I like to try to solve these before watching the video, and to think about them in my head until I see an "elegant path." In this case I hate that that "Think outside" comment is up there - I'd prefer to not have hints like that. We can use the Pythagorean theorem to solve this. First application - complete the right triangle formed by the 24-long and 7-long sides. That hypotenuse is length 25. It's mirror image in the right quarter circle will extend the 7-long line to the other side of the semicircle, so we see that that full side (including the 7-long line) is length 32. Now apply the the Pythagorean theorem again to that large triangle (with sides 24 and 32) to find that the semicircle diameter is 40. So the semicircle radius is 20. Now we have a final right triangle with hypotenuse 25, side length 20, and side length the quantity we seek. Apply the Pythagorean theorem one last time to find that H = sqrt(25^2 - 20^2) = sqrt(225) = 15. Q.E.D.
The hypotenuse of 25 calculated for the triangle shown this is also the hypotenuse of the right triangle including the base of the 1/4 circle and the unknown line segment. A 5,4,3 triangle implies if hypotenuse =25 then other sides are 20 and 15 respectively ergo line segment =15. No need to go outside the quarter circle.
No need to think outside the quater circle? O the center of the quater circle, A on the circle on the left of the diagram, B on the vertical at distance x of O, and C on the quater circle as angle ACB is 90°. R is the radius of the circle. We consider OACB inscripted in the quater circle and use the Ptolemy theorem: AB.OC = AC.OB + BC.OA, then: 25.R = 24.x + 7.R That gives: R= (24/18).x = (4/3).x Now we use the Pythagorean theorem in triangle OAB: AB^2 = OA^2 + OB^2, so 25^2 = (16/9).x^2 + x^2, giving that x^2 = 625/(25/9) = 225. Finally x = 15 (and R = 20)
i dont understand extending the line which perfectly meets the other end of the circle (B) how does this property work and what conditions must it have?
Thanks for sharing the fun power of math! Here is a video solving a similar problem using a free, browser-based modeling tool. ruclips.net/video/BLDpTQ5ediM/видео.html
I like to try to solve these before watching the video, and to think about them in my head until I see an "elegant path." In this case I hate that that "Think outside" comment is up there - I'd prefer to not have hints like that.
We can use the Pythagorean theorem to solve this. First application - complete the right triangle formed by the 24-long and 7-long sides. That hypotenuse is length 25. It's mirror image in the right quarter circle will extend the 7-long line to the other side of the semicircle, so we see that that full side (including the 7-long line) is length 32. Now apply the the Pythagorean theorem again to that large triangle (with sides 24 and 32) to find that the semicircle diameter is 40. So the semicircle radius is 20. Now we have a final right triangle with hypotenuse 25, side length 20, and side length the quantity we seek. Apply the Pythagorean theorem one last time to find that H = sqrt(25^2 - 20^2) = sqrt(225) = 15.
Q.E.D.
Simple, cler and clever. Nice.
Love your sense of humor.
Pytagorean theorem:
s²=24²+7²
s = 25 cm
d²=24²+(7+s)²
d = 40 cm, r = 20 cm
h² = s² - r²
h² = 25² - 20²
h = 15 cm ( Solved √ )
The hypotenuse of 25 calculated for the triangle shown this is also the hypotenuse of the right triangle including the base of the 1/4 circle and the unknown line segment. A 5,4,3 triangle implies if hypotenuse =25 then other sides are 20 and 15 respectively ergo line segment =15. No need to go outside the quarter circle.
so I assume your math teacher also taught you how to Psychicly guess triangls ratios
No need to think outside the quater circle? O the center of the quater circle, A on the circle on the left of the diagram, B on the vertical at distance x of O, and C on the quater circle as angle ACB is 90°. R is the radius of the circle.
We consider OACB inscripted in the quater circle and use the Ptolemy theorem: AB.OC = AC.OB + BC.OA, then: 25.R = 24.x + 7.R
That gives: R= (24/18).x = (4/3).x
Now we use the Pythagorean theorem in triangle OAB: AB^2 = OA^2 + OB^2, so 25^2 = (16/9).x^2 + x^2, giving that x^2 = 625/(25/9) = 225.
Finally x = 15 (and R = 20)
wow
I've often wondered why we weren't just taught Ptolemy's theorem in high school instead of the Pythagorean theorem. It's more general.
Love this bro! Keep going❤
i dont understand extending the line which perfectly meets the other end of the circle (B) how does this property work and what conditions must it have?
Look at Thales's theorem: en.wikipedia.org/wiki/Thales%27s_theorem
@@MindMathEnigmas thank you very much
a right angle inscribed in a circle intercepts a semicircle, because an inscribed angle has measure equal to half that of the arc it intercepts
@@jonah8891 thanks man
Nice problem, very nice solution
15
Nice
15 cm
Thanks for sharing the fun power of math! Here is a video solving a similar problem using a free, browser-based modeling tool. ruclips.net/video/BLDpTQ5ediM/видео.html