Can anyone solve this math for me?
-
The explanation don't explain why AE must be a diameter of the circle. What makes that obvious?
Thales' Theorem
-
No, but someone could
-
Thales' Theorem
Thanks. I had the same Q: https://en.m.wikipedia.org/wiki/Thales's_theorem
-
x = 15
Denote the origin of the circle O and the points A, B, C. 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) -
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 two lengths are equal and the bottom-right angle is 90 degrees.
but does it have to be a given, or can we actually prove that it has to be
-
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.
-
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.
-
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.
-
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.
-
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.
-
Take a look at this page, it'll give you not only your answer but explain how to solve it
-
Well the drawing is wrong. I measured it with a ruler and it should be 9
-
24,7 and 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.
-
No, but someone could
Correct.
-
Jesus.
Jesus is always the answer
-
OP sneakily making Lemmy do their homework, well played.
-
Tats so cool! Did you just do that or find it?
-
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.
-
Take a look at this page, it'll give you not only your answer but explain how to solve it
-
No, sorry, I'm dumb.
