The answer can be found easily without adding extra lines. I'll explain starting at 4:30 where AP and PB are found = 9 and CQ is also found to be 9. Triangle TAP area is 6*9/2 = 27. Triangle PQB area is 9*9/2 = 40.5. With the square area being 117, the total area in square and triangles is 117 + 27 + 40.5 = 184.5. The brown triangle area can be found by subtracting triangles TAB and TSQ from this total. TAB area = 6*18/2 = 54. TSQ is half the square area (117/2 = 58.5). 184.5 - 54 - 58.5 = 72.
Nice. But still... It does not require that much math to solve. Let's start with a few observations: 1. We are looking for the area of TBQ. And that's the area of TPQ (half of TPQS) + area of PBQ - area of TPB. 2. Area of TPB = half the area of TAB = area TAP. 3. You do need the lenght of AP to get the area of TAP. But you don't need the side of TPQS. It's simply sqrt of TPQS (117) - TA^2 (36) = sqrt of 81 = 9. 4. QC = AP => area PBQ = 1/2 AP^2 Therefore, area TBQ = 117/2 + 9^2/2 - 6x9/2 = (117 + (9x9 - 6x9))/2 = (117+3x9)2 = 144/2 just to show how easily you can do it in your head 🙂
Here is my version with trigonometry to share with. Let AP=sqrt(117-36)=9=PB ->AB=9+9=18 TB=sqrt(6^2+18^2)=6*sqrt(10) sin a=TA/TB=6/(6*sqrt(10))=1/sqrt(10) Applying sine law to triangle PTB sin(b)/PB=sin(a)/PT ->sin(b)/9=1/sqrt(10)/(3*sqrt(13))=3/sqrt(130) cos b=sqrt(1-sin(b)^2)=11/sqrt(130) sin(
TP²-6²=AP²→ AP=√(117-36)=9→ AB=2*9=18. Si R es la proyección ortogonal de Q sobre PQ → Los triángulos TAP y PRQ son congruentes→ QR=AP=9 y PR=TA=6 → Área TQB=ABQ+ATQ-TAB =(18*9/2)+[6*(9+6)/2]-(6*18/2) =81+45-54=72. Gracias y un saludo.
Interesting... I ended up using "functions of lines" to find crossing points, and a little Pythagoras too. -3x + 54 is line BQ Horizontal from T (not shown) intersects BQ at x = 16, y = 6, call it [M] Now there are 3 triangles and one rectangle to find area of. 6 * 16 is the rectangle ATCM area Upper triangle is 1/2 16 * 3 in area Lower right triangle is 1/2 * 2 * 6 And the big white lower triangle is 1/2 6 * 18 Brown = 6 * 16 + [1/2 2 * 6] - [1/2 6 * 18] + [1/2 3 * 16] Brown = 96 + 6 - 54 + 24 Brown = 72 and that's that.
First draw a vertical line from Pt Q to the x-axis (Pt X). Now there are 2 triangles beneath square SQPT. One is 6x9. Ther other is 9x6. That means there is a 3rd triangle on the right QBX with a height of 9 and a base of 3. The area of the Trapazoid TAXQ will be = 1/2 *(6+9)*(9+6)=112.5. The area of triangle QBX = 1/2 *3*9 =13.5. The sum of those 2 areas is 126. The area of triangle TAB = 1/2 *6*18 =54 Subtract these 2 areas = 126-54 = 72 Easy problem after you realize Triangles TAP and PQX are similar.
In triangle TAP: AP^2 = TP^2 - AT^2 = 117 - 36 = 81, so AP = PB = 9 and AB = 18. In triangle TAB: TB^2 = TA^2 + AB^2 = 36 + 324 = 360, so TB = 6.sqrt(10). Now let's use an orthonormal center A and first axis (AB). We have T(0; 6) and B(18; 0), VectorTB(18; -6). The equation of (TP) is (-6).(x) - (18).(y-6) = 0 or x +3.y -18 = 0. We have now VectorPQ(6; 9) and so Q(9+6; 0+9) or Q(15; 9). The distance from Q to (TP) is abs(15 +3.9 -18)/sqrt((1)^2 + (3)^2) = 24/sqrt(10) The area of the green triangle is then (1/2). TB .distance from Q to (TB) = (1/2).(6.sqrt(10)).(24/sqrt(10)) =72.
