A Very Nice Geometry Problem | You should be able to solve this! | 2 Methods

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[@santiagoarosam430]

(6+3)²+4²=EB²=97→ 97-5²=X²=72→ X=6√2 ud.
Gracias y un saludo cordial.

[@vaan_jee]

There is a 3rd method :
- consider O as the intersection between EB and DC
- ED // CB => use Thales in ECBD : CB = 2 ED, so OC = 2.OD and OD + OC = 4 => OC = 8/3 and OD = 4/3
- use Pythagore in BCO to find OB = sqrt(97)/3 and in EDO to find OE = 2.sqrt(97)/3 => EB = sqrt(97)
- use Pythagore in ABE : EB² = AE² + x² => x² = EB² - AE² = 97 - 25 => x = 6.sqrt(2)

[@Grizzly01-vr4pn]

You mean 'O as the intersection between EB and DC' I think.

[@vaan_jee]

​@@Grizzly01-vr4pnyes thanks, i have changed it !

[@Johanbetter2332]

A cute One, i prefer yours

[@michaeldoerr5810]

The answer is 6*sqrt(2). Apart from the fact that this is another property involving Pythagorean triples, the second method seems easier than the first method. The first method involves applying Law of Cosines twice and then make sure that you set the angle at a location opposite another right angle. That might be applicable to Pythagorean triplet sectors of regular polygons. I could be wrong. And I hope that I have gotten it almost right to the point of being people "should be able to do this". I would like to see a playlist with that title invoked so that I can be a JEE certified mathematician!!!

[@giuseppemalaguti435]

Posto BC..=α..risultano le equazioni 9sinα=5+4cosα...X=9cosα+4sinα...elevo al quadrato X^2=81+16-25=72

[@maluhadli468]

BE square is root 97 not 97

[@murdock5537]

Nice! btw: φ = 30°; AEB = τ; BEC = γ → AEC = τ + γ →
sin⁡(τ + γ) = sin⁡(τ)cos⁡(γ) + sin⁡(γ)cos⁡(τ) = (6/485)(20 + 43√2) < sin⁡(3φ) → τ + γ < 3φ

[@devondevon4366]

Answer sqrt 72 or 8.49
Draw two lines, one parallel to 4 and the other parallel to 3.
These two lines meet at a right angle: the longest line is 9 (6+3), and the
shortest is 4. Draw another line ( the hypotenuse) to form a right triangle.
The length of the hypotenuse h= sqrt ( 9^2 + 4^2) = sqrt (81 + 16) = sqrt 97
almost made an error putting 4^2 as 8 instead of 16
Since another triangle was formed, the length of its hypotenuse is also sqrt 97, and its other two sides are 5 and X
Hence, X = sqrt ( sqrt 97^2- 5^2)
= sqrt (97 - 25
= sqrt (72)

[@GiuseppeAriano]

An unbeareble sensation of deja-vu

[@RealQinnMalloryu4]

(5)^2=25 (3)^2=9 (4)^2=16 (6)^2=36 {25+916+36}=116 116/90°ABC=1.26ABC 1^1.2^13^1 2^1^1 2^1 (ABC ➖ 2ABC+1).

[@tanchong2629]

X=10

[@devondevon4366]

The length of the hypotenuse is sqrt 97 or 9.849, hence x must be less than 9.849 and hence less than 10

[@imetroangola4943]

Mais fácil do que tirar dinheiro de cego!

[@prossvay8744]

x=6√2

[@Grizzly01-vr4pn]

The first method is ridiculous.

[@danielsteve-5759]

10 ?