Calculate area of the Blue right triangle | Circle inscribed | Important Geometry skills explained
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- เผยแพร่เมื่อ 15 ก.ย. 2024
- Learn how to find the area of the Blue shaded right triangle with side length 32. Circle of radius 12 inscribed by the right triangle. Important Geometry skills are also explained: area of Triangle formula; Pythagorean Theorem; Two-tangent theorem; Circle Theorem. Step-by-step tutorial by PreMath.com
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Calculate area of the Blue right triangle | Circle inscribed | Important Geometry skills explained
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I liked how the Pythagorean formula containing one unknown resolved into a linear equation from complete cancellation of the quadratic terms. In hindsight it's obvious they must but when I saw it, I got that moment of happiness that I can only compare to the satisfaction of having a completely clean home. And yes, I have OCD.
2r=a+b-c
a²+b²=c²
You can use this formulas to fid b
Again there are several interesting ways to solve this challenge. I used the tan 2A rule to find the tan of angle BAC and from there used tan to find the length BC and so on. Love these problems, thank you.
To avoid the equation, I did a proportion :
32/x=24/x-12
x=48
base = 48 +12=60
Thanks for the lesson. Ciao
Sir it was a good question..❤❤
Glad to hear that
Thanks for your feedback! Cheers!
You are awesome, Suman. Keep it up 👍
Love and prayers from the USA! 😀
A very interesting question, I solved it because a similar question came in my recently finished board exams. It was really a fun to do it. Great work
Keep it up🙏🙏🙏
BF=12 AD=AE=32-12=20 DC=FC=x
32^2+(12+x)^2=(20+x)^2
1024+144+24x+x^2=400+40x+x^2
16x=768 x=48
32＊(12+48)/2=960
area of the Blue Triangle : 960
Excellent!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
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Thanks
You are very welcome!
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You are awesome, Pramod. Keep it up 👍
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Nice problem! It would be beneficial for me and others to work on the algebraic side in re-writing the equation for solving.
My approach used a derivative of the Pythagorean formula that involved the difference of two squares and factoring them yielded the same result.
You are providing a GREAT service!
Glad you think so!
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You are awesome, my dear friend. Keep it up 👍
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Radius of incircle of the right triangle can be calculated with formula r = (p + b - h)/2 where p is perpendicular, b is base and h is hypothenuse. Knowing r and p and the fact that h^2 = p^2 + b^2 we can easily find the value of b. And the square S = p*b/2.
Cool beans! Very systematic and logical.
I solved for the area using Tangent = Opposite/Adjacent or (TOA). Using TOA I determined that angle A = 61.9275 and angle C = 28.0724869 . I got length of line BC = 60 and the area came out to be the same which is 960
Yes, this is how I worked it out - see my comment.
Very interesting👍
Thanks for sharing😊😊
I solved it the same way, i am very happy,thanks teacher
Due to the tan properties, AD=AE, DC=CF, EB=FB, as the radius of the circle is 12, AE and BF will be 20 and 12 respectively. Then let DC and CF be x, by the formula Pr=2A , we can have : (32+12+x+20+x)(12)=2(32)(12+x)/2. After calculated the x=48, sub it into the formula of triangle area: 32(12+48)/2=960
Very nice, no calculator or trig needed.
Really interesting PROBLEMS you r bringing about sir... 🌹
Brilliant problem, I always seem to struggle to get started with the ones where you use the quadratic doohickie, I fully understand it but have a problem getting started, thinking where to stick the variable, and its frustrating the daylights out of me more practice I guess 👍🏻😃
Boa Noite MESTRE
Muit[issimo obrigado pelas aulas
Graças a DEUS e ao Sr consegui aprender Geometria
DEUS lhe Abençoe
As OF amd OE are radii of the inscribed circle, and AB and BC are tangent to the circle, ∠OEB and ∠BFO are right triangles, and BFOE is a square of side length 12, as 12 is the radius of the circle.
Let x be the length of the triangle base BC. By Two Tangent Theorem, for any point that connects to two tangents of a circle, those two line segments are congruent. As AB, BC, and CA are all tangent ti the circle, we have three such congruent pairs: EB and BF, FC and CD, and DA and AE.
As AE = 32-12 = 20, DA is also 20. As FC = x-12, CD is also x-12. This means CA = x-12+20 = x+8
Triangle ∆ABC:
a² + b² = c²
32² + x² = (x+8)²
1024 + x² = x² + 16x + 64
16x = 1024 - 64 = 960
x = 960/16 = 60
A = bh/2 = 60(32)/2 = 960
Amazing sir
Keep watching
Thanks for your feedback! Cheers!
