Can you find area of the Yellow Quadrilateral? | (Semicircle) |
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- Опубликовано: 7 ноя 2024
- Learn how to find the area of the Yellow Quadrilateral. Important Geometry and Algebra skills are also explained: Quadrilateral; Thales' theorem; Pythagorean theorem; Triangle area formula; circle area formula. Step-by-step tutorial by PreMath.com
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A = ½πR² = 578π cm² --> R= 34cm
Pytagorean theorem:
c² = (2R)² - 32² --> c= 60cm
Pytagorean theorem:
x² = c² - 36² --> x= 48cm
Quadrilateral area :
A = A₁ +A₂ = ½bh + ½bh
A = ½ 60*32 + ½36*48
A = 1824 cm² ( Solved √ )
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This video is much more helpful than the comments. And this is for future practice!!!
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Yeah, needed a paper and pen for this one to do the squares and squareroots...
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Once you have the diameter of 68 and the right triangle for which it is the hypotenuse, it can be faster to note that after a quick common factoring that 68 and 32 can be reduced to 4*17 and 4*8, and 17 and 8 are two elements of the Pythagorean Triple of (8,15,17). This gives you the BD length of 60, which can then be fed into triangle BCD, for which we have another Pythagorean triple 12*(3,4,5), which gives the missing DC length of 48.
I have found that in many of these types of problems knowing the first five or so primitive (irreducible) Pythagorean triples (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25), & (20, 21, 29) often offers shortcuts to find lengths.
You are correct; recognizing Pythagorean triples can be very useful!
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AO=OB=r r*r*π*1/2=578π r=34 AB=2r=68
DB=√[68²-32²]=60 DC=√[60²-36²]=48
Quadrilateral area = 36*48*1/2 + 60*32*1/2 = 864 + 960 = 1824
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The area of the semicircle is πr²/2=578, from which r=34 we have BD=√(68²-32²)=60, and we have CD=√(60²-36²)=48, from which the area of the quadrilateral ABCD is equal to (32*60)/2+(48*36)/2=1824.
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Radio =34---> DB=√(68²-32²)=60 ---> CD=√(60²-36²)=48 ---> Área sombreada =(32*60 +36*48)/2 =1824 ud².
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Can I ask about the graphic tablet and the application you used
MacBook pro along with Canva, Camtasia techsmith.
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Semicircle O:
A = πr²/2
578π = πr²/2
1156π = πr²
r² = 1156
r = √1156 = 34
Draw BD. As A and B are points on the diameter and D is a point on the circumference, then ∠BDA = 90°.
Triangle ∆BDA:
BD² + DA² = AB²
BD² + 32² = 68²
BD² = 4624 - 1024 = 3600
BD = √3600 = 60
Triangle ∆BCD:
BC² + CD² = DB²
36² + CD² = 60²
CD² = 3600 - 1296 = 2304
CD = √2304 = 48
The area of quadrilateral ABCD is equal to the sum of the areas of triangles ∆BDA and ∆BCD.
Quadrilateral ABCD:
A = 60(32)/2 + 48(36)/2
A = 960 + 864 = 1824 sq units
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r^2 = 578(2) = 1156 = 34
diameter = 68
68^2 - 32^2 = (100)(36) = 60^2
60^2 - 36^2 = 96(24) = 48^2
area = 60(32)/2 + 36(48)/2 = 960 + 864 = 1824
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@PreMath Thank YOU 🙂
Full circle area would be 1156pi, so r = sqrt(1156) = 34.
ADB is a right triangle with a side of 32 and a hypotenuse 68 (due to 2r).
68^2 - 32^2 = 3,600, so DB = 60.
ADB has area of 60*16 = 960 un^2.
Now for BCD: 3,600 - 36^2 = 3,600 - 1296 = 2304.
sqrt(2304) = 48
24*36 = 864 un^2
864 + 960 = 1824 un^2
I have now looked. Yes, we followed the same path.
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Let's find the area:
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First of all we calculate the diameter d=2r of the semicircle:
A = πr²/2 = π(d/2)²/2 = πd²/8
578π = πd²/8
4624 = d²
⇒ d = √4624 = 68
According to the theorem of Thales the triangle ABD is a right triangle. Therefore we can apply the Pythagorean theorem:
AB² = AD² + BD²
BD² = AB² − AD² = d² − AD² = 68² − 32² = 4624 − 1024 = 3600
⇒ BD = √3600 = 60
The triangle BCD is a also right triangle, so we can apply the Pythagorean theorem again:
BD² = BC² + CD²
CD² = BD² − BC² = 60² − 36² = 3600 − 1296 = 2304
⇒ CD = √2304 = 48
Now we are able to calculate the area of the yellow quadrilateral:
A(ABCD) = A(ABD) + A(BCD) = (1/2)*AD*BD + (1/2)*BC*CD = (1/2)*32*60 + (1/2)*36*48 = 960 + 864 = 1824
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S=1824
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Thale's ale ...the elixir of the gods! 🙂
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Solution:
Semicircle Area = 578π = πr²/2
πr² = 1156π
r² = 1156
r = 34
Thales Theorem (according to this theorem, the angle ADB is a right Angle) and applying Pythagorean Theorem in triangle ABD, at the same time, we have:
DB² + 32² = 68²
DB² = 4624 - 1024
DB² = 3600
DB = 60
In next step, applying Pythagorean Theorem in triangle BCD, we have:
DC² + 36² = 60²
DC² = 3600 - 1296
DC² = 2304
DC = 48
Quadrilateral Area (QA) = Area ABD + Area BCD
QA = ½ 32 . 60 + ½ 36 . 48
QA = 960 + 864
QA = 1824 Square Units ✅
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34 is the radius, sqrt(68^2-32^2)=60, sqrt(60^2-36^2)=48, therefore the answer is 1/2×(32×60+36×48)=1824.😅
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1824
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Plenty of computation 😢
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