There is a simple way to solve this problem as well! However, I incorporated the additional concepts... My apologies for taking a longer approach! Thank you for your continued love and support🙏
I like the solution though it may be solved in an easier and speedy method. We enjoy every right solutions whether it is short or lengthy. Every solution is beautiful for its own nature. Thanks to all.
*Simple solution:* Power at a point theorem C: DC² = EC × BC 2² = 1 × (2r + 1)→ 4 = 2r + 1 r = 3/2. Therefore, the area of the semicircle is: πr²/2 = *9π/8 square units.*
3.534 This might easier than I thought. Draw a perpendicular line from D to the circle's center to form a right triangle the sides are r + 1 , r and 2. Employed Pythagorean Theorem Hence (r+1)^2 = r^2 +4 r^2 + 2r +1 =r^2 +4 2 r+ 1 = 4 2r =4-1 (earlier I put 4 + 1 , instead of 4 -1 and got 5) 2r =3 r = 1.5 Area of semi circle = 2.25 pi/2 1.125 pi or 3.534 answer
DCO is a right triangle with sides r, 2, and r+1. r^2 + 4 = r^2 + 2r + 1 4 = 2r+1 2r=3 r = 3/2 r^2 = 9/4 For a full circle it is (9/4)pi, so for the semicircle the area is (9/8)pi un^2 (9pi)/8 Approx 3.53 un^2
Very easy! In triangle ODC we have OC^2= OD^2 + DC^2, so (R + 1)^2 = R^2 = 2^2, so R^2 + 2.R + 1 = R^2 + 4, so R = 3/2. The area of the semi circle is then (Pi/2).(R^2) = (Pi/2).(9/4) = (9/8).Pi
There is a much simpler way to solve this then any steps were taken by PreMath and the comments. We have a tangent to the circle and a crossing line to the circle from the same point This all we need.
I think this problem not need tangent circle AB and OD that you suppose by X. It's enough use triangle OCD with Pytagoras formula. Warm regards from Indonesia. 🙏🙏🙏
Equation 1 not needed to find r, just start with ODC and use Pythagorean Theorem. Equation 1 would have been helpful if you had asked for the area of the triangle ABC!👍
Bro I watch your videos more less everyone after i got your channel so don't take it as a hate comment But i think 🔺DCO is right Triangle so we can apply pythagoras on it and if radius is r then we get the equation (r+1)^2=r^2 + 2^2 and solve for r why make things complicated? Is there any reason? 🤔
Solution: Lets label the lengths below BO = DO = EO = r Applying Pythagorean Theorem in ∆ CDO, we will get: CD² + DO² = CO² (2)² + (r)² = (r + 1)² 4 + r² = r² + 2r + 1 4 = 2r + 1 2r = 3 r = 3/2 Semicircle Area (SA) = ½ π r² SA = ½ π (3/2)² SA = ½ π 9/4 SA = 9π/8 Square Units ✅ SA ≈ 3.5342 Square Units ✅
Nah, too much work. Figured it out from the thumbnail. Right triangle: sides 2 and r, hypotenuse 1+r. r=3/2. Then finally 9/4 pi for a circle and 9/8 pi for a semicircle. ✨Magic!✨
Let's find the area: . .. ... .... ..... ADC is a tangent to the circle, so we can apply the tangent-secant theorem. With r being the radius of the semicircle we obtain: CD² = CE*BC = CE*(CE + OB + OE) = CE*(CE + r + r) = CE*(CE + 2*r) CD²/CE = CE + 2*r CD²/CE − CE = 2*r ⇒ r = (CD²/CE − CE)/2 = (2²/1 − 1)/2 = (4 − 1)/2 = 3/2 Let's check this result. Since AB and ADC are tangents to the semicircle, according to the two tangent theorem we know that AB=AD. AB is a tangent to the semicircle, so we know that ∠ABC=90°. Therefore ABC is a right triangle and we can apply the Pythagorean theorem: BC = CE + 2*r = 1 + 2*(3/2) = 4 AC² = AB² + BC² (AD + CD)² = AD² + BC² AD² + 2*CD*AD + CD² = AD² + BC² 2*CD*AD = BC² − CD² ⇒ AB = AD = (BC² − CD²)/(2*CD) = (4² − 2²)/(2*2) = (16 − 4)/4 = 12/4 = 3 ⇒ AC = AD + CD = 3 + 2 = 5 We obtain that ABC is a 3-4-5 right triangle. The triangle OCD should also be a right triangle and similar to ABC: CD = 2 OD = r = 3/2 OC = OE + CE = r + CE = 3/2 + 1 = 5/2 OD:CD:OC = (3/2):2:(5/2) = 3:4:5 ✅ Everything seems to be all right. So we can finally calculate the area of the semicircle: A(semicircle) = πr²/2 = π*(3/2)²/2 = 9π/8 Best regards from Germany
MY RESOLUTION PROPOSAL : 01) Radius = R lin un 02) OD^2 + DC^2 = OC^2. Note that OC = (R + 1) lin un. 03) R^2 + 4 = (R + 1)^2 04) R^2 + 4 = R^2 + 2R + 1 ; 2R = 3 ; R = 3/2 lin un 05) Seemi Circle Area = (Pi * R^2) / 2 06) SCA = (9*Pi/4) / 2 07) SCA = 9Pi/8 sq un Therefore, MY BEST ANSWER : The Semicircle Area is equal to (9Pi/8) Square Units. Or Semicircle are is equal to approx. 3,5 Square Units.
