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The picture in the attached figure
we know that
If a tangent segment and a secant segment are drawn to a <span>circle </span><span>from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment
</span>so
DC²=BC*CA-----> CA=DC²/BC
DC=25
BC=14
CA=25²/14-----> CA=44.64
CA=BC+BA----> BA=CA-BC----> BA=44.64-14----> BA=30.64
BA is the diameter
hence
<span>the length of diameter BA is 30.64----> round to the nearest tenth---> 30.6
</span>
the answer is
<span>the length of diameter BA is 30.6</span>
we know that
If a tangent segment and a secant segment are drawn to a <span>circle </span><span>from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment
</span>so
DC²=BC*CA-----> CA=DC²/BC
DC=25
BC=14
CA=25²/14-----> CA=44.64
CA=BC+BA----> BA=CA-BC----> BA=44.64-14----> BA=30.64
BA is the diameter
hence
<span>the length of diameter BA is 30.64----> round to the nearest tenth---> 30.6
</span>
the answer is
<span>the length of diameter BA is 30.6</span>