Can you find area of the Yellow shaded triangle? | (Inscribed Circle) |
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- Опубліковано 27 чер 2024
- Learn how to find the area of the Triangle ABC. Important Geometry skills are also explained: Pythagorean Theorem; triangle area formula; Two tangent theorem; SOHCAHTOA. Step-by-step tutorial by PreMath.com
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Great
Glad to hear that!
Thanks for the feedback ❤️
Like the professor did, I determined that AE=AD = 28 , drop a perpendicular from D to AE at M. This, DM, is the height of triangle ADE. triangle AMD is similar to ABC so it is easy to use 3,4,5 ratios to calculate the height of the yellow triangle, DM, to be 22.4, because we know the hypotenuse is 28. So area of yellow triangle is 1/2 * (28*22.4) = 313.6
How I did it, but used congruent triangles ratio.
1/ Label r= the radius and AD= m and DC= n (m AC=14x5=70 (The triangle ABC is a 3-4-5 triple)--> sin theta= 4/5 and p=168/2=84
We have rp= 1176-> r= 1176/84= 14
Note that m x n = area of triangle ABC too, so
m+n= 70
m x n =1176
-> we have the equation ( By Vieta ‘s theorem😊)
sq x - 70x + 1176= 0
--> x= 28 and x’= 42
--> m= 28
Area of the yellow triangle= 1/2 sqm . 4/5=313.6 sq units
Interesting👍
a=56, b=2*1176/56=42, c^2=a^2+b^2= 3136+1764=4900, c=70, r=(a+b-c)/2=(56+42-70)/2=28/2=14, AE=AE=b-r=42-14=28,
area triangle ADE =28^2*sin angle EAD/2= 784*4/2*5=313,6,
I did this a different way, but it was far more complex than yours and needed a calculator for intermediate calculations. Yours was much better.
I made a point H on the left for right triangle ADH, and calculated HD via similar triangles. I then used HD as height for ((HD)*28)/2 (I actually used (HD)*14 for area. I previously calculated 28 by going around the tangents.
Thanks PreMath
Thanks Sir so much
That’s very nice
We are lucky then learn from you .
❤❤❤❤❤
Triangle ∆ABC:
Aᴛ = bh/2
1176 = 56h/2 = 28h
h = 1176/28 = 42
As BC = 56 = 4(14) and AB = 42 = 3(14), ∆ABC is a 14:1 ratio 3-4-5 Pythagorean triple right triangle and CA = 5(14) = 70.
Draw radii OE and OF. As AB and BC are tangent to circle O at E and F respectively, ∠OEB = ∠BFO = 90°. As ∠EBF = 90°also, then ∠FOE = 90° as well and EBFO is a square with side length r.
As DA and AE are tangents to circle O that intersect at A, DA = AE = 42-r. As FC and CD are tangents to circle O that intersect at C, FC = CD = 56-r.
CA = CD + DA
70 = (56-r) + (42-r)
70 = 98 - 2r
2r = 28
r = 28/2 = 14
DA = AE = 42 - (14) = 28
Let ∠DAE = α. sin(α) = BC/CA = 56/70 = 4/5.
Triangle ∆DAE:
Aₜ = DA(AE)sin(α)/2
Aₜ = 28(28)(4/5)/2
Aₜ = 392(4/5)
Aₜ = 1568/5 = 313.6 sq units
AB = (2.area of ABC)/56 = 42
Then AC^2 = 56^2 + 42^2 = 4900, so AC = 70
ABC is similar to a (3, 4, 5) right triangle, with lenghts multiplicated by 14. In a (3, 4, 5) right triangle the radius of the inscripted circle is 1, so here it is 14.
Then AE = 42 - 14 = 28 = AD
The yellow area is (12).(28).(28).sin(angleABAC)
with sin(angleBAC) = 4/5 Finally the yellow area is 1568/5.
Thank you!
You're welcome!
Hello Pre-Math, I'll be grade 10 after the summer and I need some help and advice on Mathematics on the future lessons. Things like solving a quadratic equation, functions and tangents of circles. Could you please make more of those kinds of videos? Thanks
Робимо всі підрахунки для трикутника 3-4-5 з коефіціентом 14. Радіус - 1, сторони жовтого трикутника -2, площа - 1/2×(2^2)×(4/5)×(14^2)
(42+56+70)/2-56=28, then the area is 1/2×28^2×(4/5)=313.6.😊
AB*56/2=1176 AB=42 AC=70
42-r+56-r=70 2r=28 r=14
AE=AD=42-14=28 28*4/5=112/5
Yellow Area = 28*112/5*1/2 = 1568/5 = 313.6
How :
42-r+56-r = 70?
Please proof it 🙏🏻🙏🏻🙏🏻
@@andryvokubadra2644
EO=BF=EB=OF=r AE=AD=42-r DC=DF=56-r
AC=AD+DC=42-r+56-r=70
@@himo3485
Please proof EB = BF = OF = EO 🙏🏻🙏🏻🙏🏻
If EO = OF or EB = BF, I agree. But if both, we must get a proof 🙏🏻🙏🏻🙏🏻
r=A/p=14...AE=AD=28...A=(1/2)28*28*sinBAC=392*(56/70)=392*(4/5)=1568/5
Excellent!
Thanks for sharing ❤️
Trying to do it all in my head, came up with 320. Close, but no cigar.
Let's find the area:
.
..
...
....
.....
