SOLUTION: 3 circle, radii of centre PQR are 3.5 cm, x cm, 7 cm. Given that angle of PQR = 90 degree.
a. show that {{{2x^2+21x-49=0}}}
b. find x
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Question 704018: 3 circle, radii of centre PQR are 3.5 cm, x cm, 7 cm. Given that angle of PQR = 90 degree.
a. show that
b. find x
Answer by sachi(548) (Show Source): You can put this solution on YOUR website!
3 circle, radii of centre PQR are 3.5 cm, x cm, 7 cm. Given that angle of PQR = 90 degree.& of course the missing part is they are touching each other
so PQ=3.5+x=7/2*x QR=7+x & PR=7+3.5=7+7/2=21/2
PQR is a right triangle
so PQ^2+QR^2=PR^2
or(7/2+x)^2+(7+x)^2=(21/2)^2=441/4
or 49/4+x^2+7x+49+x^2+14x=441/4
or 2x^2+21x=441/4-49/4-49=49
or 2x^2+21x-49=0 ANS a
now multiplying 8 on both sides
16x^2+168x-392=0
or (4x)^2+2*4x*21+441-441-392=0
or (4x+21)^2=833
or x=(-21+/-sqrt833)/4
ans
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