I did it differently, first assigning lengths along bottom. Extend & add lines to form rectangle with P & Q as corners. Lower left point is E & upper right is F. Let QC = 4K, then CD = 8K, DE = 3K. PE (side of square) = 8K. X = (12/15)(8K) = (32/5)K. Solve for K using triangle PEQ to get K = 15/17. X = (15/17)(32/5) = 96/17.
∠BPQ = ∠DQP (alternate interior angles formed by transversal PQ and parallel lines PB and QD) Therefore, right triangles APS, BPT, CQT, DQS are similar, with diagonals of length 3, 11, 4, 12 respectively. If we let AS = 3a, then we get BT = 11a, CT = 4a, DS = 12a If we let AP = 3b, then we get BP = 11b, CQ = 4b, DQ = 12b Sides of square: AD = AS + DS = 3a + 12a = 15a BC = BT + CT = 11a + 4a = 15a AB = BP − AP = 11b − 3b = 8b CD = DQ − CQ = 12b − 4b = 8b 8b = 15a → b = 15a/8 Using Pythagorean theorem in right triangle APS we get AS² + AP² = PS² (3a)² + (3b)² = 3² [divide by 9] a² + b² = 1 a² + (15a/8)² = 1 a² + 225a²/64 = 1 289a²/64 = 1 a² = 64/289 a = 8/17 x = DS = 12a = 12(8/17) = 96/17 *x = 96/17*
Месяц назад+1
from Morocco thank you for your clear wonderful explanations
X = 96/17. At the 6:00 mark, I have notcied that if complementary angles are involved and both triangles are HL similar, they BOTH have a pair of sudes that are the LAST two letters in the triangle subtended for alpha angle the common pair is set equal to the first two letters of the alpha similar triangles and for beta, the similar sides are set equal the first and last letters of beta similar triangles. Let me know if that is a sufficient explanation for a problem that I might want to practice on!!!
Let s be the side length of square ABCD. Let M be the point on DA where TM is perpendicular to DA and BC and parallel to AB and CD. Extend CD to N and drop a perpendicular from P to N. As ∠PNQ = ∠SMT = 90° and ∠MTS and ∠NQP are corresponding angles and thus congruent, ∆PNQ and ∆SMT are similar triangles. By the same token, ∆SDQ is also similar to ∆SMT and ∆PNQ. PN/QP = SM/TS s/15 = SM/8 SM = 8s/15 SM² + MT² = TS² (8s/15)² + s² = 8² 64s²/225 + s² = 64 289s²/225 = 64 s² = 64(225/289) = 8²(15²)/17² s = √(8²(15²)/17²) = 120/17 SD/QS = PN/QP x/12 = s/15 12s = 15x x = 12s/15 = 4s/5 x = 4(120/17)/5 = 4(24)/17 x = 96/17 ≈ 5.647 units
Seems overly complicated. Let t=angle PQD and draw a perpendicular to AD through T with foot U. Then 3 sin(t) + x = 15 sin(t) so x=12 sin(t) and 15 sin(t)=8cos(t). So x=12 sin(arctan(8/15))=96/17. You can get away without the trig. In the end this comes down to PRQ ~ SDQ ~ SUT ~ TCQ, where R is the foot of the perpendicular to DC through P. The rest is Pythagoras and PR=DQ-CQ.
