Question

questions attached please help me its geometry

Answer 1

erm
I'm not entirely sure how to solve for the first... sorry :/

for the second
the bottom length would be 18 - 3 - 3 = 12
so then in the triangle to the right, you'll have a base of 12-3 = 9
and a height of 3
so then use pythagorean theorem to find AB :)

Answer 2

thank you alot i did it the same way i was just not sure :) and its okay ill think about it

Answer 3

ok, I think I figured out the first :)
\[AB = \sqrt{(4x+x)^{2} + 5^{2}} = \sqrt{(5x)^{2} + 5^{2}} = \sqrt{25x^{2} + 25}\]\[CD = \sqrt{(4x)^{2} + (2x + 5)^{2}} = \sqrt{16x ^{2}+4x ^{2}+20x+25}\]AB = CD
so 25x^2 + 25 = 16x^2 + 4x^2 + 20x + 25
simplifies to 5x^2 - 20x = 0 solve for x

Answer 4

im sorry i didnt understand?

Answer 5

I used pythagorean theorem to get expressions for AB and CD
and since AB = CD
I put what I got together and simplified :)

Answer 6

im sorry ik im asking alot bt can u tell me how u solved this 25x^2 + 25 = 16x^2 + 4x^2 + 20x + 25

Answer 7

its a bit confusing

Answer 8

sure :)
sorry if it was a bit jumbled and messy
25x^2 + 25 = 16x^2 + 4x^2 + 20x + 25 add like terms
25x^2 + 25 = 20x^2 + 20x + 25 subtract 20x^2 from both sides
-20x^2 -20x^2
5x^2 + 25 = 20x + 25 subtract 20x from both sides
-20x -20x
5x^2 - 20x + 25 = 25 subtract 25 from both sides
-25 -25
5x^2 - 20x = 0 try to solve for x from here :)

Answer 9

thank you so much <3

Answer 10

you're welcome :)

Answer 11

Let the ladder be x

\( \large \begin {cases} x^2=5^2+(OA)^2 &\text {if OA>0}\\
x^2=(OC)^2+(OD)^2 &\text {if OC>0}\\
OD=5+2AC &\text {if OD>0} \\
AC=\frac{OA}{5} &\text {if AC>0} \\
OC=4AC&\text {if x>0}\\ \end {cases} \)