Can you find area of the Green shaded Rectangle? | (Quarter Circle) |

Learn how to find the area of the Green shaded Rectangle. Important Geometry and Algebra skills are also explained: Pythagorean theorem; area of the triangle formula; area of the rectangle formula. Step-by-step tutorial by PreMath.com
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Can you find area of the Green shaded Rectangle? | (Quarter Circle) | #math #maths | #geometry
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[@BBMathTutorials]

Fabulous sir🙋🏻‍♂️thank you for this great video🙏🏼👍🏼

[@PreMath]

My pleasure😀 Thanks for the feedback ❤️

[@devondevon4366]

1.46 Let the A and B be the length of the two legs of the triangle Then Area = 5= 1/2 A*B 5=1/2 AB 10 =AB and A= 10/B (solving for A) Equation F Since the hypotenuse = 5 units, then A^2 + B^2 = 5^2 A^2 + B^2 = 25 Equation G (10/B)^2 + B^2 = 25 substituting Equation F into Equation B 100/B^2 + B^2 = 25 100 + B^4 = 25B^2 (multiply both sides by B^2) Let B^2 = N (Introduce a new variable N) then 100 + N^2 = 25 N N^2 -25 + 100 = 0 (N-20)(N-5) factor N=20 , and N= 5 Hence, B = sqrt 20 and B= sqrt 5 or B =4.4721359 and B= 2.236067977 4.4721359 is the longer leg of the triangle, and 2.236067977 is its shorter leg Hence, the difference between the longer leg and 5 is the WIDTH of the small GREEN RECTANGLE, or 5 - 4.472136 = 0.527864 Hence, the difference between the shorter leg and 5 is the LENGHT of the small GREEN RECTANGLE or 5 - 2.2360678 =2.7639320 Hence, 0.527864 times 2.7639320 is the Area of the GREEN RECTANGLE. 0.527864 * 2.7639320 = 1.458 or 1.46 Answer

[@johnbrennan3372]

If you call length of green rectangle x then ed=5-x and since area of triangle eod is 5 then od=10/(5-x).Using pythag. You get an equation that gives 2 values for x,one being valid ie x=5- sqr5. da= 5-2sqroot 5.So green area= (5-sqr5)(5-2sqr5)=35-15sqr5.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@JLvatron]

Instead of labelling the triangle sides as a & b, I labelled the rectangle sides a & b. So my triangle sides were 5-a & 5-b. Theoretically, this should still work, but the math was getting nowhere. great video!

[@PreMath]

Great job! Thanks for the feedback ❤️

[@nandisaand5287]

I used eqns 1 & 2 to find Base & Height: H=10/B B²+(10/B)²=(5)² B²+(100/B²)=25 B²+(100/B²)-25=0 B⁴-25B²+100=0 (B²-5)(B²-20)=0 B=sqrt(5); 2sqrt(5) Area= (5-sqrt(5))•(5-2sqrt(5)) =25-5sqrt(5)-10sqrt(5)+10 =35-15sqrt(5)

[@PreMath]

Excellent! Thanks for sharing ❤️

[@quigonkenny]

Let the unlabeled vertex of the green rectangle on BC be M and the unlabeled vertex on AB be N. Let ∠EOD = θ. As OE is a radius of quarter circle O, OE = OC = 5. This means that OD = OEcos(θ) = 5cos(θ) and DE = OEsin(θ) = 5sin(θ). Triangle ∆ODE: Aᴛ = OE(OD)sin(θ)/2 5 = 5(5cos(θ)sin(θ)/2 5sin(θ)cos(θ) = 2 2sin(θ)cos(θ) = 4/5 sin(2θ) = 4/5 sin(2θ) = O/H ==> O = 4, H = 5 cos(2θ) = A/H = (√H²-O²)/H cos(2θ) = (√5²-4²)/5 = (√25-16)/5 = √9/5 = 3/5 sin(2θ/2) = √((1-cos(2θ))/2) sin(θ) = √((1-3/5)/2) = √((2/5)/2) = 1/√5 DE = 5sin(θ) = 5(1/√5) = √5 OD² + DE² = OE² OD² + (√5)² = 5² OD² = 25 - 5 = 20 OD = √20 = 2√5 As OABC is a square and BMEN is a rectangle, then NB = ME = MD-DE = 5-√5 and BM = EN = OA-OD = 5-2√5. Rectangle BMEN: Aᵣ = wh = (5-√5)(5-2√5) Aᵣ = 25 - 10√5 - 5√5 + 10 Aᵣ = 35 - 15√5 ≈ 1.459 cm²

[@PreMath]

Excellent! Thanks for sharing ❤️

[@jamestalbott4499]

Nice! Appreciate how the algebraic identity was used to solve the problem.

