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

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Пікірлер: 16

@pwmiles56 6 күн бұрын

The altitude of P over BC is 3/4 the altitude of A which itself is 10 sin 45 = 5 sqrt 2. Hence, CPQ = (1/2) base x height = (1/2) 8 (3/4) 5 sqrt 2 = 15 sqrt 2

@RAG981 5 күн бұрын

Excellent answer.

@quigonkenny 5 күн бұрын

Drop a perpendicular from A to E on BC. As AB = 10 and ∠ABE = 45°, AE = 10sin(45°) = 10(1/√2) = 10/√2 = 5√2. Draw AC. We are given that PQ = 3AP. Let AP = x, so PQ = 3x. Drop a perpendicular from C to F on AQ. Let CF = h. The area of ∆QCP is equal to PQ(CF)/2 = 3xh/2. Similarly the area of ∆ACQ = (3x+x)h/2 = 4xh/2 = 2xh. So [∆QCP]/[∆ACQ] = (3xh/2)/2xh = 3/4. Triangle ∆ACQ: [ACQ] = QC(AE)/2 = 8(5√2)/2 = 20√2 Red Triangle ∆QCP: [QCP] = (3/4)20√2 = 15√2 sq units

@santiagoarosam430 6 күн бұрын

La horizontal por P corta a BA por los ¾ de su longitud; la distancia vertical de esa intersección a QC es ¾*10/√2=30√2/8 ---> Área QCP =½*8*30√2/8 =15√2 ud². Gracias y saludos.

@ABhaim 6 күн бұрын

let angle AQP = a, AP = x, PQ = 3x, and AQ = 4x. according to the sine theorem: 4x/sin45 = 10/sin(a) 2x/sin45 = 5/sin(a) 2*sqrt(2)*x = 5/sin(a) x = 5*sqrt(2)/(4*sin(a)) 3x = PQ = 15*sqrt(2)/(4*sin(a)) [PQC] = PQ * QC * sin(180-a) /2 = 15*sqrt(2)/(4*sin(a)) * 8 * sin(a) /2 = 15*sqrt(2)

@olesgaiduk5159 5 күн бұрын

Let AC = AB and ∠ACV = 45º then AM = 5√2; ⇒ BC = 10√2; ⇒ ВQ = ВС - QC = 10√2 - 8. Place the origin of coordinates at point B(0; 0); where Q((10√2 -8); 0); A (5√2; 5√2); Since the line division coefficient QA ⇒ k= QP / PA = 3/1 =3 then Yр is the coordinate of point P ⇒ Yр = (Yq + k* Ya )/(1 + k) = (0 +3*5√2) /(1+ 3) = 15√2/4 ⇒ S_QPC = 0.5Yp *CQ = 0.5 * (15√2/4) *8 = 15√2 .

@michaeldoerr5810 6 күн бұрын

The area of CPQ is 15*sqrt(2) units squared. Looks like another sequence of clever geometry that relies on which points are collinear. I could be wrong. Then again there had to be a jusitified pairs of right angles.

@imetroangola4943 6 күн бұрын

*_Solução:_* Sejam AP=x, BQ=y e ∠AQB=α. [CPQ]=(PQ×QC×sin(180°-α))/2 *[CPQ]=3x •8•sinα/2= 12x sinα* A área [ABQ] é dada por: BQ×AP×sinα/2=AB×BQ×sin 45°/2 BQ×AP×sinα=AB×BQ×sin 45° 4x•y•sinα=10•y•√2/2 4x•sinα=5√2 ×(3) 12x•sinα=15√2. Portanto, *[CPQ]=15√2*

@Zina308 4 күн бұрын

Join AC. Drop a perp. from Q to AB at point N. Let QN= BN= y. In issoc. right triangle NBQ, BQ= Y V2.Let [APC]= m, [PCQ]= 3m.[ABQ]/ [AQC]= y V2/8. 5y/ 4m= y V2/8. m= 5 V2. [PCQ]= 3m= 15 V2.

@RealQinnMalloryu4 6 күн бұрын

(10)^2=100H/ (8)^2=64 A/ {100H/64A/}=164H/A {45°B+45°P+130°Q}=220°BPQ {220°BPQ/64H/A= 30.40PBQH/A 30.4^10 3^10.4^10 3^2^5.4^2^5 3^1^1.4^1^1 3^1.2^2 3^1.1^2 3.2 (PBQH/A ➖ 3PBQHH/A+2).