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

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

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[santokhsidhuatla]

Area of circle= pi *r^2=49pi
r=49^1/2=7
Half chord=(8+5)/2=6.5
Perpendicular =7^2-6.5^2=p^2
P=2.598
Green area=1/2 x 2.598 x 1.5=
=1.9485 square units

[Uber]

AO = 7, AT=6.5 then TO=2.6 no need to find OP.

[quigonkenny]

Circle O:
Aₒ = πr²
49π = πr²
r² = 49
r = √49 = 7

As ∠OTP = 90° and AB is a chord, OT bisects AB amd AT = TB = (8+5)/2 = 13/2. As TB = 13/2 and PB = 5, TP = 13/2-5 = 3/2.

Draw radius OB.

Triangle ∆OTB:
OT² + TB² = OB²
OT² + (13/2)² = 7²
OT² = 49 - 169/4 = 27/4
OT = √(27/4) = (3√3)/2

Triangle ∆OTP:
Aᴛ = bh/2 = (3/2)((3√3)/2)/2 = (9√3)/8 ≈ 1.95 sq units

[Ibrahimfamilyvlogl]

First comments mashallah very nice sharing sir❤❤

[egillandersson]

Another approach : consider the triangle ABO, whose sides are all known (7, 7 and 13) and calculate his area by Heron's formula. Then, you get the height OT, as you know the area and the correspondant base (13).

[skoijlg]

∆ AOB равнобедренный, значит высота является медианой. BT=(8+5)/2=6, 5, PT=6, 5-5=1.5. OR=√7^2-6.5^2=√6, 75. S=1.5*√6.75/2=1, 948

[marioalb]

Radious of circle:
A = πR² = 49π cm²
R= 7 cm
Chord:
c = 8+5 = 13 cm
½c = 6, 5 cm
Intersecting chords theorem:
(R+h)(R-h) = (½c)²
R² - h² = (½c)²
h² = 7² - 6, 5² = 6, 75
h = 2, 598 cm
Area of green triangle:
A = ½ b.h
A = ½ (6, 5-5) . 2, 598
A = 1, 9486 cm² ( Solved √ )

[giuseppemalaguti]

OP^2=h^2+1, 5^2=5^2+r^2-2*5*r*cos(arccos6, 5/r)..legge del coseno .=25+49-70*6, 5/7=9..h=9-2, 25=6, 75..h=√6, 75...Agreen=√6, 75*1, 5/2=9√3/8

[marcgriselhubert]

The radius of the circle is R = OA = 7
AT = TB = (AP +PB)/2 = 13/2
In triangle OAT: OT^2 = OA^2 - AT^2 = 49 - 169/4 = 27/4,
so OT = (3/2).sqrt(3)
TP = TB - PB = 13/2 - 5 = 3/2
The green area is (1/2).OT.TP = (1/2).((3/2).sqrt(3)).(3/2)
= (9/8).sqrt(3).
That was very easy.

[alscents]

T is the midpoint of AB, so [TP] = [TB] - [PB] = (8 + 5) / 2 - 5 = 3/2
By Pythagoras [TO]^2 = [AO]^2 - [TP]^2 = 7^2 - (13/2)^2
Area of triangle TOP = (1/2) [TP] [TO] = (1/2) (3/2) (sqrt(49 - 169/4)) = 3 * sqrt(27/4) / 4 = 9 sqrt(3) / 8
Area of triangle TOP = 1.948557 square units

[santiagoarosam]

49π=πr²→ r=7 → 7²-[(8+5)/2]²=h²→ h=3√3/2 ; b=(13/2)-5=3/2 → Área triángulo verde =bh/2 =(1/2)(3/2)(3√3/2)=9√3/8
Gracias y un saludo cordial.

[maxhagenauer]

First, the area of the circle is 49*pi so radius is 7.

Now call the base "b" and height "h" of the triangle. Construct line from O to A and a line from O to B so you have 2 right triangles. Then we have:

(8 - b)^2 + h^2 = 7^2

(5 + b)^2 + h^2 = 7^2

So we get b = 3/2, h = 3 * sqrt(3) / 2. Area of the triangle is 1/2 * b * h = 1/2 * ( 3/2 ) * ( 3 * sqrt(3) / 2 ) = 9 * sqrt(3) / 8 or about 1.949.

[MaximeDube-whzr]

8= √49-(7-x)^2
5=√49-(7-x)^2
13/2= √49-(7-x)^2
x=4.401
7- 4.401=2.598
PT=8-6.5=1.5
A=(1.5)(2.598)/2 = 1.9485

[ashutoshkumardalei]

Fantabulous hands up to your efforts ❤

[yalchingedikgedik]

That’s very good
Nice and wonderful method
Thanks Sir
❤❤❤❤❤

[montynorth]

Radius OB = 7 as calculated.
AB length = 8 + 5 = 13.
T is the centre of line AB.
TB = ( 8 + 5 ) / 2 = 6.5.
Therefore TP = 6.5 - 5 = 1.5.
Joining points O & B.
OT^2 = OB^2 - TB^2 by Pythagoras.
OT^2 = 7^2 - 6.5^2.
OT = 2.598.
Area of green triangle = 1/2 x 1.5 x 2.598.
1.95.( correct to 2 decimal places).

[MrPaulc]

r=7
TP is side a
TO is side b
PO is side c
Chord AB is 13, so AT is 13/2
49 - (13/2)^2 = b^2
49 - 169/4 = b^2, so b^2 = 196/4 - 169/4 = 27/4
b = (3*sqrt(3))/2
a = 3/2, because AT = TB
((3/2)*3*sqrt(3))/4 un^2 is the area of the green triangle.
((9/2)*sqrt(3))/4 = (9*sqrt(3))/8 un^2 approximates to 1.949 un^2
Having now looked at your video, I notice that the green triangle is a 30, 60, 90.
Thanks once again.

[DB-lgsq]

شكرا لكم على المجهودات
يمكن استعمال TA=TB
TP=3
OT^2=OA^2-8^2

S=3/2 ×racine(49/pi -64)

[gaylespencer]

Since we know the radius is 7, I used two Pythagorean formulas to find x (= the distance from T to P).

(8-x) and (5+x), where TO is the common side to both formulas.

[yxkrurq]

Let's solve this using an intersecting chords approach. If the circle radius is r then pi*r^2 = 49*pi so r = 7. Now chord AB = 13 and since OT is perpendicular to AB, T is the midpoint of AB and AT = 6.5 and TP = AP - AT = 8 - 6.5 = 1.5. Next we use the intersecting chords formula AT*TB = (r + TO)(2r -(r +TO)). We substitute r =7 to get (13/2)^2 = 49 - TO^2 so TO is 3*sqrt(3)/2 and the area of the triangle is (1/2)*TP*TO = (1/2)*(3/2)*(3*sqrt(3)/2) = (9/8)*sqrt(3) = 1.9485 square units.