A Nice Geometry Problem 

As second easier solution : triangles BDO and ABC are similar(angle angle), so R/3=2/4 => R=3/2

Should be pretty straightforward. 1. Notice that segment AD equals segment AC (3). 2. Therefore, segment AB is 5. So, with a right triangle with one leg = 3 and the hypotenuse = 5 makes the other leg 4. 3. So, now we can write 2 equations in 2 unknowns: a. 2r + a = 4 (where a is the length of segment BE), and b. r² + 2² = (r + a)² 4. Solve for 'a' in equation a, so a = (4 - 2r). Then substitute for a in equation b, which gives us an equation in only r: r² + 2² = (r + 4 - 2r)², or r² + 4 = (4 - r)², r² + 4 = 16 - 8r + r² 5. Canceling the r²s and combining terms yields: r = 1.5 6. Therefore the area of the semicircle is: πr²/2 = π•1.5²/2 = *1.125π* And that is the answer. Now to watch the vid. Cheers! - s.west

Llamamos "O" al centro del semicírculo. AC=AD=3→ AB=2+3=5→ BC=4 → Razón de semejanza entre los triángulos ODB y ACB: s=2/4=1/2→ OD=3s=3/2 → Área del semicírculo =π(3/2)²/2 =9π/8. Gracias y un saludo cordial.

AD=AC=3 ( tangent line segments drawn from point A are equal ) So AB=AD+BD=3+2=5 By Pythagoras theorem in orthogonal triangle ABC you find BC=4. So OB=BC-OC=> OB=4-R In orthogonal triangle BOD => OB²=BD²+0D² => (4-R)²=2²+R² => 16-8R+R²=4+R² => R=3/2 At last area of semicircle = πR²/2=9π/8

Solution by finding radius of circle by areas: Using the two tangents theorem, we find AD = AC = 3. AB = AD + BD = 3 + 2 = 5. Using the Pythagorean theorem, we find BC = 4. Area of ΔABC = (1/2)bh = (1/2)(BC)(AC) = (1/2)(4)(3) = 6. Designate the center of the semicircle as O, as in the video. Construct OD and OA. OD = OC = R + radius of circle. Area of ΔABO = (1/2)bh. Let AB be the base, then OD is the height and area = (1/2)(5)(R) = 2.5R. Area of ΔACO = (1/2)bh. Let AC be the base, then ODCs the height and area = (1/2)(3)(R) = 1.5R. C = area of ΔABO + area of ΔACO = 2.5R + 1.5R = 4R. However, area of ΔABC = 6, so 4R = 6 and R = 6/4 = 3/2. Area of semicircle = (1/2)πR² = (1/2)π(3/2)² = (1/2)π(9/4) = 9π/8, as Math Booster also found..

Let O be the center of the semicircle and r be the radius. By two tangents theorem, CA = AD = 3, a both tangents converge at A. This means that AB = 2+3 = 5, which means ∆BCA is a 3-4-5 Pythagorean triple right triangle, and BC = 4. Draw OD. As ∠ODB = ∠BCA = 90° and ∠DBO = ∠ABC, ∆BCA and ∆ODB are similar triangles. Triangle ∆ODB: OD/DB = CA/BC r/2 = 3/4 r = 2(3/4) = 3/2 A = πr²/2 = π(3/2)²/2 A = (9π/4)/2 = 9π/8 ≈ 3.534

1.125 pi or 3.534 Answer line AD = 3 (circle tangent theorem) Hence AB = 5 (2+3) Hence, BC = 4 (3-4-5 right triangle) Let 0 = center of the semi-circle, then line D0 = r, and it is a right triangle (tangent circle theorem) Hence, line 0B = 4-r Hence, 2^2 + r^2 = (4-r)^2 4 + r^2 = 16 + r^2 -8r 8 r = 12 2r =3 r = 1.5 Area of circle = 2.25 pi Hence, area of semi-circle = 2.25 pi/2 = 1.125 pi or 3.534 Answer

S=9π/8≈3,5325≈3,54