Even my teacher can't do this, let's see who's smart

Even my teacher can't do this, let's see who's smart

kys brainlet

Tell your teacher to stop being lazy

9 + 25 + sqrt((9+25)^2 + (50-18)^2 )

Ok, I see now how tried to solve it. You add three horizontal sections together which sum to the length. You know the distance from the edge of the board to the center of all circles so all you need is the horizontal component of the line connecting the two radii of the small and large circle.
Draw a triangle which has two of its corners at the radii and the third corner in the middle of the two smaller circles radii. This is a right triangle so just use Pythagoras but you need to solve for a non-hypotenuse side so it should be - not +.
9 + 25 + sqrt( (9+25)^2 - (25-9)^2 ) should be the correct formula for that component. 64 is the answer which is 4 cm less than the sum of diameters of a large and small circle.

>let's see who's smart
Not the guy who failed to rotate his picture.

68cm^3

Trisect the right side:
radius of big circle + the "distance" between the centers + radius of small circle.
The two radii are obvious, but now the distance between the two circles:
imagine three lines:
1. connecting the two centers
2. a parallel line to the right side passing through center of small circle
3. parallel to bottom side passing through big circles center

these 3 lines form a triangle with hypotenuse: 25cm+9cm
and the short side: 25cm
now use pythagoras theorem: b = sqrt((25cm)^2+(34cm)^2) = sqrt(1781cm^2)
length l = 25cm + 9cm + sqrt(1781cm^2)
easy.

ololo
answer is 64

The answer is obviously between the diameter of the large circle and the sum of the diameters of a large and a small circle. This gives your answer a range of 50-68. Your answer is 76.
You forgot to minus 9cm from the short side which is needed as the line should start at the center of a small circle, not at the bottom of the board. Also you solved Pythagoras for c not b.

>Even my teacher can't do this

Stop lying and do your own homework, kid

Where are more difficult problems? Why so simple...

I got one for you :

We have a cup shaped like a cube with height of 12cm and we fit 4 spheres inside it,each sphere has a radius of 3cm

How much volume of liquid we xan fit inside the cup?

12*12*12-4*4/3*pi*3^3 = 1275.61 ?

school level of difficulty

[math] 1577.2 cm^3[/math]

oh shit wait, I mean [math]15,77*10^-4 cm^3[/math]

wtf am I doing [math] *10^-2 [/math] instead of [math] 10^-4 [/math]

The centres of the circles form an isosceles triangle with 25+9 cm (sum of radii) as two side lengths and (50-9*2)=16cm as the last one.

Length of rectangle=25cm+9cm+height of isosceles triangle

Height of isosceles triangle=17*cot(arcsin(17/36))

Therefore l=25+9+17cot(arcsin(17/36))

boobs?

>do my homework for me

There's a board for this:

68
50 +9 +9
Easy lol

64 bruh, easy.

>cm^3

Here's my attempt.

That's ridiculously easy though. Join the centers of the circles. You know the length of the two non-vertical sides because it's just 9+25, and the length of the vertical side is easy because it's just 50-9-9. You know the distance from the vertical edge to the right edge of the rectangle is 9 and you know the distance from the left point to the left side is 25cm. So all you need to do is work out the perpendicular height relative to the left point which is easy because you know all three side lengths.

How are you too dumb to do this, this is early high school level

this was so fucking easy that I got a decently clean solution/sketch while figuring it out

forgot