First of all we calculate the missing side lengths of the triangle ABC. Since this triangle is a right triangle, we can conclude:
A(ABC) = (1/2)*AB*BC ⇒ AB = 2*A(ABC)/BC = 2*1176/56 = 42
AC² = AB² + BC² = 42² + 56² = (3*14)² + (4*14)² = (5*14)² = 70² ⇒ AC = 70
The radius R of the inscribed circle can be calculated from the area of the triangle and its perimeter P:
P = AB + AC + BC = 42 + 70 + 56 = 168
A(ABC) = (1/2)*P*R ⇒ R = 2*A(ABC)/P = 2*1176/168 = 14
AB, AC and BC are tangents to the circle. According to the two tangent theorem we known that AD=AE and BE=BF. Therefore we obtain:
BE = BF = R = 14
AD = AE = AB − BE = 42 − 14 = 28
Now we are able to calculate the area of the yellow triangle:
A(ADE)
= (1/2)*AD*AE*sin(∠DAE)
= (1/2)*AD²*sin(∠DAE)
= (1/2)*AD²*sin(∠BAC)
= (1/2)*AD²*(BC/AC)
= (1/2)*28²*(56/70)
= 313.6
Best regards from Germany
1/ Label the radius of the inscribed circle=r AD=m and CD= n
2/ AB= 42=14x3, BC=56=14x4--> BC=14x5=70 (the triangle ABC is a 3-4-5 triple)
3/ Calculating r
Let p be half of the perimeter: p=168/2=84
We have r.p=area of the triangle ABC=1176-> r=14
4/ Calculating m and n (m
S=313,6 square units
Excellent!
Thanks for sharing ❤️
My way of solution ▶
A(ΔABC)= 1176
⇒
1176= AB*BC/2
BC= 56
⇒
1176= AB*56/2
AB= 42
according to the Pythagorean theorem:
AB²+BC²= CA²
42²+56²= CA²
CA= √4900
CA= 70
according to the incircle equation:
r= 2A/U
U= AB+BC+CA
U= 42+56+70
U= 168 length units
⇒
r= 2*1176/168
r= 14 length units
∠ CAB= α
tan(α)= 56/42
if we consider the deltoid D(CDOF):
tan(α/2)= r/a
tan(α)= 2tan(α/2)/[1-tan(α/2)²]
tan(α/2)= x
⇒
56/42= 2x/(1-x²)
4/3= 2x/(1-x²)
4x²+6x-4=0
x₁= 1/2
x₂= -2
⇒
tan(α/2)= 1/2
tan(α/2)= 14/a
1/2= 14/a
a= 28
A(ΔAED)= 1/2*a*a*sin(α)
sin(α)= 56/70
= 1/2*28²*56/70
= 2*28²/5
A(ΔAED)= 1568/5
A(ΔAED)= 313,6 square units
we could also find a,b and c in the way by drawing the 3 deltoids, we can see easily that:
a+b= 42
b+c= 56
a+c= 70
⇒
a-b= 14
a+b= 42
⇒
a= 28
b= 14
c= 42
.....
398
1176 × 7/22 = r^2
..
R= ....×7/22 r
P = 2 × 1176 r
X + y + z = 1176 r
Z =1176 r - y
Z = y 1175
X + 1176 y=1176y
X=0
1176 y = 56
294 y =14
4 2 y = 2
Y = 1/21 = r
1/2 × 56 × 1176 y = 1176
Y = 1 /28
R = 4 /3
Yellow area...will be ...!!!!
STEP-BY-STEP RESOLUTION PROPOSAL :
01) AB = (1.176 * 2) / 56
02) AB = 2.352 / 56
03) AB = 42
04) AC = sqrt(1.764 + 3.136)
05) AC = sqrt(4.900)
06) AC = 70
07) Incircle Radius (R) = (56 + 42 - 70) / 2
08) R = (98 - 70) / 2
09) R = 28 / 2
10) R = 14
11) AE = AD = 42 - 14 = 28
12) sin(BÂC) = 56 / 70 = 28 / 35 = 4 / 5 (4 * 7 = 28 and 5 * 7 = 35) = 0,8
13) Yellow Area = [(28 * 28 * sin(BÂC)] / 2
14) Yellow Area = (784 * 0,8) / 2
15) Yellow Area = 627,2 / 2
16) Yellow Area = 313,6
17) ANSWER : Yellow Area equal to 313,6 Square Units.
This the Answer from the Department of Mathematics of The Islamic International Institute for the Study of Ancient Knowledge, Thinking and Wisdom, locate in Cordoba Caliphate - Al Andalus.
You seemed to make the assumption that chord ED formed the base of isosceles triangle AED. There is no proof that AE =AD.
Thank you for your excellent work.
If you go to around the 4:30 mark, he discusses the Two-Tangent Theorem and why AE=AD as both E and D are tangent points between the triangle ABC and the inscribed circle.
@@trumpetbob15 no, he implies that the tangents are at 90 degrees. Which is true for any chord.
There is no proof that the chord is perpendicular to the height of the triangle.
@@briancherdak788 I'm confused by your response, but let me try explaining it a different way. Points E and D are both tangent points for triangle ABC and the inscribed circle (definition of an inscribed circle). Thus, under Two-Tangent Theorem, a point outside the circle (point A), is equal distance from from both tangents - AE=AD. (That is the theorem at the 4:30 mark.) After finding the lengths of the sides, we now need to know the angle in the middle, called theta. When PreMath did the trigonometry section, he first started by calculating sin of theta for the right triangle ABC (opposite AB divided by hypotenuse AC). Then, to calculate the area of the yellow triangle, he used that sin value already calculated in the area formula that does not require a right triangle = 1/2 * side * side * sin(angle in the middle). The two sides are both 28 and he found the sin of the angle in the middle, theta, in the first step, and then completed the problem. The chord ED MUST be the base of a triangle ADE since the shortest distance between two points is a straight line and so the only way to connect lines AD and AE is by connecting line ED, which also happens to be a chord of the circle. The length of chord ED is actually not used in solving this problem; it is simply there to create the yellow triangle ADE. Does that help explain it a little more?
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