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).
Yes, you’re right and more precise. I don’t like to use approximate figures and carry on in future computing. Here, OT = 3 sqrt (3) /2 and finally S = 9 sqrt (3) /4 as you mentioned.
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 √ )
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.
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.
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).
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.
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.
No need to find OP! We can find OT because OA = 7 and OTA is a right triangle. OT * OT = 7*7 - 6.5*6.5 = 6.75 => OT = ~2.6 => The area of the green triangle = OT(2.6) * (TP)1.5 * 0.5 = ~1.95
OT = a TP = 3/2 Then: (7 + a) . (7 - a) = 13/2 . 13/2 49 - a² = 169/4 196 - 4a² = 169 4a² = 27 a = √(27/4) a = 3√3/2 A = ½ 3/2 . 3√3/2 A = 9√3/8 Square Units A ~= 1,948 Square Units
I'm writing an updated version of Euclid's Elements. Gonna plagiarize the whole damn thing...chapter 1, Intersecting Chord Theorem. What could possibly go wrong? 🙂
I wondered how you got from sqrt(6.75) to 2.6. You apparently rounded. I've never known a mathematician to round before the final answer. Couldn't you more exactly get the square root of 6.75 by reverting it back into a fraction? 6.75 = 6 + 3/4 = 24/4 + 3/4 = 27/4. And IIRC, sqrt(27/4) = sqrt(27)/sqrt(4), meaning sqrt(6.75) = 3(sqrt(3))/2. Right? Therefore, the final exact answer would be 1/2 x 3sqrt(3)/2 x 3/2 = 9sqrt(3)/8.
Circle area=49π so πr^2=49π r=7 Connect O to A In ∆ OAT OT^2+AT^2=OA^2 OT^2+(6.5)^2=7^2 So OT=3√3/2 PT=BT-BP=6.5-5=1.5 So Green triangle area=1/2(3√3/2)(1.5)=9√3/8=1.95 square units.❤❤❤
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.
If you use Pythagoras' theorem on triangle OTA you get OT^2+TA^2=AO^2 where AO=r=7,TA=13/2, OT is therefore√(r^2-13/2^2). =√(49-169/4) = √(196/4-169/4) =√(27/4) =(3√3)/2 ≈ 2.59808. Using that then the area of the green triangle =1/2(OT)(TP) ≈ 1.94856. Not the exact number 1.95 using OT=2.6 Ok, it's about 0.99926% out but almost 1% could be a problem in other areas - you would come up short of, or overshot Mars by almost 1.5 million miles.😂
Let's find the area: . .. ... .... ..... First of all we calculate the radius R of the circle: A = πR² 49π = πR² 49 = R² ⇒ R = 7 Since OT is perpendicular to AB, we can conclude that T devides the chord AB into two parts with the same length. Therefore we obtain: AT = BT = AB/2 = (AP + BP)/2 = (8 + 5)/2 = 13/2 Now let's add the points Q and R on the circle such that QR is the diameter where OT is located on. Then we can apply the intersecting chords theorem: QT*RT = AT*BT (OQ + OT)*(OR − OT) = AT² (R + OT)*(R − OT) = AT² R² − OT² = AT² OT² = R² − AT² = 7² − (13/2)² = 49 − 169/4 = (196 − 169)/4 = 27/4 ⇒ OT = 3√3/2 Now we are able to calculate the area of the green triangle: PT = AP − AT = 8 − 13/2 = 16/2 − 13/2 = 3/2 A(OPT) = (1/2)*OT*PT = (1/2)*(3√3/2)*(3/2) = 9√3/8 ≈ 1.949 Best regards from Germany
That was over complicated. Short leg is 1.5. Longer leg is the short leg of a right triangle with hypotenuse of 7 and long leg of 6.5. So that length is sqrt(27)/2. Area = bh/2= (3sqrt(3))/8
STEP-BY-STEP RESOLUTION PROPOSAL : 01) AB =13 lin un 02) AP = 8 lin un 03) BP = 5 04) AT = BT = 13 / 2 lin un = 6,5 lin un 05) TP = 6,5 - 5 = 1,5 lin un 06) OA = 7 lin un 07) OT^2 = OA^2 - AT^2 08) OT^2 = 49 - 42,25 09) OT^2 = 6,75 10) OT = sqrt(6,75) lin un ; OT ~ 2,5981 lin un 11) Green Triangle Area (GTA) = (OT * TP) / 2 12) GTA = (sqrt(6,75) * 1,5) / 2 13) GTA ~1,95 sq un Thus, OUR ANSWER : The Green Triangle Area is approx. equal to 1,95 Square Units.
Geometry is beautiful.
Glad to hear that!
Thanks for the feedback ❤️
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
Excellent!
Thanks for sharing ❤️
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
AO = 7, AT=6.5 then TO=2.6 no need to find OP.
Thanks for the feedback ❤️
Ok
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).
S=9√3/8≈1,95
Excellent!
Thanks for sharing ❤️
Yes, you’re right and more precise. I don’t like to use approximate figures and carry on in future computing. Here, OT = 3 sqrt (3) /2 and finally S = 9 sqrt (3) /4 as you mentioned.
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 √ )
∆ 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
Excellent!
Thanks for sharing ❤️
Да Филиповна
First comments mashallah very nice sharing sir❤❤
Thanks for liking
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.
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.
That’s very good
Nice and wonderful method
Thanks Sir
❤❤❤❤❤
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
Fantabulous hands up to your efforts ❤
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
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).
Excellent!
Thanks for sharing ❤️
Thank you!
You're welcome!
Thanks for the feedback ❤️
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.
Excellent!
Thanks for sharing ❤️
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 =1,948557......ud².
Gracias y un saludo cordial.
Excellent!
Thanks for sharing ❤️
Area = 49*Pi = Pi*R^2 ->R=7. Now find OP=x using intersecting chord theorem at P. [7+x] * [7-x] = 8*5 = 40 = 49-x^2 -> x^2 = 9 -> x=3. Now find OT=y and TP=z. y^2 + (8-z) ^2 = 7^2 -> y^2 + 8^2 - 16*z + z^2 = 8^2 + 16*z + 3^2 (since y^2 + z^2 = x^2 = 3^2) -> 49 = 64 + 9 - 16*z -> 16z = 64+9-49 = 24 -> z=3/2. Now y is found by Pythagoras (30-60-90 triangle) to be 3*sqrt (3)/2. Thus [OTP] = OT*TP/2 = y*z/2 = 3*sqrt (3)/2 * 3/2/2 = 9*sqrt (3) / 8 = 1.95 sq. units
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.
Excellent!
You are very welcome!
Thanks for the feedback ❤️
No need to find OP! We can find OT because OA = 7 and OTA is a right triangle. OT * OT = 7*7 - 6.5*6.5 = 6.75 => OT = ~2.6 => The area of the green triangle = OT(2.6) * (TP)1.5 * 0.5 = ~1.95
OT = a
TP = 3/2
Then:
(7 + a) . (7 - a) = 13/2 . 13/2
49 - a² = 169/4
196 - 4a² = 169
4a² = 27
a = √(27/4)
a = 3√3/2
A = ½ 3/2 . 3√3/2
A = 9√3/8 Square Units
A ~= 1,948 Square Units
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
I'm writing an updated version of Euclid's Elements. Gonna plagiarize the whole damn thing...chapter 1, Intersecting Chord Theorem. What could possibly go wrong? 🙂
😀
Thanks for the feedback ❤️
👏👏👏👏👏👏👏👏
I wondered how you got from sqrt(6.75) to 2.6. You apparently rounded. I've never known a mathematician to round before the final answer. Couldn't you more exactly get the square root of 6.75 by reverting it back into a fraction? 6.75 = 6 + 3/4 = 24/4 + 3/4 = 27/4. And IIRC, sqrt(27/4) = sqrt(27)/sqrt(4), meaning sqrt(6.75) = 3(sqrt(3))/2. Right?
Therefore, the final exact answer would be 1/2 x 3sqrt(3)/2 x 3/2 = 9sqrt(3)/8.
green area is more precise (9*sqrt3)/8 = 1,9486
TP=1.5 and from triangle ATO where AO=7 and AT = 6.5 you can get length of OT etc.
