Circle a and circle b are congruent. cd is a chord of both circles. if ab = 8 ft and cd = 6 ft, how long is a radius?

Let us say that the intersection point of lines AB and CD is called point E. The lines AB and CD are perpendicular to each other which also means that the triangle CEB is a right triangle.

Where the line CB is the radius of the circle while the side lengths are half of the whole line segment:

EB = 0.5 AB = 0.5 (8 ft) = 4 ft

CE = 0.5 CD = 0.5 (6 ft) = 3 ft

Now using the hypotenuse formula since the triangle is right triangle, we can find for the radius or line CB:

CB^2 = EB^2 + CE^2

CB^2 = (4 ft)^2 + (3 ft)^2

CB^2 = 16 ft^2 + 9 ft^2

CB^2 = 25 ft^2

CB = 5 ft = radius

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐

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find the upper bound of the range.

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step-by-step explanation: