A Very Nice Geometry Problem | You should be able to solve this!

[@Ramkabharosa]

Area(ΔBPQ) = d = √{(a+b+c)² - 4.a.b} = √{(9+5+4)² - 4.(9).(5)} = √{2².(9)² - 4.(45} = √{4.(81 - 45)} = √{4.(36)} = 2.(6) =12. Here a = area(ΔABP) (a side = width of rectangle), b = area(ΔBCQ) (a side = height of rectangle), & c = area(ΔDPQ)(the other rt Δ).

[@jimlocke9320]

Let AB = CD = b (base of rectangle), BC = AD = h (height of rectangle), AP = m, therefore PD = h - m, CQ = n, therefore DQ = b - n. Area ΔABP = (1/2)(AB)(AP) = (1/2)(b)(m) but is given as 9, so (1/2)bm = 9 and bm = 18. Area ΔBCQ = (1/2)(CQ)(BC) = (1/2)(n)(h) but is given as 5, so (1/2)nh = 5 and nh = 10. Area ΔDPQ = (1/2)(DQ)(PD) = (1/2)(b - n)(h - m) but is given as 4 so (1/2)(b - n)(h - m) = 4 and, expanding, bh -bm - nh + mn = 8. Substitute 18 for bm and 10 for nh and simplify: bh -18 - 10 + mn = 8, bh + mn = 36. Multiply the first 2 equations together: bmnh = (18)(10) = 180 and mn = 180/bh. In bh + mn = 36, replace mn by 180/bh: bh + 180/bh = 36 Let bh = x: x + 180/x = 36. multiply both sides by x and consolidate on left side: x² - 36x + 180 = 0. Note that x = bh is the area of the square. We have the same equation that Math Booster found at 8:20. Follow the steps that Math Booster took starting at 8:20 to solve for the area of the rectangle and then deduct the given areas of 3 triangles to find that the area of ΔPBQ = 12. Note that we can not solve for the base or the height of the rectangle. We can only solve for the product of base and height. Base and height can have any pair of positive values as long as their product is 30.

[@davidellis1929]

A quick check for simple dimensions that satisfy the given conditions yields AB=6, AP=3, PD=2, DQ=4, QC=2, BC=5. Then [ABCD]=30, and [BPQ]=30-9-4-5=12. This is not rigorous; a complete and proper solution does seem to require setting up and solving a quadratic equation whose variable is [ABCD], as shown in the video.

[@prossvay8744]

Let a is Length and b is Width Area of rectangle=ab AP=18/a and CQ=10/b DP=b-18/a and DP=a-10/b Area of the DPQ=1/2(b-18/a)(a-10/b)=4 So ab=30 So area og the BOQ triangle=30-(9+5+4)=12.❤

[@oscarcastaneda5310]

Wow, I went off on such a tangent with this one ! Thank you so much for the Elegant Solution : )

[@timeonly1401]

Guess-&-check is a fine technique here only because the answer happen to turn out to be integral. But one shouldn't assume or expect that the final answer (an area) is an integer, any more than that any particular side is integer, either. If I were setting up a competition problem, I'd not make the answer easily guessable. Involving interesting fractions is likely sufficient. Use radicals if someone is especially clever. JMO.

[@giuseppemalaguti435]

QC=x,DC=b,PD=y,AD=h..risultano xh=10,(b-x)y=8,(h-y)b=18..da quest'ultima bh=18+by=18+b*8/(b-10/h)=18+8bh/(bh-10)...bh(bh-10)=18(bh-10)+8bh..questa equazione di 2 grado dà ( bh)^2-36bh+180=0..bh=30...=>A=30-9-5-4=12

[@quigonkenny]

Let DA = BC = y and CD = AB = x, so the area of the rectangle will be xy. Let PA = z and CQ = w, so DP = y-z and QD = x-w. Triangle ∆BAP: 9 = xz/2 xz = 18 --- [1] Triangle ∆QCB: 5 = yw/2 yw = 10 --- [2] xz(yw) = 18(10)

[@michaeldoerr5810]

The area is 30-18 or just 12 unita square. And by golly near the end there was probably enough more elegant way of eliminating the wrong answer. And looks like I must use that for practice!!!

[@santiagoarosam430]

Trazamos una recta horizontal por P y otra vertical por Q y ABCD queda dividido en cuatro celdas A, B, D y C, de superficies: (2*9 -a=18-a) ; (a) ; (2*4=8) y (2*5 -a=10-a)→ (18-a)/8=a/(10-a)→ a=6 = Área ABCD =18-6+6+8+10-6=30→ Área PBQ =ABCD-9-4-5=30-18=12 ud². Gracias y saludos.

