Can you find area of the triangle? | (with and without Heron's Formula) |

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PreMath

Күн бұрын

Пікірлер: 31

@LuisdeBritoCamacho 5 ай бұрын

🙂🙂🙂🙂🙂🙂🙂

@PreMath 5 ай бұрын

Thanks dear❤️

@johnspathonis1078 5 ай бұрын

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.

@billferrol4202 5 ай бұрын

Cosine rule to get one angle, then use that angle in area=1/2a.b.sinangle

@PrithwirajSen-nj6qq 5 ай бұрын

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.

@prossvay8744 5 ай бұрын

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.❤❤❤

@SirKaftar_Requiem 5 ай бұрын

We Can use cos Law

@PrithwirajSen-nj6qq 5 ай бұрын

The second method may not be called a separate method. It is a repetition of the first method .

@jamestalbott4499 5 ай бұрын

Thank you!

@unknownidentity2846 5 ай бұрын

Let's do it: . .. ... .... ..... 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 ✓

@vinijr_best5925 5 ай бұрын

It is good idea to mention in the description which method will be used. so visitors can try that.

@jamesrocket5616 5 ай бұрын

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

@santiagoarosam430 5 ай бұрын

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.

@alster724 5 ай бұрын

Very easy even without Heron's

@MrPaulc222 2 ай бұрын

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.

@murdock5537 5 ай бұрын

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

@soli9mana-soli4953 5 ай бұрын

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

@Marcus-y1m 5 ай бұрын

😮😮😮😮😮😮😮

@PrithwirajSen-nj6qq 5 ай бұрын

,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 ++

@paulbrower 5 ай бұрын

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.

@giuseppemalaguti435 5 ай бұрын

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

@devondevon4366 5 ай бұрын

84 9* 12= 108/2 =54 5* 12= 60/2 =30 ------- 84

@LuisdeBritoCamacho 5 ай бұрын

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.