Japanese College Test Geometry Problem | Find the shaded area of the triangle

[@pk2712]

Draw an altitude from P to point R on the base QC of triangle CPQ . Now draw another altitude from point P to S on line QC . It is then obvious that triangles PQR and AQS are similar . Since these triangles are similar , we can write : PR/AS=PQ/AQ ---- (1) AS in this equation is simple to solve from right triangle BAS : AS=BAsin45=10(sqrt(2)/2)=5sqrt(2) . AQ=PQ+AP=3AP+AP=4AP and AQ=3AP . Substituting all these values into equation (1) gives : PR/(5sqrt(2))=(3AP/4AP)=3/4 --- (2) Multiply equation through by 5sqrt(2) , and you get the height of triangle CPQ : h=PR=15sqrt(2)/4 . We know that the base of triangle CPQ is b=QC=8 . We then can calculate the area of triangle CPQ as : A=1/2bh=1/2(8)(15sqrt(2)/4)=15sqrt(2) .

[@ناصريناصر-س4ب]

Let H and E be the perpendicular projections of points A and P on BC respectively. We have AH = 5√2 and according to Thales' theorem we have PE/AH=PQ/AQ = 3/4. From this, PE = (3/4)*(5√2) = (15√2)/4. Therefore, the area of triangle PQC is equal to (8*15√2/4)/2 = 15√2.

[@matematicafacilcomprof.jua9231]

I did this way and I was about to write the same comment. It is easier and faster.

[@五十嵐特許事務所]

Connecting points A and C creates ⊿ABC. If BQ＝a, then the area X of ⊿ABC is X=AB×BC×sin45°/2＝5(a+8)/√2. If the area of ​​⊿AQC is [AQC], then [AQC]=(8/(8+a))[ABC]=8X/(8+a)=20√2. ∴[PQC]=(3/4)[AQC]=(3/4)×20√2=15√2.

[@imetroangola17]

*Solução:* Seja ∠AQB = α. Usando a lei do seno no ∆ AQB: sen 45°/AQ = sen α/AB (√2/2)/4AP = sen α/10 *sen α = (5√2)/4AP* Por outro lado, [PCQ] = [PQ × QC × sen (180° -α)]/2 [PCQ] = [3AP × 8 × sen α)]/2 [PCQ] = 3AP × 4 × 5√2/4AP *_[PCQ] = 15√2 U. Q._*

[@kateknowles8055]

Thank you for the geometry problem and the solution. I only got this far with this one Let X = [CPQ] [CAQ] = 4 X / 3 because AQ = 4 AP If BC =a [ABC] = (a/8 )(4X/3) because QC=8 AC is expressible in terms of a : cos (45) = ( a.a+ 10.10 - AC.AC )/ (20a) AC.AC = a.a +10.10 - 20a (1/root(2)) = 100 +a.a - 10a(root(2)) It would be too convenient if BC were 10 root(2) ? [ABC] =50? (10root(2)/8)(4/3)X =50 X = 50/10 (8)(3) /(4root(2)) = 15.root(2)? There seems to be an information gap , when that happens, wrong assumptions are easily made. It is called guessing. Assumptions might be insight, only if they lead to proven answers. Angle PCQ looks like 45 degrees but it probably is not . Is it angle ACQ that is 45 degrees? Now I am watching the video and reading other comments. (And I came back afterwards, and edited🤭)

[@fhffhff]

π-arcsin(5√2/AP) 8/sin/_QPC= 3/4AP/sin(180°-/_QPC-arcsin(5√2/AP)) arctg(5√2/AP/(3/32AP-√(1-50/AP²)))=/_QPC S∆=3AP*5√2/AP=15 √2

[@murdock5537]

φ=30° → sin⁡(3φ) = 1; ∆ ABC → ABC = 3φ/2; AB = 10; BC = BQ + CQ = BQ + 8 → CQ = CM + QM → sin⁡(BMA) = 1; AQ = AP + QP = k + 3k; AM = h = 5√2 area ∆ AQC = 4a → 4(5√2) = 4a → 3a = 15√2 = area ∆ CPQ

[@nexen1041]

That is absolutely amazing 👍

[@santiagoarosam430]

Las proyecciones ortogonales respectivas de A y P sobre QC son D y H---> PH=3*AD/4 ---> AD=AB/√2 =10/√2=5√2---> PH=15√2/4---> Área CPQ =PH*QC/2 =(15√2/4)*8/2 =15√2 u². Gracias y saludos

[@oscarcastaneda5310]

Red height =(3/4) height from A. and height from A is 10/sqrt(2) = 5sqrt(2) so Red Area is (1/2)(3/4)(5sqrt(2))(8) = 15sqrt(2).

[@matematicafacilcomprof.jua9231]

I did it the same way. Easier and faster. Greetings from Brazil.

[@giuseppemalaguti435]

h=(3/4)10sin45=30√2/8=...= 15√2/4...Ablue=(8/2)15√2/4=15√2

[@marcelomatos2584]

Height of the ΔABC is 15√︎2, |ΔACQ|=20√︎2, as PQ=3AP ⇒︎ |ΔCPQ|= ¾.|ΔACQ| so ︎|ΔCPQ|= ¾. 20√︎2= 15√︎2. 😉

[@sakurayayoi-p2r]

別解法高さが等しい三角形は底辺の比で分割される。したがって８＊√２*10/2の１/４が求める面積となる。

[@稲次将人]

1/2×8×5√2×3/4=15√2

[@roniakter5957]

Sir,where do you live

[@nenetstree914]

15*(2^0.5)

[@JerrySu-k3n]

this is a pretty ez one

[@RealQinnMalloryu4]

(10)^2 (8)^2={100+64}=164 {45°A45°B+90°C}=180°ABC/164=1.16 2^3 (ABC ➖ 3ABC+2).