Can you find area of the Triangle? | (Simple Rule) |#math #maths | #geometry | #trigonometry

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Can you find area of the Triangle? | (Simple Rule) |#math #maths | #geometry | #trigonometry

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[sergioaiex]

Thanks Sir for your solution, quite different than usual, but efficient

[cristidan]

as other viewers pointed, it can be solved without trigonometry, wich I find it easier. thank you for the video!

[himo]

BD=x DC=√3x AB+BD=DC 2+x=√3x 　　　2=(√3-1)x
x=2/(√3-1)=√3+1 DC=3+√3

area of the Triangle : 2*(3+√3)*1/2=3+√3

[kalavenkataraman]

Draw CD perpendicular to AB produced. BD = x, BC = 2x, CD = 2+x. Use Pythagoras theorem and find x . X= 1+ root 3. Area of ABC = 1/2(3+roor3)×2=3+root3

[blobfish]

Let the corner of the right angle be D, and let BD be "a". h = 2+a = a(sqrt3). a = 1 + sqrt3. h = 3 + sqrt3. A = 0.5 x 2 x h = 3 + sqrt3.

[montynorth]

Let base of dropped perpendicular h be point D.
Then triangle ADC is iscoseles 45, 45, 90 degree.
So AD = DC.
Thus BD = h - 2.
Triangle BDC.
Tan 60 = h / h - 2.
1.732 = h / h - 2.
1.732h - 3.464 = h.
0.732h = 3.464.
h = 4.732.
Area = 1/2 * 2 * 4.732.
4.732.

[quigonkenny]

Extend AB to D and drop a perpendicular from C to D. As ∠CAD = 45° and ∠ADC = 90°, ∠ECA = 90°-45° = 45°. As ∠CAE = ∠ECA = 45°, ∆AEC is an isosceles right triangle and EC = AE.

As ∠ECA = 45° and ∠BCA = 15°, ∠ECB = 45°-15° = 30°. As ∠BEC = 90°, ∠CBE = 90°-30° = 60°, and ∆BEC is a 30-60-90 special right triangle. This means that CB = 2BE and EC = √3BE. Let BE = x.

AE = EC
AB + BE = EC
2 + x = √3x
√3x - x = 2
x(√3-1) = 2
x = 2/(√3-1)
x = 2(√3+1)/(√3-1)(√3+1)
x = 2(√3+1)/(3-1)
x = 2(√3+1)/2 = √3 + 1

EC = √3x = √3(√3+1) = 3 + √3

Triangle ∆ABC:
Aᴛ = bh/2 = 2(3+√3)/2 = 3 + √3 ≈ 4.732 sq units

[wackojacko]

The fun begins when Trigonometric Formulas are divided into 2 major systems:
●Trigonometric Identities
●Trigonometric Ratios

☠ Caution Triangulation Formulas are the leading cause of Strangulation. 🙂

[jimlocke]

Drop a perpendicular from B to AC and label the intersection as point D. ΔABD is an isosceles right triangle. AD and BD have lengths √2. ΔBCD has a given 15° angle, <BCD, and a right angle by construction, so <BDC = 75°, making ΔBCD a 15°-75°-90° right triangle. The long side of a 15°-75°-90° right triangle, CD in this case, is (2 + √3) times as long as short side, BD in this case, so length CD = (2 + √3)(√2) = 2√2 + √6. Length AC = AD + CD = √2 + 2√2 + √6 = 3√2 + √6. Let AC be the base of ΔABC and BD the height. Area = (1/2)bh = (1/2)(AC)(BD) = (1/2)(3√2 + √6)(√2) = (1/2)(6 + √(12)) = 3 + √3, as PreMath also found.

[prossvay]

BD Right CD
CAD=45°=ACD
So BCF=30°
In ∆ BCD
(30°;60°;90°)
Let BD=x ; BC=2x; x√3
AD=AB+BD=2+x
DAC=ACD=45°
So 2+x=x√3
x=√3+1
So CD=√3(√3+1)=3+√3
So area of the triangle=1/2(2)(3+√3)=3+√3 square units=4.73 square units.❤❤❤

[devondevon]

4.732051

A different approach
Made labeling error by using 2 (instead of sqrt 2) but will use the same approach
Draw a 45-degree line from B to form a 45-45-90 right triangle ABP
Since, the hypotenuse = 2, and sine in a 45-45-90 triangle the sides are x, x and x sqrt 2
then 2 = x sqrt 2
2/sqrt2 =x
sqrt 2 = x
Hence, BP= sqrt 2 and AP= sqrt 2. Hence, the area of triangle ABP= sqrt 2* sqrt 2*1/2 = 1 AREA OF ABP

Let's find the area of triangle BCP.
For triangle BCP (which was formed when triangle ABP was formed), which is a 15, 75, and 90-degree right triangle (since angle B =
120 and 120-45=7),
In a 15-75-90 right triangle, the ratio of the angles are sqrt 3-1, sqrt 3 + 1, and 2 sqrt 2
The 15-degree angle corresponds to sqrt 3-1 or 0.732051
and the 75-degree angle corresponds to sqrt 3+1 2.732051
Since in triangle BCP, BP is opposite 15 degrees, and BP length is sqrt 2 (the value for x)
then CP (which is opposite the 75 degrees) can be found by
dividing sqrt 2 by 0.732051, then multiply the result by 2.73205
Hence, sqrt 2/0.73205 * 2.732051 or sqrt 2/ sqrt 3-1 * sqrt 3+ 1
1.93815114 * 2732051
BP= 5.27791
AP= sqrt 2
Area of BCP = 5.27791 * sqrt 2 * 1/2

