As soon as we know x =1 unit and r =√2 units Semiperimeter Of DPOA=(2+√2) units & that of CPOB =(√2+1) unit we may use Brahmagupta formula to find the area of Trapezium =area of cyclic quadrilateral DPOA + area of cyclic quadrilateral CPOB = √[(2+√2-2)(2+√2-2)(2+√2-√2)(2+√2-√2)] +√[(√2+1-1)(√2+1-1)(√2+1-√2)(√2+1-√2)] =2√2+√2=3√2 Area of shaded portion = (3√2 -π) sq units
A = (πr²)/2 π = (πr²)/2 2π = πr² r² = 2 r = √2 So, AB = AO + BO = √2 + √2 = 2√2. By the Two-Tangent Theorem, AD = DP = 2x & BC = CP = x. So, CD = CP + DP = x + 2x = 3x. Draw a segment thru point C and a point N on base AD such that it is perpendicular to the bases of trapezoid ABCD. This is the height of the trapezoid and forms a rectangle ABCN and a right △CND. Therefore, by the Parallelogram Opposite Sides Theorem, AN = BC = x & AB = CN = 2√2. Thus, DN = AD - AN = 2x - x = x. Apply the Pythagorean Theorem on △CND. a² + b² = c² x² + (2√2)² = (3x)² x² + 8 = 9x² 8x² = 8 x² = 1 x = 1 So, AD = 2 & BC = 1. Find the area of trapezoid ABCD. A = [(a + b)/2](h) = 1/2 * (1 + 2) * 2√2 = √2 * 3 = 3√2 Purple Region Area = Trapezoid ABCD Area - Semicircle Area = 3√2 - π So, the area of the purple shaded region is 3√2 - π square units (exact), or about 1.10 square units (approximation).
Semicircle O: A = πr²/2 π = πr²/2 r² = 2 r = √2 As CP and BC are tangents to semicircle O that intersect at C, CP = BC = x. As PF and DA are tangents to semicircle O that intersect at D, PD = DA = 2x. CD = CP + PD = x + 2x = 3x Draw CF, where F is the point on DA where CF is parallel to AB and perpendicular to BC and DA. As DA = 2x and FA = BC = x, DF = 2x-x = x. CF = AB = 2r = 2√2. Triangle ∆DFC: DF² + FC² = CD² x² + (2√2)² = (3x)² x² + 8 = 9x² 8x² = 8 x² = 8/8 = 1 x = 1 The shaded area will be equal to the area of trapezoid ABCD minus the area of semicircle O. Purple shaded area: A = h(a+b)/2 - πr²/2 A = 2√2(1+2)/2 - π A = 3√2 - π ≈ 1.101 sq units
((3x)/2 )*2r - pi 3xr - pi A full circle would have area 2pi, indicating that r = sqrt(2). 3*sqrt(2)*x - pi Find x: DC = x + 2x = 3x CM = 2*sqrt(2) where M is AD's midpoint. Right triangle: x^2 + (2*sqrt(2))^2 = (3x)^2 x^2 + 8 = 9x^2 x = 1 Back to my first line: (3/2)*(2*sqrt(2)) - pi = (3/2)*2*sqrt(2) - pi = 3*sqrt(2) - pi = 1.101un^2 (rounded).
Be R the radius of the semi circle. We have (Pi.(R^2))/2 = Pi, so R = sqrt(2). We now use an orthonormal center O and first axis (OB) We have C(sqrt(2); x) and D(-sqrt(2); 2.x), then VectorCD(-2.sqrt(2); a). The equation of (CD) is (X -sqrt(2)).(x) - (Y - a).(-2.sqrt(2)) = 0 or x.X +2.sqrt(2).Y -3.x.sqrt(2) = 0. The distance from O to (CD) is abs(x.0 + 2.sqrt(2).0 -3.x.sqrt(2))/sqrt(x^2 + 8) = (3.x.sqrt(2))/sqrt(x^2 + 8) This distance is also R = sqrt(2), so we square and obtain: 18.(x^2) = 2.(x^2) +8), which gives that x^2 = 1 and x = 1 The area of the trapezoïd is then AB.((AD + BC)/2) = (2.sqrt(2).((1 + 2)/2) = 3.sqrt(2) anf finally the purple area is 3.sqrt(2) - Pi.