Let's find the area: . .. ... .... ..... First of all we calculate the side length s of the blue square from its given area: A = s² 117 = s² ⇒ s = √117 = 3√13 The triangle APT is a right triangle, so we can apply the Pythagorean theorem: PT² = AP² + AT² ⇒ AP² = PT² − AT² = s² − AT² = 117 − 6² = 117 − 36 = 81 ⇒ AP = √81 = 9 Let A be the center of the coordinate system and let B be located on the x-axis. Then we know the coordinates of two of the three corners of the brown triangle: B: ( 18 ; 0 ) T: ( 0 ; 6 ) Now we can calculate the coordinates of the third point. Since PT and PQ are perpendicular to each other and have the same length, we can conclude: xQ − xP = −(yP − yT) ⇒ xQ = xP − (yP − yT) = xP − yP + yT = 9 − 0 + 6 = 15 yQ − yP = +(xP − xT) ⇒ yQ = yP + (xP − xT) = yP + xP − xT = 0 + 9 − 0 = 9 From the known coordinates we can calculate the area of the brown triangle by calculating the vector product of the two vectors v(TB) and v(TQ): v(TB) = ( xB−xT ; yB−yT ; 0 ) = ( 18−0 ; 0−6 ; 0 ) = ( 18 ; −6 ; 0 ) v(TQ) = ( xQ−xT ; yQ−yT ; 0 ) = ( 15−0 ; 9−6 ; 0 ) = ( 15 ; 3 ; 0 ) A(PTQ) = (1/2) * | v(TP) x v(TQ) | = (1/2) * | ( 9 ; −6 ; 0 ) x (15 ; 3 ; 0) | = (1/2) * | ( 0 ; 0 ; 18*3−15*(−6) ) | = (1/2) * | ( 0 ; 0 ; 54+90) | = (1/2) * | ( 0 ; 0 ; 144) | = (1/2) * 144 = 72 Best regards from Germany
1/ AP=PB=9 and the two triangles TAP and PCQ are congruent so, QC=AP= 9 2/ Area of TQB=Area of quadrilateral ATQB- Area of TAB =Area of(TAP+TPQ+QPB) - area of TAB) =1/2(6x9+117+9x9) -54 126-54= 72 sq units😅😅😅
Using a Plane (XX ; YY) Carthesian Frame of Reference : 01) Point T = (0 ; 6) 02) Point B = (18 ; 0) 03) Point Q = (15 ; 9) 04) d (T , B) = sqrt(360) lin un 05) d (T , Q) = sqrt(234) lin un. NOTE : Line TQ is the Diagonal of Blue Square with Side = sqrt(117) lin un. Diagonal Formula = sqrt(2) * Side. 06) d (B , Q) = sqrt(90) 07) Using Heron's Formula : 08) Brown Triangle Area = 72 sq un Therefore, OUR ANSWER : Brown Triangle Area equal 72 Square Units.
The answer is 72 units squared. Golly I think that at the 5:00 mark this shows that the ASA postulate works for vertical angles and also require a common side, P. Correct??? The other side just simply follow simple logic. Easier than it looks right???
I used Heron's formula. The diagonal of a square is the side length multiplied by sqrt(2). So TQ is sqrt(117)*sqrt(2) = sqrt(234). Sides TB and QB were found the same way you did. Once I had all three side lengths, just pop 'em in to Heron's formula and Bob's yer uncle.