[@PreMath]

Glad you liked it! Thanks for the feedback ❤️

[@yalchingedikgedik8007]

That’s very nice and understandable Thanks Sir for this explain With my respects ❤❤❤❤❤ Good luck

[@santiagoarosam430]

Sistema ecuaciones: b*h=10 ; b²+h²=5²→ b=√5 ; h=2√5→ Área verde = (5-√5)*(5-2√5)=35-15√5 =1,45898.... Gracias y saludos.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@prossvay8744]

Let width=b ; Length=a Area of rectangle=ab Connect E to F OD=EF=5-a DE=OF= 5-b Area of triangle=1/2((5-a)(5-b)=5 25-5a-5b+ab=10 ab-5(a+b)+15=0 ab=5a+5b-15 (1) In ∆OEF (5-a)^2+(5-b)^2=5^2 a^2+b^2-10(a+b)+50=25 (a+b)^2-2ab-10(a+b)+25=0 Let ab=x ; a+b=y So (1) x=5y-15 And (2) y^2-2x-10y+25=0 so y^2-2(5y-15)-10y+25=0 So y=10+3√5 >5 reject ; y=10-3√5 x=50-15√5-15=35-15√5 ab=35-15√5 a+b=10-3√5 b=10-3√5-a a(10-3√5-a)=35-15√5 a=5-2√5 ; a=5-√5 b=10-3√5-5+2√5=5-√5 b=5-2√5 So Green rectangle area=(5-√5)(5-2√5)=35-15√5 cm^2=1.46 cm^2❤❤❤

[@PreMath]

Excellent! Thanks for sharing ❤️

[@devondevon4366]

1.46 It would have been much easier if the area of the triangle was 6 (instead of 5) as we would avoid having to deal with decimals as it would be a 3-4-5 right triangle, and the area of the triangle (at the top of the circle ) would be 2 * 1 =2 since 5-4=1 and 5-3 = 2. You still would deal with the same concept and must understand it, to solve the problem, but avoid dealing with radical or decimals

[@marcgriselhubert3915]

No other idea. It's very simple. We can add that a =sqrt(5) and b = 2.sqrt(5), but it is of no real need for the final result.

[@PreMath]

Thanks for the feedback ❤️

[@giuseppemalaguti435]

Agreen=xy..x+b=5,y+h=5,bh=10...A=(5-b)(5-h)=25-5(b+h)+10=35-5(b+h)...5(b+h)=35-A...(^2)..25(b^2+h^2+2bh)=1225-70A+A^2...25(25+20)=1225-70A+A^2..A^2-70A+100=0...A=35-√1125=35-15√5

[@PreMath]

Excellent! Thanks for sharing ❤️

[@gelbkehlchen]

Solution: a = horizontal side of the right-angled blue triangle, b = vertical side of the right-angled blue triangle. Because of the area of ​​the right-angled blue triangle, the following must apply: (1) a*b/2 = 5 |*2/a ⟹ (1a) b = 10/a |in (2) ⟹ In addition, the Pythagorean theorem applies to the right-angled blue triangle: (2) a²+b² = 5² (2a) a²+100/a² = 25 |with a² = x ⟹ (2b) x+100/x = 25 |*x ⟹ (2c) x²+100 = 25*x |-25x ⟹ (2d) x²-25x+100 = 0 |p-q formula ⟹ (2e) x1/2 = 25/2±√(25²/4-100) = 25/2±1/2*√(625-400) = 25/2±1/2*√225 = 25/2±1/2*15 ⟹ (2f) x1 = 25/2+1/2*15 = 20 and (2g) x2 = 25/2-1/2*15 = 5 ⟹ 1. Case: a1² = x1 = 20 |√() ⟹ a1 = √20 = 2*√5 |in (1a) ⟹ (1b) b1 = 10/(2*√5) = √5 2nd case: a2² = x2 = 5 |√() ⟹ a2 = √5 |in (1a) ⟹ (1c) b2 = 10/√5 = 2*5/√5 = 2*√5 1st case: Area of ​​the green rectangle = (5-2*√5)*(5-√5) = 25-5*√5-10*√5+10 = 35-15*√5 ≈ 1.4590 2nd case: Area of ​​the green rectangle = (5-√5)*(5-2*√5) = 25-10*√5-5*√5+10 = 35-15*√5 ≈ 1.4590

