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[jimlocke]

There is a second solution, a = 21 and h = 28. In this solution, BD = AB - a = 35 - 21 = 14. The area of ΔBCD is (1/2)bh = (1/2)(14)(28) = 196. As a check, tan(x) = BD/CD = 14/28 = 1/2. tan(2x) = 28/21 = 4/3. Applying the tangent double angle formula, this is correct. In PreMath's solution, tan(x) = 7/21 = 1/3 and tan(2x) = 21/28 = 3/4, which also is correct.

[juanalfaro]

The solution is NOT unique, because it also works with a=21 and h=28. This is because (a-h) ^2 = 49, thus a-h = +/- 7. if a-h= -7, then 2a=42 -> a=21, h=28. In that case DB=14 and [BDC] = 14*28/2 = 196 cm^2. To test the solution, a=21 --> 2x=53.13 --> x=26.565 -> tan (x) =1/2 = 14/28 works fine too.

[cyruschang]

base of the white triangle = b
height of the white triangle = h
bh/2 = 294 => h = 2(294)/b
b^2 + h^2 = 35^2 => b^2 + (2(294)/b)^2 = 35^2
let b^2 = y
y^2 - (35^2)y + (2(294))^2 = 0
2y = (35^2) +/- ✓((35^4) - 4(2(294))^2)
294 = (7)(7)(6), 35 = (7)(5)
2y = (35^2) +/- 49✓((5^4) - 4(2(6))^2) = (35^2) +/- 49✓(25^2 - 24^2) = (35^2) +/- 49✓((25 + 24)(26 - 24)) = (35^2) +/- 49(7) = 49 (25 +/- 7) = 49(32) or 49(18)
y = 49(16) or 49(9)
b = 28 or 21
The big triangle (white + green) is an isosceles triangle (2x, 90 - x, 90 - x)
The sum of the white base and green base = 35, the green base = 35 - 28 or 35 - 21 => 7 or 14
the green area = (7/28)(294) cm^2 or (14/21)(294) cm^2 = 73.5 cm^2 or 196 cm^2

[marcgriselhubert]

I propose something else.
*Area of ACD = 294 = (1/2).h.AD = (h^2)/(2.tan(2.x)) with h = CD. Unknown area = A = (1/2).h.DB = (1/2).h.(h.tan(x)) = (1/2).(h^2).tan(x)
So A/294 = tan(x).tan(2.x) and A = 294.tan(x).tan(2.x).
*We have: h/AD = tan(2.x) and h.AD = 2.294 = 588. Let's multiply: h^2 = 588.tan(2.x). Let's divide: AD^2 = 588/tan(2.x)
In ADC: 35^2 = h^2 + AD^2, so 35^2 = 588.tan(2.x) + 588/tan(2.x), we simplify: 12.(tan(2.x))^2 -25.tan(2.x) +12 = 0. Delta = 625 -576 = 49
Two possibilities. tan(2.x) = 4/3 or tan(2.x) = 3/4.
*If tan(2.x) = 4/3, then tan(x) = 1/2 (easy), and then A = 294.(4/3).(1/2) = 196.
If tan(2.x) = 3/4, then tan(x) = 1/3, and then A = 294.(3/4).(1/3) = 73.5

[dariosilva]

There is another solution. When you took square root of (a-h)^2, you can get +7 or -7. The difference between two lengths can actually be negative.

[ABhaim]

why is 196 not a solution too?
I also realized this is an isosceles triangle, so if AD=a, then DB=35-a and h=588/a
according to Pythagorean Theorem, CB^2 = (35-a)^2 + (588/a)^2
and according to Al Kashi's Law of Cosines = CB^2 = 35^2 + 35^2 - 2*35*35cos(2x) = 2*35^2(1 - cos(2x)) = 2*35^2(1 - a/35)
putting these 2 equations together,
(35-a)^2 + (588/a)^2 = 2*35^2(1 - a/35)
1225 -70a + a^2 + 588^2/a^2 = 2550 - 70a
-1225 + a^2 + 588^2/a^2 = 0
solving for a^2 we get either 784 or 441
therefore, a is equal to their (positive) square root 28 or 21, making h 21 or 28 respectively, but here is the tricky part.
remember that DB = 35-a, it means it can be either 7 or 14 and that matters because is means the area can be either 0.5*7*21 = 73.5 OR 0.5*14*28 = 196!