Did it the same way. I think it turned out less messy than the solution PreMath gave. It turns out the terms inside the square root were in the form (a+b)(a-b)(c+d)(c-d)/4 so easy to calculate (a² - b²)(c²-d²)/4 = 5184 = 72²
@ 4:38 if I cut two slices of bread from the same loaf and flipped one for grilled Ham and Cheese , one would think my sandwich is incongruent. ...but no it's not! So harmony at the dinner table is maintained. 🙂
Let me name the Square as A B C D.From B draw a perpendicular line to meet at E of the line CE so that the angle BEC is Rt.Angle. BE is given as 6 units. EC we don't know. The line EC extended to meet at F. So that the perpendicular dropped from D meet EC extension at F. We have the area of Square ABCD as 117 Sq. Units.It means AB BC CD DA all are of length √117. NowTake the Triangle BCE you have BE as 6 units. BCas √117. So EC is =√117^2-36.Which is 117-36=81.√81. =9 units. EC is 9 so EC=CF=9 units. To find the area of BDF BD is known as √√234. DF is 6 given. BF we can find out . Take the triangle BEF whose sides are BE as 6 units EF as 9+9as 18 units .So BF^2 =6^2+18^2. So √6^2+18^2. Contd
The answer can be found easily without adding extra lines. I'll explain starting at 4:30 where AP and PB are found = 9 and CQ is also found to be 9. Triangle TAP area is 6*9/2 = 27. Triangle PQB area is 9*9/2 = 40.5. With the square area being 117, the total area in square and triangles is 117 + 27 + 40.5 = 184.5. The brown triangle area can be found by subtracting triangles TAB and TSQ from this total. TAB area = 6*18/2 = 54. TSQ is half the square area (117/2 = 58.5). 184.5 - 54 - 58.5 = 72.
You still need the extra lines.
@@d-8664 Lines DT and DQ aren't needed for the approach I described.
Nice. But still... It does not require that much math to solve.
Let's start with a few observations:
1. We are looking for the area of TBQ. And that's the area of TPQ (half of TPQS) + area of PBQ - area of TPB.
2. Area of TPB = half the area of TAB = area TAP.
3. You do need the lenght of AP to get the area of TAP. But you don't need the side of TPQS. It's simply sqrt of TPQS (117) - TA^2 (36) = sqrt of 81 = 9.
4. QC = AP => area PBQ = 1/2 AP^2
Therefore, area TBQ = 117/2 + 9^2/2 - 6x9/2 = (117 + (9x9 - 6x9))/2 = (117+3x9)2 = 144/2 just to show how easily you can do it in your head 🙂
Very helpful step by step. Thanks.
TQ = 3sqrt(26)
TP = 3sqrt(13)
AP^2+36 = 117 by Pythagerean Theorem
4AP^2 +36 = TB^2
cos(BTA) = 6/TB
cos(PTA) = 6/TP
angle BTP = angle BTA - angle PTA
angle BTP + angle QTB + angle STQ = 90 degrees
angle BTP + angle QTB = 45 degrees
Area = 1/2*TQ*TB*sin(angle QTB) squared units
Here is my version with trigonometry to share with.
Let AP=sqrt(117-36)=9=PB
->AB=9+9=18
TB=sqrt(6^2+18^2)=6*sqrt(10)
sin a=TA/TB=6/(6*sqrt(10))=1/sqrt(10)
Applying sine law to triangle PTB
sin(b)/PB=sin(a)/PT ->sin(b)/9=1/sqrt(10)/(3*sqrt(13))=3/sqrt(130)
cos b=sqrt(1-sin(b)^2)=11/sqrt(130)
sin(
S² = 117cm² --> S= 3√13cm
d²= 2 S² --> d = 3√26cm
a² = S²-6² --> a = 9cm
c² = (2a)²+6² --> c = 3√40cm
α = atan(2a/6)-atan(a/6)
β = 45° - α = 29,7449°
A = ½ c.d.sinβ
A = 72 cm² ( Solved √ )
TP²-6²=AP²→ AP=√(117-36)=9→ AB=2*9=18.
Si R es la proyección ortogonal de Q sobre PQ → Los triángulos TAP y PRQ son congruentes→ QR=AP=9 y PR=TA=6 → Área TQB=ABQ+ATQ-TAB =(18*9/2)+[6*(9+6)/2]-(6*18/2) =81+45-54=72.
Gracias y un saludo.
TQ=Sqrt(117)•Sqrt(2)=3•Sqrt(26)
TB=Sqrt[(6)²+(18)²]=19
Angle PTA=arctan(9/6)=56.3
Angle ATB=arctan(18/6)=71.6
Angle BTP=71.6-56.3=15.3
Angle BTQ=45-15.3=29.7
Area BTQ=½(3•Sqrt(26))•(19)•sin(29.7)
Area=72
This is awesome, many thanks, Sir!
