Olympiad Mathematics | A Very Nice Geometry Problem

[@raghvendrasingh1289]

Diameter= 5×2/√3 hence radius is 5/√3 if we join centre to upper vertex of triangle we have to find sum of areas of a triangle and a sector area of triangle = (1/2)(1/2)(25/√3) = 25√3/12 area of sector = π× ( 25/3) × (1/6) = 25π/18 Total area 25 (2π+3√3 )/36

[@jimlocke9320]

[@saronohandoyo4958]

I try to find the diameter first since AB is diameter, so if point C is touching the circle then ABC will be right triangle. AC=5. Right triangle with 30 degree will be x:x√3:2x for base:height:hypotenuse from this we get height = x√3 = 5 so we get base of triangle x=5/√3 hypotenuse = 2x =10/√3 which also this is a diameter.

[@IllllllllllIIlIIIIlIlllI]

I was able to solve the problem before seeing your solution! I'm really happy. What a great feeling

[@soli9mana-soli4953]

Being angle CAB = 30°, we can also draw an equilateral triangle with side=5 inscribed in the circle, the area over the chord AC (segment) is 1/3 the difference between tha area of the circle and the area of the equilateral triangle. Area equilateral triangle = s²/4*√ 3 = 25/4* √ 3 radius circle*2 = side/sin60°= 5/√ 3/2 = 5/3√ 3 Area circle = 25/3pi yellow area = 1/2*area circle - (area circle - area triangle)*1/3 yellow area = 1/6*area circle + 1/3*area triangle yellow area = 1/6*25/3pi + 1/3*25/4√ 3 yellow area = 25/18pi + 25/12√ 3 = 25/36*(2pi +3√ 3)

[@lasalleman6792]

I just set up a 30 , 90 60 degree triangle. Going Pythagorean, line AB comes out to 5.78. Total area of the triangle is 7.22. The line from C to B Is the sine of angle 30 and has a distance of 2.89. Then simply find the segment area between the arc CB and line CB. Use centroid of circle at O (60 degrees) as bases for angle. Use formula-- area of segment = R^2/2*(pi/360 *Angle at O - Sine of O) Area of segment turns out to be .7566. Total purple shaded area is 7.97. Right on the money!

[@prossvay8744]

Purple area=25√3/12+25π/18.❤

[@Latronibus]

This is a relatively easy one by this channel's standards. Draw BC, ABC is 30 60 90, the longer leg is 5, so the shorter leg is 5/sqrt(3), so that area is 25/(2sqrt(3))=25sqrt(3)/6. Also, AB is the hypotenuse of ABC, so the diameter is 10/sqrt(3). Now draw OC. COB is 60 degrees, and OC=OB, so COB is equilateral by SAS, and its side length is the radius, which is 5/sqrt(3). Therefore the area of COB is sqrt(3)/4 25/3=25 sqrt(3)/12, and the area of that sector is 25 pi/18. So now we have ABC + sector - COB, to avoid double counting COB. So that adds up to 25 sqrt(3)/12 + 25 pi/18.

[@marcelowanderleycorreia8876]

Today is sunday!! 😎😂👍

[@ritwikgupta3655]

Nice method. Nice problem. So many methods. Another: AO=OC, ∆AOC is isosceles, ACO=30°. So the external angle COB=30°+30°=60° Cosine formula gives R=5*5/2*5/cos30° So we get area of segment COB Area of ∆AOC=Ht of ∆AOC*R/2 Ht of ∆AOC=5*sin30° All methods are connected.

[@devondevon4366]

7.97333 Easy Let's label the shaded region ABC draw a line from B to C to form a right triangle (Thales Theorem) This triangle is 30-60-90 Hence, the sides are 5, 2.88675, and 5.7735.' Hence, the radius also = 2.88675 since it is one-half the diameter. Draw a line from the circle's center to C to form an isosceles triangle, (30 , 120, 30 degrees) . Its sides are 5, 2.88675, and 2.88675 Use Heron's formula to calculate the area = 3.61 Since the angle to the left of the triangle is 60 degrees ( 180 - 120 degrees), and since the radius of the circle is 2.88675, then 1/6 (60/360) of the circle is area = remaining sector. The circle area = 26.16 and hence , the area of the sector = 26.16/6 = 4.36 Hence, the area of the shaded region = 3.61 + 4.36 = 7.9733 answer

[@michaeldoerr5810]

The shaded area is 25[(3sqrt(3)+2pi)/36] units square. Looks like I need to know that if a circle theorem is used to show why there does need to be a right angle constructed in a sector ever, this is the reason why. And I shall use this for practice!!!

[@RealQinnMalloryu4]

(5)^2=25{60°A+60°B+60°C}=180°ABC/25=7.5 ABC(ABC ➖ 7ABC+5).

[@marioalb9726]

2R = 5 / cos 30° --> R=5/√3 cm A = A₁ +A₂= ½cRsin30° + ½.⅓πR² A = 3,6084 + 4,3634 A = 7,97176 cm² ( Solved √ )

[@alexniklas8777]

5^2+r^2=(2r)^2; 25+r^2=4r^2; r= 5/√3; S=πr^2×30°/360°= π×25/36 Thanks sir!

[@Latronibus]

No, that doesn't work, because we don't have a sector with central angle 30 degrees. In fact our region *contains* a sector with central angle 60 degrees, plus a bunch more area. Or, just eyeballing it, you can tell that the region is around a third or so of the circle, and the work you did already shows the circle is 25 pi/3, so this region should be a lot more than just 1/12 of that.

[@pojuellavid]

АСB=90°, BC=x, AB=2*x. x^2+5^2=(2*x)^2. --> x=5/√3 S=((5/√3)*5)/2=25/√12

[@anikdolui1018]

Let O be the centre and AB = 2R Join O and C & B and C. Hence, AO = BO = CO = R. Angle ACB = 90 deg bcz AB is diameter of the circle. Hence, applying Pythagoras theorem in right angled triangle ABC we get R = 5/✓3. Area of ABC = 1/2.AC.BC = 1/2.5.5/✓3 = 25/2✓3 Please comment if i have committed any mistake.

[@luiscostacarlos]

anikdolui Essa área é a área do triângulo faltou a área do segmento.

[@rabotaakk-nw9nm]

[purple]=(25/36)(2π+3vʼ3)≈7.972 sq.un.

[@riyatmoko8241]

4

[@YardenVokerol]

O=½[AB], H=½[AC], R²=|AO|²=|AH|²+|HO|² |AH|=½|AC|=½•5=5/2 ∠A=30°⇒|HO|=|AH|/√3 ⇒R²=(5/2)²+(5/2)²/3=(5/2)²•4/3=(5/√3)² R=5/√3 ∠A=∠ACO⇒∠CON=30°+30°=60°=π/3 S(OC͡B)=R²•(π/3)/2 S=½|AC|•|HO|+R²•π/6=½•5•(5/2)/√3+(5/√3)²•π/6 S=25/(4√3)+25π/18 S=25√3/12+25π/18