no idea because this hasn't been taught yet and I cant understand the book.

1. A 20 ft. ladder leans against a vertical wall. If the bottom of the ladder is pulled away at the rate of 2 ft per sec, write a formula f(t), the distance of the top of the ladder from the floor as a function of t.

2. Let P be the perimeter of an equilateral triangle. Write a formula A(P), the area of the triangle as function of the perimeter.

3. One of the legs of a right triangle has a length of 4cm. Express the length of the altitude perpendicular to the hypotenuse as a function of the length of the hypotenuse.

2. Originally Posted by reiward
Can anyone help me with this ? Thank you in advance. I really need the answers because I will report this. Sorry to be of bother, I really have no idea because this hasn't been taught yet and I cant understand the book.

1. A 20 ft. ladder leans against a vertical wall. If the bottom of the ladder is pulled away at the rate of 2 ft per sec, write a formula f(t), the distance of the top of the ladder from the floor as a function of t.

2. Let P be the perimeter of an equilateral triangle. Write a formula A(P), the area of the triangle as function of the perimeter.

3. One of the legs of a right triangle has a length of 4cm. Express the length of the altitude perpendicular to the hypotenuse as a function of the length of the hypotenuse.
2. The perimeter of an equilateral triangle of length $l$ will be $3l$.

So $P = 3l$ or $l = \frac{P}{3}$.

For any triangle, the area can be found using Heron's formula

$A = \sqrt{s(s - a)(s - b)(s - c)}$, where $s = \frac{a + b + c}{2}$.

Since it's an equilateral triangle, $a = b = c = l$.

So $s = \frac{3l}{2}$ and the area is

$A = \sqrt{\frac{3l}{2}\left(\frac{3l}{2} - l\right)\left(\frac{3l}{2} - l\right)\left(\frac{3l}{2} - l\right)}$

$= \sqrt{\frac{3l}{2}\left(\frac{l}{2}\right)^3}$

$= \sqrt{\frac{3l}{2}\left(\frac{l^3}{8}\right)}$

$= \sqrt{\frac{3l^4}{16}}$

$= \frac{\sqrt{3}l^2}{4}$

$= \frac{\sqrt{3}\left(\frac{P}{3}\right)^2}{4}$

$= \frac{\frac{\sqrt{3}P^2}{9}}{4}$

$= \frac{\sqrt{3}P^2}{36}$.

3. Originally Posted by reiward
Can anyone help me with this ? Thank you in advance. I really need the answers because I will report this. Sorry to be of bother, I really have no idea because this hasn't been taught yet and I cant understand the book.

1. A 20 ft. ladder leans against a vertical wall. If the bottom of the ladder is pulled away at the rate of 2 ft per sec, write a formula f(t), the distance of the top of the ladder from the floor as a function of t.
The floor, wall, and ladder form a right triangle with the length of the ladder as hypotenuse. Use the Pythagorean theorem to write an equation connecting those. If then the bottom leg of the triangle is given by $a= a_0+ 2t$ where $a_0$ is the initial distance from the wall to the bottom of the ladder and t is the time in seconds.

2. Let P be the perimeter of an equilateral triangle. Write a formula A(P), the area of the triangle as function of the perimeter.

3. One of the legs of a right triangle has a length of 4cm. Express the length of the altitude perpendicular to the hypotenuse as a function of the length of the hypotenuse.
Again, a right triangle. Dropping the altitude from the right angle to the hypotenuse divides that right trangle into two right triangles, both of which are similar to the original triangle. Pythagorean theorem and similar triangles.

For all of these, the first thing you should have done was draw a picture. Then use what you know about triangles from geometry.

4. dHello, reiward!

HallsofIvy gave you an excellent game plan for Problem #3.

3. One of the legs of a right triangle has a length of 4cm.
Express the length of the altitude perpendicular to the hypotenuse
as a function of the length of the hypotenuse.
Code:
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We have right triangle $ABC,\;C = 90^o,\;AC = 4.$

Let $h = AB$, the hypotenuse.

Let $y = CD$, atitude to hypotenuse.

Let $x = AD,\;\text{then: }h-x \:=\:DB$

We find that: . $\Delta ADC \:\sim \: \Delta CDB \:\sim\:\Delta ACB$

Hence: . $\dfrac{y}{x} \:=\:\dfrac{h-x}{y} \quad\Rightarrow\quad y^2 \:=\:hx-x^2$ .[1]

In $\Delta ADC\!:\;x^2+y^2 \:=\:4^2 \quad\Rightarrow\quad x \:=\:\sqrt{16-x^2}$

Substitute into [1]: . $y^2 \:=\:h\sqrt{16-y^2} - (16-y^2)$

. . $y^2 \:=\:h\sqrt{16-y^2} - 16 + y^2 \quad\Rightarrow\quad h\sqrt{16-y^2} \:=\: 16$

. . $\sqrt{16-y^2} \:=\:\dfrac{16}{h} \quad\Rightarrow\quad 16 - y^2 \:=\:\dfrac{256}{h^2}$

. . $y^2 \:=\:16 - \dfrac{256}{h^2} \:=\:\dfrac{16h^2 - 256}{h^2} \:=\:\dfrac{16(h^2-16)}{h^2}$

Therefore: . $y \;=\;\dfrac{4\sqrt{h^2-16}}{h}$