Let's find the area: . .. ... .... ..... First of all we calculate the radius of the semicircle: A = πr²/2 π = πr²/2 2 = r² ⇒ r = √2 According to the two tangent theorem we know: CP = BC = x DP = AD = 2x Now let's add point E on AD such that ABCE is a rectangle. In this case CDE is a right triangle and we can apply the Pythagorean theorem: CE² + DE² = CD² AB² + (AD − AE)² = (CP + DP)² (OA + OB)² + (AD − BC)² = (CP + DP)² (r + r)² + (2x − x)² = (x + 2x)² (2r)² + x² = (3x)² 4r² + x² = 9x² 4r² = 8x² r²/2 = x² ⇒ x = r/√2 = √2/√2 = 1 Now we are able to calculate the area of the purple region: A(purple) = A(ABCD) − A(semicircle) = (1/2)*(BC + AD)*AB − π = (1/2)*(x + 2x)*(2r) − π = 3xr − π = 3√2 − π ≈ 1.101 Best regards from Germany
My way of solution ▶ First, we draw the radius r, which is equal to [OP] : [OP]= r Let's consider the Deltoid OBCP, the two triangles in this Deltoid are equal to each other: ΔOBC= ΔOCP ⇒ [OB]= [OP]= r [OC]= [CO] the hypotenuse of both triangles ⇒ [BC]= [CP] [CP]= x b) Let's consider the Deltoid AOPD, the two triangles in this Deltoid are equal to each other: ΔAOD= ΔOPD ⇒ [AO]= [OP]= r [DO]= [OD] the hypotenuse of both triangles ⇒ [DA]= [PD] [PD]= 2x c) Asemicircle= π π= πr²/2 r²= 2 r= √2 d) Let's consider the right triangle ΔKCD [DK]= 2x -x [DK]= x [KC]= 2r [KC]= 2√2 [CD]= x+2x [CD]= 3x By applying the Pythagorean theorem we can write: [DK]² + [KC]²= [CD]² x²+ (2√2)²= (3x)² 8x²= 8 x²= 1 x= 1 e) Apurple= A(ABCD) - Asemicircle A(ABCD)= (2x+x)*2r/2 A(ABCD)= 3*√2 ⇒ Apurple= 3√2 - π Apurple ≈ 1,10 square units
Solution: Let's calculate the radius and the diameter Semicircle Area = θ/360° π r² π = 180°/360° π r² π/π = 1/2 r² 1 = 1/2 r² r² = 2 r = √2 d = 2√2 Let's suppose a point K in the segment AD, such that, the segment KC is parallel to AB. Therefore, we will project a right triangle CDK, and we will calculate its 3 sides AD = DP = 2x BC = CP = x CD = DP + CP = 2x + x CD = 3x AK = BC = x KD = AD - AK KD = 2x - x KD = x KC = AB = 2√2 Let's applying the Pythagorean Theorem to calculate "x" KC² + KD² = CD² (2√2)² + (x)² = (3x)² 8 + x² = 9x² 8x² = 8 x² = 1 x = 1 Purple Shaded Area (PSA) = Trapezoid Area - Semicircle Area ... ¹ Trapezoid Area = ½ h (a + b) Trapezoid Area = ½ 2√2 (2 + 1) Trapezoid Area = √2 (3) Trapezoid Area = 3√2 ... ² Replacing ² in equation ¹ PSA = 3√2 - π square units ✅ PSA ≈ 1,1010 square units ✅
Nice Work! This Equation needed Tan functions to estimate !! THANKS for your Daily videos again Sir !
You are very welcome!
Thanks for watching! ❤️
Obrigado espetacular muito bom não sei inglês mas compreendo oque vc faz!
Área of semicircle:
A₁ = ½πR² = π cm² --> R = √2 cm
Pytagorean theorem:
x² + (2R)² = (2x+x)²
9x² - x² = 4R²
x²= 4R²/8= ½R²= ½.2= 1 --> x=1cm
Área of trapezoid :
A₂ = 2R.x + ½.2R.x = 3.R.x = 3*√2*1
A₂ = 3√2 cm²
Shaded Area :
A = A₂ - A₁ = 3√2 - π
A = 1,101 cm² ( Solved √ )
As soon as we know x =1 unit and r =√2 units
Semiperimeter Of DPOA=(2+√2) units
& that of CPOB =(√2+1) unit
we may use Brahmagupta formula to find the area of Trapezium
=area of cyclic quadrilateral DPOA + area of cyclic quadrilateral CPOB
= √[(2+√2-2)(2+√2-2)(2+√2-√2)(2+√2-√2)]
+√[(√2+1-1)(√2+1-1)(√2+1-√2)(√2+1-√2)]
=2√2+√2=3√2
Area of shaded portion
= (3√2 -π) sq units
Purple shaded area=1/2(1+2)(2√2)-π=3√2-π square units.❤❤❤
A = (πr²)/2
π = (πr²)/2
2π = πr²
r² = 2
r = √2
So, AB = AO + BO = √2 + √2 = 2√2.