[@misterenter-iz7rz]

a^2+b^2=25, a×b=10, to find (5-a)(5-b)=25-5(a+b)+10=35-5(a+b), (a+b)^2=25+20=45, thus the answer is 35-5sqrt(45)=5(7-3sqrt(5)).😊

[@PreMath]

Excellent! Thanks for sharing ❤️

[@marcelowanderleycorreia8876]

Great question as always! 👍👍

[@PreMath]

Glad you think so! Thanks for the feedback ❤️

[@sanjoystore9799]

Great explanation.thank you sir.

[@PreMath]

Glad you liked it You are very welcome! Thanks for the feedback ❤️

[@murdock5537]

ab/2 = 5 → ab = 10 → b = 10/a → a^2 + (a/10)^2 = 25 → k = a^2 = (1/2)(25 ± 15) → a = √5 → b = 2√5 → green rectangle area = (5 - a)(5 - b) = 5(7 - 3√5)

[@PreMath]

Excellent! Thanks for sharing ❤️

[@h.g.buddne]

I decided to take Euclids theory since it was a no brainer that the hight is 2. I determined that the hight divided c in 4 and 1. Now I could calculate both sides of the triangle to be sqr of 5 and sqr of 20. So 5 - root 5 times 5 minus root 20 (2 times root 5) is the area were looking for and results in 35 minus 15 times root5😊

[@cyruschang1904]

xy = 10 x^2 + y^2 = 5^2 = x^2 + (10/x)^2 = 25 x^4 - 25x^2 + 100 = 0 (x^2 - 5)(x^2 - 20) = 0 (x, y) = (✓5, 2✓5) or (2✓5, ✓5) green area = (5 - ✓5)(5 - 2✓5) = 35 - 15✓5

[@user-wv1td1rp6j]

If you set the area of the right triagle as (ODE)= 6 cm^2 then the anser is more simple (2 cm^2)

[@himo3485]

OE=5 5=(√5)² 1 : 2 : √5 OD=x DE=2x x*2x/2=5 2x²=10 x=√5 Green Rectangle area = (5-√5)*(5-2√5) = (35 - 15√5)cm²

[@PreMath]

Excellent! Thanks for sharing ❤️

[@AmirgabYT2185]

S=5(7-3√5)≈1,4

[@rabotaakk-nw9nm]

A≈1.45898 !!! 😬

[@PreMath]

Excellent! Thanks for sharing ❤️

[@rubenvela44]

3² + 4² = 5² The sides of the rectangle are 1 and 2 with a perimeter of 6 because 4 plus 1 equals 5 and 3 plus 2 equals 5 so (5-4)(5-3) is the area 1 times 2

[@PreMath]

Thanks for the feedback ❤️

[@unknownidentity2846]

