Can you find area of the Semicircle? | (Triangles) | #math #maths | #geometry

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Can you find area of the Semicircle? | (Triangles) | #math #maths | #geometry

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[abeonthehill]

Another clear and succinct demonstration Professor ………Thanks for sharing Man !

[laxmikantbondre]

Steps
1. Assume CD = h
2. AD = 34/h, DB = 136/h ... Area of both ∆s is goven.
3. By Thales theorum ang ACB is 90
So CD is Altitude on the Hypotuneous of a Rt. traingle
4. So CD is Geometric Mean of AD and DB
So h Sqr = 34/h * 168/h
5. So solving this CD = h = 2√17
Hence AD = 34/h = √17 and DB = 136/h = 4√17. So Diameter = AB = 5√17

[allanflippin]

I took an approach more like the way Mr. PreMath usually does! That's similar triangles. The key is Thales theorem: angle ABC is a right angle. We can work out alpha and beta angles to show that yellow and green triangles are similar. Also, as pointed out in the video, AD and DB have the proportion 17:68. Adding them up, 2r = 5*AD or 5/4*DB. I'll call the height CD "h". The total of both triangles area is height * base / 2. Or h*2r/2 = (17+68). Simplified to h*r = 85.
The ratio CD:AB equals DB:CD. Filling in the numbers, h * (2r/5) = (8r/5) / h. Cross multiplying, h^2 = 16 * r^2 / 25. Taking square root, h = 4r/5. Now consider h*r = 85, or r = 85/h.
h = 4*85/h*5. or h = 68/h. Multiplying by h, h^2 = 68 or h = sqrt(68) or h = 2 * sqrt(17). From above, r = 5h/4. That means r = 10 * sqrt(17) / 4 or r = 5 * sqrt(17) / 2.
Squaring, r^2 = 25 * 17 / 4. or r^2 = 425/4. The semicircle area is pi * r^2 / 2 or 425/8 pi.

[PrithwirajSen-njqq]

(1/2*h*AD)/(1/2*h*BD)
=17/68=1/4
>AD/BD=1/4
If AD =x BD =4x
Diameter is 5x
radius = 2.5 x
h =√(x *4x ) geometric mean theorem
h = 2x
Take the 17 sq cm triangle
1/2*x*2x=17
> x =√17
Radius = 2.5*√17
Area of semicircle
=1/2(π*17*25/4) sq cms

[anatoliy]

Simple but pleasant task. 👍💯Thank you so much, Professor

[santiagoarosam]

Razón entre áreas s²=68/17=4→ Razón de semejanza =s=√4=2→ Si AD=a→ CD=2a→DB=4a→ 2a*4a/2=4a²=68→ a=√17→ AB=AD+DB=2r =a+4a=5a=5√17 → r=5√17/2→ Área del semicírculo =πr²/2 =(π*25*17)/(4*2) =425π/8 =166, 8971...
Gracias y un saludo.

[Waldlaeufer]

17 * 4 = 68, this means, that the ratio of the legs is 1 : √4 = 1 : 2.

x * 2x = 2 * 17
x² = 17
x = √17

d = AB = 1x + 4x = 5x = 5√17
r = 5/2 √17

A(semicircle) = 1/2 * r² * π = 1/2 * (5/2 √17)² π = 1/2 * 25/4 * 17 * π = (25 * 17) / 8 π = 53.125 π = 166.9 square units

[himo]

17 : 68 = 1: 4 = 1² : 2²
AD=x CD=2x BD=4x (x+4x)*2x/2=17+68 5x²=85 x=√17
AO=BO=5x/2=5√17/2

Semicircle area = 5√17/2*5√17/2*π*1/2 = (425π/8)cm²

[cyruschang]

yellow hypotenuse = x
gree hypotenuse = x✓(68/17) = 2x
x(2x)/2 = 17 + 68 = 85
x = ✓85
circle diameter = ✓(x^2 + 4x^2) = x✓5 = 5✓17
semi circle area = π[(5✓17)/2]^2 ÷ 2 = 25(17)π/8 = 425π/8 (cm^2)

[murdock]

∆ ABC → AB = AD + BD = 2r/5 + 8r/5 = 2r; CD = h; sin⁡(ADC) = 1; hr = 85
h^2 = (2r/5)(8r/5) → h = 4r/5 → hr = (4r^2)/5 = 85 → r^2 = (25/4)17 → πr^2/2 = (25/8)17π

[prossvay]

AD/BD=17/68=1/4
So AD=1a ; BD=4a
Let CD=x
So x°2=(1a)(4a)
So x=2a
Area of triangleABC=1/2(x)(5a)=85
1/2(2a)(5a)=85
So a=√17
Diameter of semisecle=1a+4a=5a=5√17
So Radius=5√17/2
Area of

[misterenter-izrz]

Quite creative. but not too difficult 🎉. a, b=2a, 2b=4a, then 34=2a^2, a^2=17, 2r=5a, 4r^2=25a^2, then the answer is 1/2 r^2 pi=1/2 × 25/4×17pi=425/8 pi.😅
::: '

[phungpham]

1/ Label CD= h, AD=a, BD= b and the diameter AB= d
Consider the two right triangle ACD and BCD
We have h. a=2x17 (1)

and h.b=2x68=2x4x17(2)
—> (1)x(2)
sq h xaxb= sq 2xsq2xsq17
Because axb= sq h ( the right triangle altitude theorem)
—> sqhxsqh= sq2xsq2sq17
h=2sqrt17
2/ hxd=2x85=170
—> d=170/(2sqrt17)=85/(sqrt17)
Area= 1/2 pi . sq(85/sqrt17)/4 =1/8 pi . sq(85/sqrt17) = 166.9 sq cm