[laxmikantbondre]

There is another solution where a =21 and h =28. This can be done by in previous step you took sqr(a-h) = a2+h2-2ah, instead do Sqr(h-a)=a2+h2-2ah.
Then values will be interchanged.

[highlyeducatedtrucker]

Add me to the pile of people who got a 2nd solution. I just used Equation 1 (a*h=588) and Equation 2 (a^2 + h^2 = 35^2) as a system of simultaneous equations, and got two results. (Well, actually I got four results, but two of those involved negative lengths so they were discarded.)

[quigonkenny]

As ∠CAD = 2x and ∠ADC = 90°, ∠DCA = 90°-2x. As ∠BCD = x and ∠CDB = 90°, ∠DBC = 90°-x. As ∠DCA = 90°-2x and ∠BCD = x, ∠BCA = 90°-x. As ∠ABC = ∠BCA = 90°-x, ∆CAB is an isosceles triangle and AB = CA = 35.

Let AD = a and DC = b.

Triangle ∆ADC:
Aᴛ = bh/2
294 = AD(DC)/2 = ab/2
ab = 2(294) = 588 --- [1]

AD² + DC² = CA²
a² + b² = 35² = 1225 --- [2]

(a+b)² = a² + b² + 2ab
(a+b)² = 1225 + 2(588) = 1225 + 1176
(a+b)² = 2401
a + b = ±√2401 = ±49
a + b = 49 --- [3] (a, b > 0)

(a-b)² = a² + b² - 2ab
(a-b)² = 1225 - 2(688) = 1225 - 1176
(a-b)² = 49
a - b = ±√49 = ±7 --- [4]
(will need to check for a > b and b > a)

a + b = 49 <-- [3]
a + (a-7) = 49 <-- [4+] (for a > b) **
2a = 56
a = 28

b = 49 - 28 = 21

a + b = 49 <-- [3]
a + (a+7) = 49 <-- [4-] (for b > a) **
2a = 42
a = 21

b = 49 - 21 = 28

So (a, b) = (28, 21) and (a, b) = (21, 28) both appear to be valid. We'll solve for both instances.

(a, b) = (28, 21):

AB = AD + DB
35 = 28 + DB
DB = 35 - 28 = 7

Aₜ = bh/2 = 7(21)/2 = 147/2 = 73.5 cm²

(a, b) = (21, 28):

AB = AD + DB
35 = 21 + DB
DB = 35 - 21 = 14

Aₜ = bh/2 = 14(28)/2 = 14(14) = 196 cm²

So the green area equals either 73.5 cm² or 196 cm².

[santiagoarosam]

Ángulos: A=2Xº→ C=90º-2Xº+Xº=90º-Xº=B→ ABC es triángulo isósceles→ AC=AB=35→ AD=a→ DB=35-a →
a*b=294*2→ a=294*2/b → Si a²+b²=35²→ (294*2/b)²+b²=35²→ b=28→ a=21→ DB=35-21=14→ Área verde =14*28/2=196 cm².
Otra solución válida sería: b=21→ a=28→ 35-28=7→ Área verde =7*21/2=73, 5 cm²
Una comprobación previa hubiese facilitado los cálculos: Si CDA fuese triángulo 3/4/5→ 35=7*5→ a=7*3=21 ; b=7*4=28 y 21*28/2=294 cm²→ Hipótesis correcta→ Solución trivial.
Gracias y un saludo cordial.

[devondevon]

answer 196
The triangle is an isosceles since angle B is 90-x (180- 90 + x)
and angle C is also 90-x ( 180- (90 + 2x ) + x)
Hence, line AB = 35
Let's label the triangle to the left n and p
hence np =588 ( 298 *2)
Since the hypotenuse of the triangle to the left = 35, it is possibly a 3-4-5 scaled up by 7
Hence, one of the base = 3 x 7 = 21
and the other 4 x 7 = 28
And since 21 x 28 = 588, then it is a 3-4-5 scaled up by 7
In which the height is 28
Since AB = 35
then the length of the other base = 14 (35 - minus 21)
Hence the area = 28 x 14 x 1/2 or 28 x 7 = 196 Answer

[rabotaakk-nwnm]

… AB=35…a+h=49; a-h=7 => h=21
[ABC]=½35•21=367.5; [BCD]=367.5-294=73.5

[unknownidentity]