φ = 30° → sin(3φ) = 1; ∎PQST → PQ = QS = ST = PT = 3√13 → QT = PS = 3√26; AP = BP = a
TA = 6; sin(TAP) = 1; APT = δ; ABT = θ; BTP = β; PTA = α → sin(δ) = cos(α); QTB = γ = 3φ/2 - β
a = √(117 - 36) = 9; sin(δ) = 2√13/13 = cos(α) → cos(δ) = 3√13/13 = sin(α); α + β = ϑ
BT = √(36 + 324) = 6√10; sin(θ) = cos(ϑ) = √10/10 → cos(θ) = sin(ϑ) = 3√10)/10
sin(β) = sin(ϑ - α) = sin(ϑ)cos(α)-sin(α)cos(ϑ) = 3√130)/130 → cos(β) = 11√130/130 →
sin(γ) = sin(3φ/2 - β) = sin(3φ/2)cos(β) - sin(β)cos(3φ/2) = (√2/2)(cos(β) - sin(β)) =
(√2/2)(11√130/130 - 3√130/130) = 4√65/65 →
area ∆ BQT = (1/2)sin(γ)(QT)(BT) = (1/2)(4√65/65)(3√26)(6√10) = 72
or:
CQP = APT = δ → sin(δ) = 2√13/13 → cos(δ) = 3√13/13 → tan(δ) = sin(δ)/cos(δ) = 2/3
BQ = k; sin(δ) = 2√13/13 = CP/3√13 → CP = 6 → BC = 3 → k = 3√10; QT = 3√26; BT = 3√26;
QTB = γ → 90 = 324 + 360 - (36√260)cos(γ) → cos(γ) = 7√65/65 →
sin(γ) = 4√65/65 → area ∆ BQT = (1/2)sin(γ)(3√26)(6√10) = 72
Interesting... I ended up using "functions of lines" to find crossing points, and a little Pythagoras too.
-3x + 54 is line BQ
Horizontal from T (not shown) intersects BQ at x = 16, y = 6, call it [M]
Now there are 3 triangles and one rectangle to find area of.
6 * 16 is the rectangle ATCM area
Upper triangle is 1/2 16 * 3 in area
Lower right triangle is 1/2 * 2 * 6
And the big white lower triangle is 1/2 6 * 18
Brown = 6 * 16 + [1/2 2 * 6] - [1/2 6 * 18] + [1/2 3 * 16]
Brown = 96 + 6 - 54 + 24
Brown = 72
and that's that.
I was afraid we were going to get the coordinates, then calculate lengths of each side, then use Heron's formula 😂😂😂
You already had a trapezoid when you put the vertical line in. Then add the triangle on the far right, and subtract the triangle at the bottom left.
Square PQST:
A = s²
117 = s²
s = √117 = 3√13
Triangle ∆TAP:
TA² + AP² = PT²
6² + AP² = (3√13)²
AP² = 117 - 36 = 81
PB = AP = √81 = 9
Triangle ∆TAB:
TA² + AB² = BT²
6² + 18² = BT²
BT² = 36 + 324 = 360
BT = √360 = 6√10
Triangle ∆QST:
QS² + ST² = TQ²
(3√13)² + (3√13)² = TQ²
TQ² = 117 + 117 = 234
TQ = √234 = 3√26
Let ∠BTP = α and ∠QTB = θ. As ∠TPQ = 90° and TP = PQ, ∆TPQ is an isosceles right triangle and α+β = ∠QTP = (180°-90°)/2 = 45°. By the law of cosines:
Triangle ∆BTP:
cos(α) = (TP²+BT²-PB²)/(2(TP)BT)
cos(45°-θ) = (117+360-81)/(2(3√13)6√10)
cos(45°-θ) = 396/36√130 = 11/√130
cos(45°)cos(θ) + sin(45°)sin(θ) = 11/√130
cos(θ)/√2 + sin(θ)/√2 = 11/√130
cos(θ) + sin(θ) = 11/√65
√(1-sin²(θ)) + sin(θ) = 11/√65
√(1-sin²(θ)) = 11/√65 - sin(θ)
1 - sin²(θ) = 121/65 - 22sin(θ)/√65 + sin²(θ)
2sin²(θ) - 22sin(θ)/√65 + 56/65 = 0
sin²(θ) - 11sin(θ)/√65 + 28/65 = 0
√65sin²(θ) - 11sin(θ) + 28/√65 = 0
√65u² - 11u + 28/√65 = 0 --- u = sin(θ)
u = [-(-11)±√((-11)²-4(√65)(28/√65))]/2(√65)
u = (11±√(121-112))/2√65
u = (11±3)/2√65 = (5.5±1.5)/√65
u = 7/√65 | u = 4/√65
sin(θ) = 7/√65 | sin(θ) = 4/√65
θ = sin⁻¹(7/√65) ≈ 60.26° ❌ θ ≤ 45°
θ = sin⁻¹(4/√65) ≈ 29.74° ✓
Brown triangle ∆QTB:
A = QT(TB)sin(θ)/2
A = 3√26(6√10)(4/√65)/2
A = 3√2(6√2)2 = 18(4) = 72 sq units
First draw a vertical line from Pt Q to the x-axis (Pt X). Now there are 2 triangles beneath square SQPT. One is 6x9. Ther other is 9x6. That means there is a 3rd triangle on the right QBX with a height of 9 and a base of 3.