There is a simple way to solve this problem as well!
However, I incorporated the additional concepts...
My apologies for taking a longer approach!
Thank you for your continued love and support🙏
OK, I was going to point that out, but you and a plethora of your viewers have already done so.
I am not the one who is going to criticize your approach. All Problems are good to revisit Mathematical Concepts.
I was wondering if working with ODC would have been quicker. It’s always good to see multiple approaches to a problem. It keeps the mind nimble
Good brother
The small circle alone can handle the problem
Much simpler and quicker to just use triangle ODC
OD² + DC² = OC²
r² + 4 = (r + 1)²
r = 3/2
π . r ² / 2 = 8 π / 9
9× pi/8.
Yes, I just did that, but I did it after a long day at work, so I excuse myself :)
Correct I was going to say the same. No need to complete the problem further. But yeah we get to know one more way to solve the problem
Join OD and apply Pythagoras theorem in triangle ODC
C'est plus pratique et rapide .
I did that, thanks for saying that is the fastest way
Essa, sem dúvidas, é a forma mais rápida de resolver o problema. Mas o interessante é ver as propriedades geométricas que estão envolvidas. 👍👍
That’s what I did.
Exactly
By Tangent secant theorem
(1+2r)*1=2^2
By solving it we'll get the radius
∆ODC, OD=r, DC=2, OC=r+1. r^2+4=(r+1) ^2. r=1,5. S=3.14*1,5^2/2=3.14*1,125.
That's how I did it
The improvement i got is because of such a great teacher like you , thanks a lot sr
I did it .
Very good! 👍👍
Radius can be obtained by in triangle odc
4+r²=(r+1)². Раскрываем, r² ушёл, 4-1=3, остаётся 3=2r, откуда r=3/2. 1,5;2;2,5 это дробный Египет, (3;4;5)/2. S=πr²/2=π(3/2)²/2=π*3²/2³=(9/8)π=1¹/₈π.
I like the solution though it may be solved in an easier and speedy method.
We enjoy every right solutions whether it is short or lengthy. Every solution is beautiful for its own nature.
Thanks to all.
*Simple solution:*
Power at a point theorem C:
DC² = EC × BC
2² = 1 × (2r + 1)→ 4 = 2r + 1
r = 3/2. Therefore, the area of the semicircle is:
πr²/2 = *9π/8 square units.*
Pythagoras inΔODC
rr+4=(r+1)^2=rr+2r+1
2r=3, r=3/2
Area =πr^2/2= 9π/8
NICE breakdown sir & THANKU for daily Solutions !
Thank you sir
DC²=BC*EC (tangent secant theorem). (2r+1)*1=2², so r=3/2, and S=1/2*(3/2)² *π=9π/8.
3.534
This might easier than I thought. Draw a perpendicular line from D to the circle's center to form a right triangle
the sides are r + 1 , r and 2.
Employed Pythagorean Theorem
Hence (r+1)^2 = r^2 +4
r^2 + 2r +1 =r^2 +4
2 r+ 1 = 4
2r =4-1 (earlier I put 4 + 1 , instead of 4 -1 and got 5)
2r =3
r = 1.5
Area of semi circle = 2.25 pi/2
1.125 pi or 3.534 answer
(r+1)^2-r^2= 4; r^2+2r+1-r^2=4;
2r=3; r=3/2.
S= π×r^2/2= 9π/8
That was a simple question sir
I had just joined OD and applied pythagoras theorem
We will get radios
I solved it 😊😊😊
Radius
Interessante o exercício. Gostei.
Thank you!
Well done sir ,
OD=OE=r r²+2²=(r+1)² r²+4=r²+2r+1 2r=3 r=3/2
Semicircle area = 3/2*3/2*π*1/2 = 9π/8
Nice! BE = 2r → 4 = (1)(2r + r) → r = 3/2 → πr^2/2 = 9π/8
r^2 + 2^2 = (r + 1)^2 = r^2 + 2r + 1 => r = 3/2
Semicircle area = (3/2)(3/2)(π)/2 = 9π/8
Thanks easy
Please directly apply Pythagoras Theorem in triangle ODC ,
( r+1)* = r* +2*,
or , r* +2r +1 = r* + 4
whence 2r = 3
& r = 3
DCO is a right triangle with sides r, 2, and r+1.
r^2 + 4 = r^2 + 2r + 1
4 = 2r+1
2r=3
r = 3/2
r^2 = 9/4
For a full circle it is (9/4)pi, so for the semicircle the area is (9/8)pi un^2
(9pi)/8
Approx 3.53 un^2
1/Label the diameter BE= d
By tangent secant theorem:
Sq CD= CExCB
-> 4= 1x(1+d) --> d=3
Area= pi x sqd/8= pi x 9/8 sq units😅😅😅
Easy way triangle ODC (r+1)^2 = r^2+2^2. r^2+2r+1=r^2+4. 2r=3. r=3/2
A= 3,14x9/4
S=1,125π≈3,532
Use the small triangle without referring to the similarity of triangles
And who used the Pythagorean theorem in the small triangle
@ 7:17 , equivalent equations. I love the math! 😊
There is no need to calculate x.
Join OD, OCD is right triangle.
Now
4+r^=(r+1)^
4= (r+1+r) (r+1-r)
4= 2r+1
r=1.5
Very easy! In triangle ODC we have OC^2= OD^2 + DC^2, so (R + 1)^2 = R^2 = 2^2, so R^2 + 2.R + 1 = R^2 + 4, so R = 3/2.
The area of the semi circle is then (Pi/2).(R^2) = (Pi/2).(9/4) = (9/8).Pi
(R + 1)^2 = R^2 + 4
R = 1.5
1 (1 + 2r) = 2^2
1 + 2r = 4
2r = 3
r = 3/2
...
Sol
In ∆ODC
2^2+r^2=(r+1)^2
=>r=3/2
∆=πr^2/2=9π/8
Tangent Secant theorem:
(2R+1)*1=2²
R = (4-1)/2 = 1,5 cm
A = ½πR² = 3,5343 cm² ( Solved √ )
Joining oD and applying py theoram than r=3/2
full circle area is =πr²
So semicircle area is = 1/2 πr²
= 11/7×9/4
= 3.53
There is a much simpler way to solve this then any steps were taken by PreMath and the comments. We have a tangent to the circle and a crossing line to the circle from the same point
This all we need.
Use power of point theorem to find r then use (pi*r^2)/2
I think this problem not need tangent circle AB and OD that you suppose by X.
It's enough use triangle OCD with Pytagoras formula.
Warm regards from Indonesia.
🙏🙏🙏
r/(r+1)=h/(h+2)
r(h+2)=h(r+1)
2r=h
(2r+1)^2+(2r)^2=(2r+2)^2
4r^2+4r+1+4r^2=4r^2+8r+4
4r^2-4r-3=0
r^2-r-(3/4)=0
r=[1+-sqrt(1+3)]/2
r=1.5
A=pi(1.5)^2/2=9pi/8
There is overhead in video... r^2 + 2^2 = (r + 1)^2 --- it is enough to solve this issue.
R^2 + 4 = (r + 1)^2
And solve for r. Easy Peasy!
