making the BOLD ASSUMPTION that the angle of the arch is 90deg (the bottom right corner of your diagram), then the dashed lines will lead you to the value of the bold line.
If the original assumption is correct, then the answer is 15.
Spent too long trying to figure out if this was loss or not.
Well? Is it?
I'ma go with 8 because it's slightly longer than 7
X=42
Shouldn't the person who to lazy to measure x solve this?
Take a look at this page, it'll give you not only your answer but explain how to solve it
https://mathematicsart.com/solved-exercises/solution-find-the-distance-bc-quarter-circle/
Thanks a lot
Dude so smart i was already breaking out the angles and testing everything out
This is the worst way to go about doing it, because you should never assume a drawing is made to scale unless it is specifically marked as such. A protractor would be useless if the drawing isn’t to scale. Generally speaking, if a problem isn’t drawn to scale, it’s because all of the info you need to solve it is already present in the drawing. You don’t need to bust out the protractor to measure angles, because the angles can either be calculated from the available info, or aren’t needed in the first place.
I meant more like trying to use cos tan sin and the pythogoras thing not actually measuring it by hand haha
Note that the problem states that the outer shape is a quarter circle, information not provided in OP's question.
Knowing it is a quarter circle is important because it allows us to validate that the bottom-right angle is 90 degrees.
I actually came to the comments to see if we had this information! Thanks.
but does it have to be a given, or can we actually prove that it has to be
It has to be given, otherwise there would be infinitely many solutions.
You would need some other information to link the line segment X to the rest of the figure.
The explanation don't explain why AE must be a diameter of the circle. What makes that obvious?
Thales' Theorem
How have I had years of math and not seen this? I mean, it’s not super useful for me, but I would have thought I would have seen something like this in pre calculus at least.
Geometry, class six or seven.
I teach this to my grade 9 class in Canada. It’s on the curriculum.
Nice. I have no recollection of seeing this before.
Tbf most Canadian grade 9 teachers skip it.
Wow, that's cool
Ooh clever
Tats so cool! Did you just do that or find it?
OP sneakily making Lemmy do their homework, well played.
Jesus.
Jesus is always the answer
if we assume the bottom right corner is a right angle and is the center of the arc, then it is solvable in the manners that others here have already described. if either of those is not the case, and the image itself doesn't state, then there is insufficient information to solve it.
24 and 7 make a pythagorean triple with 25 as the hypotenuse. If the problem uses one pythagorean triple, it probably uses another, so I assume x is 15, and the radius is 20.
Not the most complete answer, but definitely the fastest one^^
Well the drawing is wrong. I measured it with a ruler and it should be 9
Ok, but what does a well have to do with that?
I assume you need to calculate the red triangle's hypotenuse but it seems like there are too many degrees of freedom to lock down any of the other sides or angles of the triangle including X unless I'm missing some hack involving chords and reflected angles.
Try to draw a full semicircle and extend the 7 units long red line, you will notice it falls on the other corner of the semicircle. In fact, every way of drawing two segments from a semicircle corner to the same point if the circumference forms a right triangle.
Now, on the original figure, draw the hypotenuse of the red triangle, you will notice the hypotenuse is as long as the extension you draw earlier, because both start from the same height and fall on a corner of the same semicircle. That means that you can find the extension by calculating the hypotenuse.
Now, you can calculate 7+extension to get the cathetes of the extended triangle, and it's hypotenuse is the diameter of the semicircle. You can divide the diameter by two to get the radius.
Now, you notice that: X, the radius, and the red hypotenuse form a right triangle, and you know the length of the red hypotenuse and of the radius, so you can find X.
No, sorry, I'm dumb.
Hello dumb! I'm dad
Try to draw a full semicircle and extend the 7 units long red line, you will notice it falls on the other corner of the semicircle. In fact, every way of drawing two segments from a semicircle corner to the same point if the circumference forms a right triangle.
Now, on the original figure, draw the hypotenuse of the red triangle, you will notice the hypotenuse is as long as the extension you draw earlier, because both start from the same height and fall on a corner of the same semicircle. That means that you can find the extension by calculating the hypotenuse.
Now, you can calculate 7+extension to get the cathetes of the extended triangle, and it's hypotenuse is the diameter of the semicircle. You can divide the diameter by two to get the radius.
Now, you notice that: X, the radius, and the red hypotenuse form a right triangle, and you know the length of the red hypotenuse and of the radius, so you can find X.
No, but someone could
Correct.
x = 15
Denote the origin of the circle O and the points A, B, C clockwise starting from the left. From the isosceles triangle OAB we get 2 r sin(alpha/2) = 24, where alpha is the angle between OA and OB.
Construct the line orthogonal to OB that goes through C. The length of the line, h, between C and the intersection is h = 7 sin(beta) = x sin(90 - alpha). Denote the lengths of the parts of OB a and b, where a is connected to B. We have a + b = r
Use Thales circle theorem to find that the triangle ABA' completes the red shape, with A' on the circle opposite to A. That means that the angle between A'A and A'B is alpha/2, but A'OB is also an isosceles triangle. So the angle on the other side, beta, has to be the same. Thus, beta = alpha/2.
Now, put everything together: a = 7 cos (alpha/2), b = h cot(90 - alpha) = 7 sin(alpha/2) tan(alpha), r = 12 / sin(alpha/2).
a + b = r <=> cos(alpha/2) sin(alpha/2) + sin^2(alpha/2) tan(alpha) = 12 / 7
1/2 sin(alpha) + 1/2(1 - cos(alpha)) tan(alpha) = 12/7 <=> tan(alpha) = 24/7
From the identity for h we know that x = 7 sin(alpha/2) / cos(alpha). Insert alpha = arctan(24/7)
Draw a symmetrical thingy, the calculations are then simple. I'll let you figure out on your own what calculation is connected with what geometry:
sqrt(24^2+7^2) = 25
25 + 7 = 32
sqrt(32^2+24^2)=40
40/2=20
x = sqrt(25^2-20^2)=15
The two red lines are at a right angle, you can connect them and solve for the hypotenuse using the normal a^2 + b^2.
At this point you now have a second triangle that contains the X you want. Also, since the outer shape is a quarter-circle (assumed) you know that the corner in the bottom right is 90 degrees which makes the two side equal in length.
Since it’s an equilateral triangle, and you know the hypotenuse, you can back it out with c^2 = 2a^2 and solve it that way.
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