[@KipIngram]

Well, it's laughably obvious using the "theory of KZbin math questions" that the answer is 12. But... let's try to do it right. Let H and W be the height and width of the rectangle, and let x be the length DQ and y be the length DP. Let A be the unknown triangle area. Then... (1) H*W = A + 18 (2) x*y/2 = 4 --> x*y = 8 (3) (W-x)*H/2 = 5 --> W*H - x*H = 10 --> x*H = H*W - 10 (4) W*(H-y)/2 = 9 --> W*H - y*W = 18 --> y*W = H*W - 18 Insert (1) into (3) and (4): (5) A + 18 - x*H = 10 --> A - x*H = -8 --> x*H = A + 8 (6) A + 18 - y*W = 18 --> A - y*W = 0 --> y*W = A Here's two equations in three unknowns. Now form the product of the two equations: (A - x*H)*(A - y*W) = 0 A^2 - (x*H+y*W)*A + x*y*H*W = 0 A^2 - (A+8+A)*A + 8*(A+18) = 0 -- Substitute from (5) and (6) as well as (1) and (2) A^2 - (2*A+8)*A + 8*A + 144 = 0 A^2 - 2*A^2 - 8*A + 8*A + 144 = 0 -A^2 + 144 = 0 A^2 = 144 A = 12 Q.E.D

[@SGuerra]

A questão é muito bonita. Parabéns pela escolha! Brasil Setembro 2024. The question is very beautiful. Congratulations on your choice! Brazil September 2024.

[@RealQinnMalloryu4]

(9)^=81(5)^^2=25 (4)^2=16 {81+25+16}=122H/AA/0/360°ABCD/122H/AA/O/= 2.116 2.10^10^4^4 2.2^5^2^5^2^2^2^2 1.1^1^1^1^1^1^1^2 1^2 (ABCDH/AA/O/➖ 2ABCDH/AA/O/+1).

[@devondevon4366]

12 Let's assume the sides of the rectangle are integers Since the area of the triangle to the left = 5, then the sides are 5 and 2 2 * 5 = 10 , hence BC =5 and QC =2 Hence, the width of the rectangle = 5 Since the area of the triangle to the left = 4, then the sides are 1, 8 , 2, an 4 Let's try 2 and 4 (look more reasonable) Hence, DQ =4 Hence, DC = 6 ( QC + DQ) Hence, the length of the rectangle =6, and PD =2 But if PD = 2, then AP must =3 (since the width of the rectangle is said to likely = 5) If AP =3, then AB must = 6= DC= 6 And DC= 6 Hence, the area of the rectangle = 5 * 6 = 30 and since the area of the unshaded rectangles = 18 (9 + 5 + 4) Then the area of the shaded region = 12 (30-18) Answer

[@zdrastvutye]

it seems to end up in an equation that will cancel out la (see line 20): 10 print "mathbooster-a very nice geometry problem august2024" 20 dim x(2,2),y(2,2):a1=9:a2=4:a3=5:la=4:sw=(a1+a2+a3)/10/la 30 l2=sw:goto 60 40 da=2*a2/l2:l1=la-da:db=2*a3/la:lb=db+l2:dg=l1*lb/2:dg=(a1-dg)/a1 50 return 60 gosub 40 70 l21=l2:dg1=dg:l2=l2+sw:if l2>1000*sw then stop 80 l22=l2:gosub 40:if dg1*dg>0 then 70 90 l2=(l21+l22)/2:gosub 40:if dg1*dg>0 then l21=l2 else l22=l2 100 if abs(dg)>1E-10 then 90 110 print l2:ages=la*lb-a1-a2-a3:print "die gesuchte flaeche=";ages 120 x(0,0)=0:y(0,0)=0:x(0,1)=l2:y(0,1)=0:x(0,2)=0:y(0,2)=la-l1 130 x(1,0)=l2:y(1,0)=0:x(1,1)=lb:y(1,1)=0:x(1,2)=lb:y(1,2)=la 140 x(2,0)=0:y(2,0)=la-l1:x(2,1)=lb:y(2,1)=la:x(2,2)=0:y(2,2)=la 150 masx=1200/lb:masy=850/la:if masx run in bbc basic sdl and hit ctrl tab to copy from the results window