Can you find area of the Green shaded Rectangle? | (Quarter Circle) | #math #maths | #geometry

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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🙏🏼👍🏼

[prossvay]

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❤❤❤

[devondevon]

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

[yalchingedikgedik]

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

[misterenter-izrz]

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)).😊

[johnbrennan]

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.

[nandisaand]

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)

[jamestalbott]

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

[murdock]

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)

[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

[santiagoarosam]

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.

[giuseppemalaguti]

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

[himo]

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

[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!

[marcgriselhubert]

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.

[devondevon]

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

[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

[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😊

[rubenvela]

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

[cyruschang]

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