By the Two-Tangent Theorem, AD = DP = 2x & BC = CP = x.
So, CD = CP + DP = x + 2x = 3x.
Draw a segment thru point C and a point N on base AD such that it is perpendicular to the bases of trapezoid ABCD. This is the height of the trapezoid and forms a rectangle ABCN and a right △CND. Therefore, by the Parallelogram Opposite Sides Theorem, AN = BC = x & AB = CN = 2√2.
Thus, DN = AD - AN = 2x - x = x. Apply the Pythagorean Theorem on △CND.
a² + b² = c²
x² + (2√2)² = (3x)²
x² + 8 = 9x²
8x² = 8
x² = 1
x = 1
So, AD = 2 & BC = 1.
Find the area of trapezoid ABCD.
A = [(a + b)/2](h)
= 1/2 * (1 + 2) * 2√2
= √2 * 3
= 3√2
Purple Region Area = Trapezoid ABCD Area - Semicircle Area
= 3√2 - π
So, the area of the purple shaded region is 3√2 - π square units (exact), or about 1.10 square units (approximation).
S=3√2-π≈1,105≈1,11
Semicircle O:
A = πr²/2
π = πr²/2
r² = 2
r = √2
As CP and BC are tangents to semicircle O that intersect at C, CP = BC = x. As PF and DA are tangents to semicircle O that intersect at D, PD = DA = 2x.
CD = CP + PD = x + 2x = 3x
Draw CF, where F is the point on DA where CF is parallel to AB and perpendicular to BC and DA. As DA = 2x and FA = BC = x, DF = 2x-x = x. CF = AB = 2r = 2√2.
Triangle ∆DFC:
DF² + FC² = CD²
x² + (2√2)² = (3x)²
x² + 8 = 9x²
8x² = 8
x² = 8/8 = 1
x = 1
The shaded area will be equal to the area of trapezoid ABCD minus the area of semicircle O.
Purple shaded area:
A = h(a+b)/2 - πr²/2
A = 2√2(1+2)/2 - π
A = 3√2 - π ≈ 1.101 sq units
((3x)/2 )*2r - pi
3xr - pi
A full circle would have area 2pi, indicating that r = sqrt(2).
3*sqrt(2)*x - pi
Find x:
DC = x + 2x = 3x
CM = 2*sqrt(2) where M is AD's midpoint.
Right triangle:
x^2 + (2*sqrt(2))^2 = (3x)^2
x^2 + 8 = 9x^2
x = 1
Back to my first line: (3/2)*(2*sqrt(2)) - pi
= (3/2)*2*sqrt(2) - pi = 3*sqrt(2) - pi = 1.101un^2 (rounded).
Nice, but why go back to first line once you find x when you have a simplified formula for area on fourth line?
Be R the radius of the semi circle. We have (Pi.(R^2))/2 = Pi, so R = sqrt(2). We now use an orthonormal center O and first axis (OB)
We have C(sqrt(2); x) and D(-sqrt(2); 2.x), then VectorCD(-2.sqrt(2); a). The equation of (CD) is (X -sqrt(2)).(x) - (Y - a).(-2.sqrt(2)) = 0
or x.X +2.sqrt(2).Y -3.x.sqrt(2) = 0. The distance from O to (CD) is abs(x.0 + 2.sqrt(2).0 -3.x.sqrt(2))/sqrt(x^2 + 8) = (3.x.sqrt(2))/sqrt(x^2 + 8)
This distance is also R = sqrt(2), so we square and obtain: 18.(x^2) = 2.(x^2) +8), which gives that x^2 = 1 and x = 1
The area of the trapezoïd is then AB.((AD + BC)/2) = (2.sqrt(2).((1 + 2)/2) = 3.sqrt(2) anf finally the purple area is 3.sqrt(2) - Pi.
Thank you!
x^2 + r^2 + 4x^2 + r^2 = (x + 2x)^2
5 x^2 + 2r^2 = 9x^2
2x^2 = r^2 = 2
x = 1, r = √2
Purple area = (2x + x)(2r)/2 - π = 3x(2r)/2 - π = 3√2 - π
AO=BO=r DP=2x CP=x DC=2x+x=3x
r*r*π*1/2=π r=√2 AB=√2+√2=2√2
(2x-x)²+(2√2)²=(3x)² x²+8=9x² 8x²=8 x²=1 x=1
Purple shaded area = (2+1)*2√2*1/2 - π = 3√2 - π
@ 5:06 , that's the "kicker" too solving. 🤔 ... right, dropped that perpendicular EC like Wile E. Coyote drops an anvil too snuff The Roadrunner. 😊
Bom dia Mestre
Thanks easy❤
You make every statement sound like a question
{2x+2x ➖ }+{x+x ➖ }={2x^2+x^2}=2x^4 180°/2x^4=90x^4 3^30x^4 3^10x^4 3^2^5x^5 3^2^1x^2^2 3^1^1x^1^2 3x^2 (x ➖ 3pix+2).