Equation 1 not needed to find r, just start with ODC and use Pythagorean Theorem. Equation 1 would have been helpful if you had asked for the area of the triangle ABC!👍
3.534
Bro I watch your videos more less everyone after i got your channel so don't take it as a hate comment
But i think 🔺DCO is right Triangle so we can apply pythagoras on it and if radius is r then we get the equation (r+1)^2=r^2 + 2^2 and solve for r why make things complicated? Is there any reason? 🤔
Dear sir , why you are resorting to so much of unnecessary steps ? . It speaks bad of your reputation.
Pai r square.
Omg why is this so dam difficult no wonder kids hate math
9pi/8
Solution:
Lets label the lengths below
BO = DO = EO = r
Applying Pythagorean Theorem in ∆ CDO, we will get:
CD² + DO² = CO²
(2)² + (r)² = (r + 1)²
4 + r² = r² + 2r + 1
4 = 2r + 1
2r = 3
r = 3/2
Semicircle Area (SA) = ½ π r²
SA = ½ π (3/2)²
SA = ½ π 9/4
SA = 9π/8 Square Units ✅
SA ≈ 3.5342 Square Units ✅
Very very unsatisfactory.
Nah, too much work. Figured it out from the thumbnail.
Right triangle: sides 2 and r, hypotenuse 1+r. r=3/2.
Then finally 9/4 pi for a circle and 9/8 pi for a semicircle.
✨Magic!✨
2 other way easier method.
Rafiq's classes
Let's find the area:
.
..
...
....
.....
ADC is a tangent to the circle, so we can apply the tangent-secant theorem. With r being the radius of the semicircle we obtain:
CD² = CE*BC = CE*(CE + OB + OE) = CE*(CE + r + r) = CE*(CE + 2*r)
CD²/CE = CE + 2*r
CD²/CE − CE = 2*r
⇒ r = (CD²/CE − CE)/2 = (2²/1 − 1)/2 = (4 − 1)/2 = 3/2
Let's check this result. Since AB and ADC are tangents to the semicircle, according to the two tangent theorem we know that AB=AD. AB is a tangent to the semicircle, so we know that ∠ABC=90°. Therefore ABC is a right triangle and we can apply the Pythagorean theorem:
BC = CE + 2*r = 1 + 2*(3/2) = 4
AC² = AB² + BC²
(AD + CD)² = AD² + BC²
AD² + 2*CD*AD + CD² = AD² + BC²
2*CD*AD = BC² − CD²
⇒ AB = AD = (BC² − CD²)/(2*CD) = (4² − 2²)/(2*2) = (16 − 4)/4 = 12/4 = 3
⇒ AC = AD + CD = 3 + 2 = 5
We obtain that ABC is a 3-4-5 right triangle. The triangle OCD should also be a right triangle and similar to ABC:
CD = 2
OD = r = 3/2
OC = OE + CE = r + CE = 3/2 + 1 = 5/2
OD:CD:OC = (3/2):2:(5/2) = 3:4:5 ✅
Everything seems to be all right. So we can finally calculate the area of the semicircle:
A(semicircle) = πr²/2 = π*(3/2)²/2 = 9π/8
Best regards from Germany
MY RESOLUTION PROPOSAL :
01) Radius = R lin un
02) OD^2 + DC^2 = OC^2. Note that OC = (R + 1) lin un.
03) R^2 + 4 = (R + 1)^2
04) R^2 + 4 = R^2 + 2R + 1 ; 2R = 3 ; R = 3/2 lin un
05) Seemi Circle Area = (Pi * R^2) / 2
06) SCA = (9*Pi/4) / 2
07) SCA = 9Pi/8 sq un
Therefore,
MY BEST ANSWER :
The Semicircle Area is equal to (9Pi/8) Square Units. Or Semicircle are is equal to approx. 3,5 Square Units.
Why did u go thru all that nonsense with the x. Why y just don’t do. R squared plus 4 equal to (r+2) squared
𝕊𝕀𝕄ℙ𝕃𝔼 𝕎𝔸𝕐
𝕊𝕝𝕠𝕧𝕖 𝕓𝕪 𝕦𝕤𝕚𝕟𝕘 𝕋𝕣𝕚𝕒𝕟𝕘𝕝𝕖 𝕆𝔻ℂ