Calculate the length AF in a rectangle | Important Geometry and Trigonometry skills explained

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Calculate the length AF in a rectangle | Important Geometry and Trigonometry skills explained

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[dieternagel]

I like your videos. I always have a peace of paper and a pen beside my bed. I watch your videos on the TV in the sleeping room. Before I start a video I think about the strategy to solve it. With this very video I was curious what will take you 9:52 to come to the solution. Some times your video is much shorter as I expect it to be. This tells me I have not found the elegant solution yet. But today...

[mattsta]

There's a simpler way to solve this. If you drop a vertical line from point F to the baseline AB, you get a right triangle with sides 1.5 and 7.5 units. AF=SQRT(1.5^2 +7.5^2) = 7.649

[kunalchakraborty]

I would draw a perpendicular from F to AD which would cut the line into 2 equal half of length 3/2=1.5 unit , the length of the perpendicular is 3 + ( 9/2)=15/2. So AF= sqrt root ( (15/2)*2 + (3/2)*2) = 7.6485

[harikatragadda]

Cool!
Drop a perpendicular from F to AB at G.
By proportionality,
FG = ½*CB = 3/2
In the Right Triangle FGB, GB = 9/2
AG = 12 - 9/2 = 15/2
In the Right Triangle FGA,
AF² = AG² + DG²
AF = ½√234

[misterenter-izrz]

Using coordinate method, that is to determine the coordinates of F, and find the length of AF. Clearly E=(3, 3), B=(12, 0), so F=(7.5, 1.5), thus

[MrMichelX]

I found the solution more easily: F is the center of a rectangle's diagonal. The diagonals of a rectangle cut each other in their center. Thus, you can draw a line perpendicular to the base of the rectangle, by F, which cuts the rectangle by his middle.
You have now a square triangle and by using pythagorean theorem you find that AF²=(3+4.5)²+1.5², which gives you that AF nearly equals to 7.648.
Thank you for having taught to me the cosines law anyway!

[hoanquoc]

Từ điểm F vẽ 2 đường trung bình của tam giác ECB. Công việc còn lại nó đơn giản hơn nhiều bạn ơi. Cám ơn bạn đã đăng những bài toán hình học như này vì nó giúp mọi người tư duy tốt hơn, và có những cách giải tốt hơn. Mình rất thích xem kênh của bạn. Lời cảm ơn đến từ Việt Nam .

[wackojacko]

So the kicker here is the transversal line EB that give us the alternate angle pair CEB and FBA ...so cool! 🙂

[albertofernandez]

Using the Pytagorean theorem on the ECB triangle:
EB²=3²+9²
EB²=9+81
EB²=90
EB=√90
EB=3√10
So, FB=3√10/2
Using the Thales, FO=3/2
Using the Pytagorean theorem:
(3√10/2)²=(3/2)²+OB²
45/2=9/4+OB²
OB²=45/2-9/4
OB²=90/4-9/4
OB²=81/4
OB=9/2
AO=12-9/2=24/2-9/2=15/2
Finally, using the Pytagorean theorem again over AFO triangle:
AF²=(3/2)²+(15/2)²
AF²=9/4+225/4
AF²=234/4
AF²=117/2
AF=√(117/2)

[ybodoN]

For a trigonometric solution from the start:
Using the TOA formula, we instantly know that tan θ = 3/9.
Then, since _cos θ = 1/√(1+tan² θ), _ we obtain cos θ = 3/√10.
Finally, using the CAH formula, we also find that 2x = 3√10.

[Copernicusfreud]

Yay! I solved it. I also used the law of cosines, but only after drawing a line segment between points A and E. AE is 3* (sq rt 2). Angle AEF = 116.565051. EF = 3 * (sq rt 10)/2. AF is approximately 7.65

[DG_Seo]

We can apply the Appollonius Theorem for triangle AEB.

[ZahidHussain-vi]

It's very simple by geometry
Since F is the centre of circle

[murdock]

Nice, many thanks, Sir! √234/2 = 3√26/2
another way:
AB = AG + BD = (3 + 9/2) + 9/2 → FG = BC/2 → sin(AGF) = sin⁡(BGF) = 1 →
AF = √((15/2)^2 + (3/2)^2) = 3√26/2

[MrPaulc]

Going through the back catalogue here :)
Length of EB is sqrt(90) due to sqrt(3^2 + 9^2).
The vertical distance of F from AB is 3/2 as it's the midpoint.
Also, the horizontal distance of F from DA is 3 + (9/2)
This gives a right triangle with short sides of 3/2 and 3+(9/2)
3+(9/2) = 15/2
(3/2)^2 + (15/2)^2 = 9/4 + 225/4 = 234/4
AF = sqrt(234/4)
234 divisible by 9 so sqrt(9)*sqrt(26) so 3*sqrt(26)
(3*sqrt(26))/2 approximates to 7.65
It's unusual for me to take a simpler path than you, but I didn't have to explain stuff.
Thank you again.

[jphilsol]

Thales theorem : the values of the sides of ECB triangle can be divided by 2, then pythagorean theorem, and the expected value is find as well.

[spiderjump]

From F, construct a line perpendicular to AB and DC, forming points G and H on lines AB and DC respectively.
Triangles EHF and ECB are similar.(A-A)
Hence, HF=3/2 and EH=4.5
FG=HF=1.5
AG=DH=7.5
AF^2= FG^2+AG^2
=117/2
AF can then calculated

[AndreasPfizenmaier-yw]

Mattsta is right, now I see his solution, too!

[davidellis]

An entirely different approach is coordinate geometry. Let A=(0, 0), B=(12, 0), C=(12, 3) and E=(3, 3). F is the midpoint of EB, so its coordinates are (7.5, 1.5). Then AF marks the distance between (0, 0) and (7.5, 1.5), defined as sqrt(7.5^2+1.5^2), which simplifies to the value (3/2)*sqrt(26), equivalent to the stated answer of sqrt(234)/2.

[slphang]

Why not use concept of similar triangles?