Extremely complicated, video method See below an easier method: s = side of square a = horizontal proyection of segment PQ tanα = s/a = 1/(1+7/8) --> α=28.0725° x = (8+4) sin α x = 5,647 cm ( Solved √ )
Ya corregí, un signo me equivoqué.. Lo hice de 2 formas diferentes. 5x/4 +(114-x2)/3 =(144-x2) Y tambien realizando ecuaciones cuadraticas de 144=(5x2/4)2 +((144-x2)/3)2 y salieron 2 respuestas 97/17 con la resta y 8,75 con la suma
2nd method because ABCD is square so AB//CD --> SA/SD = SP/SQ ( Thales) SA/x = 3/(4+8) SA=x/4 AD = AS + SD = x+x/4 = 5x/4 Beside ABCD is square so CD=AD= 5x/4 we have TC//SD so CD/CQ = TS/TQ ( Thales ) (5x/4)/CQ = 8/4 CQ = (5x/4)(4/8) = 5x/8 So DQ = DC +CQ = 5x/4 +5x/8 = 15x/8 Because right Triangle SDQ at D so SD^2+DQ^2 = SQ^2 ( Pythagoras) x^2 +(15x/8)^2 = 12^2 x^2+(225/64)x^2 = 144 (289/64)x^2=144 x^2=144.(289/64) x=sqrt (144.(289/64))=96/17 Thank you, love from Vietnam.
I did it differently, first assigning lengths along bottom. Extend & add lines to form rectangle with P & Q as corners. Lower left point is E & upper right is F.
Let QC = 4K, then CD = 8K, DE = 3K. PE (side of square) = 8K. X = (12/15)(8K) = (32/5)K.
Solve for K using triangle PEQ to get K = 15/17. X = (15/17)(32/5) = 96/17.
∠BPQ = ∠DQP (alternate interior angles formed by transversal PQ and parallel lines PB and QD)
Therefore, right triangles APS, BPT, CQT, DQS are similar, with diagonals of length 3, 11, 4, 12 respectively.
If we let AS = 3a, then we get BT = 11a, CT = 4a, DS = 12a
If we let AP = 3b, then we get BP = 11b, CQ = 4b, DQ = 12b
Sides of square:
AD = AS + DS = 3a + 12a = 15a
BC = BT + CT = 11a + 4a = 15a
AB = BP − AP = 11b − 3b = 8b
CD = DQ − CQ = 12b − 4b = 8b
8b = 15a → b = 15a/8
Using Pythagorean theorem in right triangle APS we get
AS² + AP² = PS²
(3a)² + (3b)² = 3² [divide by 9]
a² + b² = 1
a² + (15a/8)² = 1
a² + 225a²/64 = 1
289a²/64 = 1
a² = 64/289
a = 8/17
x = DS = 12a = 12(8/17) = 96/17
*x = 96/17*
from Morocco thank you for your clear wonderful explanations
AB=a→ a²+(15a/8)²=15²→ a=120/17 → (a-x)/3=x/12→ [(120/17)-x]/3=x/12→ x=96/17=5,64705...
Gracias y un saludo.
X = 96/17. At the 6:00 mark, I have notcied that if complementary angles are involved and both triangles are HL similar, they BOTH have a pair of sudes that are the LAST two letters in the triangle subtended for alpha angle the common pair is set equal to the first two letters of the alpha similar triangles and for beta, the similar sides are set equal the first and last letters of beta similar triangles. Let me know if that is a sufficient explanation for a problem that I might want to practice on!!!
(l-x)/3=TC/4=√(64-l^2)/8..TC=x-√(64-l^2)...l=lato del quadrato ..risulta 4x^2/9=64-l^2,l=5x/4..per cui x=96/17
Let s be the side length of square ABCD. Let M be the point on DA where TM is perpendicular to DA and BC and parallel to AB and CD. Extend CD to N and drop a perpendicular from P to N.
As ∠PNQ = ∠SMT = 90° and ∠MTS and ∠NQP are corresponding angles and thus congruent, ∆PNQ and ∆SMT are similar triangles. By the same token, ∆SDQ is also similar to ∆SMT and ∆PNQ.
