Using analytical (or coordinate) geometry with origin D(0,0), and points C(10,0), B(10,5), A(0,5), P(4,5), Line DB is y=(0.5)x and line PC is y = -(5/6)x +25/3. We solve these two equations to find that the lines intersect at Q((25/4), (25/8)). Point M (25/4, 5) is where the perpendicular from Q intersects line AB. The shaded area is that of trapezium AMQD minus that of triangle PMQ: (1/2)(5 + (15/8))(25/4) - (1/2)(15/8)(9/4) or 155/8 square units.
▲BPQ~▲CDQ, BP/CD=PQ/QD=6/10=3/5. Достраиваем в т. E ▲DEQ~▲BCQ - общие отрезки, общее подобие. Тогда BC/DE=3/5=5/(5+x)⇔3(5+x)=5*5=25⇔5+x=25/3=8¹/₃⇔x=8¹/₃-5=3¹/₃. AB=DC и складывается из высот треугольников как 3y+5y=8y=9, откуда y=9/8=1¹/₈, h=5y=5⁵/₈. S(DEQ)=5⁵/₈*8¹/₃/2=5⁵/₈*4¹/₆. S(AEP)=3¹/₃*4/2=3¹/₃*2=6²/₃. Отнимем от предыдущей. 45/8*(25/6)-20/3=(45*25)/(8*6)-6²/₃. 45*25=40*25+5*25=125+100*10=1125. 1125/48=23²¹/₄₈. 23²¹/₄₈-6²/₃=17²¹/₄₈-²/₃. Общий множитель 48*3=150-6=144. 21*3=63, 48*2=100-4=96, 96-63=36-3=33. 17-33/144. 144-33=111. 16¹¹¹/₁₄₄.
Thank you for the problem and the solution. It looks easy but I did not find the easier way on my own. [ABCD] = 5.(6+4) =50 Triangle ABD is half of 50. [ABD] =25 [CDB] =25 = [DQC] +[ CQB] Triangle PBC =(1/2)5.6 [PBC] = 15 so [APCD ] =50-15 [APCD] = 35 PC = sqrt(61) by using Pythagoras' theorem in triangle PBC. [PBC] and [CBD] overlap . BD = sqrt(125) by using Pythagoras' theorem in triangle BCD (or ABD) The area [ APQD] is shaded red. It is 25 - [PBQ] The similar triangles whose areas will be useful here are BQP and DQC DC is 10 and BP is 6 so PQ/QC = BQ/QD = 6/10 =3/5 QC is (5/8)PC QD is 5/8 BD S, the semiperimeter of triangle DQC is (1/2)( QC+QD+CD) = (1/2)(5/8 sqrt(61)+ 5/8sqrt(125)+10) 10.9 approx Using Heron's formula in triangle DQC s - a = (1/2)(- (5/8)sqrt(61) +(5/8 )sqrt(125) +10) 6.05 approx s - b = (1/2)(+(5/8)sqrt(61)- (5/8)sqrt(125) +10) 3.95 approx s - c=(1/2)( +(5/8)sqrt(61) +(5/8)sqrt(125) - 10) 0.935 approx (not easy enough for me to get a right answer in practice so I worked back to here from the video answer) Area [DQC] = sqrt (s.(s - a).( s - b). (s - c) = 125/8 Area [BQP] =(36/100) [DQC] = (9/25 )( 125/8 ) = 45/8 Shaded red area = 25 - 45/8 Shaded red area = 155/8 or 19 +3/8
Pendientes: de CP =6/5 ; de BD =10/5=2 ---> Si BC=b+(5-b)---> 2b=6(5-b)/5---> b=15/8---> Área APQD=ABD-PBQ =5(4+6)/2 -(6*15/8)/2 =155/8 =19,375 u². Gracias y saludos
Here's an easier solution. [ABCD]=50, so [DCB]=25. Since triangles PBQ and QDC are similar, DQ:QB=DC:PB=10:6=5:3, so [QDC]=(5/8)*25=125/8 and [BQC]=(3/8)*25=75/8. Then [QDC] and [PBQ] are in the ratio of the squares of their sides, so [PBQ]=((3/5)^2)*(125/8)=45/8. Finally, [APQD]=[ABD]-[PBQ]=25-(45/8)=155/8.
