133k views
1 vote
I’ve been stuck on question 1 for a while, can you guys help?

I’ve been stuck on question 1 for a while, can you guys help?-example-1
User Bhavinb
by
8.3k points

1 Answer

2 votes

Answer:


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

User Blomster
by
8.0k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories