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Can you find area of the Red shaded Trapezoid? | (Trapezium) | #math #maths | #geometry

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[Mediterranean]

since 7+x is the radius
h is the geometric mean
so h= sqrt x*(14+x)
h^2=14x+x^2
h^2= 900-x^2 (from BFC)
subtract the equation to get
2x^2+14x-900=0
x^2-7x-900=0
x=24 (x>0)
h^2+x^2=900
h^2+(6*4)^2=(6*5)^2
h^2=6^2*3^2
h=6*3=18
Area (14+14+x)h/2
(28+24)*9=52*9=468

[yalchingedikgedik]

Thanks Sir
That’s very enjoyable
❤❤❤❤
Good luck with glades

[MrPaulc]

Call AD, x.
Draw a vertical down from B.
(1) (r-7)^2 + x^2 = 900.
On the left, using the midpoint of AB,
(2) 7^2 + x^2 = r^2.
It may be that I need another, more independent equation involving r, but see how this goes first.
(1) r^2 - 14r + 49 + x^2 = 900
(2) 49 + x^2 = r^2
I think some rearranging may be in order.
(1) r^2 = 14r - 49 - x^2 + 900
(2) r^2 = 49 + x^2
I might try seeing what happens if I add the two equations, as that will eliminate some of the unknowns:
2r^2 = 14r + 900 (surely it must be harder than that).
2r^2 - 14r - 900 = 0
r^2 - 7r - 450 = 0
Quadratic formula:
(7+or-sqrt(49 - 4*1*-450))/2 = r
(7+or-sqrt(1849))/2 = r
I think I might be on the right track as sqrt(1849) is an integer (43).
(7+43)/2 = r (I have removed the spurious negative result).
So r = 25.
Calculate x with right triangle ADO (much ADO about nothing).
625 - 49 = x^2
576 = x^2, so x = 24
Now for the trapezoid: (32 + 14)/2 = 23.
23*24 = 552 sq units.
I have now watched the video. I had a false start with this which I deleted from the above. We went slightly different paths but the essentials were there. I did consider making a full circle and forming a mirror image of the trapezoid. I considered employing properties of a cyclic quadrilateral - until I noticed it was a pentagon LOL.

[phungpham]

1/ Just name the diameter as CC’, the CF=a and the height BF= h.
Chord AB//CC’——> C’ABC is an isosceles trapezoid ——> C’D=a and the trangle C’BC is a right one.
2/ By using the right triangle altitude theorem, we have:
Sq h = a .(a+14) (1)
Also sqh =sq30-sqa (2) ——> a(a+14) = 900-sqa——> sqa+14a= 900 -sqa
We have the equation : 2sqa + 14a -900 =0—-> a= 18 ( negative result rejected)
3/ The triangle BFC is a 3-4-5 triple ——-> h=4x6=24
Area of the trapezoid = 1/2 (14+14+18) 24= 552 sq units

[uwelinzbauer]

Hello Professor and friends of the channel,
We have two right triangles completely consisting of integer sidelengths.
7, 24, 25 and 18, 24, 30!
Two pythagorean triples-
I like 👌 pythagorean triangles!
Thanks for the nice video,
I wish a happy Sunday 😊

[prossvay]

draw BE right CD
Let CE=x ; BE=h
x^2+h^2=30^2=900 (1)
Draw circle is complete
so h^2=x(14+x)
h^2=x^2+14x
(1) x^2+x^2+14x-900=0
2x^2+14x-900=0
So x=18
CD=14+18=32
h^2=18(14+18)
h=24
Red trapezoid area=1/2(14+32)(24)=552 square units.❤❤❤

[quigonkenny]

Drop a perpendicular from B to E on OC. As ∠ODA = ∠DAB = 90°, DC and AB are parallel. Draw radii OB and OF, where OF is perpendicular to AB at intersection point G. As AB is a chord of semicircle O and OF a radius, this means that AG = GB = 14/2 = 7. As OG and BE are both perpendicular to OC this means OE = GB = 7 and EC = OC-OE = r-7.

