Find the area of the quadrilateral | A Very Nice Geometry Problem

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[ناصريناصر-سب]

Triangle ABC has angle A equal to 30° and C equals 75°, so it is isosceles. Therefore, AC=AB=5, so AD=CD=5/√2. The area of the quadrilateral ABCD is

[jimlocke]

At 6:20, we can bisect <BAC, dividing ΔABC into two congruent 15°-75°-90° right triangles. The 15°-75°-90° right triangle appears frequently in problems, so I have suggested that all of us have its properties handy. One property is that its area is equal to the hypotenuse squared divided by 8, or 5²/8 = 25/8 in this case. Since ΔABC is composed of two of these triangles, its area is 25/8 times 2, or 25/4. Skip ahead to 8:10.

[kateknowles]

Joined AC and drew from D a perdendicular line DE to E on AC . In triangles AED and CED we have DE is common, angle E is 90º, and AD = CD which is given. So these two triangles are congruent. Therefore angle DAE equals angle DCE and they are each 45º because ADE = CDE = ½ of 90º and we have two congruent right -angled isosceles triangles here.
Now, in triangle ABC, angle A = 75º- 45º (the 75º is given) angle A is 30º (B is 75º, it is also given) so angle C is 75º because 180-75-30 = 75.
Triangle ABC is therefore isosceles with two angles equal to 75º, so AC=5

The area [ACD] is ½AD.CD AD.AD +CD.CD = AC.AC by Pythagoras' theorem so AD.CD = ½ AC.AC
The area [ACD] is AC.AC/(2x2) = 25/4 =6 +1/4

The area [ABC] is ½ AC.AB sin(30º) this is also 25/4 = 6 +1/4
Added these together
[ABCD] equals 12½

This time I have come to the solution which is also in this video independently.
Thank you Maths Booster for a good Geometry Problem.

[quigonkenny]

Draw AC. As CD = DA and ∠CDA = 90°, then ∆CDA is an isosceles right triangle and ∠DAC = ∠ACD = (180°-90°)/2 = 45°. Let CD = DA = x.

As ∠DAB = 75° and ∠DAC = 45°, then ∠CAB = 75°-45° = 30°. As ∠ABC = 75°, then ∠BCA = 180°-(30°+75°) = 75°. As ∠ABC = ∠BCA = 75°, then ∆CAB is an isosceles triangle and CA = AB = 5.

Triangle ∆CDA:
CD² + DA² = AC²
x² + x² = 5²
2x² = 25
x² = 25/2
x = √(25/2) = 5/√2

The area of quadrilateral ABCD is equal to the areas of triangles ∆CDA and ∆CAB.

Quadrilateral ABCD:
A = bh/2 + (absinC)/2
A = (5/√2)(5/√2)/2 + (5(5)sin30°)/2
A = (25/2)/2 + 25(1/2)/2
A = 25/4 + 25/4 = 25/2 = 12.5 sq units

[giuseppemalaguti]

AD=a, CB=t...risultano le due equazioni..t/sin30=√2a/sin75(t.seni)..2a^2=5^2+t^2-2*5*t*cos75(t.coseno), risulta a=5/√2..t=2, 58819...A(somma

[marioalb]

Diagonal AC is d=5 cm
A = A1 + A2
A = ½d½d + ½d²sin(75°-45°)
A = ¼5² + ¼5² = ½25
A = 12.5 cm² ( Solved √ )

[himo]

AC=5 BE=AB*1/2=5*1/2=5/2 DE=5*1/2=5/2
area of ABCD = 5*5/2*1/2 + 5*5/2*1/2 = 25/4 + 25/4 = 50/4 = 25/2

[alexniklas]

S= 5×5×sin30°+(1/2)×(5/√2)^2/2=
25/4+25/4= 25/2= 12, 5

[RealQinnMalloryu]

(5)^2=25 10^10^1 2^5^2^5^11^1^2^1^12^1(ABCD ➖ 2ABCD+1)

[papebu]

Chi e che parla, la figlia di Fantozzi. Usate la voce vostra che crea più interesse a sentire. Lo lasciato a metà il video.