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


The triangle ACD is a right triangle, so we can apply the Pythagorean theorem. Additionally we know the area of this triangle. Therefore we can conclude:

A(ACD) = (1/2)*AD*CD
AC² = AD² + CD²

A(ACD) = (1/2)*AD*CD
AC² ± 2*AD*CD = AD² ± 2*AD*CD + CD²

AC² ± 4*A(ACD) = AD² ± 2*AD*CD + CD² = (AD ± CD)²

AC² + 4*A(ACD) = (35cm)² + 4*(294cm²) = 1225cm² + 1176cm² = 2401cm² = (49cm)²
AC² − 4*A(ACD) = (35cm)² − 4*(294cm²) = 1225cm² − 1176cm² = 49cm² = (7cm)²

(AD + CD)² = (49cm)²
(AD − CD)² = (7cm)²

First case: AD > CD

AD + CD = 49cm
AD − CD = 7cm
⇒ AD = (49cm + 7cm)/2 = 28cm
∧ CD = (49cm − 7cm)/2 = 21cm

tan(2x) = CD/AD = (21cm)/(28cm) = 3/4

tan(2x) = 2*tan(x)/[1 − tan²(x)]
3/4 = 2*tan(x)/[1 − tan²(x)]
3*[1 − tan²(x)] = 8*tan(x)
3 − 3*tan²(x) − 8*tan(x) = 0
tan²(x) + (8/3)*tan(x) − 1 = 0

tan(x) = −4/3 ± √[(−4/3)² + 1] = −4/3 ± √(16/9 + 9/9) = −4/3 ± √(25/9) = −4/3 ± 5/3

Since 0<x<90°, there is only one useful solution:

tan(x) = −4/3 + 5/3 = 1/3

tan(x) = BD/CD ⇒ BD = CD*tan(x) = (21cm)*(1/3) = 7cm

Since the green triangle is also a right triangle, we obtain:

A(BCD) = (1/2)*BD*CD = (1/2)*(7cm)*(21cm) = 73.5cm²

Second case: AD < CD

CD + AD = 49cm
CD − AD = 7cm
⇒ CD = (49cm + 7cm)/2 = 28cm
∧ AD = (49cm − 7cm)/2 = 21cm

tan(2x) = CD/AD = (28cm)/(21cm) = 4/3

tan(2x) = 2*tan(x)/[1 − tan²(x)]
4/3 = 2*tan(x)/[1 − tan²(x)]
4*[1 − tan²(x)] = 6*tan(x)
4 − 4*tan²(x) − 6*tan(x) = 0
tan²(x) + (3/2)*tan(x) − 1 = 0

tan(x) = −3/4 ± √[(−3/4)² + 1] = −3/4 ± √(9/16 + 16/16) = −3/4 ± √(25/16) = −3/4 ± 5/4

Since 0<x<90°, there is only one useful solution:

tan(x) = −3/4 + 5/4 = 2/4 = 1/2

tan(x) = BD/CD ⇒ BD = CD*tan(x) = (28cm)*(1/2) = 14cm

Since the green triangle is also a right triangle, we obtain:

A(BCD) = (1/2)*BD*CD = (1/2)*(14cm)*(28cm) = 196cm²

Summary:
AD > CD ⇒ A(BCD) = 73.5cm²
AD < CD ⇒ A(BCD) = 196cm²

Best regards from Germany

[LuisdeBritoCamacho]

STEP-BY-STEP RESOLUTION PROPOSAL (considering that CD < AD as herein depicted):

01) AC = 35 cm.

02) (AD * DC) / 2 = 294 sq cm ; AD * DC = 588 sq cm

03) AD^2 + DC^2 = 35^2 ; AD^2 + DC^2 = 1.225

04) System of Two Equations :

a) AD * DC = 588
b) AD^2 + DC^2 = 1.225

05) Solutions : AD = 28 cm and CD = 21 cm. We could also see that 35 = (7 * 5). And as we have a Pythagorean Triple ( 3 ; 4 ; 5 ) : ( (3 * 7) ; (4 * 7) ; (5 * 7) ) = (21 ; 28 ; 35 ). (21 * 28) = 588

06) tan(2X) = 21 / 28 ; tan(2X) = 3/4

07) tan(2X) = (2tan(X)) / (1 - tan^2(X))

08) 3/4 = (2tan(X)) / (1 - tan^2(X))

09) 3 * (1 - tan^2(X)) = 4 * (2tan(X))

