The area of the semicircle is π/2. The area of △ABC = 1/2(AB)(OC). Now, in the △OEC if OE = 1 then OC = 2 and in the △OEA if OE = 1 then OA = (2√3)/3. Therefore AB = (4√3)/3 and the area △ABC = (4√3)/3. Yellow area = (4√3)/3 - π/2
We can also find the value of the side of the equilateral traingle by drawing the radius only .... Because after drawing the radius to side CB we will get OB= 1/sin60⁰ =2/√3 And X will be equal to 2OB =4√3 Then the area of the ABC traingle will be √3x²/4 = 4√3/3
Much easier to find OB by using 30-60-90 ratios on Triangle OBD: 2/sq root 3 = 1/OB; OB = 2/ sq root 3. Side of equilateral triangle = 2 x OB. = 4/sq root 3. Area triangle = sq root 3/4 x side (OB) ^2 = 4 x sq root 3/3 - pi/2.
O is midpoint of AB so BCO is a 30° - 60° - 90° triangle. D is tangent to the circle so CDO is also a 30° - 60° - 90° triangle. The area of the circle is π² cm² so its radius is 1 cm. Therefore OD is 1 and OC = 2 (property of a 30° - 60° - 90° triangle). Then AB = 4 / √3 ⇒ area △ABC = (4 / √3)² √3 / 4 = ⅓ 4√3. Subtract ½ π to get the area of the yellow region: ⅓ 4√3 − ½ π.
Yes, once we have the height of the equilateral ΔABC, we can apply well known formulas to determine its area. If h is the height of the equilateral triangle and a is the side length, h = (√3)(a)/2. Having determined that h = 2, we get 2 = (√3)(a)/2 and a = 4/(√3). The well known formula for equilateral triangle area is A = (√3)(a²)/4, in this case A = (√3)(4/(√3))²/4 = (√3)(16/3)/4 = (4√3)/3, from which the area of the semicircle is subtracted.
Looking at the diagram, I thought: “It would be nice if the triangle were equilateral, but how do you prove it?” And then, reading it again, that turned out to be given. Moral: always read the problem carefully first! So, to work: The area of the circle is pi, so the radius is sqrt(pi/pi) = sqrt(1) = 1; considering triangle ODC: side OD = 1 and is the side opposite the 30-degree angle, so it's half the hypotenuse OC, so OC = 2. This is the altitude of the equilateral triangle; now considering triangle OBC: letting side BC = x, as you did, and invoking Pythagoras: x^2 - 4 = (x/2)^2 = (x^2)/4; simplifying: x^2 - 4 = (x^2)/4; multiplying both sides by 4: 4(x^2) - 16 = x^2; subtract x^2 from both sides and add 16 to both swides: 3(x^2) = 16; dividing both sides by 3: x^2 = 16/3; so: x = sqrt (16/3) = 4/sqrt (3) = (4/3)(sqrt (3)). The area of the equilateral triangle is (1/2)(2)((4/3)(sqrt (3)) = (4/3)(sqrt (3)). Subtract (pi/2) for the area of the semicircle, and the shaded area equals (4/3)(sqrt (3)) - (pi/2) Voila! Coraggio.... 🤠
1/ The triangle ODB is a 30-90--60 one so DB .sqrt3 = OD= 1----> DB= 1/sqrt3------> AB = 4 DB= 4/sqrt3 2/ Area of yellow region= Area of the equilateral - Area of the semi circle = 1/4 sqAB . sqrt3 - pi/2 = 4sqrt3/3 - pi/2
As ∆ABC is equilateral, ∠CAB, ∠ABC, and ∠BCA are all 60°, and AB, BC, and CA are the same length. As AC and BC are tangent to Circle O at E and D respectively, OE and OD are equal to the radius r of the circle and ∠OEC and ∠ODC are 90°. By Two Tangent Theorem, EC and DC are equal and thus ∆AOC and ∆BOC are congruent. As ∠ACO and ∠BCO split ∠ACB equally, they are each 30°. By Complementary Angles, ∆BDO and ∆ADO are congruent with each other and similar to ∆OEC and ∆ODC, which are also congruent with each other. Circle O: A = πr² π = πr² r² = 1 r = 1 Triangle ∆ODC: O/H = sin θ r/OC = sin 30° = 1/2 OC = 2r = 2(1) = 2 Triangle ∆BDO: A/H = cos θ r/OB = cos 30° = (√3)/2 OB = 2r/√3 = 2(1)/√3 = 2/√3 Triangle ∆ABC: A = bh/2 A = [2(2/√3)]2/2 = 4/√3 = (4/3)√3 Yellow area: (4/3)√3 - π/2 ≈ 0.74 cm²
The radius of the circle is 1 cm The length of each side of the triangle is 4/✓3 The yellow area = triangle area - half the circle area = (4/✓3) - (π/2) = (4✓3)/3 - (π/2) = [8✓3 - 3π] / 6
I was on a road trip last week and was making up problems for myself and I made one almost exactly the same as this! Great explanation too. The way I originally did it was to connect the center of the circle to the point of tangency to create a 30-60-90 triangle and then solve using the radius of 1 and area of equilateral as (sqrt(3)s^2)/2 minus area of semicircle. Yours is much easier, thank you sir
Other method. Area △ABC it's 2*OB*OC/2=OB*OC. From △OCB and high OD=r -> OC=r/sin30 OB=r/cos30. If r=1 then area △ABC=1/(sin30*cos30). Or area △ABC=1/((1/2)*(√3/2))=4/√3=4*√3/3. Area of the Yellow region 4*√3/3- ½ π.