Let's find the area: . .. ... .... ..... The triangle ODE is a right triangle, so we can apply the Pythagorean theorem: OD² + DE² = OE² DE² = OE² − OD² DE = √(OE² − OD²) The area of this triangle can be calculated as follows: A(ODE) = (1/2)*OD*DE 2*A(ODE) = OD*DE 2*A(ODE) = OD*√(OE² − OD²) 4*A²(ODE) = OD²*(OE² − OD²) 4*(5cm²)² = OD²*[(5cm)² − OD²] 100cm⁴ = (25cm²)*OD² − OD⁴ OD⁴ − (25cm²)*OD² + 100cm⁴ = 0 OD² = (25/2)cm² ± √[(25/2)²cm⁴ − 100cm⁴] OD² = (25/2)cm² ± √[(625/4)cm⁴ − (400/4)cm⁴] OD² = (25/2)cm² ± √[(225/4)cm⁴] OD² = (25/2)cm² ± (15/2)cm² First solution: OD² = (25/2)cm² + (15/2)cm² = (40/2)cm² = 20cm² ⇒ OD = (2√5)cm ⇒ DE = √[(5cm)² − 20cm²] = √(25cm² − 20cm²) = (√5)cm Second solution: OD² = (25/2)cm² − (15/2)cm² = (10/2)cm² = 5cm² ⇒ OD = (√5)cm ⇒ DE = √[(5cm)² − 5cm²] = √(25cm² − 5cm²) = (2√5)cm Now we are able to calculate the area of the green rectangle. For both solutions we obtain: A(green rectangle) = (OA − OD)*(AB − DE) = [5cm − (2√5)cm]*[5cm − (√5)cm] = (25 − 5√5 − 10√5 + 10)cm² = (35 − 15√5)cm² ≈ 1.459cm² Best regards from Germany

[@PreMath]

Excellent! Thanks for sharing ❤️

[@LuisdeBritoCamacho]

STEP-BY-STEP RESOLUTION PROPOSAL : 01) OD = X cm 02) DE = Y cm 03) OD * DE = X * Y = 10; and 04) X^2 + Y^2 = R^2 ; X^2 + Y^2 = 25 05) Solving a System of Two Nonlinear Equations with Two Unknowns : 06) X * Y = 10 and X^2 + Y^2 = 25 07) Positive Solutions : 08) X = sqrt(5) ; X ~ 2,24 cm 09) Y = 2 * sqrt(5) ~ 4,45 cm 10) Sides of Green Rectangle: 11) Length (L) = (5 - sqrt(5)) cm 12) Height (h) = (5 - 2*sqrt(5)) cm 13) Green Triangle Area = (L * h) sq cm 14) Green Triangle Area = [(5 - sqrt(5)) * (5 - 2*sqrt(5))] sq cm 15) Green Triangle Area = (35 - 15*sqrt(5)) sq cm Thus, OUR BEST ANSWER : Green Rectangle Area is equal to (35 - 15*sqrt(5)) Square Centimeters or approx. equal to 1,46 Square Centimeters. Greetings from The Great Land of Islam!! A Land of Wisdom, Peace and Knowledge.

[@PreMath]

Excellent! Thanks for sharing ❤️🙏

[@emilioplentz]

The triangle is 3 x 4 x 5. 5 - 3 =2; 5 - 4 =1; area = 2 x 1 = 2

[@PreMath]

Thanks for the feedback ❤️

[@Teamstudy4595]

Ans; 15(3 - _/5) square Unit

[@Birol731]

My way of solution ▶ A(ΔODE)= 5 cm² let OD= a DE= b ⇒ ab/2= 5 ab= 10 r = 5 cm EO= 5 cm which is equal to c (a+b)²= a²+2ab+b² a²+b²= c² c= 5 c²= 25 ⇒ (a+b)²= 25+2*10 (a+b)²= 45 a+b= √45 a+b= 3√5 a*b= 10 b= 10/a ⇒ a+10/a= 3√5 a²+10= 3√5a a² - 3√5a+10=0 Δ= 45-4*1*10 Δ= 5 a₁= (3√5+√5)/2 a₁= 2√5 b₁= 10/2√5 b₁= √5 a₂= (3√5-√5)/2 a₂= √5 b₂= 10/√5 b₂= 2√5 ⇒ it is irrelevant if we take a as a₁ or a₂: a= 2√5 b= √5 Arectangle= (r-a)*(r-b) = (5-2√5)*(5-√5) = 25-5√5-10√5+2*√5*√5 = 25+10-15√5 = 35 -15√5 Arectangle= 5(7-3√5) cm² Arectangle ≈ 1,46 cm²

[@AmirgabYT2185]

I used the same method 😂

[@PreMath]

Thanks for the feedback ❤️