Calculate the Radius of the Quarter Circle | Important Geometry and Algebra skills explained

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Calculate the Radius of the Quarter Circle | Important Geometry and Algebra skills explained

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[bigm]

Another excellent stepwise explanation! Thank you Professor!❤🥂

[libertarianguy]

The squareroot of 117 is 3*(13^0.5) so this can be reduced even further to 6*(13^0.5).

[andreare]

How do you know that BCD is a line segment?
I suppose you have to state that if you extend BC and given that <ABC is 90°, then D must be on the diameter opposite A. You take for granted that when you mirror the triangle AOC, BC will be a line, but I am not sure you can say that.

[cleador]

Perhaps I miss something: why are BC and CD on the same line (i.e. BCD is 180 degrees)?

[DougNeville]

I got to BD = 26 but then used similar triangles. Triangles ABD and COD are similar and give you 2r/36 = 26/r. Rearrange to get r^2 = 36 x 13 (and leave the calculation like that since you are about to take a square root !), and so r = 6 sqrt 13.

[philipkudrna]

AC = 26 by Pythagoras. AC is also the missing leg/extension of BC to the other end of the semicircle. Thus the Diameter is sqrt(24^2+(10+26)^2) or 43, 27. The radius is half or 21, 63.

[prbmax]

Scaffolding to perfection in every lesson. New term I learned from you. Thanks.

[rickdesper]

117 = 9*13. You could simplify sqrt(117) = 3sqrt(13). You could simplify your equations by factoring right at the start.
When you have 24^2 + 36^2 = 4r^2, factor out 12. 12^2(2^2 + 3^2) = 4r^2, simplifies to 12^2(13) = 4r^2, divide by 4, (6^2)*13 = r^2, so r = 6 sqrt(13).
The general rule is that, if (a, b, c) is a Pythagorean triple, so is (ka, kb, kc) for any positive value of k. In this case, we're looking at a triangle with two legs of 12*2 and 12*3, so rather than look at the (24, 36, ??) triple, we can make our life easier by considering the (2, 3, ??) triple. And that's simple, it's (2, 3, sqrt(13)). When you use simpler numbers you'll have simpler computations. And in this case, you would not have left a square factor under the radical.

[kevinmadden]

Sqrt117 eauals3(sqrt13).Therefore 2(sqrt 117) equals 6 sqrt(13). Also, you could use the projection theorem involving the legs of a right triangle and the projection of the leg upon the hypotenuse.

[EnnioPiovesan]

Once you find that AC=26 (and so CD) you can observe that triangle ABD is similar to triangle COD. So we can write AD:CD=BD:OD, so 2r:26:36:r and from this 2r^2=26*36 = 2*13*6^2 so r^2=13*6^2 and r=6*sqrt(13).

[theoyanto]

Another master class, thanks again 👍🏻

[dannymeslier]

rtsq 117 = 3 rtsq 13, so the final answer could also be written 6 rtsq 13

[jieqixu]

We must know that to calculate a right triangle, we can first scale it down.
24：36--2：3 The common factor is 12.
So the hypotenuse is 12√13. We only need to calculate the sum of the squares of 2 and 3, don't forget the 12 after the root sign.

[gideonlapidus]

Beautiful proof, you definitely need to think outside the box with these problems.

[Vedant-Goyal]

Please tell which app do you use to write on ?

[misterenter-izrz]

A bit tricky, this puzzle. Extend BC to draw a semicircle, we get a right angle triangle, with sides 24, 10+26, the diameter, so radius is root of 24^2+36^2/2=6 root 13=21.63 approximately🙂done.

[marcelowanderleycorreia]

Beautifull question!! Congratulations professor!!

[srirajan]

Echoing the below comments: great proof and well-articulated!! An alternative to SAS congruence of triangles AOC and DOC is to think of rotating triangle ABC 180 degrees about the vertical axis to create its mirror on the right semicircle: the AC hypotenuse becomes the as-drawn CD hypotenuse, automatically resulting in 26 for CD.

[tanaysingh]

You are videos are very helpful thanks for giving so much content to us

[jimmayors]

Shouldn't the green lettered ABD triangle sides have been 24 + (36 + 10) + 2r?