[joeschmo]

I just saw them as similar triangles, ADC, CDB, and ACB. Yellow is a 1:2:sqrt5 triangle, green is 2:4:2sqrt5, and big is sqrt5:2sqrt5:5. Diameter is 5, radius is 5/2, area is 25pi/4.
*Now* scale. So 1:2:sqrt5 has an area of 1unit, but is 17cm2 so that's the scaling factor.
Circle area is (17)(25pi/4). Semicircle is half that, so (17)(25pi/8) or 425pi/8. ✨Magic!✨

[quigonkenny]

Let ∠CAB = α and ∠ABC = β, where α and β are complementary angles that sum to 90°. As ∠ADC = 90°, ∠BCA = 90°-α = β and thus ∆BCA and ∆ADC are similar triangles. Similarly, as ∠BCD = 90°-β = α, ∆CDB is also aimilar to the above triangles.

As ∆CDB and ∆ADC are similar triangles and the area of ∆CDB is 4 times the area of ∆ADC, then the lengths of the sides of ∆CDB are √4 = 2 times the length of the corresponding sides of ∆ADC. Thus DB = 2DC. And as the two triangles are similar, this means DC = 2AD. Let AD = x.

AB = AD + DB
AB = AD + 2DC
AB = AD + 2(2AD) = 5AD = 5x

Triangle ∆CDB:
Aᴛ = bh/2 = DB(DC)/2
68 = 4x(2x)/2 = 4x²
x² = 68/4 = 17
x = √17

AB = 5x
2r = 5√17
r = (5√17)/2

Semicircle O:
Aₒ = πr²/2 = π((5√17/2)²/2
Aₒ = π(25)(17)/8 = 425π/8 ≈ 166.90 cm²

[hongningsuen]

Method using base side ratio = area ratio for equal height triangles and intersecting chords theorem:
1. Triangles ADC and BDC are equal height triangles.
Hence AD:BD = 17:68 = 1:4
2. Let R be the radius.
AB = AD + BD = 2R.
Hence AD = (2/5)R and BD = (8/5)R
4. For complete circle, by intersecting chords theorem
(CD)^2 = (AD)(BD)
(CD)^2 = (2/5)(8/5)R^2 (equation 1)
5. In triangle ADC, area = 17 = (1/2)(AD)(CD)
Hence CD = (17)(2)/[(2/5)R] = (5)(17)/R
CD^2 = [(5)(17)/R]^2 (equation 2)
6. From equations (1) and (2)
(2/5)(8/5)R^2 = [(5)(17)/R]^2
R^4 = (5^4)(17^2)/(2)(8)
R^2 = (5^2)(17)/4 = 425/4
7. Area of semicircle = (1/2)(425/4)pi = (425/8)pi

[RondoCarletti]

for both triangles use A=1/2*g*h. h is equal to both triangles. So calculate 17 (and also 68) = 1/2*2*h (it works) or 1/2 *4*h (this is realistic) = Area 157, 0796...

[juanalfaro]

Let CD=h and AD=x -> BD=2R-x. Since 17=x *h/2 and 68 = (2R-x) *h/2, then 2R-x = 4x -> 2R=5x -> R=5x.2. Chord Intersection Theorem: (2R-x) *x = h^2 = 4x*x = 4x^2 -> h=2x. Now [ADC] = x *h/2 = 2x^2/2 = x^2 = 17 -> x=sqrt (17) -> R=5*sqrt (17)/2 --> A(semicircle) = PI*R^2/2 = Pi/2 * [5*sqrt (17)/2] ^2 = Pi/2 *25*17/4 = Pi*425/8 cm^2

[LuisdeBritoCamacho]

STEP-BY-STEP RESOLUTION PROPOSAL USING THE LAW OF SIMILAR TRAINGLES :

01) Let AD = X cm
02) Let BD = Y cm
03) Let AB = (X + Y) cm
04) Let CD = h cm
05) X * h = 34 ; h = 34/X
06) Y * h = 136 ; h = 136/Y
07) 34/X = 136/Y ; Y/X = 4 ; Y = 4X
08) BD = 4X cm
09) AB = 5X cm
10) Triangle (ACD) is Similar to Triangle (BCD) ; So :
11) h/X = Y/h ; h^2 = (X * Y) ; As Y = 4X ; h^2 = 4X^2 ; sqrt(h^2) = sqrt(4X^2) ; h = 2X
12) 2X * X = 34 ; 2X^2 = 34 ; X^2 = 34/2 ; X^2 = 17 ; X = sqrt(17) ; Or :
13) 2X * 4X = 136 ; 8X^2 = 136 ; X^2 = 136/8 ; X^2 = 17 ; X = sqrt(17)
14) If : X = sqrt(17) ; Then 5*X = 5 * sqrt(17)
15) AB = 5 * sqrt(17)
16) Radius (R) = AB/2 cm ; R = (5 * sqrt(17)) / 2) ; R^2 = (25 * 17) / 4 ; R^2 = 425/4
17) Semi Circle Area (SCA) = Pi * R^2 / 2
18) SCA = ((425/2) * Pi) / 2 ; SCA = 425Pi / 8 ; SCA ~ 167 sq cm

Thus,

OUR PLANE AND SIMPLE ANSWER IS : Semicircle Area is approx. equal to 167 Square Centimeters.

Greetings from Cordoba!!

[wackojacko]

All Hail Euler and Set theory ...the hierarchy of the Right triangle within is the Isoceles and Scalene...the Oblique...the Acute and Obtuse...and Isoceles. Knowing the relationships of their angles and ratios of their sides allows one too readily calculate problems such as this one. How much fun can a person have in one day? 🙂