The area of the Trapazoid TAXQ will be = 1/2 *(6+9)*(9+6)=112.5.
The area of triangle QBX = 1/2 *3*9 =13.5. The sum of those 2 areas is 126.
The area of triangle TAB = 1/2 *6*18 =54
Subtract these 2 areas = 126-54 = 72
Easy problem after you realize Triangles TAP and PQX are similar.
TP = sq.rt. 117.
AP^2 = 117 - 36 = 81.
AP = 9.
Tan ATP = 9 /6 = 1.5.
ATP = 56.310 degrees.
Tan ATB = 18 / 6 = 3.
ATB = 71.565 degrees.
Therefore PTB = 71.565 - 56.310 = 15.255 degrees.
From square, PTQ = 45 degrees.
Therefore BTQ = 45 - 15.255 = 29.745 degrees.
TP = sq.rt. 117.
Diagonal TQ = sq.rt. 234 = 15.297.
TB = sq.rt. of 6^2 + 18^2 = 18.974.
Then brown area = 1/2 x TQ x TB x sin BTQ.
1/2 x 15.297 x 18.974 x sin 29.745.
72.
In triangle TAP: AP^2 = TP^2 - AT^2 = 117 - 36 = 81, so AP = PB = 9 and AB = 18. In triangle TAB: TB^2 = TA^2 + AB^2 = 36 + 324 = 360, so TB = 6.sqrt(10).
Now let's use an orthonormal center A and first axis (AB). We have T(0; 6) and B(18; 0), VectorTB(18; -6). The equation of (TP) is (-6).(x) - (18).(y-6) = 0
or x +3.y -18 = 0.
We have now VectorPQ(6; 9) and so Q(9+6; 0+9) or Q(15; 9). The distance from Q to (TP) is abs(15 +3.9 -18)/sqrt((1)^2 + (3)^2) = 24/sqrt(10)
The area of the green triangle is then (1/2). TB .distance from Q to (TB) = (1/2).(6.sqrt(10)).(24/sqrt(10)) =72.
Let's find the area:
.
..
...
....
.....