10) 3 - 3tan^2(X) = 8tan(X)

11) 3tan^2(X) + 8tan(X) - 3 = 0

12) tan(X) = Y

13) 3Y^2 + 8Y - 3 = 0

14) Y = - 3 or Y = 1/3 ; tan(X) = - 3 or tan(X) = 1 / 3. Let's consider the Positive Solution:

15) tan(X) = 1 / 3 ; and tan(X) = BD / CD ; BD / CD = 1 / 3 ; BD / 21 = 1 / 3 ; BD = 7

16) Green Area (GA) = (BD * CD) / 2 ; GA = (7 * 21) / 2 ; GA = 147 / 2 ; GA = 73, 5


Therefore,


OUR BEST ANSWER IS:


Green Triangle Area equal to 73, 5 Square Cm.

[imetroangola]

*_Realmente existem duas soluções_*

Pois,

a+h=49 e ah=588, logo

a=28 e h=21 ou a=21 e h=28, logo:

b=7 e h=21 ou b=14 ou h=28. Portanto,

A área ABD pode ser:

7×21/2= *73, 5cm²*

ou

14×28/2= *196cm²*

[viaducjy]

rôle symétrique a et h donc 2 possibilités a=28, h =21 ou a=21, h=28
Donc il y a deux cas aire de 73, 5 ou aire de 196

[jmcxzcs]

a and h are exchangeable, so there are two answers

[murdock]

Awesome! φ = 30°; ∆ ABC → AB = AD + BD = a + b; CD = h; BC = k
AC = 35; ADC = 3φ; CAB = 2δ; BCD = δ
ah/2 = 294 → ah = 588 = 12(49) = 3(7)4(7) = 21(28)
AC = 35 = 5(7) → a = 28 → h = 21 → cos⁡(2δ) = 4/5 →
cos⁡(δ) = √((1/2)(1 + cos⁡(2δ))) = 3√10/10 = h/k → k = 7√10 →
area ∆ BCD = (1/2)sin⁡(δ)hk = (1/2)(√10/10)(21)7√10 = 147/2

[Birol]

My way of solution ▶
The area of the triangle ΔADC is given as 294 cm²
A(ΔADC)= 294 cm²
⇒
194= 1/2* [AD]*35*sin(2x)
⇒
[AD]*sin(2x)=

If we apply the sinus theorem for this triangle ΔADC, we get:
sin(90°)/35= sin(2x)/[DC]= sin(90-2x)/[AD]
let's consider the first and third term of this equation:
sin(90°)= 1
⇒
1/35= sin(90-2x)/[AD]

sin(90-2x)= sin(90°)*cos(2x) - cos(90°)*sin(2x)
sin(90-2x)= cos(2x)
⇒
1/35= cos(2x)/[AD]
[AD]= 35*cos(2x)

if we put this value ([AD]) in Equation-1 we get:
[AD]*sin(2x)= 16, 8
⇒
35cos(2x)*sin(2x)= 16, 8
cos(2x)*sin(2x)= 0, 48

if we multiply both sides with 2, we get:
2*cos(2x)*sin(2x)= 0, 96
= sin(4x)
⇒
sin(4x)= 0, 96
4x= arcsin(0, 96)
4x= 73, 74°
x= 18, 435°

[AD]= 35*cos(2x)
[AD]= 35*cos(2*18, 435°)
[AD]= 35*cos(36, 87°)
[AD]= 28 cm

sin(2x)/[DC]= sin(90°)/35
[DC]= 35*sin(2x)
[DC]= 35*sin(36, 87°)
[DC]= 21 cm

III) if we apply the sinus theorem for the green triangle ΔDBC we get:
sin(x)/[DB]= sin(90°-x)/[CD]= sin(90°)/[BC]
let's consider the first two terms of this equation:
sin(18, 435)/[DB]= sin(90°- 18, 435°)/[CD]
0, 31622/[DB]= sin(71, 565°)/21
⇒
[DB]= 7 cm

Agreen= A(ΔDBC)
Agreen= 1/2*[CD]*[DB]
[CD]= 21 cm
[DB]= 7 cm
⇒
Agreen= 1/2*21*7
Agreen= 73, 5 cm²

[misterenter-izrz]

294×2=588=35×a sin 2x, 588+2A=35×(a+b)sin 2x, 2A=35×b sin 2x, cos 2x=a/35, tan x=b/(35 sin