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
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
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.
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
@@phungpham1725 Thanks for your solution and sorry for the delayed answer. I was just wondering about the formula area = m*n, so I tried to prove it: A(ABC) = m*n = AD*CD = AE*CF # two tangent theorem = (AB − BE)*(BC − BF) = (AB − r)*(BC − r) = AB*BC − AB*r − BC*R + r² = 2*(1/2)*AB*BC − (AB + BC)*r + r² = 2*A(ABC) − (AB + BC + AC)*r + AC*r + r² = 2*A(ABC) − 2*A(ABC) + AC*r + r² = AC*r + r² = (1/2)*AC*r + (1/2)*AC*r + r² = (1/2)*AC*r + (1/2)*(AD + CD)*r + r² = (1/2)*AC*r + (1/2)*(AE + CF)*r + r² = (1/2)*AC*r + (1/2)*AE*r + (1/2)*CF*r + r² = A(ACO) + A(AEO) + A(CFO) + A(BEOF) This is obviously true, therefore the formula must be correct. But since I used AB*BC=2*A, this formula works only for right triangles, does'nt it? Best regards from Germany
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
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.
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 .....
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.
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 ...!!!!
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?
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
Amazing
Interesting👍
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.
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,
Thank you!
You're welcome!
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
@@phungpham1725 Thanks for your solution and sorry for the delayed answer. I was just wondering about the formula area = m*n, so I tried to prove it:
A(ABC)
= m*n
= AD*CD
= AE*CF # two tangent theorem
= (AB − BE)*(BC − BF)
= (AB − r)*(BC − r)
= AB*BC − AB*r − BC*R + r²
= 2*(1/2)*AB*BC − (AB + BC)*r + r²
= 2*A(ABC) − (AB + BC + AC)*r + AC*r + r²
= 2*A(ABC) − 2*A(ABC) + AC*r + r²
= AC*r + r²
= (1/2)*AC*r + (1/2)*AC*r + r²
= (1/2)*AC*r + (1/2)*(AD + CD)*r + r²
= (1/2)*AC*r + (1/2)*(AE + CF)*r + r²
= (1/2)*AC*r + (1/2)*AE*r + (1/2)*CF*r + r²
= A(ACO) + A(AEO) + A(CFO) + A(BEOF)
This is obviously true, therefore the formula must be correct. But since I used AB*BC=2*A, this formula works only for right triangles, does'nt it?
Best regards from Germany
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 🙏🏻🙏🏻🙏🏻
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
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
Thanks PreMath
Thanks Sir so much
That’s very nice
We are lucky then learn from you .
❤❤❤❤❤
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.
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 ❤️
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)
Trying to do it all in my head, came up with 320. Close, but no cigar.
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
.....
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.
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 ...!!!!
398
Der Akzent nervt ...
quiet ...
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?