Let's find the area:
.
..
...
....
.....
First of all we calculate the radius of the semicircle:
A = πr²/2
π = πr²/2
2 = r²
⇒ r = √2
According to the two tangent theorem we know:
CP = BC = x
DP = AD = 2x
Now let's add point E on AD such that ABCE is a rectangle. In this case CDE is a right triangle and we can apply the Pythagorean theorem:
CE² + DE² = CD²
AB² + (AD − AE)² = (CP + DP)²
(OA + OB)² + (AD − BC)² = (CP + DP)²
(r + r)² + (2x − x)² = (x + 2x)²
(2r)² + x² = (3x)²
4r² + x² = 9x²
4r² = 8x²
r²/2 = x²
⇒ x = r/√2 = √2/√2 = 1
Now we are able to calculate the area of the purple region:
A(purple) = A(ABCD) − A(semicircle) = (1/2)*(BC + AD)*AB − π = (1/2)*(x + 2x)*(2r) − π = 3xr − π = 3√2 − π ≈ 1.101
Best regards from Germany
My way of solution ▶
First, we draw the radius r, which is equal to [OP] :
[OP]= r
Let's consider the Deltoid OBCP,
the two triangles in this Deltoid are equal to each other:
ΔOBC= ΔOCP
⇒
[OB]= [OP]= r
[OC]= [CO] the hypotenuse of both triangles
⇒
[BC]= [CP]
[CP]= x
b) Let's consider the Deltoid AOPD,
the two triangles in this Deltoid are equal to each other:
ΔAOD= ΔOPD
⇒
[AO]= [OP]= r
[DO]= [OD] the hypotenuse of both triangles
⇒
[DA]= [PD]
[PD]= 2x
c) Asemicircle= π
π= πr²/2
r²= 2
r= √2
d) Let's consider the right triangle ΔKCD
[DK]= 2x -x
[DK]= x
[KC]= 2r
[KC]= 2√2
[CD]= x+2x
[CD]= 3x
By applying the Pythagorean theorem we can write:
[DK]² + [KC]²= [CD]²
x²+ (2√2)²= (3x)²
8x²= 8
x²= 1
x= 1
e)
Apurple= A(ABCD) - Asemicircle
A(ABCD)= (2x+x)*2r/2
A(ABCD)= 3*√2
⇒
Apurple= 3√2 - π
Apurple ≈ 1,10 square units
Solution:
Let's calculate the radius and the diameter
Semicircle Area = θ/360° π r²
π = 180°/360° π r²
π/π = 1/2 r²
1 = 1/2 r²
r² = 2
r = √2
d = 2√2
Let's suppose a point K in the segment AD, such that, the segment KC is parallel to AB. Therefore, we will project a right triangle CDK, and we will calculate its 3 sides
AD = DP = 2x
BC = CP = x
CD = DP + CP = 2x + x
CD = 3x
AK = BC = x
KD = AD - AK
KD = 2x - x
KD = x
KC = AB = 2√2
Let's applying the Pythagorean Theorem to calculate "x"
KC² + KD² = CD²
(2√2)² + (x)² = (3x)²
8 + x² = 9x²
8x² = 8
x² = 1
x = 1
Purple Shaded Area (PSA) = Trapezoid Area - Semicircle Area ... ¹
Trapezoid Area = ½ h (a + b)
Trapezoid Area = ½ 2√2 (2 + 1)
Trapezoid Area = √2 (3)
Trapezoid Area = 3√2 ... ²
Replacing ² in equation ¹
PSA = 3√2 - π square units ✅
PSA ≈ 1,1010 square units ✅
STEP-BY-STEP RESOLUTION PROPOSAL :
01) AD = DP = 2X
02) BC = CP = X
03) AB = 2sqrt(2)
04) CD = 3X
05) 9X^2 = 8 + X^2
06) 8X^2 = 8 ; X ^2 = 1 ; X = 1
07) Trapezoid [ABCD] Area = (3sqrt(2)) sq un
08) Purple Shaded Area = [(3sqrt(2)) - Pi] sq un
09) PSA ~ 1,10 sq un
Thus,
OUR BEST ANSWER IS :
Purple Shaded Area equal to approx. 1,10 Square Units - Exact Form = [(3sqrt(2)) - Pi] Square Units.