Find Area of the Blue Triangle without Trigonometry | Important Geometry skills explained

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Find Area of the Blue Triangle without Trigonometry | Important Geometry skills explained

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[Ankitsingh-yjm]

We are thanking you sir for providing us a amazing question every day . These questions gives us capability to think about question how to solve

[Funishawsomish]

I found that 16 is the common height of both triangles so I used the pythagorean theorem to find that the total base of both triangles is 30 I then subtracted 12 to get 18 18•16•1/2 is 144 so that is the answer

[MarieAnne.]

At 5:34, I don't understand why we would even bother with negative square root. x+12 is the length of side BD, so it has to be positive.

[johankotze]

Finding CD = 16 using Pythagoras is simple.
Then finding BD = 30 by Pythagoras is also simple.
AB = 18 is a simple subtraction.
The Area = (1/2)(18)(16) = 144

[theoyanto]

Nice simple problem today, bit of a gift really, but still much appreciated, thank you 🤓👍🏻

[vidyadharjoshi]

20sq - 12 sq = CDsq. CD = 16. ( CD can also be taken from 3:4:5 triangle ). 34sq - 16sq = BDsq. BD = 30, AB = 18. Area = 144

[bhalusingh]

Sir I did this like this first I completed the triangle into a rectangle
Then by Pythagoras theorem find the length
As there was a diagonal then by heron, s formula find the area of blue triangle
Therefore
34²-16²=30²
Then, subtract 12 from 30 and get 18
The used heron formula
√36(18)(2)(16)
144 ans

[murdock]

Nice, many thanks!
∆ACD = pyth. triple (4(3 - 4 - 5)) → CD = 16 →
∆BCD = pyth. triple (2(8 - 15 - 17)) → BD = 30 → AB = 18 → 16(18)/2 = 144 = ∆ABC
btw:
AD/AC = sin⁡(φ) = 3/5 → CD/AC = cos⁡(φ) = 4/5
BD/BC = sin⁡(θ) = 15/17 → CD/BC = cos⁡(θ) = 8/17 →
sin⁡(θ - φ) = sin⁡(θ)cos⁡(φ) - sin⁡(φ)cos⁡(θ) = 36/85 → (θ - φ) ≈ 25, 057° → φ ≈ 36, 87°

[colonelbean]

First one I‘ve completed in my head!

A simple one but a good one!

[flavrt]

Fun puzzle. I took the 16-30-34 pyth triple, into the A = ½ x 16 x (30-12) route.

[procash]

Very simple problem. Solved it in no time

[kennethstevenson]

One shortcut, since 16 is the common height in both triangles I got (15x16) - (6x16) = (9x16) = 144 square units.

[alexniklas]

CD= 16; DB=(34^2-16^2)^(1/2)=30;
A=30×16/2-12×16/2=144

[Copernicusfreud]

I solved it. I used method 1 in the video. I then used Heron's formula for a second method.

[tombufford]

The height = 16 from sqrt(400-144)=sqrt(256). AB = sqrt(17*17*2*2 - 16*16)=30. From half base * height, The area of the large triangle = 15*16=240. small triangle = 8*12=96.Area of blue region = 240-96 = 144.

[quigonkenny]

As DA = 12 = 3(4) and AC = 20 = 5(4), CD = 4(4) = 16 (4:1 ratio 3-4-5 Pythagorean triple).

As CD = 16 = 8(2) and BC = 34 = 17(2), DB = 15(2) = 30 (2:1 ratio 8-15-17 Pythagorean triple).

AB = DB - DA
AB = 30 - 12 = 18

A = bh/2 = 18(16)/2 = 18(8) = 144

[Ramkabharosa]

|CD|² = |AC|² - |AD|² = 20² - 12² = 400 - 144 = 256 = 16².
So |CD| =16. |BD|² = |AC|²-|AD|² = 34²-16² = 2².(17²-8²)
= 2².(289-64) = 2².(225) = 2².15². So |BD|= 2.(15)= 30.
Area(ABC)= (½).|AB|.|CD|= (½).(30-12).16=18.(8)=144.

[ybodoN]

Let AD = a, AC = b and BC = c. Then AB = √(c² − b² + a²) − a and we could eventually apply Heron's formula.
Whatever the method, since CD = √(b² − a²), the area of the blue triangle is ½ (√(c² − b² + a²) − a) √(b² − a²).
With a = 12, b = 20 and c = 34, the area of the blue triangle is ½ (√(34² − 20² + 12²) − 12) √(20² − 12²) = 144.

[Birol]

Accoring the Pythagorean theorem, we have:
CD²+DA²= CA²
CD²+12²=20²
CD²= 400-144
CD²= 256
CD= 16 length unit
Again:
CD²+DB²= CB²
256+DB²=34²
DB²= 900
DB= √900
DB= 30 LU
AB= 30-12
AB= 18 LU
Blue area, BA
BA= AB*CD/2
= 18*16/2
= 144 square units is the answer !

[misterenter-izrz]

Let h be the height of the triangles, so h=16 clearly by pythagorean theorem, the area is hAB/2=8AB, and AB=BD-AD=sqrt(34^2-16^2)-12=30-12=18, therefore the answer is 8x18=144.🙂