Can you find area of the Blue Square? | (Semicircle) |
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- เผยแพร่เมื่อ 2 ก.ค. 2024
- Learn how to find the area of the Blue Square inscribed in a semicircle. Important Geometry and Algebra skills are also explained: Pythagorean theorem; Thales' theorem; area of the square formula; similar triangles. Step-by-step tutorial by PreMath.com
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merci sire , excellent exercice
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tan α = Rcos45°/R = 1/√2
α = 35,26°
2R = 32 / cos α
R = 19,596 cm
A = ½R²
A = 192 cm² ( Solved √ )
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In the third step, instead of considering the similar triangles, extend OP to become a diameter of the circle and then use the intersecting chords theorem: (32 - a√3)a√3 = (a√2 + a)(a√2 - a) = a² => a = 8√3
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Nice Solution for the rigorous problem. Thank you Sir.
Recognizing that AO is the same as the diagonal of the square, that is, both are radii, I took that arctan of PAO as a/(a*2^.5) which was 1/2^5. That angle was 35.26438... So the cosine of this angle is 32 divided by 2*a*2^.5 (which is the diameter). Solve for a which is 13.85640646... Then square that number to equal 192.
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3rd alternative solution:
1/ Just assume that a=1 so r= sqrt2 and AP= sqrt3 and area of blue square= 1
2/ The quadrilateral PCBO is cyclic so,
APxAC=AOxAB-->ACxsqrt3=sqrt2x2sqrt2=4--> AC=4/sqrt3
Now if AC = 32, the area of the blue square = 1x sq((32/(4/sqrt3))=64x3= 192 sq units😊
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Brilliant use of Similar triangles 📐 in the end.
Can we know the name of the app/software with which you are writing ✍️ electronically on the computer screen?
Diagonal of the square = radius of circle =a√2
AP =√[ (a√2)^2 +a^2]=a√3
DP is extended to T to get a chord.
Now intersecting chord theorem states that
DP *PT =AP *PC -- (1)
(DP =PT =a)
AP=a√3
PC= 32 - a√3
Now from equation No. 1 we get
a^2= a √3(32 - a√3)
> 4a^2=32a√3
> 4a =32√3
> a=8√3
a^2=(8√3)^2=64*3=192
Area of square = 192 sq units
Comment please
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Let the radius of semicircle O be r and the side length of square OPDE be s. Draw radius OD. As OD is also the diagonal of square OPDE, this gives us a direct relationship between s and r via Pythagoras:
Triangle ∆OPD:
OP² + PD² = OD²
s² + s² = r²
2s² = r²
s² = r²/2
s = √(r²/2) = r/√2 --- [1]
r = √2s --- [2]
Draw CB. By Thales' Theorem, as A and B are opposite ends of a diameter and C is a point on the circumference, then ∠BCA = 90°. As ∠AOP = 90° as well, and ∠PAO is common, triangles ∆AOP and ∆BCA are similar.
Triangle ∆BCA:
BC/CA = OP/OA
BC/32 = s/r = (r/√2)/r --- [1]
BC = 32/√2 = 16√2
BC² + CA² = AB²
(16√2)² + 32² = (2r)² = (2(√2s))² --- [2]
8s² = 512 + 1024 = 1536
s² = 1536/8 = 192 sq units
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Be R the radius of the circle and t = angleBAC
In triangle AOB: OP = R/sqrt(2) and AO = R
Then AP^2 = OP^2 +AO^2 = (R^2)/2 + R^2
= (3/2).R^2 and AP = sqrt(3/2).R
Then cos(t) = AO/AP = R/(sqrt(3/2).R)
=sqrt(2/3)
In triangle ABC: cos(t) = AC/AB,
so: sqrt(2/3) = 32/(2.R) and R = 16.sqrt(3/2)
The area of the square is (R/sqrt(2))^2
=(R^2)/2 = (256.(3/2))/2 = 64.3 = 192
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Good morning sir I like it
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R=x\/2, ВС/32=х/х\/2, ВС=32/\/2=16\/2, (2R)^2=32^2+(16\/2)^2, (2x\/2)^2=1024+512, 8x^2=1536, x^2=1536/8, x^2=192.