Triangle ∆BEC:
EC² + BE² = CB²
(r-7)² + BE² = 30²
r² - 14r + 49 + BE³ = 900
BE² = 851 + 14r - r²
BE = √(851+14r-r²)

Triangle ∆OEB:
OE² + EB² = OB²
7² + (851+14r-r²) = r²
49 + 851 + 14r - r² = r²
2r² - 14r - 900 = 0
r² - 7r - 450 = 0
(r-25)(r+18) = 0
r = 25 | r = -18 ❌

BE = √(851+14(25)-(25)²)
BE = √(851+350-625)
BE = √(1201-625)
BE = √576 = 24

Trapezoid ABCD:
A = h(a+b)/2
A = 24(14+(25+7))/2
A = 12(14+32)
A = 12(46) = 552 sq units

[Waldlaeufer]

i) r² = h² + 7²
h² = r² - 49

ii) (r - 7)² + h² = 30¹
r² - 14 r + 49 + h² = 900

i) into ii)
r² - 14r + 49 + r² - 49 = 900
2r² - 14r = 900
r² - 7r = 450
r² - 7r + (7/2)² = 450 + (7/2)²
(r - 7/2)² = 450 + 49/4 = 1800/4 + 49/4 = 1849/4
r - 7/2 = +- 43/2
r1 = 7/2 + 43/2 = 50/2 = 25
r2 = 7/2 - 43/2 = -36/2 = -18 (rejected since negative)

r = 25
x = 25 - 7 = 18
h² = 25² - 49 = 576
h = 24

A(red) = 1/2 * (14 + 18 + 14) (24) = 1/2 * 46 * 24 = 23 * 24 = 24² - 24 = 576 - 24 = 552 square units

[Vakyanshpro]

Please make videos on coordinate geometry.
Btw first view

[santiagoarosam]

El punto "P" es la proyección ortogonal de B sobre OC y BP=h→ h²=30²-(r-7)² → Potencia de P respecto a la circunferencia =(r+7)(r-7)=h²=30²-(r-7)²→ r=25 → h=24 →
Área ABCD =24*[14+(14/2)+25]/2 =552 ud².
Gracias y un saludo cordial.

[JSSTyger]

h^2+(r-7)^2=900
h^2+7^2=r^2
900-(r-7)^2=r^2-7^2
900-r^2-49+14r=r^2-49
900+14r=2r^2
r^2-7r-450=0

h^2+49=625
h=24

[marcgriselhubert]

Let's use an orthonormal, center O, first axis (OC).
We have B(R.cos(t); R.sin(t)) with R the radius of the circle and 0 < t < 90°. Then A(-R.cos(t); R.sin(t)) C(R; 0)
AB = 2.R.cos(t) = 14, so R.cos(t) = 7 and (R^2).(cos(t))^2) = 49. (eq 1)
VectorBC(R.(1 -cos(t); -R.sin(t)),
so BC^2 = (R^2).(2 -2.cos(t)) = 900 (eq 2)
With eq 1 and eq 2 : (1 - cos(t))/ (cos(t))^2 = 450/49
Then 450.(cos(t))^2 + 49.cos(t) -49 =0
Delta = 49^2 + 49.450.4 = 49.1849, andsqrt(Delta) = 7.43 = 301. Then cos(t) = (-49 + 301)/900 = 252/900 = 7/25
(cos(t) beeing positive). Then R = 7/cos(t) = 25
(Cos(t))^2 = 49/625 and (sin(t))^2 = 1 -(49/625) = 576/625, so sin(t) = sqrt(576/625) = 24/25 (beeing positive). Then R.sin(t) = 24 is the height of the trapezoïd.
The area of the trapezoïd is ((14 +(7 +25))/2).24 =552

[LuisdeBritoCamacho]

STEP-BY-STEP RESOLUTION PROPOSAL :