PN/QP = SM/TS
s/15 = SM/8
SM = 8s/15
SM² + MT² = TS²
(8s/15)² + s² = 8²
64s²/225 + s² = 64
289s²/225 = 64
s² = 64(225/289) = 8²(15²)/17²
s = √(8²(15²)/17²) = 120/17
SD/QS = PN/QP
x/12 = s/15
12s = 15x
x = 12s/15 = 4s/5
x = 4(120/17)/5 = 4(24)/17
x = 96/17 ≈ 5.647 units
tanα = 1/(1+7/8) --> α=28.0725°
x = (8+4) sin α
x = 5,647 cm ( Solved √ )
!!! EASIER !!!
Suggestion: using red over black isn't easy to see and read.
PS=PQ/5，PS/SQ=AS/x=1/4
AS/CT=3/4，CT/x=4AS/3x=1/3
CT=x/3，AS=x/4
CD=5x/4，CQ=5x/8
（15x/8）²+x²=144
289x²/64=144
x=12*8/17=96/17
AD² + (PA + DC + CQ)² = (PS + ST + TQ)²
a = AD = DC
a² + [(3/8)a + a + (4/8)a]² = (3 + 8 + 4)²
Seems overly complicated. Let t=angle PQD and draw a perpendicular to AD through T with foot U. Then 3 sin(t) + x = 15 sin(t) so x=12 sin(t) and 15 sin(t)=8cos(t). So x=12 sin(arctan(8/15))=96/17.
You can get away without the trig. In the end this comes down to PRQ ~ SDQ ~ SUT ~ TCQ, where R is the foot of the perpendicular to DC through P. The rest is Pythagoras and PR=DQ-CQ.
Extremely complicated, video method
See below an easier method:
s = side of square
a = horizontal proyection of segment PQ
tanα = s/a = 1/(1+7/8) --> α=28.0725°
x = (8+4) sin α
x = 5,647 cm ( Solved √ )
Extremely easy:
tanα = 1/(1+7/8) --> α=28.0725°
x = (8+4) sin α
x = 5,647 cm ( Solved √ )
3/(8+4)=AS/x 1/4=AS/x AS=x/4
4/(8+4)=TC/x 1/3=TC/x TC=x/3
x+x/4=5x/4 5x/4-x/4-x/3=2x/3
(2x/3)²+(5x/4)²=8² 4x²/9+25x²/16=64 64x²/144+225x²/144=64
289x²=144*64 (17x)²=12*12*8*8=96² 17x=96 x=96/17
Ya corregí, un signo me equivoqué..
Lo hice de 2 formas diferentes.
5x/4 +(114-x2)/3 =(144-x2)
Y tambien realizando ecuaciones cuadraticas de 144=(5x2/4)2 +((144-x2)/3)2 y salieron 2 respuestas 97/17 con la resta y 8,75 con la suma
2nd method
because ABCD is square so AB//CD
--> SA/SD = SP/SQ ( Thales)
SA/x = 3/(4+8)
SA=x/4
AD = AS + SD = x+x/4 = 5x/4
Beside ABCD is square so CD=AD= 5x/4
we have TC//SD so
CD/CQ = TS/TQ ( Thales )
(5x/4)/CQ = 8/4
CQ = (5x/4)(4/8) = 5x/8
So DQ = DC +CQ = 5x/4 +5x/8 = 15x/8
Because right Triangle SDQ at D so
SD^2+DQ^2 = SQ^2 ( Pythagoras)
x^2 +(15x/8)^2 = 12^2
x^2+(225/64)x^2 = 144
(289/64)x^2=144
x^2=144.(289/64)
x=sqrt (144.(289/64))=96/17
Thank you, love from Vietnam.
tanα = 1/(1+7/8) --> α=28.0725°
x = (8+4) sin α
x = 5,647 cm ( Solved √ )
3rd and easier method
(3)^2(8)^2={9+64}=73 360°ABCD/73x 4.68ABCDX 4.4^17 2^2.2^2^17^1 1^1.1^2^1^1 21 (ABCDX ➖ 2ABCDX+1).
X=96/17.
Difficult
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Oxford new syllabus
D1, D2, D3 Math solutions are availabe.