First of all we calculate the side length s of the blue square from its given area:
A = s²
117 = s²
⇒ s = √117 = 3√13
The triangle APT is a right triangle, so we can apply the Pythagorean theorem:
PT² = AP² + AT² ⇒ AP² = PT² − AT² = s² − AT² = 117 − 6² = 117 − 36 = 81 ⇒ AP = √81 = 9
Let A be the center of the coordinate system and let B be located on the x-axis. Then we know the coordinates of two of the three corners of the brown triangle:
B: ( 18 ; 0 )
T: ( 0 ; 6 )
Now we can calculate the coordinates of the third point. Since PT and PQ are perpendicular to each other and have the same length, we can conclude:
xQ − xP = −(yP − yT) ⇒ xQ = xP − (yP − yT) = xP − yP + yT = 9 − 0 + 6 = 15
yQ − yP = +(xP − xT) ⇒ yQ = yP + (xP − xT) = yP + xP − xT = 0 + 9 − 0 = 9
From the known coordinates we can calculate the area of the brown triangle by calculating the vector product of the two vectors v(TB) and v(TQ):
v(TB) = ( xB−xT ; yB−yT ; 0 ) = ( 18−0 ; 0−6 ; 0 ) = ( 18 ; −6 ; 0 )
v(TQ) = ( xQ−xT ; yQ−yT ; 0 ) = ( 15−0 ; 9−6 ; 0 ) = ( 15 ; 3 ; 0 )
A(PTQ)
= (1/2) * | v(TP) x v(TQ) |
= (1/2) * | ( 9 ; −6 ; 0 ) x (15 ; 3 ; 0) |
= (1/2) * | ( 0 ; 0 ; 18*3−15*(−6) ) |
= (1/2) * | ( 0 ; 0 ; 54+90) |
= (1/2) * | ( 0 ; 0 ; 144) |
= (1/2) * 144 = 72
Best regards from Germany
1/ AP=PB=9
and the two triangles TAP and PCQ are congruent so, QC=AP= 9
2/ Area of TQB=Area of quadrilateral ATQB- Area of TAB
=Area of(TAP+TPQ+QPB) - area of TAB)
=1/2(6x9+117+9x9) -54
126-54= 72 sq units😅😅😅
PTQB=(PTQ+QPB)-PTB=(117/2)+(81/2)-(54/2)=72
Using a Plane (XX ; YY) Carthesian Frame of Reference :
01) Point T = (0 ; 6)
02) Point B = (18 ; 0)
03) Point Q = (15 ; 9)
04) d (T , B) = sqrt(360) lin un
05) d (T , Q) = sqrt(234) lin un. NOTE : Line TQ is the Diagonal of Blue Square with Side = sqrt(117) lin un. Diagonal Formula = sqrt(2) * Side.
06) d (B , Q) = sqrt(90)
07) Using Heron's Formula :
08) Brown Triangle Area = 72 sq un
Therefore,
OUR ANSWER :
Brown Triangle Area equal 72 Square Units.
The answer is 72 units squared. Golly I think that at the 5:00 mark this shows that the ASA postulate works for vertical angles and also require a common side, P. Correct??? The other side just simply follow simple logic. Easier than it looks right???
I lati sono TB= √36+324=√360,TQ=√2√117=√234,e l'angolo compreso è 45+arctg9/6-arctg18/6..poi A=(1/2)absinα=72..
(TQB)=(TQP)+(QPB)-(TPB),also i used a bit trigonometry to find the areas of QPB and TPB triangles and therefore the result is 72.
So BF is √360. Now we have all yhe three sides as √234. And 6 and √360. Using the formula as s(s-a)(s-b)(s-c) .square root of that gives you the area.
Thank you!
you can find the lengths of BT , QT AND QB ..then use heron's formula
The idea of construction is 👍🏻
I used Heron's formula. The diagonal of a square is the side length multiplied by sqrt(2). So TQ is sqrt(117)*sqrt(2) = sqrt(234). Sides TB and QB were found the same way you did. Once I had all three side lengths, just pop 'em in to Heron's formula and Bob's yer uncle.
Did it the same way. I think it turned out less messy than the solution PreMath gave. It turns out the terms inside the square root were in the form (a+b)(a-b)(c+d)(c-d)/4 so easy to calculate (a² - b²)(c²-d²)/4 = 5184 = 72²
S=72🔥
I got same answer but used herons formula which was messy.
Clumsy puzzle leads to a Clumsy solution 😢
72
90 square unit ?
Sancs
@ 4:38 if I cut two slices of bread from the same loaf and flipped one for grilled Ham and Cheese , one would think my sandwich is incongruent. ...but no it's not! So harmony at the dinner table is maintained. 🙂
😊😊😊
Let me name the Square as A B C D.From B draw a perpendicular line to meet at E of the line CE so that the angle BEC is Rt.Angle. BE is given as 6 units. EC we don't know. The line EC extended to meet at F. So that the perpendicular dropped from D meet EC extension at F. We have the area of Square ABCD as 117 Sq. Units.It means AB BC CD DA all are of length √117. NowTake the Triangle BCE you have BE as 6 units. BCas √117. So EC is =√117^2-36.Which is 117-36=81.√81. =9 units. EC is 9 so EC=CF=9 units.
To find the area of BDF BD is known as √√234. DF is 6 given. BF we can find out . Take the triangle BEF whose sides are BE as 6 units EF as 9+9as 18 units .So BF^2 =6^2+18^2. So √6^2+18^2. Contd