1) RADIUS : R = OA = OB = OC
2) As OA = OB = R, we can observe that Triangle [AOB] is an Isosceles Triangle with Base = 14 lin un; and Sides equal to R (R = OA = OB).
3) Let's call the Height of that Isosceles Triangle : h
4) h^2 = R^2 - 7^2 ; h^2 = (R^2 - 49) lin un
5) h^2 = 30^2 - (R - 7)^2
6) h^2 = 900 - (R^2 - 14R + 49)
7) h^2 = 900 - R^2 + 14R - 49
8) h^2 = (851 + 14R - R^2) lin un
9) As : h^2 = h^2
10) 851 + 14R - R^2 = R^2 - 49
11) 900 +14R - 2R^2 = 0
12) Two Roots : R = - 18 (Negative Root) or R = 25 (Positive Root)
13) R = 25 lin un
14) h^2 = (R^2 - 49)
15) h^2 = 625 - 49
15) h^2 = 576
16) h = sqrt(576)
17) h = 24 lin un
18) Major Base = 32 (B = 25 + 7 = 32)
19) Minor base = 14
20) h = 24
21) T = (32 + 14) * 12
22) T = 46 * 12
23) T = 552 sq un
ANSWER : Red Trapezoid Area equal to 552 Square Units.

See You Soon For Another Cartoon!!

[himo]

14/2=7 h²+7²=r² h²+(r-7)²=30²
　　　　　　　 h²=r²-7² h²=30²-(r-7)²　　

(r+18)(r-25)=0 r>0, r=25 h=24 DC=25+7=32

Trapezoid area : (14+32)*24/2=46*12=552

[misterenter-izrz]

Need some workload of computation😮, r^2=7^2+h^2, h^2+(r-7)^2=30^2, r^2-49+r^2-14r+49=900, 2r^2-14r-900=0, r^2-7r-450=(r+18)(r-25)=0, r=-18, rejected, r=25, thus h^2=25^2-49=24^2, h=24, therefore the area is

[jimlocke]

Using the Pythagorean theorem on ΔBFC generates an equation with 2 unknowns. A second equation is required. PreMath generates the second equation by applying the Pythagorean theorem to ΔOFB. Another way to get a second equation is to construct the other half of the semicircle and extend AD into a vertical chord. The horizontal chord through CD is a perpendicular diameter, so it bisects the vertical chord. We use the intersecting chords theorem. The horizontal chord is divided into lengths x + 14 and x, the vertical chord into h and h, so h² = (x + 14)(x) = x² +14x. We get the same second equation that PreMath got, but have used a different method. We proceed to solve the system of 2 equations as PreMath did in the video.

[robertlynch]

My solution was exactly the same as yours... which is rather remarkable, since it almost NEVER is! Thanks again. . . . A = 552

[DB-lgsq]

شكرا لكم على المجهودات
يمكن استعمال المثلث OBC
BC^2=OB^2+OC^2-2OB×OC
×cosBOC

x=18

[unknownidentity]

Let's find the area:
.
..
...
....


First of all we add point E on the diameter such that ABED is a rectangle. In this case the triangles BOE and BCE are both right triangles and we can apply the Pythagorean theorem. For reasons of symmetrie we have DO=EO=DE/2=AB/2. With R being the radius of the semicircle we obtain:

BE² + EO² = BO²
BE² + CE² = BC²

EO² − CE² = BO² − BC²
EO² − (CO − EO)² = BO² − BC²
(AB/2)² − [R − (AB/2)]² = R² − BC²
(14/2)² − [R − (14/2)]² = R² − 30²
7² − (R − 7)² = R² − 30²
49 − R² + 14*R − 49 = R² − 900
0 = 2*R² − 14*R − 900
0 = R² − 7*R − 450

R = 7/2 ± √[(7/2)² + 450] = 7/2 ± √(49/4² + 450) = 7/2 ± √(49/4 + 1800/4) = 7/2 ± √(1849/4) = 7/2 ± 43/2

Since R>0, there is only one useful solution:

R = 7/2 + 43/2 = 50/2 = 25

Since BE is the height of the trapezoid, we are able to calculate its area:

BE² = BO² − EO² = R² − EO² = 25² − 7² = 625 − 49 = 576
⇒ BE = 24

A(ABCD)
= (1/2)*(AB + CD)*BE
= (1/2)*(AB + DO + CO)*BE
= (1/2)*(AB + DO + R)*BE
= (1/2)*(14 + 7 + 25)*24
= 12*46 = 552

Best regards from Germany

[solimana-soli]

h² = R² - 7²
h² = 30² - (R - 7)² then comparing