Question

in the figure pqr is right triangle a circlr of radius 7 cm is drawn with the center Q and another circle of radius is 11 cm id drawn with center at P if PQ is 20 cm find the radius centered at R​

Answer 1

Answer:

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Step-by-step explanation:

so here let the radius of circle with centre R be x.cm ie let AR=RB=x.cm

so here circle with centre Q is 7.cm and that of circle with centre P is 11.cm

hence AQ=7.cm

BP=11.cm

so thus then

PR=BR+BP

QR=AQ+AR (P-B-R,Q-A-R)

ie PR=(x+11).cm

BR=(x+7).cm

so as triangle PQR is a right angled triangle

By Pythagoras Theorem

we get

QP²=PR²+QR²

ie (20)²=(x+7)²+(x+11)²

ie 400=x²+14x+49+²+22x+121

ie 400=2x²+170+36x

ie 2x²+36x-230=0

so dividing by 2 on both sides

we get now then

x²+18x-115=0

x²+23x-5x-115=0

x(x+23)-5(x+23)=0

ie (x+23)(x-5)=0

ie x=-23 OR x=5

but as radius,length,dimensions,etc are never negative,x=-23 is absurd

so x=5

thus AR=RB=5.cm

Hejce radius of circle with centred R is 5.cm