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I've learned 3 methods from the video and comments:
After r (radius) = sqrt2 x a (side of square) by Pythagoras theorem
1. Using Thales theorem and then similar triangles to get a
2. Using sides of right-angled triangle to get cosBAC then Thales theorem to get r and then a
3. Using intersecting chords theorem to get a
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One more solution:
1. Let a= OP=PD; x=AP; PC = 32-x;
2. Let us extend DP to the left to intersect with the circle at point N => PN = PD = a;
3. PN*PD = x*(32-x) => a^2 = x*(32-x); (1)
4. x^2 = r^2 + a^2; (2)
5. r^2 = 2*a^2; (3)
6. system of equations {(1),(2),(3)}
7. let us put (3) into (2): x^2 = 3*a^2 => x = a*sqrt(3); (4)
8. Let us put (4) into (1):
a^2=a*sqrt(3)*(32-a*sqrt(3));
as a!=0 lets devide both sides by a
a = 32*sqrt(3) - 3*a;
4a = 32*sqrt(3);
a= 8*sqrt(3);
Ablue = a^2 = (8*sqrt(3))^2 = 64*3 = 192 sq units.
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Cos(A)=√(2/3). Опускаем перпендикуляр из центра окружности на AC. Делит хорду пополам. а√2=16/(√2/3). а=8√3.
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with the help of you, i can❤
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Very nice! ∆ ABC → AB = 2r = AO + BO; AC = 32; ∎EDPO → PO = EO = DE = DP = a →
DO = AO = BO = a√2 = r; sin(AOP) = 1 → PAO = δ → PO = a; AO = a√2 → AP = a√3 →
cos(δ) = a√2/a√3 = √6/3 = 32/2a√2 → a = 8√3 → ∎EDPO = 64(3) = 192
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Congratulations for the question sir!!
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Okay, I'm dazzled.
Let's find the area:
.
..
...
....
.....
Since OEDP is a square, all interior angles are right angles. Therefore the triangle ODE is a right triangle and we can apply the Pythagorean theorem. With r being the radius of the semicircle and s being the side length of the square we obtain:
OD² = OE² + DE²
r² = s² + s² = 2*s²
According to Thales theorem the triangle ABC is a right triangle. The triangle OAP is also a right triangle. Now let's have a look at the interior angles of these two triangles:
∠AOP = ∠ACB = 90°
∠OAP = ∠BAC = α
∠APO = ∠ABC = β
So these two triangles are similar and we can conclude:
AO/AP = AC/AB
r/AP = 32/(2*r)
r/AP = 16/r
AP/r = r/16
⇒ AP = r²/16
Now we can apply the Pythagorean theorem to the right triangle OAP in order to obtain the area of the square:
AP² = AO² + OP²
(r²/16)² = r² + s²
(2*s²/16)² = 2*s² + s²
(s²/8)² = 3*s²
s⁴/64 = 3*s²
Since s=0 is not a useful solution, the only useful solution is:
A(square) = s² = 3*64 = 192
Best regards from Germany
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32/2r=r/√(r^2+l^2)...l,lato del quadrato..con r=√2l..sostiisco, risulta l=8√3...A=192
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PO=OE=ED=DP=x OD=AO=OB=√2x
⊿AOP∞⊿ACB BC=32/√2=16√2
AB=2√2x (16√2)²+32²=(2√2x)² 512+1024=8x² 8x²=1536
x² = Blue Square area = 192
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OBDP is a square
So OB=BD=DP=PO=x
And OD is Radius of semicircle =x√2
∆ AOP~ ∆ ACB
x/x√2=BC/32
1/√2=BC/32
So BC=16√2
In ∆ ABC
(16√2)^2+(32)^2=(2x√2)^2
So x=8√3
Blue square area =(8√3)^2=192 square units.❤❤❤
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Ok, muy bien
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PreMath likes square roots a lot. And when it should and when it shouldn't. Let's ban square roots! OD^2=2a^2. AP^2=3a^2. (AB^2)/(AC^2)=(AP^2)/(AO)^2.