Thanks for the feedback ❤️
Thanks from Morocco
Circle area=49π
so πr^2=49π
r=7
Connect O to A
In ∆ OAT
OT^2+AT^2=OA^2
OT^2+(6.5)^2=7^2
So OT=3√3/2
PT=BT-BP=6.5-5=1.5
So Green triangle area=1/2(3√3/2)(1.5)=9√3/8=1.95 square units.❤❤❤
Excellent!
Thanks for sharing ❤️
r*r*π=49π r=7
TO=x TP=y
x²+(8-y)²=7² x²+(5+y)²=7²
x²+(8-y)²=x²+(5+y)² x²+64-16y+y²=x²+25+10y+y² 26y=39 y=3/2 x²=27/4 x=3√3/2
Green Triangle area = 3√3/2*3/2*1/2 = 9√3/8
Excellent!
Thanks for sharing ❤️
If AT = 8 and AO = 7, then hypothenuse is smaller then cathetus?
AP=8, AT = 8-1.5
√3*(9/8)
Excellent!
Thanks for sharing ❤️
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.
1.95 Sq.units
Excellent!
Thanks for sharing ❤️
If you use Pythagoras' theorem on triangle OTA you get OT^2+TA^2=AO^2 where AO=r=7,TA=13/2, OT is therefore√(r^2-13/2^2). =√(49-169/4) = √(196/4-169/4) =√(27/4) =(3√3)/2 ≈ 2.59808.
Using that then the area of the green triangle =1/2(OT)(TP) ≈ 1.94856. Not the exact number 1.95 using OT=2.6
Ok, it's about 0.99926% out but almost 1% could be a problem in other areas - you would come up short of, or overshot Mars by almost 1.5 million miles.😂
Thanks for the feedback ❤️
OA=7
AT=8
AT < OA
???
This is not possible. If the area of the circle is 49π, the radius is 7. OA = r = 7, but 7 < AT = 8
I was following along with this but the blue line AP=8, not just the segment AT≠8
@@ericb5634 OK. That explains it. Thanks!
Let's find the area:
.
..
...
....
.....
First of all we calculate the radius R of the circle:
A = πR²
49π = πR²
49 = R²
⇒ R = 7
Since OT is perpendicular to AB, we can conclude that T devides the chord AB into two parts with the same length. Therefore we obtain:
AT = BT = AB/2 = (AP + BP)/2 = (8 + 5)/2 = 13/2
Now let's add the points Q and R on the circle such that QR is the diameter where OT is located on. Then we can apply the intersecting chords theorem:
QT*RT = AT*BT
(OQ + OT)*(OR − OT) = AT²
(R + OT)*(R − OT) = AT²
R² − OT² = AT²
OT² = R² − AT² = 7² − (13/2)² = 49 − 169/4 = (196 − 169)/4 = 27/4
⇒ OT = 3√3/2
Now we are able to calculate the area of the green triangle:
PT = AP − AT = 8 − 13/2 = 16/2 − 13/2 = 3/2
A(OPT) = (1/2)*OT*PT = (1/2)*(3√3/2)*(3/2) = 9√3/8 ≈ 1.949
Best regards from Germany
Excellent!
Thanks for sharing ❤️
That was over complicated. Short leg is 1.5.
Longer leg is the short leg of a right triangle with hypotenuse of 7 and long leg of 6.5.
So that length is sqrt(27)/2.
Area = bh/2= (3sqrt(3))/8
≈
STEP-BY-STEP RESOLUTION PROPOSAL :
01) AB =13 lin un
02) AP = 8 lin un
03) BP = 5
04) AT = BT = 13 / 2 lin un = 6,5 lin un
05) TP = 6,5 - 5 = 1,5 lin un
06) OA = 7 lin un
07) OT^2 = OA^2 - AT^2
08) OT^2 = 49 - 42,25
09) OT^2 = 6,75
10) OT = sqrt(6,75) lin un ; OT ~ 2,5981 lin un
11) Green Triangle Area (GTA) = (OT * TP) / 2
12) GTA = (sqrt(6,75) * 1,5) / 2
13) GTA ~1,95 sq un
Thus,
OUR ANSWER : The Green Triangle Area is approx. equal to 1,95 Square Units.
Excellent!
Thanks for sharing ❤️
АО=7, АТ=8 это невозможно
AT = 6.5, AP =8
@@Uber308 My inattention
green area is more precise (9*sqrt3)/8 = 1,9486