You are awesome, Ravi. Keep it up 👍
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by watching your other videos I was able to apply what you have taught to easily solve this one, many thanks
Tricky at first but was able to get it midway
Excellent!
Thanks for your feedback! Cheers!
You are awesome, Kevin. Keep it up 👍
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From Pythagorean identity express |AC| in terms of |BC|
if a = |BC| , b = h = |AB| = 32
c = sqrt(1024+a^2)
Comparing two formulas for Area of triangle
1/2a*ha and 1/2(a+b+c)r
1/2*32*a = 1/2(32+a+sqrt(1024+a^2))*12
16a = 6*32 +6a + 6sqrt(1024+a^2)
10a - 192 = 6sqrt(1024+a^2)
5a - 96 = 3sqrt(1024+a^2)
(5a-96)^2-9(1024+a^2)=0
25a^2-960a+9216-9216-9a^2 = 0
16a^2 - 960a=0
a^2 - 60a = 0
a(a-60)=0
a = 60
Area = 1/2*60*32
Area = 960
Great explanation
Καλησπέρα σας κύριε συνάδελφε. Σήμερα είδα την άσκησή σας και η λύση μου βασίζεται στην σχέση: (ΑΒC)=τ.ρ όπου τ=(α+β+γ)/2 και ρ η ακτίνα του εγγεγραμμένου κύκλου, σχέση που μαθαίνουν οι μαθητές μας στην 2η Τάξη του Λυκείου. Τα προκαταρτικά (BE=BF=12, AE=BD=20,BC=12+x, AC=20+x) οπότε 1/2*(32)*(12+x)={[32+(20+x)+(12+x)]/2}*12=>...x=48 οπότε ΒC=60 και μετά (ABC)=(32*60)/2=960 sq.units. Ευχαριστώ.
Yay! I solved it.
I solved this problem in 2' !!
Exactly same method.
(20+x)" = (12+x)" + 32"
x"+40x+400 = x"+24x+144 +1024
16x = 768 ➡ x = 48
A = (1/2)•32•60* = 960 < 31"
"That make You Larger Than Life" 🎶 BSB^ 😉
Nice working.
Thanks sir❤
Can you give us lessons to prpare to iranian geometry olympiad
Pleeeeeaaaaase😢
Solved it under 3 m😎
Excellent!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
Love and prayers from the USA! 😀
960
Some people may have wondered, as I did, whether this might be possible without Pythagoras. It turns out that it is. All we need is two different but related ways of calculating the triangle's area.
Spoiler alert.
The area, A, of the triangle:
A = S ∙ r, where S is the semi-perimeter of the triangle and r is the radius of the incircle.
A = p ∙ q, where p and q are the lengths of the hypotenuse on either side of its point of tangency with the incircle.
The semi-perimeter, S, of the triangle:
S = p + q + r.
Equating the two formulae for the area of the triangle:
S ∙ r = r (p + q + r).
S ∙ r = p ∙ q.
p ∙ q = r (p + q + r).
Deriving q from p and r:
(p − r) q = r (p + r).
q = r (p + r) / (p − r).
In our example:
r = 12;
p = 32 − r = 20.
p + r = 32;
p − r = 8.
So:
q = 12 (32) / 8 = 48.
p = 20, q = 48, r = 12.
Area of our triangle:
A = 12 (20 + 48 + 12) = 12 ∙ 80 = 960.
Or:
A = 20 ∙ 48 = 10 ∙ 96 = 960.
I get the different results by triangle similarities. 32/(2r+x)=24/(r+x). I dont't know what is wrong. I got x equal to 24
Are you from USA sir?it was a nice question
The area of the circle
is r×r×3.14
12×12×3.14
So the area of triangle is (32×h)/2
if you want I have Turkish olympic standart geometry tests
S=960
Area of blue right triangle=960 square units
👍
Sir are you from USA?
It was a nice question
I love following your strategies but you're moving forward too slowly for med.
Solved with formula S=pr 😅
This is a lot quicker than using Pythagoras and a lot of algebra...
Other contributors have said they used trigonometry, without giving a full, step-by-step solution.
Pause the video three minutes in, and draw OD = 12
Half of angle BAC = atn(OD/DA) = atn(12/20)
Angle BAC = 2*atn(12/20) = 61.9 degrees
Length of side BC = 32 * tan BAC = 60 (base)
Area of triangle ABC = (half of base times height) = 60*32/2 = 960
أهلاً وسهلاً-يتعذر الاشتراك ولا يوجد جرس-يتعذر التعليق-لا أعلم .
Very easy😂
❤😂👍🍺
So nice of you.
Thanks for your continued love and support!
You are the best. Keep it up 👍
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@@PreMath 🥂