= 3.73205

Recall area of ABP = 1

Hence, the area of both or the triangle above = 1 + 3.72305 = 4.72305. Answer

The area of a 75-15-90 right triangle can also be found by knowing the length of the hypotenuse.

recall that 90 degrees correspond to 2 sqrt 2
So divide sqrt 2 / 0.73205 = 1.93815114

then multiply the result by 2sqrt 2 = 5.4641
In 15, 75, and 90 degrees, the area is = the hypotenuse square divided by 8

Hence, 5.4641^2/8 = 29.85639= 3.73205, then add the 1 ( the area of ABP) = 4.73205 again the same answer

[marcgriselhubert]

Other method:
H the orthogonal projection of C on (AB) and x = BH
AHC is right isosceles and AH = HC = 2 + x
In triangle HBC: tan(angleHBC) = tan(60°) = sqrt(3) = HC/BC,
so sqrt(3) = (2 + x)/x, giving x = 1 + sqrt(3) when simplified.
So HC = 2 + x = 3 + sqrt(3) and the area of ABC is (1/2).AB.BC
= (1/2).(2).(3 +sqrt(3)) = 3 +sqrt(3).

[Mediterranean]

Drop a line from B to AC in D such as the line is perpendicular to AB
ABD is a isocèles right triangle so BD= 2
AD= 2*sqrt 2 by the Pythagorean theorem
Angle BDC is 180-45=135
So Angle CBD is 15 degrees
So we get that BDC is isoceles
DC=BD=2
AC=2(1+sqrt2)
AB=2
Area of the triangle
1/2 ab sin c
1/2*2*2(1+sqrt2)*sin 45
=sqrt 2+2

[Birol]

My way of solution ▶
we have a triangle Δ ABC
if we draw the height of this triangle, to point D, we would have a right triangle: ΔADC
there is also a smal right triangle ΔDBC, in this triangle ΔADC.
[AB]= 2
[BD]= x

let's consider the triangle ΔBDC
[BD]= x
∠CBD= 60°
⇒
tan(60°)= [DC]/[BD]
√3= [DC]/x
[DC]= √3x

Let's consider the right triangle ΔADC:
sin(45°)= [DC]/[CA]
√2/2= √3x / [CA]
[CA]= √6x

according to the Pythagorean theorem:
[AD]²+[DC]²= [CA]²
[AD]= 2+x
[DC]= √3x
[CA]= √6x
⇒
(2+x)²+(√3x)²= (√6x)²
4+4x+x²+3x²= 6x²
2x²-4x-4=0
x²-2x-2=0
Δ= 4+4*1*2
Δ= 12
√Δ= 2√3

x₁= (2+ 2√3)/2
x₁= 1+ √3

x₂= (2-2√3)/2
x₂= 1-√3
x₂ < 0 ❌
⇒
x= 1+ √3
⇒
[CA]= √6x
[CA]= √6*(1+ √3)
[CA]= √6+ 3√2

A(ΔABC)= 1/2* [CA]*[AB]*sin(45°)
A(ΔABC)= 1/2* (√6+ 3√2)*2*√2/2
A(ΔABC)= 3+√3
A(ΔABC)= 4, 732 square units

[pijanV]

u can reflect trianle across AC to form right angle triangle, it will give 30 60 90 trianglee, rest is easy

[LuisdeBritoCamacho]

The Answer is (3 + sqrt(3)) Square Units. Why?

Because the Height of the Triangle [AOC] = (h) = (2 + a) is equal to the Base of an Isosceles Triangle [AOC] with Angles (45º ; 45º ; 90º). tan(45º) = 1.

As : tan(60º) = sqrt(3) then : tan(60º) = (a + 2) / a

So : (a + 2) / a = sqrt(3)

Single Solution : a = (sqrt(3) + 3)

Base = 2
height = (sqrt(3) + 3)

Area = (sqrt(3) + 3) sq un

And that's all folks.

[santiagoarosam]

Inscribimos el triángulo ABC en el cuadrado ADCE de lado AD=a y cuya diagonal es AC→ El triángulo BDC tiene ángulos 60º/90º/30º→ BD*√3=DC→ (a-2)√3=a→ a=3+√3→ Área ABC=AB*DC/2=2*a/2=a =3+√3.
Gracias y un saludo cordial.

[misterenter-izrz]

2+x=sqrt(3)x, x=2/(sqrt(3)-1), then the area is

[MrPaulc]

i suppose we need the perpendicular height.
Make a point D to the right of AB, such that ADC is a right triangle.
AD is x + 2
<ACD is 45deg so the triangle is a right isosceles
BCD is a 30, 60, 90 with short sides of x and x + 2
x*sqrt(3) = x + 2
Square both sides: 3x^2 = x^2 + 4x + 4
Therefore, 2x^2 = 4x + 4
2x^2 - 4x - 4 = 0 is the quadratic I need
4+or-sqrt(16 - 4*2*-4))/4 = x
(4+or-sqrt(48))/4 = x
Discard minus as it will give a negative result
(4+4*sqrt(3))/4 = x
x = 1 + sqrt(3) so the height is 3 + sqrt(3)
Halve the base to 1, and the area is 3 + sqrt(3) un^2
Decimal approximation is 4.732 un^2
I see you went for Exterior Angle Theorem, and I went for calculating the extension to the base. via 30, 60, 90 plus quadratic formula. No problem, as there are often several ways to solve.
Thanks once again.

[rey-dqnx]

2+x=√3x
x=2/(√3-1)
x=√3+1
AC=√2(3+√3)
Area= ½(2)√2(3+√3)√2/2
Area=3+√3