Find the height h of the right triangle | Inscribed circle | Important Geometry skills explained

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Find the height h of the right triangle | Inscribed circle | Important Geometry skills explained

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[arghadeepdas]

First comment please pin my comment sir

[FromTheMountain]

Here's how I solved it. The area of ABC is 40h/2 = 20h. Next, split ABC into three quadrilaterals: AEOF, EBDO, and ODCF. The areas are, respectively, 4(h - 4), 16, and 4*36. Therefore, 20h = 4(h - 4) + 16 + 4*36. It follows that h = 9. No need to solve a quadratic equation!

[alexkirchoff]

I used trigonometry to find out the angle BCA/2 = θ, 4 = 36 * tan θ, BCA = 2 * θ and finally h = b * tan(BCA).Many less calculations by me(and many more by calculator 😉).I know your policy is to use always the lightest tool set necessary for the task and it's probably the most correct one for a teacher.Anyway kind regards and congrats from Italy on the excellent work you do on this channel, sir👏

[harikatragadda]

If H is the hypotenuse and r is the Inradius, then we know for a Right Triangle,
r = ½(Altitude+Base -Hypotenuse)
Here, 4= ½(h+40-H)
H-h=32

Also, r=Area/SemiPerimeter
Here 4=20h/½(h+40+H)
9h-H=40

Adding equations, h = 9

[ecksdeedeedee]

i solved it like this, using 2 formulas for area:
A=40h/2
A=rs (<- Where "r" is incircle radius and "s" is half of the perimeter")
40h/2=4s ==> 40h=8s ==> 5h=s
then multiply by 2 to get not just half but the whole perimeter:
10h=40+h+x(<- Where x is the hypotenuse)
9h=40+x ==> x=9h-40
Now substitute in pythagoras: 40^2+h^2=(9h-40)^2 ==> 1600+h^2=81h^2-720h+1600
==>h^2=81h^2-720h ==> h=81h-720 ==> 80h=720 h=9. done :)

[theoyanto]

You make look so simple, I always understand the problem and the solution but frequently I don't get off to a good start, so just more experience needed I guess... Keep 'em comin 👍🏻

[sulimanibra]

What I like with your channel the the details and accuracy leaving nothing to revise it to guess...no ways to forget ...thanks

[Fred-yqfs]

Using the tangent definition, we have a small square of side 4.
The circle is the inscribed circle, therefore its centre is the intersection of the bisector lines (Classic geometry theorem)
B being the point on the right of the triangle, we have:
t=Tan(B/2) = 4/(40-4) = 1/9
Using the double tangent formula:
Tan(B)=2t/(1-t^2)
We get: Tan(B) = 9/40
Besides: Tan(B) = (h+4)/40
So h = 5

[montynorth]

Using alternative Trig. method.
Tan DCO = 4/36
DCO = 6.34 degrees. (tan-1).
DCF = 12.68 degrees. (doubled)
Tan 12.68 = h/40. (triangle ABC)
h = 0.225 x 40 = 9.

[AnonimityAssured]

This yields an interesting result. I wonder if it's related to the largest circles in 3-4-5, 5-12-13, 7-24-25 and 8-15-17 triangles, each of which has a whole-number radius. I suspect there's a general theorem for Pythagorean triples. If I remember, I'll investigate. Thanks, as always, for the stimulation to curiosity you so often give us.

[santiagoarosam]

The two tangents to the circle drawn from each of the vertices A, B, C, have the same length and are collinear with the sides of the triangle ABC → BC=BD+BC=4+(40-4)=4+36 → AB=AE+EB=X+4 → AC=AF+FC=X+36 → We can set up two equal expressions to find the area of ∆ABC→ 40h/2 = 4²+(4X)+(4*36) → 40 (X+4)/2=160+4X → 20X+80=160+4X → 16X=80 → X=5 → h=4+5=9
Cheers

[peterdevos]

It really helps watching your video's and trying it out. A few weeks ago I hadn't a clue where to start. Then I started to get the hang of it and this one I could solve in 2 minutes. Thanks!

[soniamariadasilveira]

Gostei muito de sua explicação, sempre clara e detalhada para que compreendamos com mais facilidade. Deus o abençoe sempre! Um grande abraço!

[plamenpenchev]

Let 2a and 2b be left upper and right lower angles. Then a + b = 45 grad. Then tan(a+b) = 1,
but tan(a+b) = (tan(a)+tan(b))/(1-tan(a)tan(b)), i.e.
(4/36+4/x)/(1-(4/36)(4/x))=1
The obtained eq. is 8x = 40.

[bigm]

Another very patient stepwise explanation. Thank you!👍🥤

[toninhorosa]

I solve similar to you teachear. The only diference were how I make my Pytagora equation:
AB = h
EB = 4
AE = h - 4
BC = 40
BD = 4
DC = 40 - 4 = 36
CF = 36
AF = h - 4
Pythagora:
(36 + h - 4)^2 = 40^2 + h^2
(h + 32)^2 = 1600 + h^2
h^2 + 64h + 1024 = 1600 + h^2
Simplifying:
64h = 1600 - 1024
64h = 576
h = 576/64
h = 9

[KevinAPamwar]

Nice...
In general

Area of any triangle = 1/2*Perimeter*(Centroid radius)
Here Perimeter = 40+ h +(40*40+h*h)^1/2
Centroid radius = 4

Also, Area =1/2*h*40 =1/2*Perimeter*(Centroid radius)
h*40 =Perimeter*4
10h = Perimeter
10h =40+ h +(40*40+h*h)^1/2
9h-40 = Hypotenuse = (40*40+h*h)^1/2
81h*h -2*9h*40= h*h
80h*h= 2*9h*40
Solving for h gives 9

[brinzanalexandru]

Radius of the inscribed circle in an right angle triangle is (a+b-c)/2 where c is the hypothenuse and then use Pythagoras and you got a system in h and c which leads to h=9

[fafat]

Great
Thanks for making videos like this. these videos are really helpful for me and make me better in geometry💌❤

[jimlocke]

Method using trigonometry: We construct OC to form ΔODC and ΔOFC. Let <DCO = Θ. We note that tan(Θ)= 4/36 = 1/9. ΔODC and ΔOFC are congruent by side angle side. By congruency, <DCO = <FCO. Furthermore, <BCA = <DCO + <FCO = Θ + Θ = 2Θ. We use the formula for tangent of sum of angles: tan(a + b) = (tan(a) + tan(b))(1 - ((tan(a))(tan(b))). In this case, the 2 angles are equal. So, the formula becomes tan(<BCA) = tan(2Θ) = (tan(Θ) + tan(Θ))/(1 - ((tan(Θ))(tan(Θ))) = (2/9)/(1 - ((1/9)(1/9))) = (2/9)/(80/81) = 18/80 = 9/40. However, tan(<BCA) = h/BC = h/40 and tan(<BCA) = tan(2Θ) = 9/40. So h/40 = 9/40 and h = 9, as PreMath also found.

Checking our work (as well as PreMath's): By the Pythagorean theorem, length AC = √(h² + 40²) = √(9² + 40²) = √(81 + 1600) = √(1681) = 41 which must equal x + 36. Because x = h - 4 = 5, x + 36 = 41 and our calculation checks out.