Can you find area of the triangle? | (with and without Heron's Formula) |
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- Опубліковано 30 вер 2024
- Learn how to find the area of the scalene triangle. Important Geometry and Algebra skills are also explained: Heron's formula; Area of a triangle formula. Step-by-step tutorial by PreMath.com
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Can you find area of the triangle? | (with and without Heron's Formula) | #math #maths | #geometry
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Thanks dear❤️
One part of the problem can be done by mental arithmetic - no mathmatical manipulations required. The triangle can be made up of two standard right angle triangles - a 5:12:13 and a 9:12:15 ( 3x scaled up 3:4:5 triangle) From inspection the common perp height is 12 so area is (12 x 14)/2 =84.
The second method may not be called a separate method.
It is a repetition of the first method .
Cosine rule to get one angle, then use that angle in area=1/2a.b.sinangle
I think it is not required to know any angle.
Just find the value of Cos C.
Then sinC =√(1-cos^2C)
Put this value in 1/2absinC to get the area.
We Can use cos Law
Let's do it:
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With Heron:
s = (13 + 14 + 15)/2 = 42/2 = 21
A = √[21*(21 − 13)*(21 − 14)*(21 − 15)] = √(21*8*7*6) = √(3*7*2³*7*2*3) = 84
Without Heron:
May h be the height of the triangle according to base AB. By applying the Pythagorean theorem we obtain:
h² + x² = 13²
h² + (14 − x)² = 15²
h² + x² = 169
h² + 196 − 28*x + x² = 225
h² + x² = 169
h² + x² − 28*x = 29
28*x = 140
⇒ x = 5
⇒ h = √(169 − x²) = √(169 − 5²) = √(169 − 25) = √144 = 12
A = (1/2)*AB*h(AB) = (1/2)*14*12 = 84 ✓
1st method
Connect C to D (D on AB)
CD right AB
Let CD=h ; AD=x
so x^2+h^2=13^2=169 (1)
CD=AB-AD=14-x
h^2+(14-x)^2=15^2=225
h^2+x^2+196-28x=225 (2)
(2) 169+296-28x=225
So x=5
(1) h^2+25=169
h=12
Area of triangle=1/2(14)(12)=84 square units.
2nd method
heron's formula
√s(s-a)(s-b)(s-c)
s=(a+b+c)/2=(13+14+15)/2=21
So area of triangle√21(21-13)(21-14)(21-15)=84 square units.
3rd method
14^2=13^2+15^2-2(13)(15)cos(x)
Cos(x)=33/65
Sin(x)=√1-(33/65)^2=56/65
Area of triangle=1/2(13)(15)(56/65)=84 square units.Thanks sir.❤❤❤
Here's my cheat: there is only one perpendicular to a line from a point not on that line, and the "13" side suggests a 5-12-13 right triangle x is 5 and b is 12, so the area is 84, or half bh
(12x14 = 168). It checks out, because 14-h = 9, and 9-12-15 is also a right triangle, the basic 3-4-5.
В этом волшебном треугольнике, площадь равна двум периметрам. Р=42, А=84!
84
9* 12= 108/2 =54
5* 12= 60/2 =30
-------
84
Via Heron's Formula:
Area = √s(s-a)(s-b)(s-c)
s= (a+b+c)/2
s= (13+14+15)/2
s= 21
Area = √21(21-13)(21-14)(21-15)
Area = 84 sq.units
S=(3×14)/2=21, so A^2=21×8×7×6=3^2×7^2×2^4=(3×4×7)^2, A=84😮13^2-s^2=14^2-(15-s)^2, 30s=198, s=6.6, h^2=13^2-6.6^2, h=11.2, area=1/2×11.2×15=84.😅
Although Heron's is fine for this, I tried another way. I overcomplicated this, so messed it up a bit.
Using your labelling, I had a point D on the base. I also had a point (M) for the base's midpoint.
AM = BM = 7
DM = x
CD = h
To find x I then went:
13^2 - (7-x)^2 = 15^2 - (7+x)^2
169 - (49 - 14x + x^2) = 225 - (49 + 14x + x^2)
Remove brackets and change signs where needed:
120 + 14x - x^2 = 176 - 14x - x^2
120 + 14x = 176 - 14x
Therefore, the difference between 120 and 176 (a difference of 56) is 28x, so x = 2 as in
14x = 56 - 14x, so 28x = 56 so x = 2.
This splits the base into 5 and 9 (7-x and 7+x)
Okay, it works fine this time. I will leave this up there as an alternative method, but I messed it up first attempt. I think I forgot to change one of the signs. The 5,12,13 and 9,12,15 (multiple of 3,4,5) takes care of the rest.
Thanks once again.
Here I go again on my own!!
1) 14 = X + (14 - X)
2) Let h be the Height
3) h^2 = 15^2 - (14 - X)^2
4) h^2 = 13^2 - X^2
5) 225 - (14 - X)^2 = 169 - X^2
6) Solution : X = 5
7) h^2 = 144
8) h = 12
9) A = (14 * 12) / 2
10) A = 168 / 2
11) A = 84
12) ANSWER : Area of Triangle equal 84 Square Units.
,we may first find CosC
CosC=(169+225-196)/2*13*15
From here sinC=√(1-cos^2C)
Then 🔺 =1/2*13*15*sinC
++no need to evaluate the magnitude of angle ++
An other method could be using the Briggs formula with a,b,c the 3 sides of the triangle finding sin alpha/2 and cos alpha/2. And in this way finding sin alpha as 2*sin alpha/2*cos alpha/2. Once known sin alpha we can find area with trigonometry
Con un simple y rápido cálculo mental podemos suponer que si 15=3*15 → h=3*4=12 → Se confirma la hipótesis puesto que AB=5+9 y la ecuación 5²+12²=13² es verdadera → Área ABC=14*12/2=84 ud².
Utilizando la fórmula de Herón obtenemos el mismo resultado: Perímetro =13+14+15=42→ Semiperímetro =21→ Área ABC=√ (21*6*7*8)=84 ud².
Gracias y un saludo cordial.
It's been a while and I did not remember Heron's formula. I figured this out using the second method.
It is good idea to mention in the description which method will be used. so visitors can try that.
cos(α/2)=√21*8/210=√4/5...A=(1/2)14*15sinα=210cos(α/2)sin(α/2)=210√4/5√1/5=420/5=84
Very easy even without Heron's
Nice! φ = 30°; ∆ ABC → AB = 14; BC = 15; AC = 13; ABC = ϑ →
252 = 14(30)cos(ϑ) → cos(ϑ) = 3/5 → sin(ϑ) = √(1 - cos(ϑ)^2) = 4/5 → area ∆ ABC = (1/2)sin(ϑ)(14)15 = 84
4th method ???
Thank you!
😮😮😮😮😮😮😮
Dude, was that your phone?
Yes! 😀