Can you find area of the triangle PQR? | (Fun Geometry Problem) |

[@juanalfaro7522]

This problem had been posted a few days ago. Let BRA=90-a=b. Clearly PBA and RBA are congruent triangles, thus BP=BR=30m and AP=AR. Now AQ=24 -> AP=AR=24 -> PR=48. Now BPA=b and QPR=90-b = a. Now cos (b) = 24/30 = 0.8 and cos (a) = sin (b) = 0.6 -> AB = 30*0.6 = 18. [PQR] = PQ*QR/2 = PR^2 *0.8*0.6/2 = 48^2 * 0.24 = {24^2*2^2] *24/100 = [576*4] * 24/100 = (2304*24)/100 = (2300+4) *24/100 = (55200+96)/100 = 552.96 m^2

[@phungpham1725]

1/ AQ= 24 -> AP=AR= 24
Focus on the triangle BAR:
AR= 4x6= 24, BR=5x6=30 so, AB = 3x6= 18 ( triangle BAR is a 3-4-5 triple)
--> Area of BAR= A1=1/2 x18x24=9x24
2/ The triangle PQR is similar to the triangle BAR .
Label Area of PQR as A2
We have:
A2/A1= sq (PR/BR)=sq(48/30)=sq(8/5)
-> A2= 9x24x64/25
A2=552.96 sq units😅😅😅

[@xualain3129]

Here is my version with trigonometry to share with.
Let cos a=sqrt(1-(4/5)^2)=3/5
PQ=PB*sin b=30*sin(180-2a)=30*sin(2a)=30*2*sin a*cos a=144/5
QB=PA*cos b=30*cos(180-2a)=-30*cos(2a)=-30*(1-2*sin(a)^2)=42/5
PR=PB+BR=30+42/5=192/5
Area PQR=1/2*144/5*192/5=27648/50=55296/100=552.96

[@MathandEngineering]

Thank you so much friend this is a very interesting Trigonometric one