Question

Just need a quick check on a geometry proof:

Let O be the inscribed circle in the isosceles trapezoid ABCD where AB is parallel to CD, AB=a, and CD=b. Find the radius of the inscribed circle.

Pf: \(r=\frac{h}{2}=\frac{\sqrt {ab}} {2}\)

In order for there to be an inscribed circle, we know that \(AD=BD=\frac{a+b}{2}\). Then we can get h by the pythagorian thm as \(h=\sqrt{ab}\). Then because the inscribed circle must be tangent to both the top and the bottom of the trapezoid, we know that a diameter is parallel to the height, so our radius is h/2.

What do you think?

Answer 1

Here it is since the math won't show:
\(r=\frac{h}{2}=\frac{\sqrt {ab}} {2}\)

In order for there to be an inscribed circle, we know that \(AD=BD=\frac{a+b}{2}\). Then we can get h by the pythagorian thm as \(h=\sqrt{ab}\). Then because the inscribed circle must be tangent to both the top and the bottom of the trapezoid, we know that a diameter is parallel to the height, so our radius is h/2.

Answer 2

My picture as well:

Answer 3

I did not mark the angles because I didn't use that they are congruent

Answer 4

@ganeshie8 , do you mind taking a quick look?

Answer 5

or could you take a look @mathmale ?

Answer 6

More than anything, I just need to know if I need to expand on any points.

Answer 7

Your approach looks quite thorough, and you're to be congratulated for that.

Regarding "we know that a diameter is parallel to the height, so our radius is h/2": I agree.

Would you mind demonstrating how you'd use the Pyth. Thm. to support your statement,

"In order for there to be an inscribed circle, we know that AD=BD=a+b2. Then we can get h by the pythagorian thm as h=ab−−√. "

Answer 8

\(AD=BD=\frac{a+b}{2} \)

Answer 9

thats a typo right ?

Answer 10

it should be :
\(AD=BC=\frac{a+b}{2}\)

Answer 11

overall it looks good to me !

Answer 12

Thank you, I just didn't show pythag here, but I did list it on the paper I'm using:

\[h^2=AD^2-DP^2\]Forgot to name P on my pic, but AP is the perpendicular.
\[h^2=[\frac{a+b}{2}]^2- [\frac{b-a}{2}]^2\]
\[=\frac{a^2+2ab+b^2-b^2+2ab-a^2}{4}\]
\[=\frac{4ab}{4}\]
\[h^2=ab\]
\[h=\sqrt{ab}\]


@ganeshie8 it just looks like a minus, it is a plus

Answer 13

ohhh yea type o

Answer 14

It should be BC

Answer 15

nice work !

Answer 16

Thank you, I just needed to check it. I am absolutely awful at knowing if a proof is detailed enough

Answer 17

Thanks to both of you, I really appreciate the help!

Answer 18

yes, it is not detailed enough
thats the reason u got many questions.... guess u need to justify few things in the proof

Answer 19

unless they're not proven/used everyday in ur life

Answer 20

yea, I actually have both of the questions written on my paper as side work, I'll just include those as well

Answer 21

for example, nobody remembers on top of their heads :
AD = BC = (a+b)/2

Answer 22

ok, so I'll elaborate on why there

Answer 23

okay !

Answer 24

Thanks again!

Answer 25

np :)

Answer 26

I think I would draw it like this