Conosco tutti gli angoli,con le arctangenti...PQ=a=3√61/8 col teorema dei seni,quindi risulta A=10+(1/2)√41(3√61/8)sin(arctg5/4+arctg5/6)=10+(3/16)√41√61(50/√41√61)=10+150/16=10+75/8=155/8
Let the area of triangle PQB be equal to x, then [PQD]=(5x)/3, [BQC]=(5x)/3, [DQC]=(25x)/9, and we have [APD]=10, from which x+(5x)/3+(5x)/3+(25x)/9+10=50, from which x=45/8, so the area of APQD is equal to 10+(5/3)*(45/8)=155/8
Method using side-area-ratio of trapezium divided by diagonals & equal height triangles: 1. Join DP to form trapezium DCBP and triangle ADP. 2. Trapezium DCBP is divided by its diagonals into 4 trianlges. Let top side be T = 6 and bottom side be B = 4+6 = 10. The ratio area of top triangle : left triangle:right triange: bottom triangle = T^2:TB:TB:B^2 = 36:60:60:100 Ratio area of triangle BDP = top triangle + left triangle = 36 + 60 = 96 3. Base side ratio of triangle ADP: triangle BDP = 4:6. Hence ratio area of triangle ADP = 96 (4/6) = 64 4. Ratio area of shaded region = left triangle + triangle ADP = 60 + 64 = 124 5. Ratio area of rectangle = sum of all triangles = 36 + 60 + 60 + 100 + 64 = 320 Actual area of rectangle = 5 (4 + 6) = 50 Hence actual area : ratio area is 50:320 = 5/32 6. Actual area of shaded region = 124 (5/32) = 155/8. This is a typical exam side-area-ratio MCQ to be solved in 90 seconds.
Using analytical (or coordinate) geometry with origin D(0,0), and points C(10,0), B(10,5), A(0,5), P(4,5), Line DB is y=(0.5)x and line PC is y = -(5/6)x +25/3. We solve these two equations to find that the lines intersect at Q((25/4), (25/8)). Point M (25/4, 5) is where the perpendicular from Q intersects line AB. The shaded area is that of trapezium AMQD minus that of triangle PMQ: (1/2)(5 + (15/8))(25/4) - (1/2)(15/8)(9/4) or 155/8 square units.
▲BPQ~▲CDQ, BP/CD=PQ/QD=6/10=3/5. Достраиваем в т. E ▲DEQ~▲BCQ - общие отрезки, общее подобие. Тогда BC/DE=3/5=5/(5+x)⇔3(5+x)=5*5=25⇔5+x=25/3=8¹/₃⇔x=8¹/₃-5=3¹/₃. AB=DC и складывается из высот треугольников как 3y+5y=8y=9, откуда y=9/8=1¹/₈, h=5y=5⁵/₈. S(DEQ)=5⁵/₈*8¹/₃/2=5⁵/₈*4¹/₆. S(AEP)=3¹/₃*4/2=3¹/₃*2=6²/₃. Отнимем от предыдущей. 45/8*(25/6)-20/3=(45*25)/(8*6)-6²/₃. 45*25=40*25+5*25=125+100*10=1125. 1125/48=23²¹/₄₈. 23²¹/₄₈-6²/₃=17²¹/₄₈-²/₃. Общий множитель 48*3=150-6=144. 21*3=63, 48*2=100-4=96, 96-63=36-3=33. 17-33/144. 144-33=111. 16¹¹¹/₁₄₄.
Почему 2/3?
@AmirgabYT2185 Ой. Точно! 6/10, пардон. Сейчас поправлю.
@AmirgabYT2185 поправил, но где-то снова обсчитался, кажется.
Thank you for the problem and the solution.
It looks easy but I did not find the easier way on my own.
[ABCD] = 5.(6+4) =50 Triangle ABD is half of 50. [ABD] =25 [CDB] =25 = [DQC] +[ CQB]
Triangle PBC =(1/2)5.6 [PBC] = 15 so [APCD ] =50-15 [APCD] = 35 PC = sqrt(61) by using Pythagoras' theorem in triangle PBC.
[PBC] and [CBD] overlap . BD = sqrt(125) by using Pythagoras' theorem in triangle BCD (or ABD)
The area [ APQD] is shaded red. It is 25 - [PBQ]
The similar triangles whose areas will be useful here are BQP and DQC
DC is 10 and BP is 6 so PQ/QC = BQ/QD = 6/10 =3/5
QC is (5/8)PC QD is 5/8 BD
S, the semiperimeter of triangle DQC is (1/2)( QC+QD+CD) = (1/2)(5/8 sqrt(61)+ 5/8sqrt(125)+10) 10.9 approx
Using Heron's formula in triangle DQC s - a = (1/2)(- (5/8)sqrt(61) +(5/8 )sqrt(125) +10) 6.05 approx
s - b = (1/2)(+(5/8)sqrt(61)- (5/8)sqrt(125) +10) 3.95 approx
s - c=(1/2)( +(5/8)sqrt(61) +(5/8)sqrt(125) - 10) 0.935 approx
(not easy enough for me to get a right answer in practice so I worked back to here from the video answer)
Area [DQC] = sqrt (s.(s - a).( s - b). (s - c) = 125/8
Area [BQP] =(36/100) [DQC] = (9/25 )( 125/8 ) = 45/8
Shaded red area = 25 - 45/8
Shaded red area = 155/8 or 19 +3/8
Pendientes: de CP =6/5 ; de BD =10/5=2 ---> Si BC=b+(5-b)---> 2b=6(5-b)/5---> b=15/8---> Área APQD=ABD-PBQ =5(4+6)/2 -(6*15/8)/2 =155/8 =19,375 u².
Gracias y saludos
Here's an easier solution. [ABCD]=50, so [DCB]=25. Since triangles PBQ and QDC are similar, DQ:QB=DC:PB=10:6=5:3, so [QDC]=(5/8)*25=125/8 and [BQC]=(3/8)*25=75/8. Then [QDC] and [PBQ] are in the ratio of the squares of their sides, so [PBQ]=((3/5)^2)*(125/8)=45/8. Finally, [APQD]=[ABD]-[PBQ]=25-(45/8)=155/8.
*Solução:*
É fácil mostrar que os triângulos ∆PBQ e ∆CDQ são semelhantes, logo:
[CDQ]/[PBQ] = (10/6)² = (5/3)² = 25/9. Assim,
[CDQ] = 25/9 × [PBQ].
Note que:
[BCD] - [PBC] =[CDQ] - [PBQ]
10×5/2 - 6×5/2 = (25/9 -1)[PBQ]
10 = 16/9 × [PBQ]
[PBQ] = 45/8. Logo,
[APCD] = [ABD] - [PBQ]
[APCD] = 10×5/2 - 45/8
[APCD] = 200/8 - 45/8
*[APCD] = 155/8.*
Nice and Easy Question : )
We think Alike !
I wonder if there's a solution that uses trigonometry since the tangent of angle ABD = 1/2.
BE=5√5, sin( B)=1/√5. ∆DQC подобен ∆BQP, 6/10=x/(5√5-x). x=15√5/8. S(∆BQP)=0.5*6*15√5/8*sin(B)=45/8=5,625. S=25-5,625=19,375.
Conosco tutti gli angoli,con le arctangenti...PQ=a=3√61/8 col teorema dei seni,quindi risulta A=10+(1/2)√41(3√61/8)sin(arctg5/4+arctg5/6)=10+(3/16)√41√61(50/√41√61)=10+150/16=10+75/8=155/8
Let the area of triangle PQB be equal to x, then [PQD]=(5x)/3, [BQC]=(5x)/3, [DQC]=(25x)/9, and we have [APD]=10, from which x+(5x)/3+(5x)/3+(25x)/9+10=50, from which x=45/8, so the area of APQD is equal to 10+(5/3)*(45/8)=155/8
S=19,375 square units
(4)^2(6)^2(5)^2={16+36+25}=77 {90°A+90°B+90°C+90°D}=360°ABCD/77=50.10ABCD 5^10.10 2^3 (ABCD ➖ 3ABCD+2).
Solution:
Triangle PQB is similar to triangle DCQ because all angles are equal.
h = height of triangle PQB,
i = height of triangle DCQ.
(1) h+i = 5
Due to the similarity:
(2) i/(4+6) = h/6 |*10 ⟹ (2a) i = h*10/6 |in (1) ⟹
(1a) h+h*10/6 = 5 ⟹
(1b) h*(1+10/6) = 5 ⟹
(1c) h*16/6 = 5 |*6/16 ⟹
(1d) h = 5*6/16 = 30/16 = 15/8 ⟹
Colored area = 5*10/2-6/2*15/8 = 25-45/8 = (200-45)/8 = 155/8 = 19.375
PM=⑥、ＭＱ=⑤、MB=⑩ 1/2×⑯×⑤
155/8
Method using side-area-ratio of trapezium divided by diagonals & equal height triangles:
1. Join DP to form trapezium DCBP and triangle ADP.
2. Trapezium DCBP is divided by its diagonals into 4 trianlges. Let top side be T = 6 and bottom side be B = 4+6 = 10.
The ratio area of top triangle : left triangle:right triange: bottom triangle = T^2:TB:TB:B^2 = 36:60:60:100
Ratio area of triangle BDP = top triangle + left triangle = 36 + 60 = 96
3. Base side ratio of triangle ADP: triangle BDP = 4:6. Hence ratio area of triangle ADP = 96 (4/6) = 64
4. Ratio area of shaded region = left triangle + triangle ADP = 60 + 64 = 124
5. Ratio area of rectangle = sum of all triangles = 36 + 60 + 60 + 100 + 64 = 320
Actual area of rectangle = 5 (4 + 6) = 50
Hence actual area : ratio area is 50:320 = 5/32
6. Actual area of shaded region = 124 (5/32) = 155/8.
This is a typical exam side-area-ratio MCQ to be solved in 90 seconds.