Question

rectangle ABCD in which ab = 8 cm and BC = 20 cm is a point on AD such that angle BPC equals to 90 cm if R1 R2 and R3 are incircles of the triangles APB CPD and BPC find r1+r2+r3

Answer 1

Let AP=x cmAP=x cm

Then PD=(20−x) cmPD=(20−x) cm

In ΔABPΔABP

BP2=82+x2BP2=82+x2

In ΔPCDΔPCD

PC2=82+(20−x)2PC2=82+(20−x)2

In ΔBPCΔBPC

82+x2+82+(20−x)22x2+128–40x+400x2–20x+64(x−4)(x−16)x=202=400=0=0=4,1682+x2+82+(20−x)2=2022x2+128–40x+400=400x2–20x+64=0(x−4)(x−16)=0x=4,16

And yes, this simply means that the point PP can be nearer to AA or DD. You know sometimes there are more than one correct answer to a construction depending on the given conditions :)

Now, lets recall the formula…

Radius of inscribed circle=A12(a+b+c)=2Aa+b+cRadius of inscribed circle=A12(a+b+c)=2Aa+b+c

As, always I will try to use some simple rules to make the final calculations as easier as I can make.

AP=4,AB=8⟹BP=45‾√,PC=85‾√⟹PD=16,CD=8⟹r1+r2+r3r1=4×84+8+85‾√=83+5‾√r2=45‾√×85‾√45‾√+85‾√+20=405+35‾√r3=16×816+8+85‾√=163+5‾√=83+5‾√+405+35‾√+163+5‾√=243+5‾√+405+35‾√=24(3−5‾√)9−5+40(5−35‾√)25−45[Rationalize the denominator]=6(3−5‾√)−2(5−35‾√)=18−65‾√−10+65‾√=8AP=4,AB=8⟹r1=4×84+8+85=83+5BP=45,PC=85⟹r2=45×8545+85+20=405+35PD=16,CD=8⟹r3=16×816+8+85=163+5r1+r2+r3=83+5+405+35+163+5=243+5+405+35=24(3−5)9−5+40(5−35)25−45[Rationalize the denominator]=6(3−5)−2(5−35)=18−65−10+65=8