Find the shaded area inside the quarter circle | A Very Nice Geometry Problem

[@quigonkenny]

Square OPQS: A = s² 80 = s² s = √80 = 4√5 Triangle ∆OSQ: OS² + SQ² = OQ² s² + s² = r² r² = 2s² = 2(4√5)² = 160 r = √160 = 4√10 Red shaded area: A = [AOB] - [OPQS] A = πr²/4 - 80 A = π(4√10)²/4 - 80 A = 160π/4 - 80 A = 40π - 80 ≈ 45.664 sq units

[@marioalb9726]

R² =2*80 cm² = 160 cm² A = ¼πR² - 80 = 40π -80 A = 45,66 cm² ( Solved √ )

[@francisdelpuech6415]

That’s the way I did it too, very quick! 😊

[@AmirgabYT2185]

S=40π-80=40(π-2)≈45,71

[@PrithwirajSen-nj6qq]

The area of the square is 80 sq units Hence its side =√80=4√5 And its diagonal =√[2(side)^2]=√(2*80= 4√10 Radius = 4√10 units Area of quarter circle =π/4*16*10=40π Area of red region = (40π - 80) sq units

[@RealQinnMalloryu4]

90ABPQ/80=1.10 1.5^5ABP1.2^3^2^3 1^1^1.2^3 2^3 (ABPQ ➖ 3ABPQ+2).

[@PrithwirajSen-nj6qq]

Let us solve this problem using Co ordinate geometry. The bottom left vertex at (0,0) Then the vertex on circumference is at (a, a,) [a is the side of the square ] Equation of circle with center at (0,0) and radius r is x^2 +y^2=r^2 As the circle passes through (a, a), this point will satisfy the above equation. Hence a^2 +a^2=r^2 > r^2=2a^2 Red region = πr^2 /4 -a^2 =(π*2*a^2)/4 -a^2 =a^2(π/2-1) = 80(π/2 -1) = 45.663 sq units (approx.)

[@yuuppcc]

Ｘ×Ｘ=PS×OQ=80×2=160

[@wasimahmad-t6c]

45.6637

[@kateknowles8055]

Writing R for radius: (R+OP)(R-OP) = R^2- OP^2 and this is equal to OPx OP . .>>>> This is the way chords of a circle intersect each other.

[@kateknowles8055]

Thank you (spelling correction)

[@ناصريناصر-س4ب]

The area of the square POSQ is (Rl√2)²=80, hence R=4√10, hence the area of the red region is (π*(4√10)²/4)-80=40π-80.

[@santiagoarosam430]

r²/2=80→ r²=160→ Área sombreada roja =(160π/4)-80 =40(π-2). Gracias y un saludo.

[@adamhanna9940]

160pi - 80

[@prossvay8744]

Blue area=1/4(π)(4√10)^2-80=40π-80.❤❤❤

[@michaeldoerr5810]

That is a literal example of easier than it looks. All that you had to do was simply to calculate the area to get the x value and make use of what the square area is. And the answer is 40(1.14) units square.

[@cyruschang1904]

r = ✓(80 + 80) = 4✓10 area sought = (π/4)(4✓10)^2 - 80 = (4π)(✓10)^2 - 80 = 40π - 80 = 40(π - 2)

[@jimlocke9320]

You really do not have to compute variable a. Take the square and another square congruent to it and divide them into two triangles each by cutting along the diagonal of each. Assemble the 4 triangles into a square with the diagonals as sides. The total area is twice that of the original square, 2 x 80 = 160, so x² = 160. Area of quarter circle = (1/4)πx² = (1/4)π(160) = 40π. Shaded area = 40π - 80.

[@perskancke7960]

Too simple!

[@himo3485]

80=√80*√80 OA=OB=OQ=r r²=(√80)²+(√80)² r=√160 Shaded area = √160*√160*π*1/4 - 80 = 40π - 80 = 45.6

[@maxsalvati3626]

Circa 45

[@murdock5537]

a = 4√5 → r = a√2 = 4√10 → green area = 40(π - 2)

[@GeoVars]

40π-80

[@oscarcastaneda5310]

40(pi - 2) : ) Did this one w/o paper and pen ! It was refreshingly easier than some of the recent challenges : ) Thanks, I needed this : )

[@gikasmith5511]

Bad rounding! You take pi, not 3.14, and the area turns out to be 45.6637..... which rounds to 45.7 :) Otherwise, a very handsome problem, and as a Math teacher I am super thankful for this channel.

[@befactfull525]

your previous voice is way much better than this

[@MathBooster]

KZbin is trying a new feature to translate the voice in different languages by using AI. So, if you are getting an AI voice then you can select the original language (english) in the setting.

[@noralealmarchena]

Muy complicada la explicación. Podría ser más directa

[@kateknowles8055]

For you más directa Writing R for radius: (R+OP)(R-OP) = R^2- OP^2 and this is equal to OPx OP . .>>>> This is the way chords of a circle intersect each other.