OB=2×2/sqrt(3)=4/sqrt(3), the area of the triangle is (1/2)(16/3)sqrt(3)/2=(4/3)sqrt(3), therefore the answer is (4/3)sqrt(3)-pi/2=0.738605approximately. 😊
Radius of circle: A = π R² = π cm² R = 1 cm Side of equilateral triangle: S = 2R/cos30° = 4/√3 cm S = 2,31 cm Area of equilateral triangle: At = √3/4 . S² At = √3/4 . (4/√3)² At = 4/√3 cm² Yellow shaded area: A = At - Asc A = 4/√3 - ½π A = 0,7386 cm² ( Solved √ )
Fantastic sir
So nice of you ❤️
Thanks for your feedback! Cheers! 😀
The area of the semicircle is π/2. The area of △ABC = 1/2(AB)(OC). Now, in the △OEC if OE = 1 then OC = 2 and in the △OEA if OE = 1 then OA = (2√3)/3. Therefore AB = (4√3)/3 and the area △ABC = (4√3)/3. Yellow area = (4√3)/3 - π/2
We can also find the value of the side of the equilateral traingle by drawing the radius only ....
Because after drawing the radius to side CB we will get OB= 1/sin60⁰ =2/√3
And X will be equal to 2OB =4√3
Then the area of the ABC traingle will be √3x²/4 = 4√3/3
Thanks for your feedback! Cheers! 😀
You are awesome. Keep it up 👍
Love this stuff. Part of my regular regimen to keep me from becoming a drooler as I get older. Thank you sir for your puzzles, I enjoy them very much.
Very nice sharing 🌹🌹
Thank you! Cheers! ❤️
Much easier to find OB by using 30-60-90 ratios on Triangle OBD: 2/sq root 3 = 1/OB; OB = 2/ sq root 3. Side of equilateral triangle = 2 x OB. = 4/sq root 3. Area triangle = sq root 3/4 x side (OB) ^2 = 4 x sq root 3/3 - pi/2.
Correction last sentence: Yellow area = area triangle less area semicircle, = sq root 3/4 x side (OB) ^2 = 4 x sq root3/3 - pi/2. Hope that’s clear!
O is midpoint of AB so BCO is a 30° - 60° - 90° triangle. D is tangent to the circle so CDO is also a 30° - 60° - 90° triangle.
The area of the circle is π² cm² so its radius is 1 cm. Therefore OD is 1 and OC = 2 (property of a 30° - 60° - 90° triangle).
Then AB = 4 / √3 ⇒ area △ABC = (4 / √3)² √3 / 4 = ⅓ 4√3. Subtract ½ π to get the area of the yellow region: ⅓ 4√3 − ½ π.
Excellent!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
Yes, once we have the height of the equilateral ΔABC, we can apply well known formulas to determine its area. If h is the height of the equilateral triangle and a is the side length, h = (√3)(a)/2. Having determined that h = 2, we get 2 = (√3)(a)/2 and a = 4/(√3). The well known formula for equilateral triangle area is A = (√3)(a²)/4, in this case A = (√3)(4/(√3))²/4 = (√3)(16/3)/4 = (4√3)/3, from which the area of the semicircle is subtracted.
Looking at the diagram, I thought: “It would be nice if the triangle were equilateral, but how do you prove it?” And then, reading it again, that turned out to be given. Moral: always read the problem carefully first!
So, to work:
The area of the circle is pi, so the radius is sqrt(pi/pi) = sqrt(1) = 1;
considering triangle ODC: side OD = 1 and is the side opposite the 30-degree angle, so it's half the hypotenuse OC, so OC = 2. This is the altitude of the equilateral triangle;
now considering triangle OBC: letting side BC = x, as you did, and invoking Pythagoras:
x^2 - 4 = (x/2)^2 = (x^2)/4; simplifying:
x^2 - 4 = (x^2)/4; multiplying both sides by 4:
4(x^2) - 16 = x^2; subtract x^2 from both sides and add 16 to both swides:
3(x^2) = 16; dividing both sides by 3:
x^2 = 16/3; so:
x = sqrt (16/3) = 4/sqrt (3) = (4/3)(sqrt (3)).
The area of the equilateral triangle is (1/2)(2)((4/3)(sqrt (3)) = (4/3)(sqrt (3)). Subtract (pi/2) for the area of the semicircle, and the shaded area equals
(4/3)(sqrt (3)) - (pi/2) Voila!