(8a^2)/32^2=(3a^2)/2a^2. a^2/64=2. a^2=192 square units.😎
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Thank you so much!
My first attempt was almost the same as yours😅!
Here is my 2nd approach:
Let a and r be the side of the square and the radius of the circle respectively.
We have: r= a. sqrt2
So tan angle CAB= OP/OA=a/r=1/(sqrt2)
Note that the angle ACB= 90 degrees so, BC/AC=tan(CAB)=1/sqrt2
--> BC=16sqrt2
By using the Pythagorean theorem
sqAC+sqBC=sqAB
-> sq32+sq(16sqrt2)=sq(2r)=sq(2.a.sqrt2)
--> 8 sqa=512+1024
Area= sq a=192 sq units
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I guess when I think outside the box I'm really looking for a distant function I project onto the space in question. 🤔
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AO = OB is not given. I think that should be clarified.
tan a=1/sqrt(2), so cos a=1/sqrt(3/2), chord length=2 r cos a=2sqrt(2/3)r=32, r=16sqrt(3/2), the area is r^2/2=3×128=384.😊
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=
You need to complete this. Where is final division by 2.
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My way of solution ▶
Since this is a semicircle, if we consider the triangle ΔABC, then ∠ BCA= 90°
OE=ED=DP=PO= a
AD= r
⇒
a√2= r
a= √2 r/2
ΔABC ~ ΔAPO
AP= x
OP= a
OP= √2 r/2
BC= y
CA= 32
⇒
OP/BC= AO/CA= PA/AB
a/y= x/2r= r/32
⇒
√2 r/2 / y= r/32
y= 16√2
BC= 16√2
BC²+CA²= AB²
BC= 16√2
CA= 32
AB= 2r
⇒
(16√2)²+32²= (2r)²
256*2+1024= 4r²
1536= 4r²
r²= 384
Ablue= a²
= (√2 r/2)²
= 2r²/4
= r²/2
Ablue= 384/2
Ablue= 192 square units
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Nice problem and solution! I solved it differently, using analytic geometry, but your way was much more elegant.
You pronounced the Greek mathematician as "thails" but it should be pronounced "thay-leez". That is two syllables. For confirmation, see th-cam.com/video/8jCY1GZmdx4/w-d-xo.html.
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Прошу прощения, но эту задачу я уже решал на этой самой русской олимпиаде у Казакова!
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STEP-BY-STEP TRIGONOMETRICAL RESOLUTION PROPOSAL :
01) Square Side = X
02) Square Diagonal = X*sqrt(2)
03) Radius of Semicircle = Square Diagonal = X*sqrt(2)
04) Angle (OAP) = a and Angle (OPA) = b. aº + bº = 90º
05) tan(a) = X / X*sqrt(2) ; tan(a) = 1/sqrt(2) ; tan(a) = sqrt(2)/2
06) tan(b) = X*sqrt(2)/X ; tan(b) = sqrt(2)
07) sin^2 (a) = 1/3 and cos^2 (a) = 2/3
08) sin(a) = sqrt(3)/3 and cos(a) = sqrt(6)/3
08) BC / 32 = tan(a) ; BC / 32 = sqrt(2)/2 ; BC = (32*sqrt(2))/2 ; BC = 16*sqrt(2) ; BC ~ 22,63
09) Finding Diameter AB (2R) :
10) AB^2 = AC^2 + BC^2 ; AB^2 = 1.024 + 512 ; AB^2 = 1.536 ; AB = sqrt(1.536) ; AB (2R) = 16*sqrt(6)
11) Radius = 8*sqrt(6)
12) As stated above : Radius of Semicircle = Square Diagonal = X*sqrt(2)
13) R = X*sqrt(2)
14) So : X*sqrt(2) = 8*sqrt(6)
15) X = (8*sqrt(6)) / sqrt(2))
16) Blue Square Area = X^2
18) X^2 = 64*6 / 2 ; X^2 = 384 / 2 ; X^2 = 192
ANSWER : The Blue Square Area equal to 192 Square Units.
Best Regards from The Universal Islamic Institute for Study of Ancient Mathematical Thinking, Knowledge and Wisdom in Cordoba Caliphate.
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