Question

I’ve been stuck on question 1 for a while, can you guys help?

Answer 1

Answer:

$h=20$**

**The picture is not accurate for this problem.

Step-by-step explanation:

I would use Pythagorean Theorem to setup two equations since there are two triangles with no angle information.

Let $BC=x+(99-x)$ where $x$ is equal to the first partition of $BC$ (reading from left to right) and $(99-x)$ is equal to the second partition of $BC$ (reading from left to right).

We have the following system to solve:

$20^2=h^2+x^2$

$101^2=h^2+(99-x)^2$

I will use elimination to first solve for $x$.

Subtract the equations:

$20^2-101^2=x^2-(99-x)^2$

Factor both sides using $a^2-b^2=(a-b)(a+b)$:

$(20-101)(20+101)=(x-(99-x))(x+(99-x))$

Simplify inside the $($ $)$.

$(-81)(121)=(2x-99)(99)$

Divide both sides by 9:

$(-9)(121)=(2x-99)(11)$

Divide both sides by 11:

$(-9)(11)=(2x-99)(1)$

Simplify both sides:

$-99=2x-99$

Add 99 on both sides:

$0=2x$

Divide both sides by 2:

$x=0$

Now go to either equation we had in the beginning to find $h$.

$20^2=h^2+x^2$ with $x=0$

$20^2=h^2$

$20=h$