Coraggio.... 🤠
😉👍
❤️🌹
1/ The triangle ODB is a 30-90--60 one so DB .sqrt3 = OD= 1----> DB= 1/sqrt3------> AB = 4 DB= 4/sqrt3
2/ Area of yellow region= Area of the equilateral - Area of the semi circle = 1/4 sqAB . sqrt3 - pi/2 = 4sqrt3/3 - pi/2
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
As ∆ABC is equilateral, ∠CAB, ∠ABC, and ∠BCA are all 60°, and AB, BC, and CA are the same length. As AC and BC are tangent to Circle O at E and D respectively, OE and OD are equal to the radius r of the circle and ∠OEC and ∠ODC are 90°. By Two Tangent Theorem, EC and DC are equal and thus ∆AOC and ∆BOC are congruent. As ∠ACO and ∠BCO split ∠ACB equally, they are each 30°. By Complementary Angles, ∆BDO and ∆ADO are congruent with each other and similar to ∆OEC and ∆ODC, which are also congruent with each other.
Circle O:
A = πr²
π = πr²
r² = 1
r = 1
Triangle ∆ODC:
O/H = sin θ
r/OC = sin 30° = 1/2
OC = 2r = 2(1) = 2
Triangle ∆BDO:
A/H = cos θ
r/OB = cos 30° = (√3)/2
OB = 2r/√3 = 2(1)/√3 = 2/√3
Triangle ∆ABC:
A = bh/2
A = [2(2/√3)]2/2 = 4/√3 = (4/3)√3
Yellow area:
(4/3)√3 - π/2 ≈ 0.74 cm²
Alternatively line OD =1
BD = OB/2 and OD = ~1.732BD
AB = 4BD
OC = 1.732OB
(OB*OC) -(pi/2) = yellow
Área círculo =Pi》Radio r=1 =AE(sqrt3)》AE=1/sqrt3》AO=2AE=2sqrt3/3》OC=AO(sqrt3)=2》Área ABC=AO×OC=4sqrt3/3》》Área amarilla =ABC - (Pi/2) =(4sqrt3/3) - (Pi/2)
Gracias y un saludo cordial.
The radius of the circle is 1 cm
The length of each side of the triangle is 4/✓3
The yellow area = triangle area - half the circle area = (4/✓3) - (π/2) = (4✓3)/3 - (π/2) = [8✓3 - 3π] / 6
I was on a road trip last week and was making up problems for myself and I made one almost exactly the same as this! Great explanation too. The way I originally did it was to connect the center of the circle to the point of tangency to create a 30-60-90 triangle and then solve using the radius of 1 and area of equilateral as (sqrt(3)s^2)/2 minus area of semicircle. Yours is much easier, thank you sir
Other method. Area △ABC it's 2*OB*OC/2=OB*OC. From △OCB and high OD=r -> OC=r/sin30 OB=r/cos30. If r=1 then area △ABC=1/(sin30*cos30). Or area △ABC=1/((1/2)*(√3/2))=4/√3=4*√3/3.
Area of the Yellow region 4*√3/3- ½ π.
Excellent working 👍
Many many thanks dear ❤️
Thanks Professor!😀🥂
very nice sum
Thanks a lot dear ❤️
Wow nice puzzle!
Solve it in my head. ...Well not the decimal but the 1st answer, lol
Yay! I solved the problem.
Bravo! ❤️
Nice! DO = EO = r = 1; sin(φ) = BD/BO = 1/2 → cos(φ) = √3/2 →
tan(φ) = √3/3 = BD/1 → 3BD = √3 → BD = √3/3 → BO = 2√3/3 = BC/2 →
BC = 4√3/3 → CO = → (2√3/3)(√3) = 2 →
yellow area = (4√3)/3 - π/2 = (1/6)(8√3 - 3π)
OB=2×2/sqrt(3)=4/sqrt(3), the area of the triangle is (1/2)(16/3)sqrt(3)/2=(4/3)sqrt(3), therefore the answer is (4/3)sqrt(3)-pi/2=0.738605approximately. 😊
Great! ❤️
Thanks for sharing! Cheers!
We can easily find a solution working with the similarity between AOE and AOC right triangle of 30°60°90° type
Thanks dear ❤️
A=√(4/3)*2-π/2=4/√3-π/2
Super!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
S=(8√3-3π)/6≈0,73
Thank you
Radius of circle:
A = π R² = π cm²
R = 1 cm
Side of equilateral triangle:
S = 2R/cos30° = 4/√3 cm
S = 2,31 cm
Area of equilateral triangle:
At = √3/4 . S²
At = √3/4 . (4/√3)²
At = 4/√3 cm²
Yellow shaded area:
A = At - Asc
A = 4/√3 - ½π
A = 0,7386 cm² ( Solved √ )
1/6(8√3-3π)
Great! ❤️
There is a mistake in calculating x , when time is 7:46 , Sin60 = 2/X this is wrong sin60 =X/2 and X=sqrt3 Also can be found by Pythagorean theorem.
The sides of the equilateral triangle ABC were labeled x at 1:09. Also, at 7:04 we are told to focus on △AOC, not on △EOC 🧐
I've solved it but without using trigonometry : )
My answer is 3sqrt(3)/4-π/2
Oh wow I need to redo trig class. I used sin(60) = hypotenuse over opposite. Whoops. 4sqrt(3)/3-π/2