Trapezoid

[spark kestrel]

In trapezoid ABCD, the sum of the angles at the larger base AB is 90∘, and points M and N are the midpoints of AB and CD, respectively.

Given that:
AB=36 cm,CD=14 cm

find the length of MN

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[green kettle]

comrade, ill help u give me some time

[spark kestrel]

sure thing

[green kettle]

i forgot mid pt therorem

one sec

nvm

its for traingle i was dumb

anyways

yk AB and CN are paralle

CD*

[spark kestrel]

It's (AB+CD)/2, but since MN connects the bases and not the legs i don't think i can use that

also, i am not sure what do do with the punctuated lines gives, they mist be for something

[green kettle]

idr much but yk midsegment

theorem?

[green kettle]

paralle.

l

[green kettle]

@spark kestrel

[spark kestrel]

yes

[green kettle]

wait however the mid point is on the parallel sides

[toxic axle]

Hint: what is this angle going to be?

[spark kestrel]

90?

since the sum of the angles on the base is 90

so this one must be 90

[toxic axle]

Yeah, nice!

[green kettle]

u cud use midpt theorem

[green kettle]

damn bro smart

[toxic axle]

And now try looking at triangles OAB and ODC, as well as recall a property of a median in right triangle that falls onto the hypotenuse.

[green kettle]

im cofnused tho

angle A + angle B gotta be = 180

or else it aint a trapezoid right

thats a property

[toxic axle]

No.

[green kettle]

adj angles sum upto 180 right

[toxic axle]

Yeah, but not those.

[green kettle]

great 8th std level maths down the drain 💀

[toxic axle]

A + D and B + C are both 180°.

[green kettle]

Ohhh

[toxic axle]

A and B are independent.

[green kettle]

i see

[spark kestrel]

well, we know that AM=MB and DN=NC, and now we know that those 2 triangles are similar

[toxic axle]

Right.

Now, ON is the median of a right triangle ODC from its right angle.

So, what property does it have?

[spark kestrel]

that it splits the side into two equal parts?

[toxic axle]

Well, yes, but I mean its length.

[spark kestrel]

hmm

[somber adder]

||Similar triangles?||

[spark kestrel]

Yes, but I am not surw how to use this information

[drowsy furnace]

if two triangles ABC and XYZ are similar, then the following relation between their sides can be made:
AB/XY = BC/YZ = AC/XZ

meaning the ratio of AB to XY is the same as the ratio of BC to YZ

AB is n times bigger than XY, and BC also is n times bigger than YZ

where n is a specific coefficient that can range from 0 (not including 0) to infinity

I need to make myself russian keycaps or at least find sticker for cyrillic because copying letters one by one from the wikipedia page is quite annoying

[somber adder]

Pythagorean and similar triangles is what I was thinking

[drowsy furnace]

i think we could stay with similar triangles: ||OM/ON = MB/NC; OM = MB so ON = NC||

[toxic axle]

What I meant that ||in a right triangle the median from the right angle is half the length of the hypotenuse.|| That allows us to easily find ON and OM, and clearly NM = OM - ON.

[drowsy furnace]

correct, that i what i based my calculations on, i didn't want to go too deep into it to let catxrine get there herself

[spark kestrel]

so MN=11?

[toxic axle]

Yes.

[spark kestrel]

and still, what am i supposed to do with the punctuated part? Isn't this there for a reason?

[toxic axle]

You could also use a right triangle with those sides, I guess.

[drowsy furnace]

i think it's to help show you that the triangles are similar

[spark kestrel]

thank you all for the help!

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