Homework Geometry Problem From China | You should be able to solve this!

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[@quigonkenny]

Let BQ = PQ = QC = x. As PQ = QC, ∆PQC is an isosceles right triangle and ∠QCP = ∠CPQ = (180°-90°)/2 = 45°. As ∠ABC = 90°, ∠CAB = 90°-45° = 45° and ∆ABC is an isosceles right triangle similar to ∆PQC and AB = BC = 2x. Triangle ∆ABQ: AB² + BQ² = QA² (2x)² + x² = 5² 4x² + x² = 25 5x² = 25 x² = 25/5 = 5 x = √5 The area of the red shaded triangle ∆QPA is equal to the area of triangle ∆ABC minus the areas of triangles ∆ABQ and ∆PQC. Triangle ∆QPA: [QPA] = [ABC] - [ABQ] - [PQC] [QPA] = BC(AB)/2 - BQ(AB)/2 - QC(PQ)/2 [QPA] = 2√5(2√5)/2 - √5(2√5)/2 - √5(√5)/2 [QPA] = 20/2 - 10/2 - 5/2 = 5/2 sq units

[@prossvay8744]

CQ=PQ=BQ=a So ∆ CPQ is the isosceles right triangle So angle PCQ=CPQ=45° So BC=AB=2a In ∆ ABQ AB^2+BQ^2°AQ^2 (2a)^2+a^2=5^2 4a^2+a^2=5^2 5a^2=25 a^2=5 So a=√5 So area of triangle =1/2(2√5)^2-1/2(√5)(2√5)-1/2(√5)^2=5/2.❤❤❤

[@himo3485]

BQ=QC=PQ=x AB=2x x²+(2x)²=5² 5x²=25 x²=5 x=√5 area of triangle APQ = √5*√5*1/2 = 5/2

[@marioalb9726]

x² + (2x)² = 5² 5x² = 25 --> x = √5 cm A = ½b.h = ½x² = ½ 5 A = 2,5 cm² ( Solved √ )

[@marioalb9726]

x = 5 cos(atan 2) = √5 cm A = ½b.h = ½x² = ½ 5 A = 2,5 cm² ( Solved √ )

[@devondevon4366]

Answer 5/2 or 2.5 Triangle PQC is similar to triangle ABC since both are right triangles and share angle C. Since PQC is an isosceles, then ABC is also an isoceles. Hence, AB = BC Let BQ = n, then the bases of triangle ABC are 2n and 2n. Hence, its area = 4n^2/2 or 2n^2 Hence, the area of triangle ABQ = 2n*n*1/2 = n^2 And the area of CPQ = n*n*1/2 = n^2/2 Hence, the area of SHADED triangle in terms of n is the difference between triangle ABC and ABQ + CPQ Hence, 2n^2 - ( n^2 +n^2/2) 4n^2/2 - (2n^2/2 + n^2/2) n^2/2 Notice that triangle CPQ has the same area, n^2/2, as the SHADED triangle Let's go back to triangle, ABQ with sides 2n, n, and 5 and employ the Pythagorean Theorem (2n)^2 + n^2 = 25 5n^2= 25 n^2 = 5 n^2/2 = 5/2 Answer

[@johnbrennan3372]

Can be done without establishing that |AB|= 2x. |AB| also equals sqroot (25- x^2) . You can add the areas of the 3 smaller triangles (all in terms of x) and put the result equal to the area of the big triangle(also written in terms of x).Out of that you get the value of x and you then find the value of 1/2( x^2) for the answer.

[@michaeldoerr5810]

This is genuinely more helpful than the comments. I shall use that for practice!!! And the area for APQ is 5/2 units square. I did not know that BQ is also the base of that triangle. Good to know.

[@murdock5537]

φ = 30° → sin⁡(3φ) = 1; ∆ ABC → AB = BC = 2a = BQ + CQ = AT + BT = a + a; AQ = 5 PT = BQ = a; sin⁡(BQP) = 1; AQP = α = QAB; BQA = δ → cos⁡(δ) = sin⁡(α) = √5/5 → a = √5 → area ∆ AQP0 = (1/2)sin⁡(α)5a = 5/2

[@santiagoarosam430]

Pendiente de AC =PQ/QC=a/a=1---> BC=2a=BA --->(2a)²+a²=5²---> a=√5---> Área sombreada QPA=PQ*QB/2 =a²/2 =5/2. Gracias y saludos

[@oscarcastaneda5310]

A fun problem, Once I saw the height of the triangle I was able to solve this one without pencil and paper : )

[@RealQinnMalloryu4]

(5)^2=25 {45°ASPQ+45°C+90°B}=180°ASPQCB/26=7.5ASPQCB(ASPQCB ➖ ASPQCB+5)

[@MdShahriarHossain-l1n]

I like this problem. At first I thought I cannot solve it,but after trying I can! Now, I am feeling confident. 🎉

[@Zina308]

x= V5. [ABC]= (2V5)(2V5)/2= 10. [PQC]= [APQ], [PQC]= 1/4 [ABC], [PQC]= 10/4= 2.5. [APQ]= 2.5.

[@giuseppemalaguti435]

PQ=a ..teorema dei seni su APQ a/sin(45-arcsin(a/5))=5/sin135..a=√5...A=(1/2)5√5sin(135+45-arcsin(1/√5))=(1/2)5√5(1/√5)=5/2

[@AuriEsperanzaSQ]

Yo lo resolví sacando el area total del triángulo grande y luego le resté las áreas de los triángulos rectangulos pequeños interiores y la diferencia era el área del triángulo obstusángulo. Con la fórmula de Herón también la podía realizar ya que había encontrado todos los lados del triángulo obstusángulo pero demoraba más, terminaba mañana😂

[@SGuerra]

Que questão bonita. Parabéns pela escolha!!! Brasil Setembro 2024.

[@ademahmed1687]

What if you subtract area of triangle PQC from area of triangle ABC ???????/

[@MdShahriarHossain-l1n]

Which grade's homework is this problem?

[@sorourhashemi3249]

Thanks.easy

[@yakupbuyankara5903]

A=5/2

[@رایان-ل4ح]

با احترام لطفاً مسأله المپیاد

[@devondevon4366]

5/2 or 2.5

[@AuriEsperanzaSQ]

5/2

[@professorrogeriocesar]

Semelhança.