Question

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

Answer 1

`h=20`**

**The picture is not accurate for this problem.

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`